问:
How can we come up with the definition of natural logarithm?
I learned calculus for 2 years, but still don't understand the definition of $\ln(x)$

$$\ln(x) = \int_1^x \frac{\mathrm d t}{t}$$

I can't make sense of this definition. How can people find it? Do you have any intuition?

答:
The study of the exponential function and the logarithm is motivated by the desire to find a way of modelling a particular kind of growth, where the rate of change of a quantity is proportional to the value of that quantity at any given time. Consider how the population of a colony of bacteria might change over time—perhaps it doubles every hour. Or how your wealth compounds when you put it into a bank. In both cases, we are dealing with exponential growth. Later, when Calculus makes rigorous the notion of 'instantaneous change', we may define an exponential function as a solution to the differential equation
$$
\frac{dy}{dx}=ky \, .
$$
Solving this equation, we find that
$$
kx+C = \int \frac{1}{y} \, dy \, . \tag{*}\label{*}
$$
It seems like we have hit a roadblock, since we cannot integrate $1/y$ using the rule
$$
\int y^n \, dy = \frac{y^{n+1}}{n+1}+C \, ,
$$
since that would lead to division by zero. However, the fundamental theorem of calculus tells us that the set of antiderivatives of $1/y$ must be
$$
\int_{a}^{y} \frac{1}{t} \, dt \, .
$$
This is a set of functions that differ by a constant as we vary the value of $a$. However, since there is already a constant on the LHS of $\eqref{*}$, we can assume that $a=1$ without loss of generality. This yields
$$
kx+C = \int_{1}^{y} \frac{1}{t} \, dt \, .
$$
Let $f(y)=\int_{1}^{y} \frac{1}{t} \, dt$. This function represents the area of the region bounded by the hyperbola $1/t$, the horizontal axis, and the vertical lines $t=1$ and $t=y$:

As the value of $y$ gets larger, the area of this region also gets larger. We can then verify that $f$ is strictly increasing, meaning that $f^{-1}$ exists. Hence,
$$
kx+C = f(y) \implies y = f^{-1}(kx+C) \, .
$$
Since growth of this kind is called exponential growth, it only seems right to bestow upon the function $f^{-1}$ the new name $\exp$. The statement reads more nicely as
$$
y=\exp(kx+C) \, .
$$
Of all of the types of exponential growth, the most pleasing is that in which the rate of change of a quantity is equal to the value of that quantity (not just proportional to it). Since the set of solutions to $dy/dx = ky$ is $y=\exp(kx+C)$, the set of solutions to $dy/dx = y$ is
$$
y=\exp(x+C) \, .
$$
And of these solutions, you will no doubt agree that the simplest and most aesthetic is
$$
y=\exp(x) \, ,
$$
where $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x} \frac{1}{t} \, dt$. For obvious reasons, this function is called the natural exponentional function. There seems to be clear parallels between $\exp$ and functions of the form $g(x)=a^x$. If we return to our examples of exponential modelling—the colony of bacteria that doubles in size every hour—it certainly seems that $2^x$ is an exponential function. After all, the growth of a bacterial colony is proportional to its population at any given point in time. This suggests that
$$
2^x = \exp(kx+C)
$$
for some values of $k$ and $C$. (So far, we haven't actually defined what $2^x$ means when $x$ equals $\sqrt{2}$, say, but we'll leave to the side for the moment, and use an 'intuitive' definition of irrational exponents, where $2^{\sqrt{2}} \approx 2^{1.41421}$.) Suspecting that $a^x$ and $\exp(x)$ are somehow linked, we try to recall the essential properties of $a^x$, hoping that $\exp(x)$ shares these properties. The most important of these properties is that
$$
a^x \cdot a^y = a^{x+y} \, .
$$
And it turns out that $\exp(x)$ also has this property! In other words,
$$
\exp(p) \cdot \exp(q) = \exp(p+q) \, .
$$
However, proving this takes a little work. Since $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, it seems sensible to use this integral in our proof. Showing that $\exp(p) \cdot \exp(q) = \exp(p+q)$ is actually equivalent to showing that $f(r) + f(s) = f(rs)$, since
\begin{align}
&f(r) + f(s) = f(rs) \\
\iff &f^{-1}(f(r)+f(s)) = rs \\
\iff &f^{-1}(f(r)+f(s)) = f^{-1}(f(r)) \cdot f^{-1}(f(s)) \\
\iff &\exp(p+q) = \exp(p) \cdot \exp(q)
\end{align}
(And this should make intuitive sense, too. Notice how $f$ changes multiplication into addition, and so it is only natural that $f^{-1}$ changes addition into multiplication.) There is also a slick way to prove that $f(r)+f(s)=f(rs)$:
$$
\int_{1}^{rs}\frac{1}{t} \, dt = \int_{1}^{r}\frac{1}{t} \, dt + \int_{r}^{rs}\frac{1}{t} \, dt
$$
Let $z=t/r$, meaning that $dz=\frac{1}{r}dt=\frac{z}{t}dt$. Then we have
\begin{align}
\int_{1}^{rs}\frac{1}{t} \, dt &= \int_{1}^{r}\frac{1}{t} \, dt + \int_{1}^{s}\frac{1}{z} \, dz \\
f(rs) &= f(r) + f(s) \, .
\end{align}
Now that we have established that $\exp(p) \cdot \exp(q)=\exp(p+q)$, it seems reasonable to conjecture that $\exp(x)$ is actually equal to $a^x$ for some base $a$. To find that base, we plug in $x=1$, yielding $\exp(1)$, which we will abbreviate as $e$. This means that
$$
e^x = \exp(x) \, .
$$
This notation, apart from being very convenient, is also entirely valid. To reiterate, $\exp(p) \cdot \exp(q)=\exp(p+q)$, and so
$$
e^p \cdot e^q = e^{p+q}
$$
as expected. Geometrically, $e$ represents the number $a$ for which $\int_{1}^{a}\frac{1}{t} \, dt = 1$. This property follows directly if one recalls the definition of $\exp$. And since $\log_a(x)$ is defined as the inverse of $a^x$, it is evident that $\log_e(x)$ is the inverse of $e^x$. But we know that the inverse of $\exp(x)$ is $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, and putting these statements together, we find that
$$
\log_{e}(x) = \int_{1}^{x}\frac{1}{t} \, dt \, .
$$
In fact, this logarithm is so important, so natural, that we often dispense with the base $e$ entirely and write
$$
\log(x) = \int_{1}^{x}\frac{1}{t} \, dt \, .
$$
Earlier, we mentioned the difficulty of defining $a^x$ when $x$ is an irrational number. The logarithm solves this problem. For rational $x$, we know that
$$
a^x = e^{x\log(a)} \, .
$$
But the RHS makes sense for all $x$, and so we may define
$$
a^x = e^{x\log(a)} \, ,
$$
for $a>0$. And defining exponents in this way means that
$$
a^x \cdot a^y = a^{x+y} \, ,
$$
still holds even for irrational exponents! The proof follows smoothly if we unravel the definition of $a^x$:
$$
a^x = e^{x\log(a)} = \exp(x\log(a)) \, ,
$$
where $\exp(x)=\log^{-1}(x)$ and $\log(x) = \int_{1}^{x}\frac{1}{t} \, dt$. Armed with this knowledge, we return to the differential equation we were trying to solve earlier, and finally obtain the solution we were expecting.
\begin{align}
\frac{dy}{dx} &= ky \\
\int k \, dx &= \int\frac{1}{y} \, dy \\
kx + C &= \log(y) \\
y&=e^{kx+C} = e^{kx} \cdot e^{C} = Ae^{kx} \text{ where $A=e^C$}
\end{align}
Since $a^x$ simply represents $e$ being raised to another base, functions such as $2^x$ are also included in the solution set. In particular,
$$
\frac{d}{dx}(a^x) = \frac{d}{dx}(e^{x\log(a)}) = e^{x\log(a)} \cdot \log(a) = a^x \cdot \log(a) \, ,
$$
and so $k=\log(a)$. Hopefully, this should provide the motivation behind defining the logarithm as $\int_{1}^{x}\frac{1}{t} \, dt$.


问:
How to check if the quadratic congruence has solution or not
How to check if the quadratic congruence has solution or not, for example check:
$$x^2\equiv 1141 \pmod{2020}$$

I use Euler's Criterion but it doesn't work:
$$1141^\dfrac{2020-1}{2} \equiv 1 \pmod{2020}$$ Because $\dfrac{2020-1}{2}$ is a decimal number. Any hints would be great!

ContentLicense: CC BY-SA 4.0

Body:
$x^2\equiv 1141 mod (2020)$

$1141=163\times 7$

$2020=20\times 101$

$163^{100}=(163^{50})^2\equiv 1 mod (101)$

$7^{100}\equiv 1 mod (101)$

$1141^{100} \equiv 1 mod (101)$

$\phi (20)= 8$

$1141^8=(1141^4)^2 \equiv 1 mod (20)$

So $x^2\equiv 1 mod (2020)$

Which means there is a solution.


问:
help with meromorphic doubly periodic function
Given a non constant meromorphic doubly periodic function $f$ with real-independent periodes $f(z+\omega_1)=f(z+\omega_2)=f(z)$. Consider $F:=\{\lambda_1\omega_1+\lambda_2 \omega_2, \lambda_1, \lambda_2 \in [0,1]\}$. Prove that $f$ has the same number of poles and zeros on $F$ with multiplicities counted.

I tried basically everything (Laurent-series etc.) but what I'm missing is that I need to prove the hint: $f$ has no poles and zeroes on the boundary of $F$.

I know that $\partial F$ is a contour and for the rest of the exercises I can use the Argument Principle but I really don't know how to prove the hint.

答:
Use the Argument principle. Integration of $f'/f$ on the contour $\partial F$ is possible, and the result is

the number of zeros minus the number of poles, multiplicities counted, on the one side, and

zero, on the other side, because contour integrals on opposite sides of the fundamental parallelogram $F$ cancel each other, same values, opposite direction of the contour paths.


问:
Cubing a simple thing
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.

ContentLicense: CC BY-SA 3.0

Title: Cubing a simple thing

Body:
Given: $(x+2)^3$ we can rewrite $(x+2)^3$ as:  $(x+2)(x+2)(x+2)$

When expanding 3 terms we must first calculate $(x+2)(x+2)$ before we can multiply by the third $(x+2)$.

= $(x+2)(x+2)\implies x^2+2x+2x+4\implies x^2+4x+4$

= $ (x+2)(x^2+4x+4)\implies x(x^2+4x+4)$ + $2(x^2+4x+4)$

= $ x^3+4x^2+4x+2x^2+8x+8$

= $x^3+6x^2+12x+8$


问:
Intuition on the quality of the approximation of the sample average in the central limit theorem
Let's say that a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ is $\mathcal {o}(Y_n)$ for some other sequence of random variables $(Y_n)_{n \in \mathbb{N}}$, if $\lim_{n \to \infty}\frac{X_n}{Y_n}=0$ in distribution.

We can use this to interpret the central limit theorem for such a sequence of random variables of mean $\mu$ and variance $\sigma^2$ as an asymptotic expansion (of the distribution)
$$ \frac{1}{n}\sum_{j=1}^{n} X_j = \mu + \frac{\xi}{\sqrt{n}} + \mathcal{o}\left(\frac{1}{\sqrt{n}}\right), $$
where $\xi$ is a random variable distributed as $\mathcal{N}(0,\sigma^2)$.

Intuitively the first term comes from the law of large numbers, and the second term is a correction of order $\frac{1}{\sqrt{n}}$ coming from the central limit theorem.
Is there some intuition in why this term should be of order $\frac{1}{\sqrt{n}}$? Why isn't the expansion in some other power of $n$, e.g. $\frac{1}{n}$ as in the definition of differentiability.

Assuming that it is intuitively clear that for large $n$, the sample average should be approximately distributed as a normal distribution, why should the quality of this approximation be exactly $\frac{1}{\sqrt{n}}$ and not some other numerical value?

答:
At first let us notice that $X_i$ are independent identically distributed random variables (or vectors).

According to SLLN (strong law of large numbers, under assumption $EX_1 < \infty$) we have

$$\frac1{n}\sum_{i=1}^n X_i \to E X_1$$

almost sure. In other words, we have convergence

$$\frac{S_n - \mathbf{E}S_n}{n} \to 0,$$
where $S_n = \sum_{i=1}^n X_i$.

So, after division by $n$ the value $S_n - \mathbf{E}S_n$ tends to $0$.
It's easy to see that if we will not devide it by $n$, it will not tend to $0$. It's natural: $D \Bigl(S_n - \mathbf{E}S_n  \Bigr) = nDX_1$ and is "big" for "big" n. More precisely, the behavior of $S_n$ we may get from LIL (law of iterated logarithm).

Hence, we know that $$ \frac{S_n - \mathbf{E}S_n}{f(n)}$$
diverges when $f(n) =1$ and "vanishes" when $f(n) = n$.

In order to make the value $ \frac{S_n - \mathbf{E}S_n}{f(n)}$ "not too big and not too small" let us notice that $$D \frac{S_n - \mathbf{E}S_n}{f(n)} = \frac{n DX_1}{f^2(n)}.$$
Hence it's natural to consider the case when $f^2(n)$ has the same rate of convergence as $n$, because only in this case the dispersion of  $ \frac{S_n - \mathbf{E}S_n}{f(n)}$ is not too big and not too small. And CLT says that this is a right approach.


问:
Dot products in SVM
I was watching a lecture on Support Vector Machine by Patrick Winston and here he does this thing

Where he projects the distance vector $x_+$ $-$ $x_-$ (in red) on the unit vector (in blue) and tell that this is the width between the 2 hyperplanes (in white). That is, he performs $dot$ $product$ on  $x_+$ $-$ $x_-$ (in red) and an unit vector perpendicular to the hyperplane to find out the distance between the lines. (The equation is written at the right). How is this thing working? Can anyone please explain. Thank you.

答:
There are many ways to see this. Here is one:

Adding to $x_-$ anything parallel to the hyperplane -- that is orthogonal to $w$ -- does not change the dot product. We can use this to move $x_-$ to the projection of $x_+$ to the hyperplane through $x_-$, which we will call $x_{p}$. Then $x_+- x_{p}$ is parallel to $w$, and $(x_{p}-x_-)\cdot w=0$.

Now the distance between the hyperplanes is $|x_+- x_{p}|=(x_+- x_{p})\cdot \frac{w}{|w|}$, which is the same as $(x_+- x_-)\cdot \frac{w}{|w|}$ -- their difference being $(x_{p}-x_-)\cdot \frac{w}{|w|}=0$. This is what we wanted.

Title: Dot products in SVM


问:
Integration $\int \left(x-\frac{1}{2x} \right)^2\,dx $
Evaluate $$\int\!\left(x-\frac{1}{2x} \right)^2\,dx. $$

Using integrating by substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$. In the end, I came up with the answer to the integral as :
$$\left(\frac{1}{3}+\frac{2x^2}{3}\right)\left(x-\frac{1}{2x}\right)^3.$$

Any mistake ?

答:
Just expand
$$
\left(x-\frac{1}{2x}\right)^2 = x^2 - 1 +\frac{1}{4x^2}
$$
and integrate term by term.</p>


问:
Combinatorics - N families, each family consists of {dad, mom, child, dog} sitting on a circular table
N families, each family consists of {dad, mom, child, dog} sitting on a circular table.

there are (4n-1)! ways to sit them

in how many the parents always sit together and the dogs never together?

What i tried to think of is calculating, all possibilities ( (4n-1)! ) - possibilities of dogs sitting together

but calculating the possibilies of all the dogs sitting together is kinda impossible, because assuming n=5 then 5 dogs and i have to find number of ways for 5 dogs sitting together then 4 dogs sitting together , and 3, etc...

答:
Your original estimate of $(4n-1)!$ provides a good starting point for the analysis

The issue of the parents always sitting together is easily dealt with by presuming that each pair of parents is glued together into one unit.  This means that instead of $(4n)$ units, you have only $(3n)$ units.

However, you have to adjust that within each internal parent-unit, there are (2!) ways of internally permuting the components of the parent-unit.

Therefore, absent the issue re the dogs, the (new starting) computation is

$$[(3n-1)!] \times [(2!)^n].$$

Fortunately, positioning the dogs also admits elegance.  Remembering that each pair of parents is an inseparable unit, you actually have (not counting the dogs), a dogless computation of

$$T_1 ~=~[(2n-1)!] \times [(2!)^n].$$

There are exactly $(2n)$ slots between these $(2n)$ units.  The $n$ dogs must be assigned into these slots, with no more than 1 dog assigned to any one slot.  This can be done in the following number of ways:

$$T_2 = \binom{2n}{n} \times n!.$$

Therefore, the final answer is

$$T_1 \times T_2.$$

Addendum

Responding to the comment/question of Light Lamps.

Explanation best given with a thought-experiment.  Suppose that :

You have a round table with $4n$ chairs.

Each child sits in one chair.

Each set of two parents sit together in only one chair.  That is, each parent-unit is sharing only one chair, with one parent using the left-half of the chair, and one parent using the right-half of the chair.

The $2n$ chairs now occupied are each separated by exactly 1 empty chair.

This means that you have $2n$ empty chairs alternating with $2n$ occupied chairs.

In this thought experiment, suppose that you focus on where one specific child is seated at the circular table.  Since the occupied chairs [i.e. the $2n$ chairs assigned to humans] alternate with the empty chairs, the position of the occupied chairs is fixed (in relation to the specific child being focused on).

With respect to that child, the $(2n-1)$ other units [$(n-1)$ children and $n$ parent-units] may be permuted in $(2n-1)!$ ways, among the $2n$ chairs assigned to humans.

Note that for each parent-unit, there are $(2!)$ ways that the parent-unit may be sharing a chair, with one parent on the left and one parent on the right.  This explains my computation of

$$T_1 ~=~[(2n-1)!] \times [(2!)^n].$$

Now, the constraint that each set of parents sit together has been satisfied.  The dogs are going to be assigned seats from the $2n$ remaining (unoccupied) chairs.  This satisfies the constraint that no two dogs sit next to each other.

There are $\binom{2n}{n}$ ways of selecting which of the $2n$ currently unoccupied chairs will be used by the dogs.

Having selected the $n$ chairs to be used by the dogs, there are $(n!)$ ways of permuting the dogs in the $n$ chairs assigned to them.

This explains my computation of

$$T_2 = \binom{2n}{n} \times n!.$$

This means that the computation of

$$T_1 \times T_2$$

is appropriate for the thought experiment.

What about the actual seating of the $4n$ (people + dogs) around the table, where

$4n$ chairs are used

No chairs are left unoccupied.

The parents sit together.

No two dogs sit next to each other.

There is a bijection between the thought experiment and the actual seating.  That is, each thought experiment seating corresponds to exactly 1 actual seating:

In the actual seating, with one parent in a chair to the left of the other parent, it corresponds to the thought experiment where both parents share a chair, with one parent using the left side of the chair.

In the actual seating, where a child does not have a dog on his immediate left (for example), this corresponds to the thought experiment where the child has an empty chair to his left.

In the actual seating, where the set of parents do not have a dog on their immediate left, this corresponds to the thought experiment where the parents have an empty chair on their left.

Because of the bijection between the thought experiment and the actual seating, the computation of $T_1 \times T_2$ is also appropriate for the actual seating.


问:
Distribution of $8n\operatorname{ln}(\sqrt{0.2\hat{p}} + \sqrt{0.8(1-\hat{p})} )$

Let $X_1, ..., X_n$ be a random sample taken from a Bernoulli$(p)$ distribution. Consider $\hat{p} = (1/n)\sum_i X_i = \bar{X}_n$ as the estimator of $p$. We wish to test $H_0: p=0.2$ against $H_1: p\neq 0.2$ using the statistic
$$
8nB = - 8n\operatorname{ln}(\sqrt{0.2\hat{p}} + \sqrt{0.8(1-\hat{p})} )
$$
where $B = -\operatorname{ln}(\sqrt{0.2\hat{p}} + \sqrt{0.8(1-\hat{p})} )$ is the Bhattacharya's distance.

The Problem: Determine the asymptotic distribution of $8nB$.

I haven't been able to create any analytical solution to the problem. I have verified that $\hat{p}$ satisfies the WLLN, that is, $\hat{p} \stackrel{P}{\longrightarrow} p$. Simulation (using R) suggests that $8nB$ converges in distribution to a $\chi^2_1$:

```<code>N <- 10^4
n <- 10^4

B_distr <- c()

for(i in 1:N){
  X <- sample(0:1, n, prob=c(0.8, 0.2),replace=TRUE)
  p_hat <- mean(X)

  B <- -8*n*log(sqrt(0.2*p_hat) + sqrt(0.8*(1 - p_hat)))
  B_distr <- c(B_distr, B)
}

hist(B_distr)
psych::describe(B_distr)
ks.test(B_distr, p="pchisq", 1)
</code>```

答: Assuming $H_0$ is true we can say $$\sqrt{n}\Big(\hat{P}-\frac{1}{5}\Big) \longrightarrow N\Big(0,\frac{4}{25}\Big)$$ where $\longrightarrow$ denotes convergence in distribution. Take $$g(x)=-8\ln\Big(\sqrt{0.2x}+\sqrt{0.8(1-x)}\Big)$$ Note $g'(1/5)=0$ so we cannot use the delta$-$method. But we can, however, use the second order delta method and say that $$n \bigg(g\big(\hat{P}\big)-g(1/5)\bigg) \longrightarrow \frac{4}{25}\cdot \frac{g''(1/5)}{2} \cdot \chi^2_1$$ Since $g(1/5)=0$ and $g''(1/5)=\frac{25}{2}$ we have $$ng\big(\hat{P}\big) \longrightarrow \chi^2_1$$ This is what you're looking for.


问:
Set-up a triple integral in spherical coordinates of a solid bounded by a hemisphere and cylinder
Given $S$ as the solid that is outside the cylinder $x^{2}+y^{2}=1,$ inside the sphere $x^{2}+y^{2}+z^{2}=4,$ and above the $x y$ -plane. Set-up an iterated triple integral in spherical coordinates that is equal to $\iiint_{G} f(x, y, z) d V$ where $f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$.

How I approached this problem is that since

$f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$ and we know that from rectangular to spherical we have $\rho=\sqrt{x^2+y^2+z^2}$ and $x=\rho sin\phi cos\theta$

I know have  $\iiint_{G} f(x, y, z) d V = \iiint_{G} \rho sin\phi cos\theta*\rho^2 sin\phi d\rho  d\theta d\phi $

I've also determined the bounds for $\theta = 0$,$ 2\pi$ as well as for $\phi = 0$,$\pi/2$

My problem is determining the bounds for $\rho$ as the inner figure is a cylinder. My guess is that the upper bound of $\rho=2$ but I can't determine the lower bound.

Any help would be appreciated. Thanks.

答:
You are right about the outer bound, the inner bound will have to be the cylinder:

$$x^2+y^2=1 \implies \rho\sin\phi = 1$$

In the $\rho\phi$-plane ($\rho$ being the vertical axis) the region of integration will be shaded area to the left of the green line:

You can see from the photo that $\phi$ doesn't actually make it all the way back to $0$ (the $z$ axis in Cartesian), it stops at

$$2\sin\phi = 1 \implies \phi = \frac{\pi}{6}$$

and $\rho$ stops at

$$\rho\sin \frac{\pi}{2} = 1 \implies \rho = 1$$

From here it's easier to see how to set up the integral in both orders.

$$\int_0^{2\pi}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int_{\csc\phi}^{2} f\cdot \rho^2\sin\phi \: d\rho d\phi d\theta$$

$$\int_0^{2\pi}\int_{1}^{2}\int_{\sin^{-1}\left(\frac{1}{\rho}\right)}^{\frac{\pi}{2}} f\cdot \rho^2\sin\phi \: d\phi d\rho d\theta$$


问:
A functional equation problem: $f(xy)f\left(\frac{x+y}{xy}\right)=f(x+y)$ and $f(x+1)=f(x)+1$
According to my friends, in the test of Moroccan team selection, no one solved the following problem :

Find all functions $f : \mathbb{R}^{+*} \to \mathbb{R}^{+*} $ such that :
$$f(x+1)=f(x)+1 \hspace{2cm} \text{and} \hspace{2cm} f(xy)f\left(\frac{x+y}{xy}\right)=f(x+y)$$
For all $x$ and $y$ in $\mathbb{R}^{+*}$

I followed the basic steps to solve generally a functional equation, but it seems that this one is very special since it needs to verify 2 conditions. Any thoughts ?

答:
If we substitute $y=1$ in the second equation we can get the expression:

$$f(x) = \frac{1}{f(1+1/x) -1} = \frac{1}{f(1/x)}.$$

Let's say we want to solve how much is $f(x)$ with $x\in \mathbb{R}, x>0$. Call the integer part $\lfloor x \rfloor = a_0$ and $b_0 = x-a_0$. Applying the first rule $a_0$ times we have that:

$$f(x) = a_0 + f(b_0) = a_0 + \frac{1}{f(1/b_0)}.$$

Now call $a_1 = \lfloor 1/b_0 \rfloor$ and $b_1 = 1/b_0-a_1$ and we apply the first rule again $a_1$ times:

$$ f(x) = a_0 + \frac{1}{a_1 + f(b_1)} = a_0 + \frac{1}{a_1 + \frac{1}{f(1/b_1)}}.$$

You can continue this indefinitely. This process is a well-known algorithm, it's the computation of the continued fraction of $x$, sometimes expressed as $[a_0; a_1, a_2, ...]$. These fractions, when you stop at a particular $n$ will get you the best rational approximation to $x$ (comparing with fractions with lower denominators) and are really handy when working with reals in any computer program.

You can check that we can bound the value of the $n$th iteration, no matter the value of $f$, between the $n-1$th and $n-2$th iteration of the usual algorithm (without $f$), and we know this sequence converges to $x$ so $f(x)$ must be exactly $x$. You can check any notes in continued fractions (for example the wikipedia page) for that.


问:
Proof a graph of order $n$ and $\delta(G)\geq \sqrt{n}$ contains a cycle at most 4
"If $G$ is a graph of order $n$ and $\delta(G)\geq \sqrt{n}$, then for every vertex $v\in V(G)$ there is a cycle of length at most four in $G$ which contains $v$."

I need to prove this for every $n\in \mathbb{N}$, I understand that it can be done using proof by induction but don't know what to have as my induction hypotheses or where to go from this.

答:
Argue by contradiction. If $v$ is not part of a cycle of length $3$ or $4$, then $v$ has $\sqrt n$ neighbors $w_1, w_2, \dots, w_{\sqrt n}$ such that:

There is no edge $w_i w_j$: such an edge would create a $3$-cycle.

Aside from $v$, all other neighbors of $w_1$ through $w_{\sqrt n}$ are distinct: if $w_i$ and $w_j$ had another common neighbor, we'd get a $4$-cycle.

Why is this a problem?

(If you're still not sure, try drawing a picture of the argument above with $n=9$ vertices and $\delta(G) \ge \sqrt9 =3$.)


问:
Does $(A = v_a \implies B = v_b) \implies P(B = v_b) \geq P(A = v_a)$ make sense?
I have a question regarding the following formula:

Let $A$ and $B$ be two random variables.

If $A = v_a \implies B = v_b$, then

$\mathcal{P}(B = v_b) \geq \mathcal{P}(A = v_a)$

I was told that the hypothesis ($A = v_a \implies B = v_b$) does not make sense mathematically speaking, specifically, that the hypothesis should not contain an implication and that I should try to define this formula in another way, perhaps defining $v_b$ as a function of $v_a$. I think the formula is correct, but I was not able to find a source in which this formula (or a similar one) appears. I want to know how to show that it makes sense to have a conditional statement ($\implies$) in the hypothesis of a formula. Ideally, the formula I am interested in could be mentioned in some source. Unfortunately, I was not able to find it. I have already checked out these sources [1][2][3][4] to no avail, in addition to a search in this forum.

My question is: does this formula make sense? and if it does, which books/papers/or other sources can I check to verify it?

[1] Giangiacomo Gerla (1994). Inferences in probability logic, Artificial Intelligence, 70(1), 33 - 52.

[2] Nils J. Nilsson (1986). Probabilistic logic, Artificial Intelligence, 28(1), 71 - 87.

[3] H. T. Nguyen, M. Mukaidono, & V. Kreinovich (2002). Probability of implication, logical version of Bayes theorem, and fuzzy logic operations. In 2002 IEEE World Congress on Computational Intelligence. 2002 IEEE International Conference on Fuzzy Systems. FUZZ-IEEE'02. Proceedings (Cat. No.02CH37291) (pp. 530-535 vol.1).

[4] https://plato.stanford.edu/entries/logic-probability/#CombLogiProbTheo

答:
I finally wrote the hypothesis without a conditional, by using probabilities. I used the following equivalent hypothesis.

$$\mathcal{P}(b = v_b | a = v_a) = 1$$

I still don't know if an implication in the hypothesis is correct or not, but at least I know that this one is correct.


答:
The intersection of cylinder and sphere is at $z = \sqrt 3$

So $\phi$ at that point is $\phi = \arccos (\frac{z}{2}) = \frac{\pi}{6}$. (radius of sphere is $2$)

So $ \frac{\pi}{6} \leq \phi \leq \frac{\pi}{2} $

Also $ \frac{1}{\sin \phi} \leq \rho \leq 2$

Take a radial line from the origin to the point on the sphere and as the radius of the cylinder is $1$ the hypotenuse from origin to the point on the cylinder will be $OT = \frac{OI}{\cos(90^0-\phi)} = \frac{1}{\sin \phi}$

Title: Set-up a triple integral in spherical coordinates of a solid bounded by a hemisphere and cylinder

Body:
Given $S$ as the solid that is outside the cylinder $x^{2}+y^{2}=1,$ inside the sphere $x^{2}+y^{2}+z^{2}=4,$ and above the $x y$ -plane. Set-up an iterated triple integral in spherical coordinates that is equal to $\iiint_{G} f(x, y, z) d V$ where $f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$.

How I approached this problem is that since

$f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$ and we know that from rectangular to spherical we have $\rho=\sqrt{x^2+y^2+z^2}$ and $x=\rho sin\phi cos\theta$

I know have  $\iiint_{G} f(x, y, z) d V = \iiint_{G} \rho sin\phi cos\theta*\rho^2 sin\phi d\rho  d\theta d\phi $

I've also determined the bounds for $\theta = 0$,$ 2\pi$ as well as for $\phi = 0$,$\pi/2$

My problem is determining the bounds for $\rho$ as the inner figure is a cylinder. My guess is that the upper bound of $\rho=2$ but I can't determine the lower bound.

Any help would be appreciated. Thanks.


问:
Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx} $
$\eqalign{
  & \int {{{\left( {\cot x - \tan x} \right)}^2}dx}   \cr
  &  = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx  \cr
  &  = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx  \cr
  &  = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx}   \cr
  &  = \int {{{\left( {2\cot 2x} \right)}^2}}   \cr
  &  = \int {4{{\cot }^2}2xdx}   \cr
  &  = \int {4\left( {{{\csc }^2}2x - 1} \right)dx}   \cr
  &  = \int {\left(4{{\csc }^2}2x - 4\right)dx}   \cr
  &  = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C  \cr
  &  =  - 2\cot 2x - 4x + C \cr} $

Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.

答:
Why do you think that you've gone wrong? Your answer is correct.

Edit. To show why the solution is the same as the one in your answers, we just need to show that $\tan x-\cot x=-2\cot 2x$.
$$
\begin{align*}
&\tan x-\cot x+2\cot 2x \\
&=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}+2\frac{\cos 2x}{\sin 2x} \\
&=\frac{2\cos x\cos 2x\sin x-\cos^2 x\sin 2x+\sin^2 x\sin 2x}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos x(\cos^2 x-\sin^2 x)\sin x-\cos^2 x(2\sin x\cos x)+\sin^2 x(2\sin x\cos x)}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos^3 x\sin x-2\cos x\sin^3 x-2\sin x\cos^3 x+2\cos x\sin^3 x}{\cos x\sin x\sin 2x} \\
&=0
\end{align*}
$$


问:
Does the function have to be differentiable in local extreme(maximum and minimum) points?
let's say we have a partially defiend function

$$
f(x)=\left\{ \begin{align} x+1 & \quad\mbox{if}\quad x<5\\ x-5 & \quad\mbox{if}\quad x\ge5 \end{align} \right. $$

The graph of that function would look like this:
the graph of function from geogebra
In this specific case,would x = 5 be local minima?

答:
As always, if we have a question like this we want to go back to the Definition. I'll rewrite it just a little bit, to make it more applicable:

A real number $x_0$ is a local minimum of a real function $f\colon\mathbb{R}\to\mathbb{R}$ if there exists $\epsilon>0$ such that for all real $x$, $|x-x_0|<\epsilon\implies f(x_0)\leq f(x)$. In other words, $x_0$ is a local minimum if $f(x_0)$ is the smallest of all $f(x)$ in some neighborhood around $x_0$.

In this case $5$ is indeed a local minimum, because $f(5+\epsilon)>f(5)$ for all positive real $\epsilon$, and $f(5-\epsilon)>f(5)$ for all positive real $\epsilon<5$.

However, it's important to note that even though local extrema do not require differentiability or even continuity, being a minimum must be "two-sided" in some sense; i.e. the function

$$\begin{cases}4-x,&x<5\\x-5,&x\geq5\end{cases}$$

does not have a local minimum at $5$, because $f(5)$ is greater than $f(4.9),\ f(4.99),\ f(4.999),\ $etc.


问:
Invariance of volume form under coordinate transformations
For my topology class I have to show the following:

Let $g = g_{ij} \; dx^i \otimes dx^j$ be a metric on a differentiable manifold. Show that the volume form $dV = \sqrt{\det g} \; dx^1 \land ... \land dx^n$ is invariant under coordinate transformations.

Now, I am not quite sure what I have to do here. If $x = (x^1, ..., x^n) $ is a set of coordinates and $\tilde{x} = (\tilde{x}^1, ..., \tilde{x}^n)$ is another set of coordinates, then do I have to show that

$$d\tilde{V}( \tilde{x} ) \enspace = \enspace dV(x) \quad ?$$

But how do I know how $d\tilde{V}(\tilde{x})$ looks like? Does anyone have a hint for me?

答:
You do not know how $dV$ looks like for a specific case. So, you have to use the general form of $dV$ which is what you have written.

You need to show that if you perform a coordinate transformation on the wedge products and on the determinant of the metric at the same time, everything will be cancelled and you will recover $dV$ again.

OwnerDisplayName: Octavius

OwnerDisplayName: Kianoush Maleki


答:
Using basic coordinate transformation rules for tensors ($g_{ij}$ and $dx^i$), as well as antisymmetric properties of the wedge product, show that the wedge product

$dx^1\wedge\dots dx^n$ transforms as follows under coordinate transformations:

$$
dx^1\wedge\dots dx^n=\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n
$$

Where $\frac{\partial(x)}{\partial(\tilde{x})}$ is the determinant of the Jacobian matrix. Then, again using coordinate transformations, show that

$$
\mbox{det}\left[\tilde{g}\right]=\left(\frac{\partial(x)}{\partial(\tilde{x})}\right)^2\,\mbox{det}\left[g\right]
$$

Conclude that, under coordinate transformations:

$$
dx^1\wedge\dots dx^n=\sqrt{\frac{\mbox{det}\left[\tilde{g}\right]}{\mbox{det}\left[g\right]}}\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n
$$

Finally conclude that:

$$
\sqrt{\mbox{det}\left[g\right]}\cdot dx^1\wedge\dots dx^n =
\sqrt{\mbox{det}\left[\tilde{g}\right]}\cdot d\tilde{x}^1\wedge\dots d\tilde{x}^n
$$

i.e. the form of this quantity does not change as you change coordinates.

Finally, finally, note that any manifold is locally a Cartesian space, and there, in Cartesian coordinates $\mbox{det}\left[g\right]=1$ and $dx^1\wedge \dots dx^n=d^nV$, the standard volume form. From this conclude that $d^n V=\sqrt{\mbox{det}\left[g\right]}dx^1\wedge \dots dx^n$ is the correct volume form in any coordinate system and at all points on the manifold

Adding more specifics. I would suggest https://www.amazon.com/Tensors-Differential-Variational-Principles-Mathematics-ebook/dp/B00A735HK8#reader_B00A735HK8. It deals with all that you want in a rigorous and hands-on-way.

To compute the determinant of the metric tensor you need a way of expressing this determinant using tensors. At this point on usually reaches for Levi-Civita, but the problem with that is that Levi-Civita is not a tensor (it is a relative tensor, in the language of Lovelock and Rund). Fortunately, there is something called generalized Kroenecker delta [https://en.wikipedia.org/wiki/Kronecker_delta#Definitions_of_the_generalized_Kronecker_delta], which is a tensor (to prove this note that generalized Kroenecker delta can be written as products of usual Kronecker delta's $\delta^{\mu}_{\nu}$, so it is sufficient to prove that latter is a tensor).

Ok, with that:

$$
\det\left[\tilde{g}\right]=\tilde{\delta}^{\alpha_1\dots\alpha_n}_{1\dots n}\tilde{g}_{1\alpha_1}\dots\tilde{g}_{n\alpha_n}
$$

To check that, note that $\tilde{\delta}^{1\dots n}_{1\dots n}=1$ and then picks up a factor of -1 on every exchange of $\alpha$-s. Now you can apply change of coordinates to the RHS, since there you only have tensors:

$$
\begin{align}
\det\left[\tilde{g}\right]=\frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} \cdot
\frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n}
\cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n}
\end{align}
$$

Firstly deal with sum over $\mu$-s. Let's look at just two terms, for example:

$$
\begin{align}
\frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} =& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} +
\frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1 \dots\alpha_n}_{2, 1\dots n}+\dots\\
=& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} -
\frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot
\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}+\dots
\end{align}
$$

Note the swap of indices on the least delta. In this way we can show that the sum over $\mu$-s consist of a product $\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}$ and a completely anti-symmetrized product of partial derivatives - a Jacobian determinant:

$$
\begin{align}
\det\left[\tilde{g}\right]=\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot
\delta^{\alpha_1\dots\alpha_n}_{1 \dots n} \cdot
\frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n}
\cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n}
\end{align}
$$

Next consider:

$$
w_{\sigma_1\dots\sigma_n}=\delta^{\alpha_1\dots\alpha_n}_{1 \dots n}  g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n}
$$

And note that:

$$
w_{1\dots n}=det\left[g\right]
$$

and any exchange of $\sigma$ indices simply adds a factor of -1 to $w_{\sigma_1\dots\sigma_n}$. We can therefore repeat the trick above to get

$$
\begin{align}
\det\left[\tilde{g}\right]=&\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot
w_{1\dots n} \cdot
\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\\
=&\left(\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\right)^2\cdot\det\left[g\right]
\end{align}
$$

Title: Invariance of volume form under coordinate transformations

Body:
For my topology class I have to show the following:

Let $g = g_{ij} \; dx^i \otimes dx^j$ be a metric on a differentiable manifold. Show that the volume form $dV = \sqrt{\det g} \; dx^1 \land ... \land dx^n$ is invariant under coordinate transformations.

Now, I am not quite sure what I have to do here. If $x = (x^1, ..., x^n) $ is a set of coordinates and $\tilde{x} = (\tilde{x}^1, ..., \tilde{x}^n)$ is another set of coordinates, then do I have to show that

$$d\tilde{V}( \tilde{x} ) \enspace = \enspace dV(x) \quad ?$$

But how do I know how $d\tilde{V}(\tilde{x})$ looks like? Does anyone have a hint for me?

OwnerDisplayName: Cryo


答:
Your idea to do this:

$$(x^2 + 4x + 4)(x + 2)$$

was correct, because $(x + 2)^2 = (x^2 + 4x + 4)$. So it follows that $(x + 2)^3 = (x^2 + 4x + 4)(x + 2)$.

Let's try just multiplying it out:

To multiply $(x^2 + 4x + 4)$ by $(x+2)$, we'll multiply every term in $(x^2 + 4x + 4)$ by every term in $(x+2)$.

That's easy to do, and easy to visualise, because $(x^2 + 4x + 4)\cdot (x + 2)$ is the same as $(x \cdot(x^2 + 4x + 4)) + (2 \cdot (x^2 + 4x + 4)).$

$$(x \cdot(x^2 + 4x + 4)) = x^3 + 4x^2 + 4x$$

$$(2 \cdot (x^2 + 4x + 4)) = 2x^2 + 8x + 8$$

Add the two:

$$x^3 + \color{red}{4x^2} + \color{blue}{4x} + \color{red}{2x^2} + \color{blue}{8x} + 8 = x^3 + 6x^2 + 12x + 8$$

Title: Cubing a simple thing

Body:
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.


问:
A basic question on martingales and filtrations
It is stated as a Corollary in a book that if $\tau$ is any stopping time and $M_t$ is a martingale with respect to $(\Omega,\mathcal{F}_t,\mathbb{P})$, then so is $M_{t\wedge\tau}$. However, I am mainly concerned with the filtrations associated with the martingales $M_{t\wedge\tau}$. I know that (from the Theorem in the book from which the Corollary is deduced without stating its proof) for any $0\leq s\leq t$ we have $$\mathbb{E}[M_{t\wedge \tau} \mid \mathcal{F}_{s\wedge \tau}] = M_{s\wedge \tau}.$$ But I think the author means that $M_{t\wedge \tau}$ is a martingale associated with the same filtration as that of $M_t$ (which is $\{\mathcal{F}_t\}$), in that case, I really didn't see how one can prove that $$\mathbb{E}[M_{t\wedge \tau} \mid \mathcal{F}_s] = M_{s\wedge \tau}$$ for any $0\leq s \leq t$. Thanks for any help!

答:
It's effectively just a computation:

\begin{align*}
\mathbb{E}[M_{t \wedge \tau} | \mathcal F_s] &= \mathbb{E}[M_{t \wedge \tau}1_{\tau \le s} | \mathcal F_s] + \mathbb{E}[M_{t \wedge \tau}1_{\tau > s} | \mathcal F_s] \\
&= M_\tau 1_{\tau \le s} + \mathbb{E}[M_{t \wedge \tau}1_{\tau > s} | \mathcal F_s]
\end{align*}

so we just need to show $\mathbb{E}[M_{t \wedge \tau}1_{\tau > s} | \mathcal F_s] = M_s 1_{\tau > s}.$  Let $A \in \mathcal F_s$, and notice $A \cap \{\tau > s \} \in \mathcal F_{\tau \wedge s}$ so we have

\begin{align*}
\int_A M_{t \wedge \tau}1_{\tau > s} dP &= \int_{A \cap \{\tau > s \}} M_{t \wedge \tau} dP \\
&= \int_{A \cap \{\tau > s \}} \mathbb{E}[M_{t \wedge \tau}|\mathcal F_{\tau \wedge s}] dP \\
&= \int_{A \cap \{\tau > s \}} M_{s \wedge \tau} dP \\
&= \int_A M_{s \wedge \tau}1_{\tau > s} dP \\
&= \int_A M_{s}1_{\tau > s} dP
\end{align*}

and by definition of conditional expectation we conclude $\mathbb{E}[M_{t \wedge \tau}1_{\tau > s} | \mathcal F_s] = M_s 1_{\tau > s}.$  Therefore

\begin{align*}
\mathbb{E}[M_{t \wedge \tau} | \mathcal F_s] &= M_\tau 1_{\tau \le s} + M_s 1_{\tau > s} \\
&= M_{s \wedge \tau}.
\end{align*}


问:
$f_0 = 0$,$f_1= 1$, and $f_n=f_{n-1}$ + $f_{n-2}$, for all $n>=2$. Use mathematical induction to prove
$\sum_{i=0}^{2n}(-1)^i f_i = f_{2n-1} - 1$, for any positive  integer n.

Let P(n) be $\sum_{i=0}^{2n}(-1)^i f_i = f_{2n-1} - 1$.

Basis step P(1) : $\sum_{i=0}^{2}(-1)^i f_i = f_{2*2-1} - 1$.
1=1 => true.

Hypothetical assumption : Assume P(k) is true : $\sum_{i=0}^{2k}(-1)^i f_i = f_{2k-1} - 1$

We need to prove that P(k+1) is true: $\sum_{i=0}^{2k+2}(-1)^i f_i = f_{2k+1} - 1$

Here's where I am lost at what to do. I have tried to look at the left hand side to transform the summation into something similar to the assumption but could not find any formula to do that. What should I do to get the proof ?

Body:
Take two terms out of the sum, so you can use the induction hypothesis:

$$\sum_{i=0}^{2k+2}(-1)^i F_i = -F_{2k+1}+F_{2k+2}+\sum_{i=0}^{2k}(-1)^iF_i$$

$$=-F_{2k+1}+ F_{2k+2}+F_{2k-1} - 1=F_{2k}+F_{2k-1}-1=F_{2k+1}-1$$


问:
How would you sketch $\sin y = - \sin x$?
I at first did $\sin^{-1}$ both sides to get $y = -x$.

This is wrong, can someone explain why?

The graph is made up of a lot of crosses and looks like a grid on desmos. I need a way to get the answer without using a graphing calculator, any help?

答:
You have,

$\sin x + \sin y = 0 \Rightarrow 2\sin\frac{x+y}2 \cos\frac{x-y}2 = 0  \Rightarrow \sin\frac{x+y}{2} = 0$ or $\cos\frac{x-y}{2} = 0$

So,

$\frac{x+y}{2} = n\pi$ or $\frac{x-y}{2} = (m+\frac12)\pi$ where $m,n \in Z$

$\Rightarrow x = 2n\pi-y$ ($2n$ is always even)

or $x = (2m+1)\pi +y$   ($2m+1$ is always odd)

So,

$x = k\pi + (-1)^{k-1}y ,\ k \in \Bbb Z$

or

$y = k\pi + (-1)^{k-1}x ,\ k \in \Bbb Z$


问:
Is my asymptotic formula for $\sum_{k=1}^{m}\log(k!)$ right?
Today I tried to derive an asymptotic formula for the sum $$\sum_{k=1}^{m}\log(k!)$$
Here is what I did: This can be rewritten as $$\sum_{k=1}^{m}\sum_{1<n\le k}\log n$$
Now I applied the Euler summation formula, and this turned to
$$\sum_{k=1}^{m}\left(\int_{1}^{k}\log t\,dt+\int_{1}^{k}\frac{t-[t]}{t}dt+\log(k)([k]-k)\right)$$
$\int_{1}^{k}\frac{t-[t]}{t}dt$ is dominated by $\int_{1}^{k}\frac{1}{t}dt=\log k$, so $\int_{1}^{k}\frac{t-[t]}{t}dt=O(\log k)$, and it is also true that $\log(k)([k]-k)=O(\log k)$
Now, the sum becomes
$$\sum_{k=1}^{m}(1-k+k\log k+O(\log k)+O(\log k))=\sum_{k=1}^{m}(1-k+k\log k+O(\log k))$$
which is equal to
$$m-\frac{m(m+1)}{2}+\log H(m)+O(\log m)$$
where $H(m)$ is hyperfactorial. Is my solution right? I could have made a mistake.

答:
Comment on the Answer in the Question

One problem is that
$$
\sum_{k=1}^mO(\log(k))=O(m\log(m))\tag1
$$
However, as the result is $\sim\frac{m^2}2\log(m)$, this does not invalidate the entire result.

Equation $(7)$ of this answer shows that
$$
\begin{align}
\log(H(m))
&=\sum_{k=1}^mk\log(k)\tag{2a}\\
&=\frac{m^2}2\log(m)-\frac{m^2}4+O(m\log(m))\tag{2b}
\end{align}
$$
so we get
$$
m-\frac{m(m+1)}2+\log(H(m))=\frac{m^2}2\log(m)-\frac{3m^2}4+O(m\log(m))\tag3
$$
Thus, as long as the error term is taken to be $O(m\log(m))$ instead of $O(\log(m))$, your answer is correct.

More Precise Asymptotic Expansion

Using a result from an earlier answer,
$$
\begin{align}
\sum_{k=1}^n\log(k!)
&=\sum_{k=1}^n\sum_{j=1}^k\log(j)\tag{4a}\\
&=\sum_{j=1}^n\sum_{k=j}^n\log(j)\tag{4b}\\
&=\sum_{j=1}^n(n-j+1)\log(j)\tag{4c}\\
&=n^2\frac{2\log(n)-3}4+n\log\left(\frac{\sqrt{2\pi}}en\right)+\frac5{12}\log(n)\\
&+\log\left(\frac{\sqrt{2\pi}}{A}\right)+\frac1{12}+\frac1{12n}+O\!\left(\frac{\log(n)}{n^2}\right)\tag{4d}\\
&=\frac{n^2}2\log(n)-\frac{3n^2}4+n\log(n)+\left(\frac12\log(2\pi)-1\right)n+\frac5{12}\log(n)\\
&+\left(\frac1{12}+\frac12\log(2\pi)-\log(A)\right)+\frac1{12n}+O\!\left(\frac{\log(n)}{n^2}\right)\tag{4e}
\end{align}
$$
Explanation:

$\text{(4a)}$: definition of $n!$

$\text{(4b)}$: change order of summation

$\text{(4c)}$: evaluate the inner sum

$\text{(4d)}$: $(8)$ from this answer, where $A$ is the Glaisher-Kinkelin constant

$\text{(4e)}$: rearranging terms

By taking more terms in the Euler-Maclaurin expansions of $(6)$ and $(7)$ from the answer cited above, we get
$$
\begin{align}
&\sum_{k=1}^n\log(k!)\\
&=\frac12n^2\log(n)-\frac34n^2+n\log(n)+\left(\frac12\log(2\pi)-1\right)n+\frac5{12}\log(n)\\
&+\left(\frac1{12}+\frac12\log(2\pi)-\log(A)\right)+\frac1{12n}-\frac1{240n^2}-\frac1{360n^3}+\frac1{1008n^4}\\
&+\frac1{1260n^5}-\frac1{1440n^6}-\frac1{1680n^7}+\frac1{1056n^8}+\frac1{1188n^9}+O\!\left(\frac1{n^{10}}\right)\tag5
\end{align}
$$


问:
Orbits in graph
Can anyone help me visualize orbits in graph.

Let $G=K_4$ with vertex set $V=\{1,2,3,4\}$ and edge set $E\left(K_4\right)=\{12,13,14,23,24,34\}$

Now if i take permutation $\sigma=(1~2~3~4)\in Sym(V)$ ($Sym(V)$ is permutation group on $V$.

https://math.stackexchange.com/a/423948/520264 in this answer he says that if there is $k$-cycle in permutation then there is $\Big\lfloor\dfrac{k}{2}\Big\rfloor$ orbits of edges between vertices of $k$- cycle.

Here we have $4$-cycle in $\sigma=(1~2~3~4)$ by above formula there is $\Big\lfloor\dfrac{4}{2}\Big\rfloor=2$ orbits of edges between vertices of $4$-cycle.

Which are these two orbits?

答:
$k=4$ is a bit too small to visualize, so here is a picture for $k=8$:

Imagine that we are looking at a permutation $\sigma$ that permutes these $8$ vertices in a cycle, and we want a graph fixed by this permutation. (Maybe there are also other vertices we're not looking at, because we don't care what they do for the moment.) Then:

The $8$ red edges are an orbit of the action of $\sigma$ on $E(K_n)$. (They are either all present or all absent, if we want $\sigma$ to leave the graph unchanged.)

The $8$ blue edges are another orbit.

The $8$ purple edges are a third orbit.

The $4$ orange edges are a fourth orbit.

Essentially, if you rotate the picture above by $45^\circ$, then each edge is moved to another edge of the same color.

In general, there are $\lfloor \frac k2\rfloor$ orbits characterized by the length of an edge, when we draw the vertices in the cycle as a regular $k$-gon.

Title: Orbits in graph


答:
As $\sin y = -\sin x = \sin ( \pi + x)$, so must have
$$
y = 2n \pi + (\pi + x) = (2n+1)\pi + x
$$
for some integer $n$.

Now as $n$ varies over the set of integers, we obtain different straight lines, each of slope $1$, with $y$-intercept equal to $(2n+1) \pi$.

Hence the grid you got.

Title: How would you sketch $\sin y = - \sin x$?

Body:
I at first did $\sin^{-1}$ both sides to get $y = -x$.

This is wrong, can someone explain why?

The graph is made up of a lot of crosses and looks like a grid on desmos. I need a way to get the answer without using a graphing calculator, any help?


问:
Cubing a simple thing
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.

答:
Why is $(x^{2} + 4x + 4)(x+2)$ wrong?

$(x^{2} + 4x + 4)(x+2) = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8 = x^{3} + 6x^{2} + 12x + 8$

And it is the right answer. If you have problem remembering the power formulas for binomials, just use pascal's triangle to calculate the coefficients:

\begin{align*}
(a + b)^{0}\to                                 &1\\
(a + b)^{1}\to                              &1     &1\\
(a + b)^{2}\to                           &1     &2     &&1\\
(a + b)^{3}\to                       &1      &3     &&3     &&1\\
(a + b)^{4}\to                    &1     &4      &&6     &&4     &&1\\
\end{align*}
(and so on)

The pattern here is that every number is the sum of the two adjacent number from the previous line. These numbers are the coefficients of the nth power. The pattern of the exponents is:
$$
\sum_{i=0}^{n} a^{n-i} b^{i}
$$
Therefore:
$$
(x + 2)^{3} = (1)(x^{3})(2^{0}) + (3)(x^{2})(2^{1}) + (3)(x^{1})(2^{2}) + (1)(x^{0})(2^{3}) = x^{3} + 6x^{2} + 12x + 8
$$
By the way:

$(x^{2}+4x+4)(x^{2}+4x+4) = (x + 2)^{2} (x + 2)^{2} = (x+2)^{4} \neq  (x+3)^{3}$

Title: Cubing a simple thing


问:
Prove that : 1. $\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$ 2. $\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$
Given the function sequence $f_n(x)=ae^{-anx}-be^{-bnx}$ where $0<a<b$ Prove that :

$\sum_{n=1}^{\infty}\int_{0}^{\infty}|f_n|dx=\infty$

$\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=0$

The second part I've managed to prove by just integrating and then it's a sum of zeroes($\sum_{n=1}^{\infty}\int_{0}^{\infty}f_ndx=\sum_{1}^{\infty}(-\frac{e^{anx}}{n}+\frac{e^{-bnx}}{n})|_{0}^{\infty}=\sum_{n=1}^{\infty}0=0$ but for the first part I've got no idea. any hint please

答:
Hint: For each $n$ find the unique $x_n>0$ such that

$$\tag 1 ae^{-anx_n} = be^{-bnx_n}.$$

On $(x_n,\infty)$ the left side of $(1)$ is greater than the right side. Thus

$$\int_0^\infty|f_n| > \int_{x_n}^\infty|f_n| =\int_{x_n}^\infty f_n.$$


答:
Since this is olympiad training I'll just sketch the solution here.

Step 1

 Substitute $y=1/x$ in $f(xy)f(\frac{x+y}{xy})=f(x+y)$, we have $f(1)=1$ and hence inductively $f(n)=n$ for all positive integer $n$.

Step 2

 Substitute $y=1$ in $f(xy)f(\frac{x+y}{xy})=f(x+y)$ and eventually you get $f\left(\frac1x\right)=\frac1{f(x)}$.

Step 3

 Establish $f(a)f(b)=f(ab)$ for all $ab\geq\sqrt[3]{16}$, by considering $x+y=ab$, $xy=a$ or $xy=b$.

Step 4

 Combining all steps above, $f(a)f(b)=f(ab)$ for all $a,b$ (considering $ma, 1/m, nb, 1/n$ for large enough $m,n$) and $f(a+b)=f(a)+f(b)$.

Step 5

 Hence $f(x)=x$ for all $x$ is the only solution.

Title: A functional equation problem: $f(xy)f\left(\frac{x+y}{xy}\right)=f(x+y)$ and $f(x+1)=f(x)+1$

Body:
According to my friends, in the test of Moroccan team selection, no one solved the following problem :

Find all functions $f : \mathbb{R}^{+*} \to \mathbb{R}^{+*} $ such that :
$$f(x+1)=f(x)+1 \hspace{2cm} \text{and} \hspace{2cm} f(xy)f\left(\frac{x+y}{xy}\right)=f(x+y)$$
For all $x$ and $y$ in $\mathbb{R}^{+*}$

I followed the basic steps to solve generally a functional equation, but it seems that this one is very special since it needs to verify 2 conditions. Any thoughts ?


问:
Hadamard theorem and length of geodesics
I read  sevral times  in articles that:

"By Hadamard theorem, there is no closed  geodesics of length $2\pi$ in a space of sectional curvature $<0$?

Can someone give me a reference of this theorem, because i don't find it !

Thank you in advance

答:
The theorem is often called the Cartan-Hadamard Theorem.  See https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem

One version of the theorem is that if a space is complete and non-positively curved, e.g. a complete Riemannian manifold with sectional curvature less than or equal to zero or a complete geodesic metric space that is locally CAT(0), then its universal cover is globally non-positively curved, e.g. CAT(0).

Without passing to the universal cover, you can have closed geodesics of length $2\pi$.  For instance, the orbit space of the Euclidean plane under the action of translation by $2\pi$ has such geodesics.  So, it's important to include the hypothesis that the space is simply connected for statement you write above.


问:
Find area of triangle ABC
BD Perpendicular AC , AB =BC=a

Find the area of triangle ABC

I have tried Googling , I used formula 1/2 (base X Height) . Used Pythagorean theorem. Anyone can suggest me solution.

答: Let $y$ represent the distance $CD$. Then $y^2+x^2=a^2$, or $y=\sqrt{a^2-x^2}$.  Now you have enough to find the area of triangle $BCD$ using your area formula.  Double this to find the area of the whole triangle.</p>

Title: Find area of triangle ABC


问:
Formula for determinant
For $n \geq 2$, let $_$ be the $n \times n$ matrix defined by

$(G_n)_{ij} = 1 \text{ if } i = j$

$(G_n)_{ij} = 2 \text{ if } j = i + 1$ (mod n)

$(G_n)_{ij} = 0$ otherwise

I am told there is a general formula for det$(G_n)$ for $n \geq 2$. This is what I'd like to find.

I find that $G_2$ = $\begin{bmatrix} 1 & 2  \\ 2 & 1 \end{bmatrix}$ with det($G_2$) = -3.
$G_3$ = $\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2  \\ 2 & 0 & 1\end{bmatrix}$ with det(G_3) = 9

$G_4$ = $\begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 2 & 0 & 0 & 1\end{bmatrix}$ with det($G_4$) = -15

I can guess the general formula det$(G_n) = (-1)^{n+1}(2^n-1)$. How might I show that this is true for all n? As hint I was told to use the permutation definition of the determinant.

Any help would be appreciated.

edit:

Attempted proof:

I will use the fact that (1) If $\pi \in S_n$ is a permutation for which either $\pi(i) = i$ or $\pi(i) = i + 1 $(mod $n)$ holds for all i, then we either have $\pi(i) = i$ for all $i$ or $\pi(i) = i + 1 $(mod $n$) for all $i$.

Let det$(G_n) = \sum_{\pi \in S_n}sgn(\pi)\prod_{i=1}^nG_{i,\pi(i)}$.

Since $\pi(i) = i$ is only true for the identity map. The other n! - 1 permamutations must be of the form $\pi(i) = i + 1$.

We get det$(G_n) = sgn(id)(1)^n$ + $\sum_{\pi \in S_n - id }sgn(\pi)2^n = 1 - 2^n$. Note that the sum will be negative since we took out an even permutation and there will be one more odd permutation than even.

What have I messed up here?. This formula is correct for $G_2$ and $G_4$ but not $G_3$

答:
We have $G_n=I+2C$, where
$$
C=\pmatrix{0&1\\ &0&1\\ &&\ddots&\ddots\\ &&&\ddots&1\\ 1&&&&0}.
$$
By inspecting the entries of the powers of $C$, it is easy to see that $I,C,C^2,\ldots,C^{n-1}$ are linearly independent. Therefore the minimal polynomial of $C$ has degree $n$ and it coincides with the characteristic polynomial. It follows from the equality $C^n=I$ that the minimal and characteristic polynomials of $C$ are $p(x)=x^n-1$. Thus
$$
\det(G_n)=\det(I+2C)=(-2)^n\det\left(-\tfrac12I-C\right)=(-2)^np(-\tfrac12)=1-(-2)^n.
$$

Title: Formula for determinant


问:
Geometric proof on three colinear points
The following question consists of a final exam, which is originally written in Spanish, so please excuse me if something is lost in traslation.
The problem goes as follows: We start of with a rhombus ABCD counterclockwise, where we define point P variable on the diagonal AC. We are then defined point Q as the intersection between the circumscribed circles of ABP and CDP. We are asked to prove that D,Q and B are colinear, in other words, that Q belongs to the diagonal CD.

Ive tried to tackle the problem through congruent triangles, or angles and paralel lines, but I cant seem to notice were to make the breakthrough to the solution. The only major conclusion until now is that triangle BPD is isosceles.

答:
(1) Will $\angle 1 = \angle 2 =\angle 3$?

(2) Wll $\angle red = 90^0 - \angle 2 = 90^0 - \angle 3 = \angle purple$?

(3) Will the green marked angles equal?

If the answers are all yes, then you have all the necessary ingredients.


问:
Regular conditional probability and absolute continuity
Let's say I have two measure spaces $(X,\mathcal{F},\mu)$ and $(Y,\mathcal{G},\nu)$. I can consider the measure $\mu\times\nu$ over the product measurable space $(X\times Y, \mathcal{F} \times \mathcal{G})$. Let us also say that there is a coupling $\pi$ of $\mu,\nu$ such that $\pi \ll \mu\times\nu$. I.e., another probability measure over the same measurable space and such that $\pi(X\times A) = \nu(A), \pi(B\times Y) = \mu(B$).
From my understanding, if the spaces are Radon, I can define a regular conditional probability and disintegrate $\pi$ using $\mu$ and this conditional measure.
I am trying to show that $\mu\times\nu \ll \pi$. Indeed, let us call $\omega$ this conditional measure. I have that for a given measurable set $E$, $$(\mu \times \nu)(E) = \int_Y \mu(E_y) d\nu(y) = \int_X \nu(E_x)d\mu(x),$$ where $E_y = \{ x \in X : (x,y)\in E\}$ and $E_x$ is similarly defined.
Similarly, for $\pi$,
$$ \pi(E) = \int_Y \mu(E_y) dw(y|E_y).$$
If $\mu(E_y)=0$ for every $y$ then we clearly satisfy the absolute continuity constraint. Assume then that there exists a $y$ s.t. $\mu(E_y)\neq  0$  and that $\pi(E)=0.$ Using the disintegration in the other direction we also have that $$\pi(E) = \int_X \nu(E_x)d\xi(x|E_x).$$
Similar reasoning for $\nu(E_x)$.
If $\pi(E)=0$ then, denoting with $\hat{y}$ the element such that $\mu(E_\hat{y})\neq 0$  we must have that $\omega(\{\hat{y}\}|E_\hat{y})=0$. Is it possible that $\nu(\{\hat{y}\})\neq 0 $?

答: Correct me if I've misunderstood, but it seems to me like you're trying to prove that $\mu \times \nu \ll \pi$ for any coupling $\pi$ of $\mu$ and $\nu$.  But this is definitely false in general.  For example, let $\mu$ and $\nu$ both be Lebesgue measure on $[0,1]$, and $\pi$ be (normalized) Lebesgue measure on the diagonal $\{ (x,x) : 0 \leq x \leq 1 \}$.  Then it's clear that $\mu \times \nu$, which is $2$-dimensional Lebesgue measure, is not absolutely continuous with respect to $\pi$.


问:
Taylor series for tetration
I would like to know what is known about Taylor series for tetration (and other hyper-exponentiations).

Surprisingly, such information is rare on internet. Numerical values for expansion in hyperexponent for tetration with basis $e$ can be found here:

http://en.citizendium.org/wiki/Tetration#Taylor_expansion_at_zero

However I am interested in expansion in the basis of tetration. I look for

$$ x \uparrow \uparrow m = \sum_{n=0}^{\infty} c_n(m)x^n $$

Wolfram mathematica gives me wired result (obviously a bug) at $x=0$ for $x \uparrow 3$:

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)

where in the output $log(x)$ appears (not a polynomial).

It is better at $x=1$

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)+at+x%3D1

I imagine the point $x=0$ may be "peculiar", but at least expansion at $x=1$ should be possible.

List of my questions:

1) Is tetration analytic at $x=0$? At $x=1$?

2) If yes, is an explicit closed-form formula for $c_n(m)$ known? (at any of these points)

3) If not, is an explicit closed-form formula for $c_n(m)$ know for some specific values of $m$?

4) Same questions for extension to higher hyper-exponentiations...

Thank you.

答:
Consider the tetration of the function $e^x$
$$^n(e^x)=(e^x)_1^{(e^x)_2^{(e^x)_3^{_..^{.e^x_n}}}}$$

For a natural number n, the taylor series of that function is

$$^n(e^x)=\sum_{k}^{\infty}\frac{1}{k!}*T(n,k)*x^k$$

where $T(n,k)$ is the OEIS A210725 (if k<n, $T(n,k)=T(k,k)$)

then, for $n\in\mathbb{N}$:

$$^n(x)=\sum_{k}^{\infty}\frac{1}{k!}*T(n,k)*(ln(x))^k$$

On the limit for $n\to\infty$, the T(n,k)=T(k,k) are the coefficients of the Lambert w function.

There are other series, that you can find here.

For code able to calculate tetrations with high precision, (for a limited set of bases, and $n\in\mathbb{R}$) check Sheldonison's fatou.gp code

In Wolframalpha, you can get the series of $^n(x+1)$, and it will produce the Stirling transform of $^n(e^x)$. The series has power of x, and no logarithms.

Title: Taylor series for tetration


答:
With reference to comments: Let's be supportive.

If BD is a perpendicular, then you can split ABC into two right-angled triangles, so it is easier to work with:

Then you can use the formula $Area = \frac{1}{2}\cdot base \cdot height$ for each triangle, and add the two. You'll need to find the size of the base of each triangle using the pythagorean theorem.

Title: Find area of triangle ABC

Body:
BD Perpendicular AC , AB =BC=a

Find the area of triangle ABC

I have tried Googling , I used formula 1/2 (base X Height) . Used Pythagorean theorem. Anyone can suggest me solution.


问:
Leibniz formula for the nth derivative of $f(x)=x^{n-1} \ln x$
Problem : Calculate the derivative of the function $f: ]0,+\infty\left[\longrightarrow \mathbb{R}\right.$ defined by $f(x)=x^{n-1} \ln x$.

Solution
Let $g_{1}(x)=x^{n-1}$ et $g_{2}(x)=\ln x .$ So we have$f=g_{1} g_{2}$
for all $k$ verifying $1 \leq k \leq n-1$ we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$
The Leibniz formula gives

$\begin{array}{c}
f^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) g_{1}^{(k)}(x) g_{2}^{(n-k)}(x) \\
=x^{n-1} \frac{(-1)^{n-1}(n-1) !}{x^{n}}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) \cdots(n-1-k+1) x^{n-1-k} \frac{(-1)^{n-k-1}(n-k-1) !}{x^{n-k}} \\
=\frac{(-1)^{n-1}(n-1) !}{x}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(-1)^{n-k-1}(n-1) !}{x}\\
=-\frac{(n-1) !}{x} \cdot \sum_{k=0}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}
\end{array}$

The sum above is the same as $\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$ (prived of $k=n,$) which is equal to -1

We obtain
$$
f^{(n)}(x)=\frac{(n-1) !}{x}
$$

What I don't get:
What's the point of this step:

we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$

and which solution is wrong?

I spent a lot of time already on this, I think it's a matter of indices I can't point, both solutions seem correct to me.

How I calculated :

we have that

$\ln ^{(n)}(x)=\frac{(-1)^{n-1}(n-1) !}{x^{n}}$

and

$(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$

\begin{aligned}
f^{(n)} &=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \\
&=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! } \frac{(n-1-k) !(-1)^{n-1-k}}{x^{n-k}} \\
=& x^{-1} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) !(-1)^{n-k-1}
\end{aligned}

$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$

which should give me zero? since

$\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k} = (1+(-1))^{n}=0 $

答:
Your solution is almost correct (and neither $g_1^{(k)}(x)$ or $g_1^{(n)}(x)$ are wrong). The only problem is that $(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$ is only true for $0 \le k \le n-1$. For $k \ge n$, the $k$th derivative will be $0$. For example, the $2$nd derivative of $x^1$ is $0$.

The step
$$f^{(n)} =\sum_{k=0}^{n}\binom{n}{k}\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \tag 1$$

is right. But then at the next step, the sum should only go from $k = 0$ to $k = n-1$ since $\left(x^{n-1}\right)^{(n)} = 0$.

Then after that, the same logic applies, so you would end up with $$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n-1}\binom{n}{k}(-1)^{n-k} = \frac{(n-1)!}{x}$$


问:
Find scalars such that equation holds?
Given the equation:

$(2,-14,8)x+(-3,21,-12)y=\vec 0$

Find the scalars x and y, not both zero, such that the above equation holds.

My attempt:

I tried solving the equation system $2x-3y=0, -14x+21y=0, 8x-12y=0$ but I only obtained trivial solutions, i.e. $x=0, y=0$

Thanks for your help.

答:
Your effort makes a lot of sense! But you not only find the trivial solution: your system of equations is correct and it is equivalent to the only equation:
$$
2x−3y=0,
$$
whose solution is
$$
y=\frac{2}{3}x.
$$
So, all the ordered couples of the kind $(x,y)=(x,\frac{2}{3}x)$ are solutions.

The trivial solution $(0,0)$ is an example, other ones are $(1,\frac{2}{3})$ and $(3,2)$ or $(-3,-2)$ but one linear equation in two variables has an infinity of such ordered couples of values as solutions.

Title: Find scalars such that equation holds?


问:
Grothendieck on (topological) Chern Classes
I have been reading through the wikipedia article about Chern classes and it currently has a section devoted to the Alexander Grothendieck axiomatic approach. The language used throughout the section ("He shows using the Leray-Hirsch theorem (...)") seems to imply that there is a paper or a book written by Grothendieck himself containing the results mentioned.

Thus, my question is: does such a paper exist? Did Grothendieck ever write about the topological Chern classes?

I started my search by reading through a paper referenced in the wikipedia article, La théorie des classes de Chern and it very well may be what I am looking for. I can't tell,  because, being written in french, it is basically unreadable to me. Still, I have a feeling that this is not the right paper, because it doesn't reference the aforementioned Leray-Hirsch theorem and the notation used throughout highly suggests that this is the algebro-geometrical case.

答: That is the right paper! The "problem" is that he doesn't explicitly use Leray Hirsch, because he is constructing Chern classes for a suitable category of non-singular varieties over any algebraically closed field. In the paper, he essentially asserts Leray-Hirsch as Axiom A1 because that's really all he needs, but if you replace varieties over $k$ with complex manifolds and the chow ring $A(X)$ with the cohomology ring $\bigoplus_{i}H^{2i}(X,\mathbb{Z})$ you should be able to prove Axiom A1 using Leray-Hirsch, and that's the main use of the theorem in his approach to Chern classes. Chapter 4 of Bott and Tu follows Grothendieck's approach quite closely and I believe they do this proof.


问:
Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx} $
$\eqalign{
  & \int {{{\left( {\cot x - \tan x} \right)}^2}dx}   \cr
  &  = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx  \cr
  &  = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx  \cr
  &  = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx}   \cr
  &  = \int {{{\left( {2\cot 2x} \right)}^2}}   \cr
  &  = \int {4{{\cot }^2}2xdx}   \cr
  &  = \int {4\left( {{{\csc }^2}2x - 1} \right)dx}   \cr
  &  = \int {\left(4{{\csc }^2}2x - 4\right)dx}   \cr
  &  = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C  \cr
  &  =  - 2\cot 2x - 4x + C \cr} $

Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.

答:
$\tan x- \cot x = \frac{\sin x}{\cos x}-\frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = \frac{-\cos 2x}{\sin x \cos x} = \frac{-2 \cos2x}{\sin 2x} = -2 \cot2x$

Therefore you answer $-\cot x -4x + \tan x +c$ is correct.


问:
I have been trying this inequality for weeks and I just can't seem to solve it
If $a,b,c>0$ prove that $\frac{{(1+a)}^{2\ }{(1+b)}^2}{1+c^2}\ +\ \frac{{(1+b)}^2{(1+c)}^2}{1+a^2}\ +\frac{{(1+c)}^2{(1+a)}^2}{1+b^2}\ \geq\ 8(a+b+c)$
I have tried factorising, using the mean inequality, Schur's inequality, Cauchy's and even tried a trigonometric approach but nothing seems to work.

答:
I put the right side on the left and typed the function into Maple and then had Maple find the gradient of the function and then find all the points where the gradient was $0$.  It gave $(1,1,1)$, $(-1,1,1)$, $(1,-1,1)$, $(1,1,-1)$, and some complex roots.

All of these values make the inequality true, and the minimum of the function must occur at one of these four points.  (The quadratic terms all have positive coefficients, so the min exists and lives inside some ball containing the origin.)


问:
Seeking references on the geometry of regular four-dimensional polytopes
I've got intrested in the geometry of higher dimensional spaces. Now I'm studying four dimensional geometry. I'm especially interested in regular four dimensional polytopes and their visalization. I've found some intresting information here and here.

I am trying to find some literature focused on the geometry of the polytopes itself but not on the visalization only. Has someone been interested in the topic? What literature can you advice me to read?

答: I advice you to take a look at "Kaleidoscopes: Selected Writings of H.S.M. Coxeter" by F. Arthur Sherk, Peter McMullen, Anthony C. Thompson and Asia Ivic Weiss. Also you can check "Introduction to Geometry" by H.S.M. Coxeter (there is a chapter about four-dimensional geometry at the very end of the book). And there is also a paper called "Regular and semi-regular polytopes" by Irene POLO-BLANCO in which one can find a brief description of regular four-dimensional polytopes and some methods of studying them.</p>


问:
Show that $f(x)=x+\frac{1}{x^2+1}$ is strictly increasing.
Show that $f(x)=x+\frac{1}{x^2+1}$ is always increasing.

I noticed that $f'(x)=1-\frac{2x}{(x^2+1)^2}$, by looking at this function graph, I noticed it must be always positive, however I can't find a way to prove that. My attempt:

If $x\le0$, then clearly $f'(x)>0$. If $x\ge1$, it is also quite evident that $\frac{2x}{(x^2+1)^2}\le1$. Then, when $0<x<1$, I'm a bit lost. I can't seem to find an algebraic trick to show that $2x<(x^2+1)^2$, I'd love some help.

答:
To complete your argument:
$$
(x^2+1)^2=x^4+2x^2+1=(x^4+x^2)+(x^2+1).
$$
Now note that $(x^4+x^2)\geq 0$ and $x^2+1\geq 2x$ but equalities cannot happen simultaneously. This argument actually shows $2x<(x^2+1)^2$ always, so you don't need to split into cases.

You can also do a calculus-free proof:
$$
f(y)-f(x)=(y-x)\frac{(x-0.5)^2+(y-0.5)^2+0.5+x^2y^2}{(1+x^2)(1+y^2)}.
$$


答: In fact, if $\frac 12\leq x <1$, then $(2x) \leq (2x)^2 \leq (x^2+1)^2$. While if $0< x < \frac 12$, we clearly have $2x \leq 1 \leq (x^2+1)^2$. This is the simplest proof that I can come up with

Title: Show that $f(x)=x+\frac{1}{x^2+1}$ is strictly increasing.

Body:
Show that $f(x)=x+\frac{1}{x^2+1}$ is always increasing.

I noticed that $f'(x)=1-\frac{2x}{(x^2+1)^2}$, by looking at this function graph, I noticed it must be always positive, however I can't find a way to prove that. My attempt:

If $x\le0$, then clearly $f'(x)>0$. If $x\ge1$, it is also quite evident that $\frac{2x}{(x^2+1)^2}\le1$. Then, when $0<x<1$, I'm a bit lost. I can't seem to find an algebraic trick to show that $2x<(x^2+1)^2$, I'd love some help.


问:
What can we say about the product of elements from subgroups of the same order
Let $G$ be a finite non-abelian group and $A$ and $B$ be order $p$ subgroups of $G$, where $p$ is a prime number greater than $2$.

Since $A$ and $B$ are subgroups with order $p$, they are isomorphic to $\mathbb{Z}_p$. For any elements $a \in A$ and $b \in B$, can we say that the product $ab=ba$ (product is commutative)?

答: The symmetric group on four symbols has several subgroups of order $3$. Try some experiments.


Title: What can we say about the product of elements from subgroups of the same order

Body:
Let $G$ be a finite non-abelian group and $A$ and $B$ be order $p$ subgroups of $G$, where $p$ is a prime number greater than $2$.

Since $A$ and $B$ are subgroups with order $p$, they are isomorphic to $\mathbb{Z}_p$. For any elements $a \in A$ and $b \in B$, can we say that the product $ab=ba$ (product is commutative)?

Body:
Consider $A_4$, which has $A = \langle 1,2,3 \rangle$ and $B = \langle 2,3,4\rangle$ as order $3$ subgroups. But we see that
$$
(1,2,3)(2,3,4) = (1,2)(3,4) \neq (1,3)(2,4)=(2,3,4)(1,2,3)
$$


问:
log-normal distribution and standard normal distribution
Why is $(\log X - \mu)/σ \sim \mathcal N(0, 1)$ if $\log X \sim \mathcal N(\mu, σ^2)$?

答: Since if $Z \sim N(\mu, \sigma^2)$ then $\frac{Z - \mu}{\sigma} \sim N(0, 1)$, simply substitute $Z = \log X$ for the expression to hold.


问:
A question about the degree of an element over a field extension.
Say $K$ is a field extension of field $F$. If element $b$ is algebraic with degree $n$ over $F$, we know that $[F(b):F]=n$.
Why is it that $[K(b):K]\leq n$?</p>

答:
Recall the following facts: if $M/L$ is a field extension, then for any $a\in M$,
$$[L(a):L]=\deg(h)$$
where $h\in L[x]$ is the minimal polynomial of $a$ over $L$. The minimal polynomial of $a$ is the monic polynomial of smallest degree having $a$ as a root. Consequently, if $p\in L[x]$ is any other polynomial having $a$ as a root, we must have $h\mid p$.

Let $f\in F[x]$ and $g\in K[x]$ be the minimal polynomials of $b$ over the fields $F$ and $K$, respectively. Because $F\subseteq K$, we have that $f\in K[x]$. Because $f(b)=0$, we must have $g\mid f$, hence
$$[K(b):K]=\deg(g)\leq\deg(f)=[K(b):K].$$


问:
Cyclic Modules over a PID

Let $R$ be a PID and $M$ be an $R$-module.

If $M$ is finitely generated then show that $M$ is cyclic if and only if $M/PM$ is cyclic for every prime ideal $P$ of $R$.

Show that the previous statement need not be true if $M$ is not finitely generated.

This problem was given in an exam I took which I couldn't really tackle. Here cyclic module means a module generated by a single element.

Now for the first part, I think the main tool to apply here is the structure theorem of finitely generated modules over a PID. According to the theorem we can write $$M\cong R^n\oplus R/(a_1)\oplus \cdots \oplus R/(a_m)$$ where $a_i\in R, a_1|a_2|\dots |a_m$. Now since $R$ is a PID $P=(p)$ for some prime element $p$ of $R$. To prove the assertion I want to show $n=1$ and $m=0$. So if there is any $a_i$'s occurring then choosing a prime $q$ dividing $a_i$ and looking at $M/(q)M$ should give a module which is not cyclic. Thus we shouldn't have any $a_i$'s occurring, and then we are left with only $R^n$ but if the quotient module is cyclic then this forces $n=1$ because the quotient of $R^n$ should again be free on $n$ generators. But I'm having trouble writing down the argument, because even though I see intuitively quotienting by a prime $q$ dividing $a_i$ should give some nonzero module given by the quotient $N/(q)N$ where $N=R/(a_i)$. But how does one write this concretely? Any help would be appreciated.

About the second part, I have no idea. I tried to think of the simplest PID, $\mathbb{Z}$. I considered $\mathbb{Z}[X]$ the polynomial ring, which is an infinitely generated module over the integers. This is not cyclic clearly. But as an example this doesn't work as $\mathbb{Z}[X]/p\mathbb{Z}[X]$ is not cyclic as well. So I'd like some help finding a counterexample. But more importantly, how does one approach such a situation? I mean the counterexample I tried to think was because I just tried to simply negate all the conditions that work for the first part, namely finite generation. And I took a module which is generated by countably many things $1,X, X^2, \dots$ but that doesn't work, so how does one think of something like that?

答: I think this should be true for every ring $R$. More precisely... If $M$  is finitely generated then we can write $$M=Rx_1+\cdots + R x_n,$$ for suitable elements $x_i\in M$. We have a map $\phi\colon R\longrightarrow M$ taking $r\longmapsto r(x_1+\cdots + x_n)$. Observe that $M$ is cyclic if and only if $\phi$ is subjective, i.e. $\mathsf{Coker}(\phi)=0$. Now, we have a SES $$R\rightarrow M\rightarrow\mathsf{Coker}(\phi)\rightarrow 0,$$ so tensorizing by $-\otimes_RR/\mathfrak p$ we get the SES$$R/\mathfrak p\rightarrow M/\mathfrak p M\rightarrow\mathsf{Coker}(\phi)/\mathfrak p\mathsf{Coker}(\phi)\rightarrow 0.$$ By assumption, $M/\mathfrak p M$ is cyclic, so the map $R/\mathfrak p \rightarrow M/\mathfrak p M$ is subjective and $\mathsf{Coker}(\phi)=\mathfrak p\mathsf{Coker}(\phi)=0$. Now, thanks to Nakayama's Lemma, we get $\mathsf{Coker}(\phi)=0$.

Title: Cyclic Modules over a PID


答:
There is a common result stating that $\quad 2ab\le a^2+b^2\quad$  (just develop $(a-b)^2\ge 0$).

In particular for $a=1$ and $b=x$ you get $\, 2x\le 1+x^2$

And since $1+x^2\ge 1$ then $1+x^2\le (1+x^2)^2$ and your result is proved.


问:
Cubing a simple thing
I am trying to expand $\quad (x + 2)^3 $

I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.

I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this

$$(x^2 + 4x + 4)(x + 2)$$

This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with

$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$

Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.

答:
There is a more intuitive way to do this that works like the FOIL method.

Write $(x+2)^3$ as $(x+2)(x+2)(x+2)$.

When we multiply these we look at all the ways we can pick one element from each of the $(x+2)$s, multiply each possibility and add them all up.

What do these look like? Well, from each $(x+2)$ you can pick either an $x$ or a $2$. Since we pick one from each, we will get three things multiplied together.

If we pick an $x$ from each we will get an $x\cdot x\cdot x$ which equals $x^3$. We can only do this one way so we get $x^3$.

What if we pick an $x$ from only two of the $(x+2)$s? Then we have to pick a $2$ from the third. How many ways are there to do this? Well, we can pick an $x$ from the first two and a number from the third; we can pick an $x$ from the first and the third and a number from the second or we can pick an $x$ from the second and the third and a number from the first. This gives us $3(2x^2)=6x^2$. So far we have $x^3+6x^2$.

What if we pick only one $x$ and a $2$ from the other two? What do we get and how many ways are there to do this? We get $2\cdot 2\cdot x=4x$ and there are three ways to do this because we can pick the $x$ from either of the three. So we get $3(4x)=12x$ which gives us $x^3+6x^2+12x$ all together.

Are there any other possibilities? Yes, we can pick a number from each and there is only one way to do that so we get $2\cdot 2\cdot 2=8$ and finally we arrive at $(x+2)^3=x^3+6x^2+12x+8$.

This approach is easier than it seems if you practice a few and it works for $(x+a)^n$ for any positive n.

Title: Cubing a simple thing


问:
find all possible vectors, with a matrix
I don't know how to approach the next problem.

they give me

that are bases of a vector space V and let f: V → V be the linear transformation such that

All v∈V such that  are

they gave me these 4 answers:

I did several accounts but none of them gave me similar, then I saw that it says f (v) = f (v), does it mean that the transform gives a transform? I'm very confused

答:
$f(2v_1 + v_2 - v_3)$ is some vector. Let us call that $\mathbf w$. It turns out that $\mathbf w = 2v_1 + 3v_2 - v_3$ but we don't actually need to know it as anything more than "$\mathbf w$" for the solution.

Anyways, so we have an equation $f(\mathbf v) = \mathbf w$ and you will hopefully recall that the solution set of such an equation is a translate of the kernel (AKA null space) of $f$.

$$\{\mathbf v : f(\mathbf v) = \mathbf w\} = \mathbf v_0 + \ker f = \{\mathbf v_0 + \mathbf u : \mathbf u \in \ker f\} $$

where $\mathbf v_0$ is any particular solution to $f(\mathbf v) = \mathbf w$.

So for this problem we need two things:

we need $\mathbf v_0$

we need $\ker f$

For 1., we know already that $f(2v_1 + v_2 - v_3) = \mathbf w$ since that was the definition of $\mathbf w$. Therefore, we can take $\mathbf v_0 = 2v_1 + v_2 - v_3$ as a particular solution. This explains also why all but one of the proposed answers includes $2v_1 + v_2 - v_3$.

So now if $\ker f = \operatorname{Span}\{\mathbf u\}$ (which I'll leave up to you to solve), then the solutions to $f(\mathbf v) = \mathbf w$ is

$$\mathbf v = 2v_1 + v_2 - v_3 + \alpha \mathbf u$$

for some $\alpha \in \mathbb{R}$.

If you need a hint on solving for $\ker f$, notice that the first two columns of $M_{B',B}(f)$ are identical.


问:
If $A$ and $X/A$ are compact and $A$ closed, then $X$ is compact.
If $(X,\mathscr{T})$ is a compact topological space and $A\subset X$ is closed then we know that $A$ and $X/A$ are both compact. My question is, does the converse also hold? If $A$ and $X/A$ are compact and $A$ closed, does it imply that $X$ is compact? I believe the answer is yes. If $\mathscr{C}$ is an open cover of $X$,  then we only have to show that there is a finite subset of $\mathscr{C}$ whose elements form a union which contains $X/A$. But I am not sure how to do this

答:
Yes, $X$ is compact.

Let $\mathscr{U}$ be an open cover of $X$. $A$ is compact,so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ that covers $A$. Let $U_0=\bigcup\mathscr{U}_0$, and let

$$\mathscr{V}=\{U_0\}\cup\{U\setminus A:U\in\mathscr{U}\setminus\mathscr{U}_0\}\,.$$

The image of $\mathscr{V}$ under the quotient map is an open cover of $X/A$, so it has a finite subcover, which pulls back to yield a finite $\mathscr{V}_0\subseteq\mathscr{V}$ covering $X$. $\mathscr{U}_0\cup(\mathscr{V}_0\setminus\{U_0\})$ is then a finite open refinement of $\mathscr{U}$, which therefore has a finite subcover.


问:
Prove that $p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\pmod{\! 4}$
Let a prime number $p$ divide $a^2+b^2$ with some  $a,b \in \left\{ 1,2, \ldots , p-1 \right\}$ Prove that  $p\equiv 1 \pmod{4}$. Is the converse true?

I know that $a^2+b^2\equiv 0 \pmod{p}$ and I don't know.

答:
Proof by exhaustion.

$$
p = a^2 + b^2 = 3 \pmod 4 \implies
$$

If

$$
p = 1^2 + 1^2 = 2, \\
p = 1^2 + 2^2 = 1, \\
p = 1^2 + 3^2 = 2, \\
p = 2^2 + 2^2 = 0, \\
p = 2^2 + 3^2 = 1, \\
p = 3^2 + 3^2 = 2, \\
\pmod 4
$$
Then we have a contradiciton.  By symmetry of variables $a,b$ (namely commutativity of $+$) we have covered all possibilities.

$\blacksquare$

The other proofs are impossible to understand easily.  This one is plain and simple.


问:
Determine if the following integral is convergent or not
I want to determine the convergence of the following improper integral
$$
I := \int_{0}^{+\infty}\frac{\sin(x)}{x^2-x}dx.
$$
What I did is to rewrite the integral as
$$
\int_{0}^{+\infty}\frac{\sin(x)}{x^2-x}dx = \int_{0}^{1}\frac{\sin(x)}{x^2-x}dx + \int_{1}^{2}\frac{\sin(x)}{x^2-x}dx + \int_{2}^{+\infty}\frac{\sin(x)}{x^2-x}dx =: I_{1} + I_{2} + I_{3}.
$$
It's easy to see that $I_{3}$ converges, $I_{1} = -\infty$, $I_{2} = +\infty$, therefore I can't deduce anything, since $I_{1}$ and $I_{2}$ diverge with different signs. Is there another way of doing it? Thanks in advance!

答:
Let $$f(x)=\frac{\sin(x)}{x^2-x}$$

$ f $ is not  locally integrable at $ (0,+\infty)$, so we cannot write $$\int_0^{+\infty} f $$

If you want

$ I_1=\int_0^{\frac 12}f $ converges because $ \lim_{x\to0^+}f(x)=-1$.

$ I_2=\int_{\frac 12}^1f $ diverges because
$$f(x)\sim \frac{\sin(1)}{x-1}\; \;(x\to 1)
$$

$ I_3=\int_2^{+\infty} f$ is absolutely convergent since, $$|f(x)|\le \frac{1}{x^{\frac 32}}$$


问:
When is it possible to interchange independent (Riemann or Lebesgue) integrals and limits?
Specifically, if we have a function of two variables $f(x,y)$, does the following identity hold?
$$\lim_{x\to c}\int_a^bf(x,y)\,dy=\int_a^b\lim_{x\to c}f(x,y)\,dy$$
Under what conditions is this true? Is it enough to say that both the integral and the limit exist for this exchange to be valid? Is this true when integrating over more general sets, not necessarily a compact 1-dimensional interval?

答:
The Dominated Convergence Theorem gives a partial answer:

Let $(X,\mathfrak X,\mu)$ be a measure space, $(Y,d)$ be a metric space, and $f:X\times Y\to \mathbb C$ a function such that $x\mapsto f(x,y)$ is intregrable for every $y\in Y$. Suppose that there is some integrable $g:X\to \mathbb C$ such that, for every $y\in Y$, $|f(x,y)|\le g(x)$ for almost every $x\in X$. Let $y_0\in Y$. Then
$$
\lim_{y\to y_0} \int_X f(x,y)d\mu(x) =\int_X f(x,y_0)d\mu(x)
$$

To prove this, note that $y\mapsto \int_X f(x,y)d\mu (x)$ defines a function $h:Y\to\mathbb C$, and we need to study $\lim_{y\to y_0} h(y)$. This can be done with sequences.

I suppose you could relax this to $(Y,\tau)$ being a topological space with some decent countability assumptions, although I've never needed something of such generality.

Going a different route:

Suppose that $(X,d_X)$ and $(Y,d_Y)$ are metric space, $f:X\times Y\to \mathbb{C}$ is uniformly continuous (with respect to the metric $d_{X\times Y}((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$), and $\mu$ is some finite Borel measure on $X$. Suppose also that $x\mapsto f(x,y)$ is integrable (on $X$) for every $y\in Y$. Let $y_0\in Y$. Then
$$
\lim_{y\to y_0} \int_X f(x,y)d\mu(x) =\int_X f(x,y_0)d\mu(x)
$$

To see this, fix $\varepsilon>0$. Pick $\delta>0$ witnessing the uniform continuity of $f$. If $d_Y(y,y_0)<\delta$, then $d_{X\times Y}((x,y),(x,y_0))=d_Y(y,y_0)<\delta$, so that $|f(x,y)-f(x,y_0)|<\varepsilon$ for all $x\in X$. But then
$$
\left| \int_X f(x,y)d\mu(x)-\int_X f(x,y_0)d\mu(x)\right|\le \int_X |f(x,y)-f(x,y_0)|d\mu(x)\le \varepsilon \mu(X)
$$

A very common occurrence of this phenomenon is in PDEs, when trying to study the limit of the integral of some function on the boundary of a bounded subset of Euclidean space, say after an application of the Gauss-Green Theorem.

A less-sharp version of the previous result is when $f$ is differentiable with bounded Jacobian. Then you can use the Mean Value Theorem to bound the differences $f(x,y)-f(x,y_0)$.


问:
Interpretation of a certain general theorem used by Gauss in his work on theta functions.
I'm trying to understand the meaning of a general proposition stated by Gauss in a posthomous paper (this paper is in pp. 470-481 of volume 3 of Gauss's werke) on theta functions, a proposition which seems to serve as a guiding and organizing principle of the vast amount of relations among theta functions that he found.

Gauss's notation and definitions

Denote by $P(x,y),Q(x,y),R(x,y)$ the following functions:

$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$
$$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$
$$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$

These functions include Jacobi theta functions in their usual meaning as special cases; if $y$ is a complex number whose absolute value is $1$, and $z$ is defined to be a real number such that $y = e^{2iz}$, then we have:

$$P(x,y)=1+2cos(2z)x+2cos(4z)x^4+2cos(6z)x^9+...=\vartheta_3(z,x)$$

which follows from the identity $cos(2nz)= \frac{e^{2inz}+e^{-2inz}}{2}$. In paticular, we have:

$$P(x,1)=1+2x+2x^4+2x^9+...=\vartheta_3(0,x)$$,
So one can understand $P(x,y),Q(x,y),R(x,y)$ as a generalization of Jacobi theta function $\vartheta(z,x)$ from purely real $z$ to a complex $z$ (non-zero imaginary part of z), so that $|y| \ne 1$.

Remark: I'm not very familiar with Jacobi's publications, so it's quite possible that Jacobi's original definition of his theta functions includes also the case when $z$ is complex, so Gauss's functions $P(x,y),Q(x,y),R(x,y)$ are nothing else than simply Jacobi's theta functions with different notation.

Gauss's theorem

On August 6, 1827, Gauss stated the following "general theorem":

$$P(x,ty)\cdot P(x,\frac{y}{t}) = P(x^2,t^2)P(x^2,y^2) + R(x^2,t^2)R(x^2,y^2) $$

and then goes on to derive a multitude of relations from it.

For more comprehensive background on this question, please look at the answer to HSM stackexchange post https://hsm.stackexchange.com/questions/6256/did-gauss-know-jacobis-four-squares-theorem.

Therefore, i'd like to know how to interpret the general theorem stated by Gauss.

答:
The definition of the Gauss theta functions can be written as

$$ P(x,y) = \sum_{n\in\mathbb{Z}} x^{n^2}y^n,\;\;
   R(x,y) = \sum_{n\in\mathbb{Z}+\frac12} x^{n^2}y^n. \tag{1} $$

Now consider the product of two theta functions

$$ S := P(x,ty)\cdot P(x,y/t) = \left(\sum_{n\in\mathbb{Z}} x^{n^2}(ty)^n\right) \!
   \left(\sum_{m\in\mathbb{Z}} x^{m^2}(y/t)^m\right). \tag{2} $$

This can be rewritten as a double sum

$$ S = \sum_{n,m\in\mathbb{Z}} x^{n^2+m^2} y^{n+m}t^{n-m}. \tag{3} $$

Rewrite this using new variables

$$ j = \frac{n+m}2,\;\; k = \frac{n-m}2 \;\; \text{ where }
  \;\; n = j+k,\;\; m = j-k \tag{4} $$

to get

$$ S = \sum_{n,m\in\mathbb{Z}} x^{2(j^2+k^2)} y^{2j}t^{2k}. \tag{5} $$

The double sum $\,S\,$ splits into two cases. One is
$\,S_0\,$ where $\,n,m\,$ have the same parity with
$\,j,k\in\mathbb{Z}.\,$
The other is
$\,S_1\,$ where $\,n,m\,$ have different parity with
$\,j,k\in\mathbb{Z}+\frac12.\,$
Rewrite the sums as products

$$ S_0 = \sum_{j,k\in\mathbb{Z}} (x^2)^{k^2}(t^2)^k \cdot
(x^2)^{j^2}(y^2)^j = P(x^2,t^2)P(x^2,y^2) \tag{6} $$

and

$$ S_1 = \sum_{j,k\in\mathbb{Z}+\frac12} (x^2)^{k^2}(t^2)^k \cdot
(x^2)^{j^2}(y^2)^j = R(x^2,t^2)R(x^2,y^2). \tag{7} $$

The end result is

$$ S = S_0+S_1 = P(x^2,t^2)P(x^2,y^2) + R(x^2,t^2)R(x^2,y^2). \tag{8} $$

I think that this is similar to what Gauss' original proof was but
I have no way to know that. This approach must be very old.


问:
The variation of a function as a function
Let $f:\mathbb{R}\rightarrow\mathbb{R^*}$ be a continuous function (when $\mathbb{R^*}=[-\infty,\infty]=2$ points compactification of $\mathbb{R}$, this is not so much necessary for the question).

Let $g(x)=T_x^b(f)$ be the total variation of $f\mid_{[x,b]}$, $g$ is monotonic function, hence $g$ is differentiable almost everywhere, and in particular g is continuous almost everywhere, which means $g$ is preserving convergence almost everywhere, but is $g$ preserving convergence $\textbf{everywhere}?$

But because $f$ is uniformly continuous (continuous on a close interval),

which means $$\forall\epsilon>0 \exists\delta>0:\mid x-y\mid<\delta\Rightarrow\mid f(x)-f(y)\mid<\epsilon$$ and because $$T_x^b(f)=\underset{P}{sup}\Sigma_{i=0}^{n_P}\mid f(x_i)-f(x_{i-1})\mid$$ when $$P_{n_P}:=\{x=t_0< t_1<...<t_{n_P}=b\}$$

It led me to think that $g$ is preserving convergence $\textbf{everywhere}.$

My attempt was to fix a small $\epsilon$, and build a partition $P$ such that the mesh of $P$ is the $\delta$ that derives from this $\epsilon$, but it is not enough to place in the definition of $T_x^b(f)$.

If $g$ is not always preserving convergence for $f\in C([a,b])$, is it preserving convergence for $f\in AC([a,b])?$

Thanks in advance.

答:
Hint

Try $f(x)=xsin(\frac{1}{x})$ with $0$ at $x=0$. Then for every $\epsilon>0$ $T_0^\epsilon(f)=\infty$ but $T_0^0(f)=0$.


问:
Another Information Theory Riddle
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.

In a magic trick, there are three participants: the magician, an assistant, and a volunteer.

The assistant, who claims to have paranormal abilities, is in a soundproof room. The magician gives the volunteer six blank cards, five white and one blue. The volunteer writes a different integer from 1 to 100 on each card, as the magician is watching. The volunteer keeps the blue card. The magician arranges the five white cards in some order and passes them to the assistant.

The assistant then announces the number on the blue card.

How does the trick work?

答:
Using @vadim123 's orientation trick, you can encode 2 bits of information per card, regardless of the number written on them and the ordering of the cards. $100<2^7$, so only 3 cards + $Boolean(Sorted) $ suffice for the encoding. :P

I wonder what is the encoding that absolutely minimizes the "resource utilization"... (I wish I had the rep. to start a bounty) :P

Title: Another Information Theory Riddle


答:
here $z$-score table I look for $\frac{106.75-100}{25}=0.27$,
$z$-score is $0.6064$

and for $\frac{95-100}{25}=-0.2$,  $z$-score is $0.4207$

So  $P(95\leq X\leq 106.75)=0.6064-0.4207=0.1857$

$P(X>92)=1-P(X\leq 92)=0.6255$

For the second part you do an inverse research on that table until you see a value close to $0.1$, that is $0.1003$ which is related to $-1.28$

Therefore $\frac{x-100}{25}=-1.28\to x = 68$

Hope this helps

Title: $X:N(100,25)$, Find: $P(95\leq X\leq 106.75)$, $P(X>92)$ and $P(X\leq x) = 10\%$

Body:
$X$ is normally distributed random variable with mean $100$ and variance $25$. $X:N(100,25)$.

a) Calculate: $P(95\leq X\leq 106.75)$ and $P(X>92)$

b) Find $x$ such that $P(X\leq x) = 10\% $

This is how I solved it:

a)

$P(95\leq X\leq 106.75)=P(\frac{95-100}{5} \leq X^*\leq \frac{106.75-100}{5})=\Phi(1.35)-\Phi(-1)=0.4115 + 0.3413=0.7528$

Z-score: $z(92)=\frac{92-100}{5}=-1.6$

$P(X>92)=\Phi(-1.6)+0.5=0.4452+0.5=0.9452$

b)

$P(X\leq x) = 10\% $

$P(X\leq x) = 0.1 $

Z-score: $z(x)=\frac{x-\mu}{\sigma}=0.1$ and from this expression I get that $x=100.5$

Would appreciate if someone could check if this is correct..


问:
find a third vector to form a set that spans $\mathbb R^{3}$
i have two vectors that i need to combine with a third unknown vector to create a spanning set, but i can't figure out how to do this. i have written the two vectors i have below. this is from a singular value decomposition problem.

$u_1 = (\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$

$u_2 = (\frac{-1}{3\sqrt{2}}, \frac{-1}{3\sqrt{2}}, \frac{-4}{3\sqrt{2}})$

i am trying to solve a problem based on an example in my e-text, but they kind of skipped over this part and i can't figure out how to get this similar result for the vectors i'm working with. i have $u_1$ and $u_2$, but i can't figure out how to get $u_3$.

any advice? thank you in advance!

答:
Since $u_1$ and $u_2$ as given are unit vectors that are perpendicular to each other

(their dot product is $0$),

their vector cross product will be a unit vector perpendicular to them, which is what is sought.


问:
Prove $\lambda$ is eigenvalue of $A$ iff $\lambda^*$ is eigenvalue of $A^\dagger$ with kernel
I don't know how to prove the statement " $\lambda$ is an eigenvalue of $A$ iff $\lambda^*$ is an eigenvalue of $A^\dagger$" being $A$ an operator in a complex, finite Hilbert space. And only using kernels, images, orthogonality, and inner product concepts to prove it (can't use dimensions).

I think it has to be something similar to the following one:

$$Ax=\lambda x \xrightarrow[]{} (A-\lambda I)x=0$$

Where $x$ is the eigenvector, and then:

$$\ker(A-\lambda I)=(Img(A-\lambda I)^{\dagger})^{\perp} = Img(A^{\dagger}-\lambda^* I)^{\perp}$$

But at that point I get stuck, any hints about how to continue?

Thank you so much!

答:
You're almost there. Think about the dimension of the space
$$\text{ker}(A - \lambda I) = \text{Img}(A^\dagger - \lambda^* I)^\perp$$
We know that it's dimension is nonzero, so in particular this means that the dimension of the $\text{Img}(A^\dagger - \lambda^* I)$ must be strictly less than $n$, which is the order of $A$ and $A^\dagger$. But from rank-nullity, this would imply that
$$\text{dim}(\text{ker}(A^\dagger - \lambda^* I)) = n - \dim(\text{Img}(A^\dagger - \lambda^* I)) > 0$$
which would imply that...