问:
Computing random numbers from an arbitrary (non-Gaussian) multivariant CDF
As part of a test harness I need to create random pairs of numbers according to a specified 2D CDF (Cumulative Distribution Function).  The CDFs change over time and are the result of some opaque (i.e., ML (Machine Learning) like) process.  The CDFs are, however, numerically well behaved, e.g., they are efficiently integrated and although differentiation has accuracy issues these are not insurmountable.  The real application may have irregular domains, but I believe this part of the problem is separable.

So lets assume what seems to be the simplest form of the problem, generate pairs $(y_0,y_1)$ over the region $(y_0,y_1) \in [0,\ 1] \times [0,\ 1]$ such that $P_y(y_0,\ y_1) = y_0 + y_1$ .  The key point is that $P_y$ is not factorable so it can't be represented as the product of independent random variables.

So it seemed that a simple yet generalizable approach was to start by defining the CDF in the standard way
$$C_y(y_0,y_1) = \int_0^{y_0} \int_0^{y_1} P_y(y_0,y_1)\ dy_0\ dy_1$$
and generate independent uniform random variables $x_0$ and $x_1$ on the interval $[0,1]$.  The first major step is solving for $Y_0$ such that $C_y(Y_0,1) = x_0$; this is, equivalent to solving for $Y_0$ such that the for the random variable $y_0$, $Prob(y_0 < Y_0) = x_0$.  The second major step is using this value to solve for $Y_1$ such that
$$\frac{C_y(Y_0,y_1)}{C_y(Y_0,1)} = x_1$$
this is equivalent to finding $Y_1$ such that $Prob(y_1 | Y_0 < Y_1) = x_1$.

This result fools the eye (it looks good), but it doesn't check numerically, which caused a subtle bug in the larger test system.  I assumed a numerical bug, which I was promptly unable to find.  Then I discovered that it doesn't check symbolically either.

For the assumed PDF (Probability Density Function) $C_y = 1/2\cdot (y_0^2\ y_1 + y_0\ y_1^2)$.  So
$$Y_0 = \frac{\sqrt{1+8\ X_0}-1}{2} $$
and
$$Y_1 = \frac{\sqrt{4x_1\ Y_0 + Y_0^2 + 4x_1} - Y_0}{2} $$

Fully expanding $Y_1$ yilds a big mess
$$
Y_1 =
  \frac{1}{4} -
  \frac{\sqrt {1+8x_0}}{4} +
  \frac{\sqrt {4x_1\, \left( -\frac{1}{2} + \frac{\sqrt {1+8x_0}}{2} \right) +
    \left( -\frac{1}{2} + \frac{\sqrt{1+8x_0}}{2} \right)^{2}+4x_1}}{2}
$$

Fortunately, this simplifies like crazy when solving for the inverse (in just a second).  Let $W(x_0,x_1) = (Y_0,Y_1)$ then
$$
W^{-1}\left( Y_0,Y_1\right) = \left(\frac{Y_0^2 + Y_0}{2},
  \frac{Y_1\ (Y_0 + Y_1)}{Y_1 + 1}\right)$$

Computing the Jacobian (not sure how much detail I should include here) yields
$$ J = \frac{(2Y_0 + 1)(Y_0+2Y_1)}{2(Y_0 + 1)} $$
and because $P_x = 1$ for all $x_0$ and $x_1$ the Jacobian is the joint PDF of $(Y_0,Y_1)$ produced by this procedure.  It turns out to be a somewhat decent approximation of the CDF (only off by 4% at the worst point), but it is not exact.

I think the procedure is theoretically flawed but:  1) I'm having trouble articulating the problem and 2) I don't know what procedure I should use instead.

答:
I'm used to write $f$ for densities and $F$ for cumulative distribution functions (cdf), I hope that won't cause confusion. As you wrote, we try to get our sample from independent random variables, uniformly distributed in $[0,1]$. For one-dimensional distributions, this is simple: if $Y$ has the (continuous) cdf $F$, then $X=F(Y)$ will be uniformly distributed in $[0,1]$. The other way round, we can find a variable $Y$ with cdf $F$ as $Y=F^{-1}(X)$.

Now, let's assume we have a two-dimensional distribution with joint density $f_{Y_0, Y_1}(y_0,y_1)$. Then, we can write it as a product of conditional densities,
$$f_{Y_0, Y_1}(y_0,y_1)=f_{Y_0}(y_0)\,f_{Y_1}(y_1|y_0),\tag{1}$$ where $f_{Y_0}$ is the density of the marginal distribution, $$f_{Y_0}(y_0)=\int^\infty_{-\infty}f_{Y_0, Y_1}(y_0,y_1)\,dy_1.\tag{2}$$
For your simple model, we have to integrate only from $0$ to $1$, since the density is $0$ outside $[0,1]$, i.e. from $f_{Y_0, Y_1}(y_0,y_1)=y_0+y_1$ for $(y_0,y_1)\in[0,1]^2$, we obtain $$f_{Y_0}(y_0)=y_0+1/2.$$ The corresponding cdf is $$F_{Y_0}(y_0)=(y^2_0+y_0)/2,$$ so our first equation is $$X_0=(Y^2_0+Y_0)/2.\tag{3}$$ It's elementary to solve for $Y_0$:
$$Y_0=\sqrt{2\,X_0+1/4}-1/2.\tag{4}$$ From (1), we obtain the conditional density
$$f_{Y_1}(y_1|y_0)=\frac{y_0+y_1}{y_0+1/2},$$  giving the conditional cdf $$F_{Y_1}(y_1|y_0)=\frac{y_0y_1+y^2_1/2}{y_0+1/2}$$ for $y_1\in[0,1]$. So our second equation is $$X_1=\frac{Y_0Y_1+Y^2_1/2}{Y_0+1/2},\tag{5}$$ which can be elementarily solved for $Y_1$: $$Y_1=\sqrt{Y^2_0+X_0(2\,Y_0+1)}-Y_0.\tag{6}$$ Looking at (4) and (6), it's not obvious that $Y_0$ and $Y_1$ have the same distribution, but it's true. The Jacobian we get from (3) and (5) is exactly $Y_0+Y_1$, that's just equation (1).

I've checked the results numerically: from your density, we conclude $\mathbb{E}Y^2_0=\mathbb{E}Y^2_1=5/12$ and $\mathbb{E}Y_0Y_1=1/3$. Generating a sample of $10^8$ pairs $(y_0,y_1)$ is obvious with code like

```<code>fun pair(): Pair<Double, Double> {
    val x0 = Random.nextDouble()
    val x1 = Random.nextDouble()
    val y0 = sqrt(2.0 * x0 + 0.25) - 0.5
    val y1 = sqrt(y0 * y0 + x1 * (2.0 * y0 + 1.0)) - y0
    return Pair(y0, y1)
}
</code>```

That gives estimates for the moments mentioned above like:

```<code>0.4166417030452688
0.4166630593865879
0.3333395345332752
</code>```

The accuracy was to be expected. In your real application, the cdf will be more complicated, and inverting them may be a computational/numerical problem, but that should be a solvable problem.

Body:
I'm used to write $f$ for densities and $F$ for cumulative distribution functions (cdf), I hope that won't cause confusion. As you wrote, we try to get our sample from independent random variables, uniformly distributed in $[0,1]$. For one-dimensional distributions, this is simple: if $Y$ has the (continuous) cdf $F$, then $X=F(Y)$ will be uniformly distributed in $[0,1]$. The other way round, we can find a variable $Y$ with cdf $F$ as $Y=F^{-1}(X)$.

Now, let's assume we have a two-dimensional distribution with joint density $f_{Y_0, Y_1}(y_0,y_1)$. Then, we can write it as a product of conditional densities,
$$f_{Y_0, Y_1}(y_0,y_1)=f_{Y_0}(y_0)\,f_{Y_1}(y_1|y_0),\tag{1}$$ where $f_{Y_0}$ is the density of the marginal distribution, $$f_{Y_0}(y_0)=\int^\infty_{-\infty}f_{Y_0, Y_1}(y_0,y_1)\,dy_1.\tag{2}$$
For your simple model, we have to integrate only from $0$ to $1$, since the density is $0$ outside $[0,1]$, i.e. from $f_{Y_0, Y_1}(y_0,y_1)=y_0+y_1$ for $(y_0,y_1)\in[0,1]^2$, we obtain $$f_{Y_0}(y_0)=y_0+1/2.$$ The corresponding cdf is $$F_{Y_0}(y_0)=(y^2_0+y_0)/2,$$ so our first equation is $$X_0=(Y^2_0+Y_0)/2.\tag{3}$$ It's elementary to solve for $Y_0$:
$$Y_0=\sqrt{2\,X_0+1/4}-1/2.\tag{4}$$ From (1), we obtain the conditional density
$$f_{Y_1}(y_1|y_0)=\frac{y_0+y_1}{y_0+1/2},$$  giving the conditional cdf $$F_{Y_1}(y_1|y_0)=\frac{y_0y_1+y^2_1/2}{y_0+1/2}$$ for $y_1\in[0,1]$. So our second equation is $$X_1=\frac{Y_0Y_1+Y^2_1/2}{Y_0+1/2},\tag{5}$$ which can be elementarily solved for $Y_1$: $$Y_1=\sqrt{Y^2_0+X_0(2\,Y_0+1)}-Y_0.\tag{6}$$ Looking at (4) and (6), it's not obvious that $Y_0$ and $Y_1$ have the same distribution, but it's true. The Jacobian we get from (3) and (5) is exactly $Y_0+Y_1$, that's just equation (1).

I've checked the results numerically: from your density, we conclude $\mathbb{E}Y^2_0=\mathbb{E}Y^2_1=5/12$ and $\mathbb{E}Y_0Y_1=1/3$. Generating a sample of $10^8$ pairs $(y_0,y_1)$ is obvious with code like

fun pair(): Pair<Double, Double> {
    val x0 = Random.nextDouble()
    val x1 = Random.nextDouble()
    val y0 = sqrt(2.0 * x0 + 0.25) - 0.5
    val y1 = sqrt(y0 * y0 + x1 * (2.0 * y0 + 1.0)) - y0
    return Pair(y0, y1)
}

That gives estimates for the moments mentioned above like:

0.4166417030452688
0.4166630593865879
0.3333395345332752

The accuracy was to be expected. In your real application, the cdf will be more complicated, and inverting them may be a computational/numerical problem, but that should be a solvable problem.


问:
inverting a percentage (basically inverting the percentages in pie chart)
This is a weird question and I am not sure if it makes sense.
Say we have 4 distances (where these are 1 dimensional distances):

```<code>5
3
8
6
</code>```

so the total distance => SUM(5,3,8,6) = 22.

So for the first distance, it represents 3/22 of the total distance, which could be converted to a percent. But in my case, the shortest distance actually means a higher percentage, and a longer distance means a lower percentage.

Originally percentages would be:

```<code>(5/22)*100
(3/22)*100
(8/22)*100
(6/22)*100
</code>```

Is there a way to invert this somehow mathematically, so that a smaller fraction means a bigger percent, something like:

```<code>(22/5)/100
(22/3)/100
(22/8)/100
(22/6)/100
</code>```

however, this is not right, since there is an obvious smell, in the case of 22/x being bigger than 100. How is this done?

答:
I do not think there is a unique way to do this. For instance, you can consider the "remaining" distance:
$$5 \quad\longrightarrow\quad 22-5=17 \quad\longrightarrow\quad \frac{17}{66}\cdot 100 = 25.76\%,$$
$$3 \quad\longrightarrow\quad 22-3=19 \quad\longrightarrow\quad \frac{19}{66}\cdot 100 = 28.78\%,$$
$$8 \quad\longrightarrow\quad 22-8=14 \quad\longrightarrow\quad \frac{14}{66}\cdot 100 = 21.21\%,$$
$$6 \quad\longrightarrow\quad 22-6=16 \quad\longrightarrow\quad \frac{16}{66}\cdot 100 = 24.24\%,$$
where $66=17+19+14+16$.

ContentLicense: CC BY-SA 4.0


问:
Prove that $\frac{d}{dx}e^x=e^x$
I would like to know how to prove that
$$\frac{d}{dx}e^x=e^x$$
An argument I've seen used is that this is the defnition of the number $e$; we can say that for some number, let's call it $a$, the derivative of $a^x$ is $a^x$ and we've just called this number $e$. But a problem I have with this approach is that we are assuming that there exists a number $a$ that satisfies this property; how do we know it exists in the first place?

I've also seen an approach used that I think is very circular in its reasoning; they use the Maclaurin expansion of $e^x$ to prove by first principles that $\frac{d}{dx}e^x=e^x$. But the derivation of this series makes use of the very fact that $\frac{d}{dx}e^x=e^x$.

Any other proof will be very appreciated; an explanation of the $2$ above proofs would also be very useful.

Thank you for your help.

答:
The first time we are acquainted with exponentiation is with the idea of it as repeated multiplication. For $n \in \mathbb{N}$, we may define
$$
a^n=\underbrace{a \cdot a \cdot a \cdot \ldots \cdot a}_{\text{$n$ times}} \, .
$$
This helps us establish an important principle about exponentiation,
$$
a^{x+y}=a^x \cdot a^y \, ,
$$
which will guide us in how we define $a^n$ when $n\not\in\mathbb{N}$. To begin with, $a^0$ should be equal to $1$, since
$$
a^0\cdot a^n=a^n \implies a^0=\frac{a^n}{a^n}=1 \, .
$$
For the same reason, $a^{-n}$ should be equal to $1/a^n$. If $n$ is a rational number $p/q$, then
$$
a^n=(\sqrt[q]{a})^p \, .
$$
It's worth noting at this point that this definiton does not apply when the base $a$ is a negative number, since $(-5)^{1/2}$ is meaningless within the context of real numbers. It is possible to partially extend the definition, so that $a^{p/q}$ is meaningful when $q$ is an odd number, but from here on in we will only define $a^n$ for when $a\geq0$.* We are still unsure about how to define $a^n$ for $n\in\mathbb{R}$, since $n$ could be an irrational number. Nevertheless, it seems natural that $a^{\sqrt{2}}$ should be approximately equal to $a^{1.41421}$, and this motivatives the below definition for irrational exponents:
$$
a^n=\lim_{m(\in\mathbb{Q})\to n} a^m \, .
$$

Now that we have formally defined $a^x$ for $a\geq0$, $x\in\mathbb{R}$, we can begin to examine its properties. You will no doubt be familiar with the phrase 'exponential growth'—growth in which the rate of change of a quantity is proportional to the quantity at any given time. This means that an exponentional function may be characteristed as any solution to
$$
y'(x) = cy(x)
$$
for some constant $c$. The term 'exponentional' can give us a clue as to which functions exhibit this property, namely the ones we have just studied! If $y(x)=a^x$, then
\begin{align}
\frac{dy}{dx} &= \lim_{\Delta x \to 0}\frac{a^{x+\Delta x}-a^x}{\Delta x} \\
&= a^x \lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}
\end{align}
Note that the final limit is simply $y'(0)$, meaning that $y'(x)$ does indeed equal $cy(x)$, with $c$ being equal to the slope of $a^x$ at the point $(0,1)$. This helps explain one of the classic definitions of the constant $e$: it is the unique number $a$ such that
$$
\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x} = 1 \, .
$$
This means that $e^x$ is its own derivative. This approach has two deficiencies, though:

We haven't proven that the limit $$\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$$ exists.

We haven't shown that there is such a number where the above limit is equal to $1$.

This can be remedied by appealing to one of the alternative definitions of $e$.

Title: Prove that $\frac{d}{dx}e^x=e^x$


问:
A square appears randomly within a square of ten time its area. What is probability that the smaller square contains the larger square's center?
The alignment relative to the larger square can be anywhere between 0° to 90° (0° or 90° being identical orientation to the larger square, 45° being all the way skewed), and the probability distribution of the alignment is based on percentage of total possibility given the constraint guaranteeing that it will fit within the larger square.

<s>The largest discrete probability, for an alignment of 0°=90°, should be simple enough to figure-out‡. The smallest discrete probability, for alignment of 45°, is a bit trickier but straightforward enough.</s>Edit:The off-kilter version would be more likely to contain the center, since it has less possible places to exist and a larger proportion of them are the center. Intuitively this should correspond directly (though not necessarily uniformly) to what I am totally clueless about, which is:

how to determine the probability  distribution for between 0° to 45°,

‡ 2. how this continuum corresponds to the probability of the original problem (e.g. {1/17?} for 0° up to {2/9?} for 45° back down to {1/17?} for 90°, along finite part of some sort of curve function presumably),

and 3. how then to apply these equations formulated into the original problem to find the final probability that a square true-randomly generated wholly within a larger square with sides √(10)-times longer would surround (or contain exactly on an edge or corner) the center of the larger square.

‡ I haven't calculated the maximum or minimum ‘discrete’ probabilities (let alone any inbetween) yet. Those guesstimate values are placeholders. I am especially interested in the processes required to solve this, and appreciate any insight or clues.

答:
Let me see if I understand you problem.

You are looking at a square $S = [-R/2, R/2]^2$ of side length $R$ (for simplicity centred at the origin. I will write $C(x;\theta)$ for the smaller square of side length $r$ centred in $x\in\mathbb{R}^2$ at rotation $\theta$. You want to know what the probability is of the event that $0\in C(x;\theta)$ under the constraint $C(x;\theta)\subseteq S$.

For simplicity, we take $r \ll R$ in a suitable sense ($=$ so that everything works just fine). The first thing I would like to do is to wrap it up mathematically. For this, we define our probability space to be
$$
\Omega := \{(x,\theta)\in \mathbb{R}^2\times [0,\pi/4]\;\vert\; C(x;\theta)\subseteq S\}.
$$
Note that I only consider angles up to 45 degrees, because the situation is symmetric for angles between 45 and 90 degrees. To model "random appearance", we take the uniform distribution on $\Omega$.

First things first: Can we describe $\Omega$? For this, fix the angle $\theta$. Write $h(\theta)$ for the height ($=$ the width) of the skewed square $C(x;\theta)$. Then
$$
h(\theta) = \cos\left(\dfrac{\pi}{4} - \theta\right)\cdot\sqrt{2}\cdot r.
$$
We conclude that
$$
\Omega = \{(x,\theta)\in \mathbb{R}^2\times [0,\pi/4]\;\vert\; \Vert x\Vert_\infty \leq (R - h(\theta))/2\}.
$$
In particular,
$$
m(\Omega) = \int_0^{\pi/4} \big{(} R - r\cos(\theta)\sqrt{2}\big{)}^2\mathrm{d}{\theta} = \dfrac{\pi}{4}R^2 - 2Rr + \dfrac{2+\pi}{4}r^2,
$$
where I wrote $m$ for the Lebesgue measure on $\mathbb{R}^2$.

Now to the event we are interested in:
$$
A = \{(x,\theta)\in \Omega\;\vert\; 0\in C(x;\theta)\}.
$$
Since we took $r\ll R$ we do not need to worry about whether the squares touching $0$ are indeed included in $S$. We further simplify the problem by noticing that the size of the set of suitable angles $\theta$ for a given centre $x$ only depends on $\rho :=  \Vert x\Vert_2$. So let's fix $0\leq \rho \leq \dfrac{\sqrt{2}}{2}r$. (The upper bound is given by the half diagonal.) To get some visual intuition, take $x = (\rho, 0)$. Let us start in the most "favorable" position, i.e. at $\theta=\pi/4$. The question is by how much we are allowed to skew the square. Note that the point closest to the origin is at distance
$$
d(\theta) = \dfrac{r}{2\cos\theta}.
$$
Hence, we need
$$
\cos\theta \leq \dfrac{r}{2\rho}.
$$
This means that we have the constraint $\cos^{-1}\left(\dfrac{r}{2\rho}\right)\leq \theta \leq \dfrac{\pi}{4}$ for $\rho \geq r/2$ and bo constraint for $\rho < r/2$. This means that
$$
m(A) = \dfrac{\pi^2r^2}{8} - 2\pi\int_{r/2}^{\sqrt{2}r/2} \rho \cos^{-1}\left(\dfrac{r}{2\rho}\right)\mathrm{d}\rho = \dfrac{\pi^2r^2}{8} - \dfrac{\pi(\pi - 2)r^2}{8}.
$$

If I have not made any mistake in any of those integrals, we should get the probability
$$
\dfrac{m(\Omega)}{m(A)} = \text{something very ugly}.
$$

Edit: The following comments try to give a bit more intuitive insight into the notation.

First, note that the uniform distribution on a interval means that the probability of the event is the "length" or "size" of the event, divided by the total length of the interval. The same is going on here: Once we have understood that we want to take $(x,\theta)$ uniformly in $\Omega$, the probability of an event $A$ is the "volume" (because we operate in 3 dimensions now) of $A$, divided by the volume of $\Omega$.

Now, to obtain the volume of a set, we usually integrate, especially if the bodies are complex as in this case. But in the end, it is only "measuring the volume". The $m$ I use is just "the volume of". This "usual" volume is called Lebsegue measure, because there other ways to make sense of the word "volume", but in this case this doesn't matter.

I don't think that there is a straightforward way to obtain the explicit expression of the result without using integrals, i.e. by simply using geometric surface/volume formula, but you can always try!

For the other notation: $\Vert (x,y)\Vert_2 = \sqrt{x^2 + y^2}$ is the usual Euclidean distance, $\Vert (x,y)\Vert_\infty = \max\{\vert x\vert, \vert y\vert\}$ is the so-called infinity norm. (There are $\Vert\cdot\Vert_p$-norms for any $p\geq 1$.) When I write $r\ll R$, I mean that $r$ is a lot smaller than $R$. Here, we need this assumption when figuring out how we may describe $A$.

Also, I mainly used WolframAlpha for those horrendous integrals.

Edit 2: There is a lot easier way to calculate $m(A)$. Indeed, mjw (https://math.stackexchange.com/q/3966261) correctly points out that, for any given angle $\theta$, the centre must be in a square of side length $r$. That means that
$$
m(A) = \dfrac{\pi}{4}r^2.
$$
I corrected the above mistake.


问:
$S_3$ in $Z_{18}$? Rings and subrings
Let $a$ be an element in the ring $R$. Show that $S_a$ is a sub ring to $R$. What is $S_3$ in $Z_{18}$?

$$S_a=\begin{Bmatrix}
x\in R|xa=0
\end{Bmatrix}$$

My work:

$$x,y\in R\\ xa=0, ya=0\\ xa-ya=0\Rightarrow a(x-y)=0\Rightarrow x-y=0\\
a(xy)=(ax)y=0$$
$S_a$ is a sub ring to R.

Does this hold, or have I completely misunderstood it?

How can I continue with $S_3$ in $Z_{18}$?

EDIT:
What is $S_3$ in $Z_{18}$?

$$S_3=\begin{Bmatrix}
x\in R|3x=0
\end{Bmatrix}\\?$$

答:
The conclusion $xa−ya=0⇒(x−y)a=0⇒x−y=0$ only holds if $a$ is not a zero divisor of $R$.

But you don't need to go that far: Let $xa=0$ and $ya=0$. Then $0 = xa-ya = (x-y)a$ and so $x-y\in S_a$.

So $S_a$ is indeed a subring of $R$.

Title: $S_3$ in $Z_{18}$? Rings and subrings


问:
How does it follow $s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$?
I have two relations:

1)$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{1}^{\infty}\frac{\Lambda(n)}{n^s}$.

2)$\psi(x)=\sum_{n\leq x}\Lambda(n)$.

From these two how does it follow that $-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$, for $s>1$.
Can anyone explain how does it follow?

答:
Let's start with the definition of the von Mangoldt function ($p$ will always mean a prime) :
$$\ \Lambda(n):=\begin{cases}
\log\, p & \text{if}\ n=p^k\ \text{and}\ k>0\\
0 & \text{else}
\end{cases}$$

We may use the Euler product in $\ \displaystyle\log \zeta(s)=-\sum_{p\ \text{prime}}\log(1-p^{-s})=\sum_p\sum_{k=1}^\infty \frac{p^{-ks}}k\ $ to obtain your first relation (after derivation) :

$$\tag{1}-\frac{\zeta'(s)}{\zeta(s)}=\sum_p\sum_{k=1}^\infty \frac{\log\,p}{p^{ks}}=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\quad\text{for}\ \ \Re(s)>1$$

After that we want to use the definition of the second Chebyshev function :
$$\tag{2} \psi(x)=\sum_{n\leq x}\Lambda(n)$$

But, using Abel's sum formula with $a(n):=\Lambda(n)$ and $\phi(n):=n^{-s}$, we get :
$$\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}=s\int_1^\infty \frac {\sum_{n\leq x}\Lambda(n)}{x^{s+1}}\;dx$$
that is, using $(1)$ and $(2)$, the wished formula :
$$-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$$

Inverting this Mellin transform to express $\psi(x)$ would produce Perron's formula as shown in this derivation of the 'explicit formula' with better handling of the discontinuities.

ContentLicense: CC BY-SA 3.0


问:
Find E(X^3) given X has the distribution Exp(2)
Find $E(X^3)$ given $X$ is in $Exp(2)$

My idea is that we can use $f_X(x)=2e^{-2x}$ and integrate $\int_{0}^{a}x^3f_X(x)dx$. But from here I don't know where to go.

答:
For an exponential distribution $f(x|\lambda)=\lambda e^{-\lambda x}$, moments can be computed according to (source Wikipedia)
$$\mathbb{E}\left ( X^n \right )=\frac{n!}{\lambda^n}$$

You can compute this equation by recurrence and integration by part
$\mathbb{E}\left ( X^n \right )=\frac{n}{\lambda}\mathbb{E}\left ( X^{n-1} \right )$. In your particular application, $\mathbb{E}\left ( X^3|\lambda=2 \right )=\frac{3!}{2^3}=\frac{3}{4}$


问:
How to sketch level curves of the form $f(x,y)= |x|+|y|$?
I'm struggling to sketch the level curves of the equation $f(x,y)= |x|+|y|$
I know for finding the level curves you have to set $f(x,y) = C$ (with c a constant). But then I have the equation $|x|+|y|=C$.

So lets say $C=0$, then $|x|+|y|=0$, but how can I sketch this. Because both x and y are greater then zero, and they have to add up to 0, therefore x and y needs to be zero, but then I can't draw anything.

So lets say $C=1$,then $|x|+|y|=1$, but how can I sketch this. Can I say that $x + y= \pm1$, therefor $y=1-x$ and $y = 1+x$ (but $x$ can't be negative, so I don't know how do this problem.)

If you know the answer, it would be very much appreciated to help me

Thank you in advance.

答:
If $|x|+|y|=1$, consider four possibilities:

$x,y\geqslant0$: then you just have $x+y=1$;

$y\leqslant0\leqslant x$: then you just have $x-y=1$;

$x,y\leqslant0$: then you just have $-x-y=1$;

$x\leqslant0\leqslant y$: then you just have $-x+y=1$.

So, your curve is a square, expressed as the union of $4$ line segments. To be more precise: it's the square whose vertices are $(\pm1,0)$ and $(0,\pm1)$.


问:
An application of Muirhead inequality
I'm having some difficulties with proving the following task using Muirhead inequality.

$$\cfrac{x_1}{x_2+x_3+...+x_n}+\cfrac{x_2}{x_1+x_3+...+x_n}+...+\cfrac{x_n}{x_1+x_2+...+x_{n-1}}\ge\cfrac{n}{n-1}$$

It looks like a generalized version of Nesbitt's inequality, however using Muirhead to prove classic Nesbitt seemed way easier. Any help is appreciated. Thank you

答:
For simplicity, let's write  $\sum x_i =S$ and $A_j= S-x_j$
We need to prove

$\sum \frac{x_i}{A_i} \geq \frac{n}{n-1}$

$ \iff \sum \frac{S}{A_i} \geq \frac{n^2}{n-1}$

$\iff S(\sum \frac{1}{A_i}) \geq \frac{n^2}{n-1}$

$\iff S(\frac{n^2}{\sum A_i}) \geq \frac{n^2}{n-1}$ by AM-HM

$\iff S(\frac{n^2}{(n-1)S}) \geq \frac{n^2}{n-1}$.

Title: An application of Muirhead inequality


答:
Let $X\sim E(\lambda)$, i.e., $f_X(x)=\lambda e^{-\lambda x}$. The $n^{th}$ moment of $X$ would be:

$E\{X^n\} = \int\limits_{0}^{+\infty} x^n e^{-\lambda x} \lambda dx$

Let $u=\lambda x$, then the integration becomes:

$E\{X^n\} = \int\limits_{0}^{+\infty} \dfrac{u^n}{\lambda^n} e^{-u} du
= \dfrac{1}{\lambda^n} \int\limits_{0}^{+\infty} u^n e^{-u} du
= \dfrac{1}{\lambda^n} \Gamma(n+1) = \dfrac{n!}{\lambda^n}$

where $\Gamma()$ is the gamma function: https://en.wikipedia.org/wiki/Gamma_function

Title: Find E(X^3) given X has the distribution Exp(2)

Body:
Find $E(X^3)$ given $X$ is in $Exp(2)$

My idea is that we can use $f_X(x)=2e^{-2x}$ and integrate $\int_{0}^{a}x^3f_X(x)dx$. But from here I don't know where to go.


问:
$(Z_{11},+,\bullet )$ subfield
I apologize for my English, I hope you understand.

Does $(Z_{11},+,\bullet )$ have any subfield?

I've tried reading in my book and searching on the internet for answers.

What is the definition of if a subfield exists?

答:
Here you  mean subring, as you consider $\Bbb Z_{11}$ with two operations. Its a field and so can equivalently ask about subfields.

But $\Bbb Z_{11}$ is a prime field as $11$ is a prime and it has no proper subrings.

Title: $(Z_{11},+,\bullet )$ subfield


答:
I just add a comment to the already proposed solutions.
Since $f(x,y) = f(-x, y) = f(x, -y) = f(-x, -y)$, the level sets of $f$ are symmetric w.r.t. the reflections about the axes $(x,y) \mapsto (-x,y)$ and $(x,y) \mapsto (x,-y)$,
and w.r.t. the symmetry about the origin $(x,y) \mapsto (-x, -y)$.

Given that, it is enough to draw the level set for example in the quadrant $x\geq 0$, $y\geq 0$, and then you get the whole picture using the above symmetries.

Title: How to sketch level curves of the form $f(x,y)= |x|+|y|$?

Body:
I'm struggling to sketch the level curves of the equation $f(x,y)= |x|+|y|$
I know for finding the level curves you have to set $f(x,y) = C$ (with c a constant). But then I have the equation $|x|+|y|=C$.

So lets say $C=0$, then $|x|+|y|=0$, but how can I sketch this. Because both x and y are greater then zero, and they have to add up to 0, therefore x and y needs to be zero, but then I can't draw anything.

So lets say $C=1$,then $|x|+|y|=1$, but how can I sketch this. Can I say that $x + y= \pm1$, therefor $y=1-x$ and $y = 1+x$ (but $x$ can't be negative, so I don't know how do this problem.)

If you know the answer, it would be very much appreciated to help me

Thank you in advance.


问:
Not sure how to go about solving this integral
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$

I've let $u=x^2+3$ but can't seem to get the right answer.

Really not sure what to do.

答:
I suggest letting $u = \frac{x}{\sqrt{3}}$. Then $du = \frac 1{\sqrt{3}} dx$ and
$$
\int \left( \frac 1{x^2+3} \right) dx = \int \left( \frac 1{(\sqrt 3 u)^2 + 3} \, \sqrt 3 \right) dx = \frac 1{\sqrt{3}} \int \left( \frac 1{u^2 + 1} \right) = \frac 1{\sqrt{3}} \arctan(u) + C.
$$
Hope that helps</p>


问:
About two equinumerous partitions of the same set.
Let $\mathcal {A,B}$ be partitions of a set $X$ into $m$ subsets. Suppose that for any $k\leq m$ and any $A_1,\ldots,A_k \in\mathcal A$ there are at most $k$ elements of $\mathcal B$ contained in $\bigcup_{i=1}^kA_i.$ Does it imply that for any $k\le m$ and any $A_1,\ldots,A_k\in\mathcal A$, there are at least $k$ elements of $\mathcal B$ whose intersection with $\bigcup_{i=1}^kA_i$ is non-empty?

I'm thinking about a problem which a positive answer to the above would solve. And the above appears plausible to me.

Both the hypothesis and the conclusion say something about any $A_1,\ldots,A_k.$ My first thought was to take any $A_1,\ldots,A_k$ and getting the conclusion from the hypothesis restricted to those particular $A_i$. But this doesn't work, and I don't have any ideas.

答: I assume that $X$ is non-empty and $m>0$. The implication does, indeed hold. In the case $k=m$, this is obvious, since $\mathcal A,\mathcal B$ are both partitions of $X$ into $m$ subsets. Now, suppose $0<k<m$, and take any $k$ distinct elements $A_1,...,A_k$ of $\mathcal A$. Put $j=m-k$ and let $A_1',...,A_j'$ be the other elements of $\mathcal A$. By hypothesis, there are at most $j$ elements of $\mathcal B$ contained in $$\bigcup_{n=1}^jA_n'=X\smallsetminus\left(\bigcup_{n=1}^kA_n\right),$$ so there are at least $m-j=k$ elements of $\mathcal B$ whose intersection with  $$ \bigcup_{n=1}^kA_n$$ is non-empty.</p>


问:
$X^3 + 22Y^3 + 3Z^3 = 0$ has no rational solution
This is Exercise 1.4 in Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory.

Write $\alpha = \sqrt[3]{22}$ and $k = \mathbb{Q}(\alpha)$. I have shown that $k$ has class number 3 and the norm form is
$$ \mathrm{norm}(a + b\alpha + c\alpha^2) = a^3 + 22b^3 + 22^2c^3 - 66abc. $$

Assuming that $(x,y,z)$ is a primitive solution of $X^3+22Y^3+3Z^3=0$ with $z > 0$, we find
$$ N(x+y\alpha) = |x^3+22y^3| = 3z^3. $$
Since $3$ is totally ramified in $k$, i.e. $3\mathcal{O}_k = \mathfrak{p}_3^3$, we can write
$$ (x+y\alpha) = \mathfrak{p}_3 \mathfrak{a}, $$
for some $\mathfrak{a}$ with $N\mathfrak{a} = z^3$.
The exercise says to show that $\mathfrak{a}$ is a cube, which would lead to a contradiciton since $\mathfrak{p}_3$ is not principal, and this is where I am having trouble.

Writing $\mathfrak{a} = \prod \mathfrak{p}^{a_{\mathfrak{p}}}$, the primes with $a_{\mathfrak{p}} \neq 0$ are above the primes dividing $z$ and I want to show that $a_{\mathfrak{p}} \equiv 0 \bmod 3$ for every $\mathfrak{p}$.
If $N\mathfrak{p} = p^{f_{\mathfrak{p}}}$, the fact that the norm of $\mathfrak{a}$ is a cube says
$$ \sum_{\mathfrak{p} \mid p} a_{\mathfrak{p}} f_{\mathfrak{p}} \equiv 0 \bmod 3.$$

If $p = 2,3$ or $11$, then $p$ is totally ramified in $k$, i.e. $p\mathcal{O}_k = \mathfrak{p}_p^3$, so $p \mid z$ implies $\mathfrak{p}_p^3 \mid \mathfrak{a}$ and the result holds.

I now consider a prime $p \neq 2,3,11$ which divides $z$.
The equation gives
$$ x^3 + 22y^3 \equiv 0 \bmod p^3, $$
and because of coprimality, we find that $22$ is a cube modulo $p$ so we cannot have $f_{\mathfrak{p}} = 3$ for any $\mathfrak{p} \mid p$. This gives $a_{\mathfrak{p}} \equiv 0 \bmod 3$ when $\mathfrak{p}$ is the only prime above $p$ dividing $\mathfrak{a}$.

The problem is that I do not see how to reach that conclusion when there are more than one prime factor of $p\mathcal{O}_k$ appearing in the decomposition of $\mathfrak{a}$. For instance, if $p\mathcal{O}_k = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$, why can we not have $\mathfrak{a} = \mathfrak{p}_1 \mathfrak{p}_2^2$?

答: See here for a treatment of the similar (and more famous) equation $3x^3 + 4y^3 + 5z^3 = 0$.  This example and the one you write about are also in Chapter 10 of Cassels' Local Fields: see Lemma 9.1 and 9.2.


答: The additive group of $\Bbb Z_{11}$ contains no proper nontrivial subgroup and so $\Bbb Z_{11}$ contains no proper subfield.

Title: $(Z_{11},+,\bullet )$ subfield

Body:
I apologize for my English, I hope you understand.

Does $(Z_{11},+,\bullet )$ have any subfield?

I've tried reading in my book and searching on the internet for answers.

What is the definition of if a subfield exists?


问:
How to find angle of rotation?
$$
A=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
$$
D=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & -\left(\frac{1+\sqrt{3} i}{2}\right) & 0 \\
0 & 0 & \left(\frac{-1+\sqrt{3} i}{2}\right)
\end{array}\right]
$$

$A$ is the form of the rotation matrix in $(x, y, z)$ basis (orthonormal) and $D$ is it's form in new basis $(x', y', z')$ in which it is diagonal. The rotation occurs about the axis in the direction$$\hat{n}=\frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}}
$$

And the transformation rule between $(x, y, z)$ and $(x', y', z')$ is given as

\begin{aligned}
x &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{-1+\sqrt{3} i}{2}\right) y^{\prime}+\left(\frac{-1-\sqrt{3} i}{2}\right) z^{\prime}\right) \\
y &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{\left.-1- \sqrt{3} i) y^{\prime}+(-1+\sqrt{3} i) z^{\prime}\right)}{2}\right)\right.\\
z &=\frac{1}{\sqrt{3}}\left(x^{\prime}+y^{\prime}+z^{\prime}\right)
\end{aligned}
So my question is how to find angle of rotation? Help!

答: As you know, your rotation is a rotaion around $\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$. Take a unitary vector $v$ orthogonal to this vector; for instance, take $v=\left(\frac1{\sqrt2},-\frac1{\sqrt2},0\right)$. It happens that$$A.v=\left(0,\frac1{\sqrt2},-\frac1{\sqrt2}\right)$$and so the angle $\theta$ that you are looking after is the angle between $v$ and $A.v$. Since these are two unit vectors, $\theta$ is such that $\cos\theta=\langle v,A.v\rangle=-\frac12$, and therefore $\theta=\frac{2\pi}3$.

Title: How to find angle of rotation?


答:
Rotations in $R^3$ are the compositions of rotations in $R^2$,

A rotation in $R^2$ is written as

$\left(\begin{matrix}
\cos(\alpha)&\sin(\alpha)\\
-\sin(\alpha)&\cos(\alpha)
\end{matrix}\right)$

where $\alpha$ is the angle of the rotation

In $R^3$ you can write it as the product of two rotations in $R^2$
$\left(\begin{matrix}
\cos(\alpha)&\sin(\alpha)&0\\
-\sin(\alpha)&\cos(\alpha)&0\\
0&0&1
\end{matrix}\right)
\left(\begin{matrix}
\cos(\beta)&0&\sin(\beta)\\
0&1&0\\
-\sin(\beta)&0&\cos(\beta)
\end{matrix}\right)$

in your example you are sending the:

$x$-axis to the $y$-axis

$y$-axis to the $z$-axis

$z$-axis to the $x$-axis

so you are doing 2 rotations (over 2 different axis) of 90º degrees each

Title: How to find angle of rotation?

Body:
$$
A=\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
$$
D=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & -\left(\frac{1+\sqrt{3} i}{2}\right) & 0 \\
0 & 0 & \left(\frac{-1+\sqrt{3} i}{2}\right)
\end{array}\right]
$$

$A$ is the form of the rotation matrix in $(x, y, z)$ basis (orthonormal) and $D$ is it's form in new basis $(x', y', z')$ in which it is diagonal. The rotation occurs about the axis in the direction$$\hat{n}=\frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}}
$$

And the transformation rule between $(x, y, z)$ and $(x', y', z')$ is given as

\begin{aligned}
x &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{-1+\sqrt{3} i}{2}\right) y^{\prime}+\left(\frac{-1-\sqrt{3} i}{2}\right) z^{\prime}\right) \\
y &=\frac{1}{\sqrt{3}}\left(x^{\prime}+\left(\frac{\left.-1- \sqrt{3} i) y^{\prime}+(-1+\sqrt{3} i) z^{\prime}\right)}{2}\right)\right.\\
z &=\frac{1}{\sqrt{3}}\left(x^{\prime}+y^{\prime}+z^{\prime}\right)
\end{aligned}
So my question is how to find angle of rotation? Help!


问:
Which real functions have their higher derivatives tending pointwise to zero?
Let $\mathrm C^\infty\!(\Bbb R)$ be the space of infinitely differentiable functions $f:\Bbb R\rightarrow\Bbb R$, and define the subspace$$A:=\{f\in\mathrm C^\infty\!(\Bbb R):(\forall x\in \Bbb R)\lim_{n\rightarrow\infty} f^{(n)}(x)=0\},$$where $f^{(n)}$ is the $n$th derivative of $f\;(n=0,1,\dots).$ Clearly all polynomial functions are in $A$. Are any others?

Edit: Alfonso has answered this question well, but is there any characterization of $A$ in terms of familiar types of function?

答:
Another example would be the reciprocal function:

$$f(x)=\frac{1}{x}$$

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

By extension, $\ln(x)$ is another example, because $\frac{d}{dx}\ln(x)=\frac{1}{x}$


答:
Looking at the columns, under the transformation the unit vectors $i$, $j$ and $k$ are transformed as follows:

$$i \rightarrow j$$
$$j\rightarrow k$$
$$k\rightarrow i$$

Doing the same transformation three times would be equivalent to the identity matrix.
Therefore the matrix is a rotation of $120^o$ about the line $x=y=z$.

Title: How to find angle of rotation?


问:
How best we can allocate?
Thank you for this opportunity to share my post here

Advanced thanks to the people who answered my post

Problem: There are $5$ items and cost of these items in dollars are $3$,  $3.2$,  $4.5$,  $2.5$,  $1.7$ respectively. How  can we allocate $100$ dollars to buy the above items in a such way in order to get maximum number of items in our hand.

All types of items should be covered

Repetition of items may be allowed

Maximum number of items should be purchased

Not necessary to spend entire $100$. but, maximum amount should spend near to $100$ or with entire $100$

We need to buy each kind of item as much as possible

How to solve this type of problem?
Kindly answer
Thanks again

答:
Let $n \in \mathbb{N}$ be the amount of dollars to spend, let $a_1, a_2, ..., a_k$ be the objects we want to buy, with prices $p_1, p_2, ..., p_k \in \mathbb{N}$. As we can reorder these items freely, we can assume that their prices are ordered increasingly, that is: $p_1 \leq p_2 \leq ... \leq p_k$.

Define $p = \sum_{i=0}^k p_k$, the total price of all items.

We can write $n = qp + r$, for some $q,r \in \mathbb{N}$ and $0 \leq r < p$. This is equivalent to dividing $n$ by $p$ with rest. As we want to buy all items an almost equal amount, we therefore by $q$ sets of all items, and then have $r$ dollars left over.

With these $r$ dollars, we can buy the items $p_1, p_2, ..., p_k$, until we run out of dollars. We always buy the least expensive item first, so we have as much cash left over as possible. In this case, the question becomes (and I don't think there is a simple formula for this):

What is the largest value of $j\in\mathbb{N}$, such that $r > \sum_{i=0}^j p_i$.

This value of $j$ is the amount of items we can buy with the remaining $r$ dollars.

Putting al this together: We can buy $q$ sets of $k$ items, and with the remaining $r$ dollars we can buy $j$ items, so in in total we can buy $qk+j$ items with our $n$ dollars.

Title: How best we can allocate?


问:
Characteristic function of logarithm of random variable
If I know the characteristic function $\phi_X(t)$ of a random variable $X>0$, how can I write the characteristic function $\phi_Y(t)$ of $Y=\log(X)$?

I know that $\phi_X(t)=E[e^{itX}]$ and $\phi_Y(t)=E[e^{it\log(X)}]$. But I can't derive one from the other. Any idea? I would like to use $\phi_X(t)$ to calculate the second moments $X$.

答:
We need to compute the log-characteristic function that is the characteristic of $Y=\ln\left(X\right)$. So, we need to express the characteristic function of $Y$ in terms of values $X$: $$E\left(\exp\left(ikY\right)\right)=\int_{l_{y}}^{u_{y}}\exp\left(iky\right)f_{Y}\left(y\right)dy$$
As $y=\ln\left(x\right)$, we have $dy=\frac{dx}{x}$ and so :
$$E\left(\exp\left(ikY\right)\right)=\int_{\exp\left(l_{y}\right)}^{\exp\left(u_{y}\right)}\exp\left(ik\ln\left(x\right)\right)f_{Y}\left(\ln\left(x\right)\right)\frac{dx}{x}$$
We need to express the density $f_{Y}$ of $Y$ in terms of the density $f_{X}$ of $X$. As $Y=g\left(X\right)=\ln(X)$, because the probabilities are invariant by change of variables, the transformation is strictly monotone and derivable in all the domain of X so that:
$$f_{Y}\left(y\right)=\left|\frac{1}{g'\left(g^{-1}\left(y\right)\right)}\right|.f_{X}\left(g^{-1}\left(y\right)\right)$$
$$\Leftrightarrow f_{Y}\left(y\right)=\exp\left(y\right).f_{X}\left(\exp\left(y\right)\right)=x.f_{X}\left(x\right)$$
In the end, the log characteristic function equals:
$$E\left(\exp\left(ikY\right)\right)=\int_{\exp\left(l_{y}\right)}^{\exp\left(u_{y}\right)}\exp\left(ik\ln\left(x\right)\right)x.f_{X}\left(x\right)\frac{dx}{x}$$
$$\Leftrightarrow E\left(\exp\left(ikY\right)\right)=\int_{\exp\left(l_{y}\right)}^{\exp\left(u_{y}\right)}\exp\left(ik\ln\left(x\right)\right)f_{X}\left(x\right)dx$$
Q.E.D.


问:
irreducible representation of $GL(V,\mathbb{C}) $ on the symmetric subspace of $V\otimes V $
GL(V) acts naturally on the tensor product space $V\otimes V$. The symmetric subspace

$$S^2 V = span \{ v \otimes v \} $$

is evidently an invariant subspace. In the book by S. Sternberg, he states that it is not hard to check directly that the representation on this subspace is irreducible.

But I have no clue at all. Can anyone provide an easy proof?

答:
The standard way to do this question is by calculating the character $\chi_{S^2V}$ of the representation, and showing  $\langle \chi_{S^2V},\chi_{S^2V}\rangle=1$. But for a more a direct proof, I think probably the easiest way is to consider the corresponding Lie algebra representation, given by
$$
X\cdot(v\otimes v) = Xv\otimes v + v\otimes Xv
$$
for $X\in\mathfrak{gl}(V)\simeq \operatorname{End}(V)$.
Since $\operatorname{GL}(V)$ is connected, it can be shown that the Lie group representation is invariant iff the Lie algebra representation is invariant.

Now picking a basis $\{e_i \mid 1\le i \le n\}$ of $V$, $S^2V$ is spanned by the $n+{n\choose 2} = \frac{n(n+1)}{2}$-dimensional basis
$$
\{e_i\otimes e_i\mid 1\le i\le n\} \cup \{e_i\otimes e_j + e_j\otimes e_i \mid 1\le i< j\le n\} .
$$
For each $1\le p,q\le n$, let $X_{pq}:V\to V$ denote the linear transformation which maps $e_q$ to $e_p$, and is zero on all other basis elements of $V$ (i.e. $X_{pq}e_i = \delta_{qi}e_p$). These span $\mathfrak{gl}(V)$.

Now suppose $W\ne\{0\}$ is an invariant subspace of $S^2V$. If one can show $W$ contains one of the basis elements listed above, then it will contain all of them, by successive applications of the $X_{pq}$ (and invariance of $W$). For example,
$$
X_{21}\cdot (e_1\otimes e_1) = e_2\otimes e_1 + e_1\otimes e_2,\qquad X_{21}\cdot X_{21}\cdot(e_1\otimes e_1) = 2e_2\otimes e_2,
$$
etc. To this end, suppose
$$
\sum_{i=1}^na_i\, e_i\otimes e_i + \sum_{i<j}b_{ij}(e_i\otimes e_j + e_j\otimes e_i) = \sum_{i=1}^na_i\, e_i\otimes e_i + \frac{1}{2}\sum_{i\ne j}b_{ij}(e_i\otimes e_j + e_j\otimes e_i)
$$
is a non-zero vector in $W$, where  in the second expression we take $b_{ij}$ to be symmetric. First suppose $a_q\ne 0$ for some $q$, and apply $X_{pq}$ for some $p\ne q$ to get
$$
a_q(e_p\otimes e_q+e_q\otimes e_p) + \frac{1}{2}\sum_{j\ne q} b_{qj}(e_p\otimes e_j+e_j\otimes e_p) + \frac{1}{2}\sum_{i\ne q}b_{iq}(e_i\otimes e_p+e_p\otimes e_i) \\= a_q(e_p\otimes e_q+e_q\otimes e_p) +\sum_{j\ne q}b_{qj}(e_p\otimes e_j+e_j\otimes e_p).
$$
Then apply $X_{qq}$ to get
$$
a_q(e_p\otimes e_q+e_q\otimes e_p).
$$
This must be in $W$, and since $a_q\ne 0$, we have $e_p\otimes e_q+e_q\otimes e_p\in W$. It follows that $W = S^2V$.

Now suppose all the $a_i$ are zero, but that $b_{pq}\ne 0$ (for $p\ne q$). Applying $X_{pp}$ to the non-zero vector gives
$$
\sum_{j\ne p}b_{pj}(e_p\otimes e_j+e_j\otimes e_p),
$$
and then applying $X_{pq}$ gives
$$
2b_{pq}\,e_p\otimes e_p.
$$
Hence $e_p\otimes e_p \in W$, from which it follows that $W=S^2V$.


问:
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$
let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$

I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove

$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$

答:
The series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}=1+\frac{1}{3}+\frac{1\cdot 2}{3\cdot 5}+\frac{1\cdot2\cdot3}{3\cdot5\cdot7}+\cdot\cdot\cdot$
causes $2-\frac{\pi}{2}$ to have nice properties with the given sequence $a_{n+1}= \frac{2n+1}{n}a_n-\frac{1}{n}.$

$a_1 = 2-\frac{\pi}{2}=1-\frac{1}{3}-\frac{1\cdot 2}{3\cdot 5}-\frac{1\cdot2\cdot3}{3\cdot5\cdot7}-\cdot\cdot\cdot$

$a_2=1-\frac{2}{5}-\frac{2\cdot 3}{5\cdot7}-\cdot \cdot \cdot$

$\vdots$

$a_n=1-\frac{n}{2n+1}-\frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$

from which $a_n >1-\frac{1}{2}-\frac{1}{2^2}-\cdot \cdot \cdot = 1-\sum\limits_{i=1}^\infty 2^{-i}=1-1=0.$

Note that if $a_n$ is a decreasing sequence implies $a_n \leq \frac{1}{n+1}$

$a_{n+1}\leq a_n \Longleftrightarrow$

$\frac{2n+1}{n}a_n -\frac{1}{n} \leq a_n \Longleftrightarrow$

$(2n+1)a_n-1\leq na_n \Longleftrightarrow$

$(n+1)a_n\leq 1 \Longleftrightarrow$

$a_n\leq \frac{1}{n+1}$

and the first inequality

$a_{n+1}=1-\frac{n+1}{2n+3}-\frac{(n+1)(n+2)}{(2n+3)(2n+5)}-\cdot \cdot \cdot \leq a_n =1-\frac{n}{2n+1}- \frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$ does hold because for each term

$\frac{(n)(n+1)(n+2)\cdot\cdot\cdot(n+m)}{(2n+1)(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)}\leq \frac{(n+1)(n+2)\cdot\cdot\cdot(n+m)(n+m+1)}{(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)(2n+2m+3)}$

and cancelling like terms it is true that

$\frac{n}{2n+1} \leq \frac{n+m+1}{2n+2m+3}$ for $m\geq0.$

Although induction is tempting, I think all the properties of the sequence are captured in the series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}$.

Note that I proved also that $a_n$ is bounded below and decreasing, and thus convergent. Instead of looking at the inequalities $0<a_n\leq \frac{1}{n+1}$ as a possible induction proof, you can look at them as squeeze inequalities, i.e. $a_n \rightarrow 0$.


答:
The trace (the sum of the entries on the diagonal) of a rotation matrix is
$$
1 + 2 \cos \theta
$$
In this case
$$
1 + 2 \cos \theta = 0 \implies \cos \theta = -0.5 \implies \theta = 2\pi/3
$$

Title: How to find angle of rotation?


问:
Families of graphs for which MAXCUT values is known?
The max-cut (https://en.wikipedia.org/wiki/Maximum_cut) is NP-hard in general. However can we enumerate the families of graphs for which the value is known explicitly? For example a bipartite graph's max-cut would be trivial.</p>

答:
I would comment, but I cannot yet.

I suggest you this website, which is basically a database of all graph classes important enough to be even defined.

Specially, you can see their complexity towards the $MAXCUT$ problem listed here.


问:
Row echelon form to solve this system
I want to use the row echelon form to solve this system:
$$
\begin{cases}
4x +2 -z= 0\\
y^2 -4 =0\\
2z -x =0
\end{cases}.
$$
I am confused by the second equation: if it was $y-4$ instead of $y^2 -4$, the matrix to reduce in row echelon form should have been
$$
\begin{pmatrix}
4 & 0 & -1 &-2\\
0 &1 & 0 & 4\\
-1& 0 & 2& 0
\end{pmatrix}
$$
but how to proceed when there is a quadratic term? Could anyone please help me?
Thank you in advance!

答:
This started out as a comment, but then it got too long.

Okay, so the comments pretty much answer the specific question directly, so allow me to address the general case, for posterity's sake.

What do we do when we see a nonlinear term in a system of linear equations?

Well, it helps to consider that "linear" is a relative term; really we ought to say that an equation is linear with respect to some particular variable. For instance $x^2-2y=-3$ is linear in $x^2$, but not $x$.

With this in mind let's look at an example. Consider the following system of equations:

$$\begin{cases}
5x + 2y^2 = 3\\
2x-4z = 0\\
-3x + y^2 +z = 0
\end{cases}$$

Notice that the variable $y$ only ever appears in the form of $y^2$, so we can treat $y^2$ as its own variable (feel free to replace '$y^2$' with '$u$', if it helps). Doing so, we get the augmented matrix

$$
\begin{bmatrix}
5&2&0&|&3\\
2&0&-4&|&0\\
-3&1&1&|&0
\end{bmatrix}
$$

Which we can reduce to get

$$
\begin{bmatrix}
1&0&0&|&\frac3{10}\\
0&1&0&|&\frac34\\
0&0&1&|&\frac3{20}
\end{bmatrix}
$$

So $x=\frac3{10}$, $y^2=\frac34$, and $z=\frac3{20}$, whence $y=\pm\sqrt\frac34$.

What's really going to bake your noodle later on is: could we still do this with a system of polynomial equations?

Title: Row echelon form to solve this system


答:
Since $y$ appears in only one equation, the fact that we have $y^2$ is not a problem. You can just note $Y=y^2$, and then you get the system :
$$\begin{cases} 4x +2 -z= 0\\ Y -4 =0\\ 2z -x =0
\end{cases}$$

Title: Row echelon form to solve this system

Body:
I want to use the row echelon form to solve this system:
$$
\begin{cases}
4x +2 -z= 0\\
y^2 -4 =0\\
2z -x =0
\end{cases}.
$$
I am confused by the second equation: if it was $y-4$ instead of $y^2 -4$, the matrix to reduce in row echelon form should have been
$$
\begin{pmatrix}
4 & 0 & -1 &-2\\
0 &1 & 0 & 4\\
-1& 0 & 2& 0
\end{pmatrix}
$$
but how to proceed when there is a quadratic term? Could anyone please help me?
Thank you in advance!


问:
Related to projection principles in a linear vector field with complex scalar product definition
In a linear vector space $\mathcal{S}$ where vectors (denoted $x, y, z, \cdots$) can be complex-valued and for a complex field of multiplier parameters (denoted $\alpha, \beta, \cdots$), we assume that a scalar product is defined with the following properties

$$ \langle x,y \rangle=\langle y,x\rangle^{*} $$
$$ \langle \alpha_{1}x_{1}+\alpha_{2}x_{2},y \rangle = \alpha_{1}^{*} \langle x_{1},y \rangle +\alpha_{2}^{*} \langle x_{2},y \rangle $$
$$ \langle x,x \rangle >0 \ \ \ \textrm{if: } \ x\neq0, $$
where $^{*}$ denotes complex conjugation.

The question here is to find the value of $\alpha$ that will minimize $|x-\alpha y|$ and then show that such value will make the vectors $y$ and $x-\alpha y$ orthogonal. This is an instructive question (from Friedman, p. 8) to introduce projection theory, but starting from the basic principles and definitions given above only.

I can prove this when the definition is for real vectors, real scalar product and real parameters, by making the length $|x-\alpha y|$ vanish, leading to a quadratic equation in $\alpha$ as

$$ \alpha^{2}-2\alpha \frac{\langle x,y \rangle}{|y|^{2}}+\frac{|x|^{2}}{|y|^{2}}=0, $$

which can be solved for a real value of $\alpha$ taking into account the Cauchy-Schwarz inequality $|\langle x,y\rangle|\leq |x|^{2}|y|^{2}$, to get the sought value: $\alpha_{min}=\langle x,y \rangle/|y|^{2}$, and the rest of the question is then easy to answer.

However, I am having difficulty proving it for the case of complex scalar product definition and complex parameters given above.

答:
Expanding the inner product, one gets (with $a=\mathrm{Re}\alpha$, $b=\mathrm{Im}\alpha$):
\begin{align*}
|x-\alpha y|^2&=|x|^2-2\mathrm{Re}(\alpha\langle x,y\rangle)+|\alpha|^2|y|^2\\
&=|x|^2-2a\mathrm{Re}\langle x,y\rangle+2b\mathrm{Im}\langle x,y\rangle+(a^2+b^2)|y|^2.
\end{align*}
For $\alpha$ to minimize this expression, the partial derivatives with respect to $a$ and $b$ need to vanish:
\begin{align*}
0&=\frac\partial{\partial a}|x-\alpha y|^2=-2\mathrm{Re}\langle x,y\rangle+2a|y|^2,\\
0&=\frac\partial{\partial b}|x-\alpha y|^2=2\mathrm{Im}\langle x,y\rangle+2b|y|^2.
\end{align*}
Thus $a=\mathrm{Re}\langle x,y\rangle/|y|^2$ and $b=-\mathrm{Im}\langle x,y\rangle/|y|^2$, that is, $\alpha=\langle y,x\rangle/|y|^2$.

I'll leave it to you to show that $\alpha$ is actually a minimizer and the orthogonality relation is satisfied.


问:
How do I prove that $2\sin(x)>x$ for every $0<x<\pi/2\,$?
I managed to prove that $\bigl|e^{2 \pi  ki/n}-1\bigr|=2\sin(\pi k/n)$, I think that using the above it is possible to prove that $2\sin(x)>x$ but I'm not sure how.

答:
We work on $[0, \pi / 2]$.

Derive $f:x\mapsto 2 \sin x - x$. You get $f':x\mapsto 2\cos x -1$, and $2\cos x \geqslant 1$ is equivalent to $x\leqslant \pi / 3$, so $f$ is increasing on $[0,\pi / 3]$ and decreasing on $[\pi / 3, \pi / 2]$. And $f(0) = 0$, and $f(\pi / 2) >0$.

So for all $x\in ]0, \pi / 2[$, $2\sin x \geqslant x$


问:
Is integration by substitution always the reverse chain rule?
I understand that integration by substitution can be justified in the following way. If $I=\int f'(g(x))g'(x) \, dx$ then $I=f(g(x))+C$. If we make the substitutions
$$
u=g(x) \text{ and } du=g'(x)dx
$$
then $I$ becomes $\int f'(u)du=f(u)+C=f(g(x))+C$, which is the same. However, I often see substitutions that don't seem to take this form. For example, a common way to evaluate
$$
I=\int \sqrt{1-x^2} \, dx
$$
is by setting $x=\sin u$ (or, to be more precise, $u=\arcsin x$). The derivative of $\arcsin x$ is
$$
\frac{1}{\sqrt{1-x^2}} \, ,
$$
which does not appear in $I$. It doesn't seem like $I$ is of the form $\int f(g(x))g'(x) \, dx$. So why is this substitution justified? Are we still doing the chain rule in reverse, or is something else going on?

答:
A more sophisticated way of making substitutions is given in Michael Spivak's Calculus. Integrals of the form
$$
\int f'(g(x))g'(x) \, dx
$$
can be easily solved by substituting $u=g(x)$. In fact, many of these integrals are so simple that you can do them in your head. However, the substitution $u=g(x)$ can still be made even if the factor $g'(x)$ does not appear. In general,
$$
\int f(g(x)) \, dx \tag{*}\label{*}
$$
can be solved in the following way, provided that $g$ is one-to-one:
\begin{align}
u &= g(x) \\
x &= g^{-1}(u) \\
\frac{dx}{du} &= (g^{-1})'(u) \\
dx &= (g^{-1})'(u)du
\end{align}
This means that \eqref{*} can be transformed to
$$
\int f(u)(g^{-1})'(u) \, du \, ,
$$
which in practice often makes the integral simpler to solve than before. Hence,
$$
\int f(g(x)) \, dx = \int f(u)(g^{-1})'(u) \, du \, .
$$
The validity of this approach can be demonstrated by differentiating $\int f(u)(g^{-1})'(u) \, du$ with respect to $x$:
\begin{align}
\frac{d}{dx}\int f(u)(g^{-1})'(u) \, du &= f(u)(g^{-1})'(u)\frac{du}{dx} \\
&= f(u) \frac{dx}{du} \frac{du}{dx} \\
&= f(g(x)) \, .
\end{align}


问:
If $\lVert f(t) \rVert:[0,T] \to \mathbb{R}$ is measurable, is $f$ measurable?
Let $f:[0,T] \to X$ be a mapping to a Banach space $X$. If its norm $\lVert f(t) \rVert$ is measurable, is $f$ itself measurable? The converse is true.</p>

答: Hint: consider cases where $f$ has just two possible values.</p>


问:
Taylor development series
I don't quite understand how to get from:

$$ 3e^{t/3} =  \sum_{n=0}^{\infty} \frac{1}{3^{n-1}n!}t^n$$

to:

$$ 3e^{t/3} -3 - t = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!}t^n $$

That is what my teacher did, but I can't figure out where did the -3 and -t go.

Title: Taylor development series

Body: The expression$$\sum_{n=0}^{\infty}\frac1{3^{n-1}n!}t^n$$means$$\frac1{3^{-1}}+t+\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots$$Its first two terms are $3+t$ and therefore, since$$3e^{t/3}=\sum_{n=0}^{\infty} \frac{1}{3^{n-1}n!}t^n=\frac1{3^{-1}}+t+\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots,$$you have$$3e^{t/3}-3-t=\frac1{3\times2}t^2+\frac1{3^2\times3!}t^3+\cdots=\sum_{n=2}^\infty\frac1{3^{n-1}n!}t^n.$$


问:
How to find the resolution of these two vectors in this case?

I am getting that how did they find Angles written in Refer(b).

Like totally I am not getting which triangle they have used or drawn to find and resolve in R in the way written

答:
The Sine Rule states that for a triangle with angles $A,B,C$ opposite sides $a,b,c$ respectively, the following relation holds. $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} $$

The triangle in your problem has two angles $\alpha,\beta$. So, the third angle is $180°-\alpha-\beta=\gamma$. The sides of the triangle are $R$ (opposite $\gamma$), $R_x$ (opposite $\beta$) and $R_y$ (opposite $\alpha$). Applying the sine rule and then taking the reciprocal gives the required relation.


问:
Derive the Rodrigues' formula for Chebyshev Polynomials

Rodrigues' formula for Chebyshev Polynomials is stated as $$T_n(x)=(-1)^n2^n\frac{n!}{(2n)!}\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$$

I understand how the Rodrigues formula for all other special functions can be derived. One that for Laguerre polynomials is asked at Derive Rodrigues’ formula for Laguerre polynomials
, but that for Chebyshev Polynomials is nowhere to be found.

The generating function for the Chebyshev polynomials is

$$
g(x,z)=\frac{1-zx}{1-2zx+z^2}=\sum_{n=0}^\infty T_n(x)z^n
$$

Can I derive it starting from the generating function or is there an easier approach ?

答:
I believe that the easiest approach is to check that both $\{T_n(x)\}_{n\geq 0}$ and $\{P_n(x)\}_{n\geq 0}$ (where $T_n$ is defined in the usual trigonometric way and $P_n$ is given by your formula involving $\frac{d^n}{dx^n}$) are sequences of orthogonal polynomials with respect to the inner product
$$ \langle a(x),b(x) \rangle = \int_{-1}^{1}\frac{a(x)b(x)}{\sqrt{1-x^2}}\,dx = \int_{0}^{2\pi}a(\cos\theta)b(\cos\theta)\,d\theta$$
then check that $\deg T_n = \deg P_n = n$ and $T_n(1)=P_n(1)=1$ for any $n\geq 0$.


问:
Bijective and smooth function $\mathbb{R} \to \mathbb{R}$
I am a tutor for complex analysis this semester and my students had to show as an exercise that every holomorphic and bijective function $f: \mathbb{C} \to \mathbb{C}$ is affine linear, i.e. there exist some $a, b \in \mathbb{C}, a \neq 0$ such that $f(z) = az+ b$.

I really like this exercise and the proof I know uses some fundamental properties of $\mathbb{C}$, i.e. the fundamental theorem of Algebra, Laurent expansion and roots of unity. So to make it clear to my students how wrong this statement is in the reals, I want to give them as many counterexamples of smooth ($C^{\infty}$) bijective functions $\mathbb{R} \to \mathbb{R}$ as possible.

One can for example take $x^n$ for $n \in \mathbb{N}$ odd and sums of this, i.e. polynomials with only odd exponents.

What I am interested in now however is if we can construct examples that are not polynomials. If we allow the function to be less smooth (e.g. $C^1$), one can construct examples such as $\frac{x|x|}{2}$ but I can not find examples that are $C^{\infty}$ so far.

So my question is: do you know any bijective and $C^{\infty}$ function $\mathbb{R} \to \mathbb{R}$ that is not a polynomial?

Edit: Thank you for all your answers, I will present these to my students. I was thinking way too complicated for some reason.

答: $\tan(x)$ is smooth bijection between $(-\pi/2,\pi/2)$ to $\Bbb R$. Now choose any smooth bijection $g$ from $\Bbb R$ to $(-\pi/2,\pi/2)$. The compostion $\tan \circ g$ is a smooth bijection from $\Bbb R$ to $\Bbb R$.


答: No. We don't even need to leave Euclidean space. Let $V$ be a nonmeasurable set in $[0,T]$. Consider the map $f: \mathbb [0,T] \rightarrow \mathbb R$ that is $1$ on $V$ and $-1$ on $V^c$. Then $|f|$ is a constant function and surely measurable, but $f$ is not. </p>

Title: If $\lVert f(t) \rVert:[0,T] \to \mathbb{R}$ is measurable, is $f$ measurable?

Body: Let $f:[0,T] \to X$ be a mapping to a Banach space $X$. If its norm $\lVert f(t) \rVert$ is measurable, is $f$ itself measurable? The converse is true.</p>


Title: Taylor development series

Body:
I don't quite understand how to get from:

$$ 3e^{t/3} =  \sum_{n=0}^{\infty} \frac{1}{3^{n-1}n!}t^n$$

to:

$$ 3e^{t/3} -3 - t = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!}t^n $$

That is what my teacher did, but I can't figure out where did the -3 and -t go.

Body:
Simply, $$\sum_{n=0}^\infty a_n=a_0+\sum_{n=1}^\infty a_n=a_0+a_1+\sum_{n=2}^\infty a_n$$ and

$$\sum_{n=0}^\infty a_n-a_0-a_1=\sum_{n=2}^\infty a_n.$$


问:
How to find the integral of $\tan x - \sec x\tan x$?
$$\int(\tan x - \sec x\tan x)dx$$

I rewrote the integral like this:

$$\int\left(\frac{\sin x}{\cos x}- \sec x\tan x\right)dx$$

and using $u$-substitution for the first term I got a final answer of:

$$-\ln|\cos x| - \sec x + c$$

Where was my mistake?

答:
$$\tan x-\sec x\tan x=\frac{\sin x}{\cos x}-\frac1{\cos x}\frac{\sin}{\cos x}$$
If we first try and do the integral of $\tan x$:
$$I_1=\int\frac{\sin x}{\cos x}dx$$
$u=\cos x\Rightarrow dx=-\frac{du}{\sin x}$ and so:
$$I_1=-\int\frac1udu=-\ln|u|+C_1=-\ln|\cos x|+C_1=\ln|\sec x|+C_1$$
and now for $\sec x\tan x$:
$$I_2=\int\frac{\sin x}{\cos^2x}dx$$
$$=-\int\frac1{u^2}du=\frac1u+C_2=\frac1{\cos x}+C_2=\sec x+C_2$$
and so:
$$\int\tan x-\sec x\tan xdx=\ln|\sec x|-\sec x+C_3$$


问:
Given a linear transformation $T:V \to \mathbb R^{2 \times 2}$
Given a linear transformation $T:V \to \mathbb R^{2 \times 2}$,where $V$ is the set of complex numbers over $\mathbb R$ and $\mathbb R^{2 \times 2}$ is the set of $2 \times 2$ matrices with entries in $\mathbb R$ such that for $z=x+iy \in \mathbb C$

$$T(z)=
\begin{bmatrix}
x+7y & 5y\\
-10y & x-7y\\
\end{bmatrix}
\quad$$

Show that $T$ is an injective linear transformation.

Describe the range of $T$.

For $\lambda \in \mathbb R$ and $z_1=a+ib$ and $z_2=c+id$ we have that:

$$T(\lambda z_1+z_2)=T((\lambda a+c)+i(\lambda b+d))$$
$$=\begin{bmatrix}
(\lambda a+c)+7(\lambda b+d)& 5(\lambda b+d)\\
-10(\lambda b+d) & (\lambda a+c)-7(\lambda b+d)\\
\end{bmatrix}
\quad$$
$$=\begin{bmatrix}
\lambda a+7\lambda b& 5\lambda b\\
-10\lambda b & \lambda a-7\lambda b\\
\end{bmatrix}
\quad +\begin{bmatrix}
c+7d& 5d\\
-10d & c-7d\\
\end{bmatrix}
\quad$$
$$=\lambda T(z_1)+ T(z_2)$$

Which shows that $T$ is a linear transformation,moreover let $T(z_1)=T(z_2)$,implies that:

$$\begin{bmatrix}
 a+7 b& 5 b\\
-10 b &  a-7 b\\
\end{bmatrix}
\quad =\begin{bmatrix}
c+7d& 5d\\
-10d & c-7d\\
\end{bmatrix}
\quad$$

$$\implies b=d \implies a=c \implies z_1=z_2$$

And so $T$ is a injection.

Now we know that $\text {ker}(T)=\left\{ 0_V \right\}$ which is helpful since bu rank-nullity theorem:

$$\text{dim}V=\text { dim ker}(T)+\text { dim Im}(T)$$

Which implies that $\text { dim Im}(T)=2$ and hence $\text {  Im}  (T) \cong V$

What else can be said about the range of $T$? I don't understand what does the author mean by describe

答: Note that$$T(1)=\begin{bmatrix}1&0\\0&1\end{bmatrix}\quad\text{and that}\quad T(i)=\begin{bmatrix}7&5\\-10&-7\end{bmatrix}.$$So, since $V=\operatorname{span}(\{1,i\})$,\begin{align}\operatorname{range}(T)&=\operatorname{span}\left(\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}7&5\\-10&-7\end{bmatrix}\right\}\right)\\&=\left\{\begin{bmatrix}x+7y&5y\\-10y&x-7y\end{bmatrix}\,\middle|\,x,y\in\Bbb R\right\}.\end{align}Note that this is simply to say that $\operatorname{range}(T)=\{T(z)\mid z\in\Bbb C\}$, which is tautological. But the range of $T$ can also be described as$$\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\in\Bbb R^{2\times2}\,\middle|\,b=\frac5{14}(a-d)\wedge c=-\frac57(a-d)\right\}.$$


答: The determinant $\det T(z)$ is $x^2+y^2$ where $x,y \in \Bbb R$. So as you vary $x,y$ over the real numbers, the determinant of the images forms a paraboloid in $\Bbb R^3$. Maybe this is one such description you're looking for

Body:
Given a linear transformation $T:V \to \mathbb R^{2 \times 2}$,where $V$ is the set of complex numbers over $\mathbb R$ and $\mathbb R^{2 \times 2}$ is the set of $2 \times 2$ matrices with entries in $\mathbb R$ such that for $z=x+iy \in \mathbb C$

$$T(z)=
\begin{bmatrix}
x+7y & 5y\\
-10y & x-7y\\
\end{bmatrix}
\quad$$

Show that $T$ is an injective linear transformation.

Describe the range of $T$.

For $\lambda \in \mathbb R$ and $z_1=a+ib$ and $z_2=c+id$ we have that:

$$T(\lambda z_1+z_2)=T((\lambda a+c)+i(\lambda b+d))$$
$$=\begin{bmatrix}
(\lambda a+c)+7(\lambda b+d)& 5(\lambda b+d)\\
-10(\lambda b+d) & (\lambda a+c)-7(\lambda b+d)\\
\end{bmatrix}
\quad$$
$$=\begin{bmatrix}
\lambda a+7\lambda b& 5\lambda b\\
-10\lambda b & \lambda a-7\lambda b\\
\end{bmatrix}
\quad +\begin{bmatrix}
c+7d& 5d\\
-10d & c-7d\\
\end{bmatrix}
\quad$$
$$=\lambda T(z_1)+ T(z_2)$$

Which shows that $T$ is a linear transformation,moreover let $T(z_1)=T(z_2)$,implies that:

$$\begin{bmatrix}
 a+7 b& 5 b\\
-10 b &  a-7 b\\
\end{bmatrix}
\quad =\begin{bmatrix}
c+7d& 5d\\
-10d & c-7d\\
\end{bmatrix}
\quad$$

$$\implies b=d \implies a=c \implies z_1=z_2$$

And so $T$ is a injection.

Now we know that $\text {ker}(T)=\left\{ 0_V \right\}$ which is helpful since bu rank-nullity theorem:

$$\text{dim}V=\text { dim ker}(T)+\text { dim Im}(T)$$

Which implies that $\text { dim Im}(T)=2$ and hence $\text {  Im}  (T) \cong V$

What else can be said about the range of $T$? I don't understand what does the author mean by describe


问:
Prove that if $\lim\limits_{x \to 0} \frac{f(x)}{x}=l$ for $b ≠ 0$ then $\lim\limits_{x \to 0} \frac{f(bx)}{x}=bl$
$\lim\limits_{x \to 0} \frac{f(x)}{x}=l$   for   $b ≠ 0$   then $\lim\limits_{x \to 0} \frac{f(bx)}{x}=bl$

答:
Hint

Use the equality

$$\left\vert \frac{f(bx)}{x}-bl \right\vert = b\left\vert \frac{f(bx)}{bx}-l \right\vert$$


问:
understand the relation between orientation , in circle test and determinant
i come from a software engineering background and I'm new to computational geometry ,
I found this picture online and can't manage to understand from where the logic comes from .
I know that determinants are used to know how a transform squishes space but i don't see the link here .

答:
When a continuous curve has the implicit equation of the form $f(x,y)=0$, the sign of the LHS tells you on what side of the curve you are (by continuity, you cannot cross the curve without making the sign change).

Here the determinants enter into play because they form the implicit equations of a line by two points or a circle by three. (Cancelling a determinant expresses lineary dependency.)

$$\begin{vmatrix}x&y&1\\x_0&y_0&1\\x_1&y_1&1\end{vmatrix}=0$$

$$\begin{vmatrix}x^2+y^2&x&y&1\\x_0^2+y_0^2&x_0&y_0&1\\x_1^2+y_1^2&x_1&y_1&1\\x_2^2+y_2^2&x_2&y_2&1\end{vmatrix}=0$$


问:
integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$
$\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$

$x=2\tan^{-1}t$

$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$

$dx=\dfrac{2}{1+t^2}\,dt$

$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$

Is this the right idea? If so what do I do next?

$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$

$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$

$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$

答:
Hint:

When $\tan \dfrac{x}{2}=t$ you have $ \cos x= \dfrac{1-t^2}{1+t^2}$


答:
$$
\log|\sec(x)|=-\log|\cos(x)|
$$
If we remove the absolute value signs, then this is obvious:
\begin{align}
\log\left(\cos(x)^{-1}\right)=-\log(\cos(x)) \, .
\end{align}
This still holds for $\log|\sec(x)|$ and $-\log|\cos(x)|$. This can be verified by noting that $\log|\sec(x)|$ is equal to $\log\left(\sec(x)\right)$ or $\log\left(-\sec(x)\right)$, and whenever $\sec$ is positive, so is $\cos$.

Title: How to find the integral of $\tan x - \sec x\tan x$?

Body:
$$\int(\tan x - \sec x\tan x)dx$$

I rewrote the integral like this:

$$\int\left(\frac{\sin x}{\cos x}- \sec x\tan x\right)dx$$

and using $u$-substitution for the first term I got a final answer of:

$$-\ln|\cos x| - \sec x + c$$

Where was my mistake?


答:
You can use the change of variable $bx = y$, getting: $$\lim_{y \rightarrow 0} \frac{f(y)}{y/b} = bl$$

Note that everything is well defined (with $b \in \mathbb{R}-\{0\}$).

Title: Prove that if $\lim\limits_{x \to 0} \frac{f(x)}{x}=l$ for $b ≠ 0$ then $\lim\limits_{x \to 0} \frac{f(bx)}{x}=bl$


问:
Combining and simplifying equations
I've come across a derivation in a journal article where one of the steps is to substitute one expression into another, but I cannot simplify to their solution. The Expressions are:

Expression 1

$$
\frac{d\theta_{t}}{dt} = k_{a}q_{max}X\theta_{t}^2 -(k_{a}C_{0} + k_{a}q_{max}X+k_{d})\theta_{t} +k_{a}C_{0}
$$

Expression 2

$$
\theta_{e} = \frac{k_{a} (q_{max}X + C_{0})+k_{d} - \sqrt{\Delta}}{2k_{a}q_{max}X}
$$

In the article, it says that Expression 2 is substituted into Expression 1, to obtain Expression 3:

Expression 3

$$
\frac{d\theta_{t}}{dt} = k_{1}(\theta_{e}-\theta_{t})+k_{2}(\theta_{e}-\theta_{t})^2
$$

With the definitions

$$
k_{1}=\sqrt{\Delta}
$$
$$
k_{2} = k_{a}q_{max}X
$$

Unfortunately the authors omitted the steps in between, so to check I initially simplified Expression 1 and Expression 2 using the definitions:

$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(k_{a}C_{0} + k_{2}+k_{d})\theta_{t} +k_{a}C_{0}
$$
$$
\theta_{e} = \frac{k_{2}+ k_{a}C_{0}+k_{d} - k_{1}}{2k_{2}}
$$

Rearranging the second expression I get:

$$
k_{a}C_{0} =2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Which I substituted into the first expression to get

$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(2k_{2}\theta_{e}+k_{1})\theta_{t} +2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Multiplying out the brackets and completing the square, I get:

$$
\frac{d\theta_{t}}{dt} =k_{2}(\theta_{t}-\theta_{e})^2 - k_{2}\theta_{e}^2 -k_{1}\theta_{t}+2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Which is where I become stuck. Is there a mathematical trick other than completing the square that I am missing which will allow me to simplify this expression to obtain Expression 3?

The paper that I have obtained this derivation from is by Lui and Shen, titled "From Langmuir Kinetics to First and Second Order Rate Equations for Adsorption"

Edit

In response to a comment, Delta is defined as:

$$
\Delta=k_{a}^2(C_{0}-q_{max}X)^2+2k_{a}k_{d}(C_{0}+q_{max}X)+k_{d}^2
$$

答:
To reduce visual clutter, I'm making a few definitions:
$$\theta:=\theta_t \qquad \theta':=\frac{d\theta_t}{dt} \qquad
\phi:=\theta_e \qquad q := q_{max} \qquad a := k_a \qquad d := k_d \qquad p := k_aC_0$$
Together with the provided definitions
$$k_1 := \sqrt{\Delta} \qquad k_2:=aqX$$
the equations become
$$\begin{align}
\theta' &= k_2\theta^2-(p+d+k_2)\theta+p \tag1\\[4pt]
2k_2\phi &= p+d-k_1+k_2 \tag2 \\[4pt]
\theta' &= k_1(\phi-\theta)+k_2(\phi-\theta)^2 \tag3
\end{align}$$
As OP has encountered, simple algebraic manipulation won't get from $(1)$ and $(2)$ to $(3)$. It's required to know that
$$\begin{align}
\Delta &:= a^2(C_0-qX)^2+2ad(C_0+qX)+d^2 \tag{4} \\[4pt]
\quad\to\quad k_1^2 &=(p-k_2)^2+2d(p+k_2)+d^2 \tag{5}
\end{align}$$
There's probably a clever way to incorporate this relation, but a brute force way is to treat $(1)$ and $(2)$ as linear equations in $p$ and $d$, solve for these values, substitute into $(5)$, and simplify (dividing-through by $4k_2$). When the dust settles, we arrive at the target relation $(3)$.

Note. It's partially clever to rewrite $(5)$ as
$$\begin{align}
k_1^2 &= -4k_2p+(p+k_2)^2+2d(p+k_2)+d^2 \tag{6}\\[4pt]
\quad\to\quad 4k_2p &=(p+d+k_2)^2-k_1^2 \tag{7} \\[4pt]
&=(p+d-k_1+k_2)(p+d+k_1+k_2) \tag{8} \\[4pt]
\quad\to\quad 2p&=\phi(p+d+k_1+k_2) \tag{9}
\end{align}$$
(where we invoke $(2)$ to simplify the right-hand side of $(8)$). If we further define
$$m:=p+d+k_2 \tag{10}$$
then we have
$$\begin{align}
\theta' &= k_2\theta^2-m\theta+p\tag{1'} \\[4pt]
2k_2\phi &= m-k_1 \tag{2'} \\[4pt]
2p &= \phi(m+k_1) \tag{9'}
\end{align}$$
Eliminating $m$ and $p$ from the system to arrive at $(3)$ is then relatively straightforward.

Title: Combining and simplifying equations


问:
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$
let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$

I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove

$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$

Body:
We write the recurrence as
$$
a_{n+1}=\frac{2n+1}n a_n - \frac1n.
$$
Then we divide both sides by $\prod_{j=1}^n \frac{2j+1}j$. We obtain
$$
\frac{a_{n+1}}{\prod_{j=1}^n \frac{2j+1}j} = \frac{a_n}{\prod_{j=1}^{n-1}\frac{2j+1}j}-\frac1{(2n+1)\prod_{j=1}^{n-1} \frac{2j+1}j}.
$$
Repeatedly applying this, we have
$$
\frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=\frac{a_2}3-\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$
Note that $a_2=3a_1-1$ and
$$
\frac13+\sum_{k=2}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}=2-\frac{\pi}2,
$$
as commented on Tesla Daybreak's answer.

Then we need $a_1=2-\frac{\pi}2$ to have a bounded sequence $\{a_n\}$.

$$
\frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=2-\frac{\pi}2-(\frac13+\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j})
$$

The RHS is
$$
\sum_{k=n+1}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$

Thus,
$$
a_{n+1}=\sum_{k=n+1}^{\infty} \frac{\prod_{j=1}^n \frac{2j+1}j}{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$

After some cancellations,
$$
a_{n+1}=\frac1{2n+3} + \frac1{(2n+5)\frac{2n+3}{n+1}}+\frac1{(2n+7)\frac{2n+5}{n+2}\frac{2n+3}{n+1}}+\cdots.
$$
$$
\leq \frac1{2n+3}+\frac1{(2n+5)2}+\frac1{(2n+7)2}\leq \frac1{n+2}.
$$


问:
Explanation of proof of the transendence of $\alpha_1^{\beta_1} \dotsm \alpha_n^{\beta_n}$
I'm following this proof in the book "Transcendental Numbers" by M. Ram Murty and Purusottam Rath.

The result is a corollary of Baker's theorem.

There are a couple of things I don't understand. Why is it sufficient to show that $$\beta_1\log\alpha_1+\ldots+\beta_m\log\alpha_m=0$$ for any $\alpha_i$ and $\beta_i$ satisfying the given conditions? I can see that $$\log(\alpha_1^{\beta_1}\dotsm\alpha_m^{\beta_m})$$ is equal to the left hand side of the above, but I'm not sure why/how this is useful, or where the contradiction comes in.

Secondly, I understand that the linear independence of $\beta_1, \dotsc, \beta_m$ implies that
$$c_mA_1=\dotso=c_mA_{m-1}=A_1c_1+\dotsb+A_{m-1}c_{m-1}=0$$
But how does this imply that $A_1=\dotso=A_{m-1}=0$?

Any help would be appreciated.

Edit: added missing "$=0$" on the last line

答:
Suppose $\alpha_1^{\beta_1}\cdots\alpha_m^{\beta_m}$ is algebraic. Call it $\alpha_{m+1}$; then $$\beta_1\log\alpha_1+\cdots+\beta_m\log\alpha_m+\beta_{m+1}\log\alpha_{m+1}=0$$ with $\beta_{m+1}=-1$ (as is written in the text), contrary to what's being shown.

It's supposed that $c_m\neq 0$ (see the text). Thus, $A_1=\ldots=A_{m-1}=0$.


问:
Does the branch of the initial value determine the branch for the whole curve?
Let $\Omega$ be a connected and simply-connected open subset of $\mathbf{C}$ which does not contain the origin. Let $\gamma :[0,1] \to \Omega$ be a simple smooth curve and let $\zeta := \gamma(0)$. Let $\phi:[0,1] \to \mathbf{C}$ be a smooth map such that $e^{\phi(t)} = \gamma(t)$ for all $t \in [0,1]$ (i.e., at each point $\phi$ is a logarithm of $\gamma$). Since $\Omega$ is connected and simply-connected, there exists a unique single-valued branch $"f"$ of the logarithm defined on $\Omega$ such that $f(\zeta) = \phi(0)$. Does it necessarily follow that $\phi$ is the restriction of $f$ to the curve $\gamma$?

答:
Hint

The answer is positive and follows from:

The fact that a branch of the logarithm is uniquely defined on an open disk $D(a,r)$ such that $0 \notin D(a,r)$ by its value $f(a)$ at $a$.

The image of $[0,1]$ under $\gamma$ is path connected, hence connected.

Moreover this image is included in the simply connected open subset $\Omega$ for which $0 \notin \Omega$.


问:
Integration and periodic divergent series
Consider the divergent infinite series

$$ \sum_{n=1}^\infty 5^n \cos(nt) $$

where $t \in \mathbb{R}$.  I don't believe it has a closed form.  However, if I take the integral with respect to $t$:

$$ \int_{-\pi}^\pi \sum_{n=1}^\infty 5^n \cos(nt) \, dt = \sum_{n=1}^\infty 5^n \int_{-\pi}^\pi \cos(nt) \, dt$$
$$ = \sum_{n=1}^\infty 5^n \cdot 0 = 0$$

So even though the series is divergent it still has a finite integral.  This seems to be a consequence of the fact that it's periodic over the integration domain: the infinities sort of cancel each other out.

This can extend also to the product of series:
$$ \int_{-\pi}^\pi \left( \sum_{n=1}^\infty 5^n \cos(nt) \right) \left(\sum_{n=1}^\infty \frac 1 {7^n} \cos(nt) \right) \,dt$$
$$ = \pi \sum_{n=1}^\infty \left( \frac 5 7 \right)^n = \frac 5 2 \pi$$

I'm trying to understand the underlying properties at work here.  Specifically, is there a way to come up with a "closed form" for a periodic infinite divergent series which behaves the same under integration (for instance, something I could use with integration by parts) without actually being the limit of the partial sums of $f(t)$?  That is, something like a proxy function with the infinities sort of preemptively normalized out?

答:
If we denote the series:
$$S_N(t)=\sum_{n=1}^N5^n\cos(nt)$$
$$s_N(t)=\sum_{n=1}^N5^n\sin(nt)$$
then we can see that:
$$S_N(t)+is_N(t)=\sum_{n=1}^N5^ne^{int}=\sum_{n=1}^N\left(5e^{it}\right)^n$$
Now we have a geometric series so we can say:
$$A_N=S_N(t)+is_N(t)=\frac{\left(5e^{it}\right)^{N+1}-1}{5e^{it}}$$
now we can try and split this up into real and imaginary parts:
$$A_N=e^{-it}\frac{(5e^{it})^{N+1}-1}{5}=(5e^{it})^N-\frac{e^{-it}}{5}$$
$$A_N=5^N\left[\cos(Nt)+i\sin(Nt)\right]-\frac15\left[\cos(t)-i\sin(t)\right]$$
and so we can deduce:
$$S_N(t)=5^N\cos(Nt)-5^{-1}\cos(t)$$
so for certain values of $t$ it is possible that $S_{N\to\infty}(t)$ converges


问:
$A\rightarrow(\neg\neg B\rightarrow B)$ in intuitionistic logic
I have a basic question about intuitionistic logic. Here are two propositions in intuitionistic logic, where $B$ is not $\neg\neg$-stable (that is, it is not the case that $\neg\neg B\rightarrow B$):
$$(1). A\rightarrow(\neg\neg B\rightarrow B),\quad\quad\quad (2). A\rightarrow B.$$
By weakening rule, (2) implies (1). This is trivial. But what about the other direction? It appears to me that if we assume (1), since $B$ is not $\neg\neg$-stable, then (1) holds only if $A\rightarrow B$. Consequently, we shall infer (2) from (1)? Is that right? Thanks a lot!

答:
This does not constitute a valid inference. Choose a proposition $B$ for which neither $(\neg\neg B \rightarrow B) \rightarrow B$ nor $\neg\neg B \rightarrow B$ are provable, then set $A$ to $\neg\neg B \rightarrow B$.

You then have (1) $(\neg\neg B \rightarrow B) \rightarrow \neg\neg B \rightarrow B$, but not (2) $(\neg\neg B \rightarrow B) \rightarrow B$.

edit: How do we know that a suitable $B$ exists? Just take the empty theory over a language with one nullary relation symbol $R$, and verify that $R$ already gives rise to a suitable proposition.


问:
{$\frac{1}{x}$} , $0<x\leq1$ in terms of {$x$}, $x\geq 1$, {$x$} is fractional part of $x$
$f(x)$ $=$ {$\frac{1}{x}$}, $0<x\leq1$ where {$x$} denotes the fractional part of $x$.

$g(x) =$ {$x$}, $x\geq 1$

I want an expression for $f(x)$ in terms of x and $g(x)$.

My try-

If $x\in \mathbb{Z} , x\geq 1$ , g(x)={x}=0

f(x)={$\frac{1}{x}$}=$\frac{1}{x}$
$$f(x)=\frac{1}{x}+g(x)$$
What happens when x is a fraction?
Jacobi theta function $\theta(x)= \sum_{n=-\infty}^\infty e^{-n^2\pi x}$ This function satisfies the functional equation $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$. I want a similar functional equation of f(x) in terms of g(x)

答: If $f(x)=\left\{\frac{1}{x}\right\}$ and $g(x)=\{x\}$, then $f(x)=g(\frac{1}{x})$, i.e. $f=g\circ h$ where $h(x)=\frac{1}{x}$. Perhaps this is what the author of the problem had in mind, and you have been overthinking it?


问:
Calculate an integral using differentiating under the integral sign (Feynman's Trick)

Calculate $$\int_{0}^{441}{\frac{\pi\sin(\pi \sqrt{x})}{\sqrt{x}}}{dx}.$$

My approach: I know that
\begin{eqnarray*}
\int{\frac{\pi\sin(\pi \sqrt{x})}{\sqrt{x}}}dx&=&\pi \int \frac{\sin(\pi\sqrt{x})}{\sqrt{x}}\\
&=&2\int \sin(u)du,\quad u=\pi \sqrt{x}\\
&=&-2\cos(u)+C\\
&=&-2\cos(\pi \sqrt{x})+C
\end{eqnarray*}
so $$\int_{0}^{441}{\frac{\pi\sin(\pi \sqrt{x})}{\sqrt{x}}}{dx}.=4$$
but, how can solve this problem using differentiating under the integral sign?

答:
Let

$$ I(a)=\int_0^c dx\ \frac{\sin(ax^{1/2})}{x^{1/2}}$$

We've chosen $a$ so that taking the derivative will simplify$^\dagger$ the expression.

$$ \partial_aI(a)=\int_0^c dx \ \cos(ax^{1/2})$$

Now we perform the $x$ integral

$$ \partial_aI(a)= \frac{2}{a^2} \left( a c^{1/2} \sin(ac^{1/2})+\cos(ac^{1/2})-1 \right) $$

Then the $a$ integral

$$ I(a)=\frac{2}{a} \left( 1- \cos(ac^{1/2})\right) +C$$

To find $C$ note that $I(0)=0$ from the first equation. In this case $C=0$ and we have

$$ I(a)=\frac{2}{a} \left( 1- \cos(ac^{1/2})\right)$$

Setting $a=\pi, c=441$, and multiplying by $\pi$ gives exactly $4$.

$\dagger$ In this case, it's not clear if the 'simplified' integral is actually simpler to do on paper. Still, this is the general idea behind Feynman's trick.


问:
Curvature and Torsion problem
Calculate the curvature and torsion of
$$x= e^t\sin(t),\quad y= e^t\cos(t),\quad z= e^t$$

I'm not sure if I am doing this correctly since I am getting quite complicated results.
But I understand that curvature $$\kappa= \frac{\left|X' \times X"\right|}{v^3}$$ and $$\tau=\frac{(X' \times X")\cdot X'''}{v^3\kappa^2}.$$

Please confirm that these formulas are correct and how to use them properly.
Thank you!

答:
In these formulas, we hav that $X = (x(t), y(t), z(t))$. Therefore, we begin by calculating the derivatives of the vectors component-wise with respect to $t$ and have

$$X' = (x'(t), y'(t),z'(t)) = \left( e^t \sin(t) + e^t \cos(t) , e^t \cos(t)  - e^t \sin(t), e^t\right)$$
$$X^{\prime\prime} = \left(2e^t \cos(t) ,  - 2e^t \sin(t), e^t\right)$$
$$X^{\prime\prime\prime} = \left( 2e^t \cos(t)  - 2e^t \sin(t),  -2e^t \sin(t) - 2e^t \cos(t) , e^t\right)$$

To compute the curvature, we start by computing

\begin{align} \left| X' \times X'' \right|
& = \left| (e^{2 t}\cos(t)+e^{2 t} \sin(t), e^{2 t} \cos(t)-e^{2 t} \sin(t), -2 e^{2 t} \cos^2(t)-2 e^{2 t} \sin^2(t)) \right|  \\
&= \sqrt{6}e^{2t}\end{align}

Also, $$v^3 = \left| X' \right|^3 = \sqrt{(e^t \sin(t) + e^t \cos(t))^2 + (e^t \cos(t)  - e^t \sin(t))^2 + (e^t)^2}^3 = 3\sqrt{3}e^{3t}$$

Therefore we have that the curvature is given by $$\kappa = \frac{\left| X' \times X'' \right| }{v^3 } = \frac{\sqrt{6}e^{2t}}{3\sqrt{3}e^{3t}} = \frac{\sqrt{2}}{3}e^{-t}$$

For the torsion, we have a few more calculations. Now

$$ (X' \times X^{\prime\prime} ) \cdot X^{\prime\prime\prime} = (e^{2 t}\cos(t)+e^{2 t} \sin(t), e^{2 t} \cos(t)-e^{2 t} \sin(t), -2 e^{2 t} \cos^2(t)-2 e^{2 t} \sin^2(t)) \cdot  ( 2e^t \cos(t)  - 2e^t \sin(t),  -2e^t \sin(t) - 2e^t \cos(t) , e^t ) = -2e^{3t}$$

Also

$$v^3 \kappa^2 = 3\sqrt{3}e^{3t} \left( \frac{\sqrt{2}}{3}e^{-t} \right)^2 = \frac{2\sqrt{3}}{3} e^t$$

Therefore the torsion is $$\tau = \frac{-2e^{3t}}{\frac{2\sqrt{3}}{3} e^t} = -\sqrt{3}e^{2t}$$

Title: Curvature and Torsion problem


问:
The hypotenuse of a triangle is shortest when the other two legs are equal. Prove this using the Calculus of Variations.
I started off my proof by of course stating that the different right triangles I would be comparing should have the same area A. I was able to show that what the question is asking is true visually and computationally using the Pythagorean Theorem, and even using the triangle inequality, but I don't really know how to set it up in such a way that I can use the Euler-Lagrange equation in order to prove that this is true.

Equations:

$$\frac{1}{2}xy=A$$

So

$$y(x) = 2A/x \Longrightarrow y'(x) = -\frac{2A}{x^2}$$

答:
Let $A=\frac{xy}{2}$ be constant. Then we have $x^2 = (\frac{2A}{y})^2$.

The hypothenus is $h(y)=\sqrt{x^2 + y^2} = \sqrt{y^2 + \frac{4A^2}{y^2}}$, so $h'(y)=\frac{2y-\frac{8A^2}{y^3}}{2\sqrt{y^2 + \frac{4A^2}{y^2}}}$.

Then $h$ has got a minimum when $h'(y)=0$, so when $2y=\frac{8A^2}{y^3}$, so when $y^4 = 4\cdot \frac{x^2y^2}{4}$, so when $y=x$.


问:
Is there any software which does Ring Computations?
Are there any software that is able to compute the following problems:

Verifying if or not a subset of a ring is an ideal.

Generating all ideals of a given finite ring.

Finding maximal multiplicative closed subset of a finite ring, etc.

I found this question previously posted, but this deals with Groups.

答: Macaulay 2 is good for these problems. For example for the first one, after defining the Ring, you can check is Ideal or not with "isIdeal" command.</p>


问:
Find constants in system of equations
In the below system of equations, find the values of constants $q_1, q_2, q_3 \in \mathbb R$ so that the system has no solution.

$$\begin{cases}
2x&+y-z&+2s &= q_1\\
3x&+y-2z&+s &= q_2\\
x&+y+3z&-s &= q_3
\end{cases}$$

In order to eliminate some of the unknowns, we add the 1st with the 3rd and subtract the 2nd, so we get:
$y+4z = q1+q3-q2$
By similar more ways (multiplying, adding and subtracting) we get some more such equations.

I don't see how this system can have no solution; plus that we are one equation short (3 equations - 4 unknowns).

Any assistance is much appreciated.

EDIT:

(Apologies but I am not familiar with linear algebra!!)

What I believe I must do is the following:

Reduce the unknowns by one (don't see how), so as to have a 3x3 system. Calculate the determinant, for which it must be $D \neq 0$.
Then calculate $D_x, D_y, D_z$ assuming we have eliminated s.
Then, for the system NOT to have a solution, it must be $D=0$ and $D_x, D,y, D_z \neq 0$.

Any further help?

答:
The system can be written as$$\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} \begin{pmatrix}
x\\ y \\ z \\ s\end{pmatrix} = \begin{pmatrix}
q_1\\ q_2 \\ q_3\end{pmatrix}.$$ Now since we have that $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =3$$ the system must be consistent, independently of the values of $q_1,q_2,q_3$. (Obviously it is $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2 &q_1\\
3&1 &-2& 1 & q_2\\
1&1& 3&-1 & q_3\end{pmatrix}=3.)$$

EDIT

$$\begin{cases}
2x&+y&-z&+2s &= q_1\\
3x&+y&-2z&+s &= q_2\\
x&+y&+3z&-s &= q_3
\end{cases} \iff \begin{cases}
2x&+y&-z &= q_1-2s\\
3x&+y&-2z&= q_2-s\\
x&+y&+3z&= q_3+s
\end{cases} $$

$$\iff  \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
-x&&+4z&= q_3-q_1+3s
\end{cases}$$

$$\iff  \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
&&3z&= q_3+q_2-2q_1+4s
\end{cases}$$

From the last equation we get

$$z=\dfrac{q_3+q_2-2q_1+4s}{3}.$$

Now $$x=z+q_2-q_1+s=\dfrac{q_3+4q_2-5q_1+7s}{3}.$$

Finally we get $$y=z-2x+q_1-2s=\dfrac{-q_3-7q_2+11q_1-16s}{3}.$$ That is, the system is consistent independently of the values of $q_1,q_2,q_3.$

Title: Find constants in system of equations


答:
Yes. For all $t \in [0, 1]$ is
$$
 e^{\phi(t) - f(\gamma(t))} = \frac{e^{\phi(t)}}{e^{f(\gamma(t))}} = \frac{\gamma(t)}{\gamma(t)} = 1
$$
because $e^{f(w)} = w$ holds for all $w \in \Omega$. It follows that the continuous function
$$
 [0, 1] \ni t \mapsto \frac{1}{2 \pi i }(\phi(t) - f(\gamma(t)))
$$
takes only integer values on the connected interval $[0, 1]$, and is therefore constant. But $\phi(0) = f(\gamma(0))$ per the construction, so that $\phi(t) = f(\gamma(t))$ holds for all $t$.

Title: Does the branch of the initial value determine the branch for the whole curve?

Body: Let $\Omega$ be a connected and simply-connected open subset of $\mathbf{C}$ which does not contain the origin. Let $\gamma :[0,1] \to \Omega$ be a simple smooth curve and let $\zeta := \gamma(0)$. Let $\phi:[0,1] \to \mathbf{C}$ be a smooth map such that $e^{\phi(t)} = \gamma(t)$ for all $t \in [0,1]$ (i.e., at each point $\phi$ is a logarithm of $\gamma$). Since $\Omega$ is connected and simply-connected, there exists a unique single-valued branch $"f"$ of the logarithm defined on $\Omega$ such that $f(\zeta) = \phi(0)$. Does it necessarily follow that $\phi$ is the restriction of $f$ to the curve $\gamma$?


问:
Which real functions have their higher derivatives tending pointwise to zero?
Let $\mathrm C^\infty\!(\Bbb R)$ be the space of infinitely differentiable functions $f:\Bbb R\rightarrow\Bbb R$, and define the subspace$$A:=\{f\in\mathrm C^\infty\!(\Bbb R):(\forall x\in \Bbb R)\lim_{n\rightarrow\infty} f^{(n)}(x)=0\},$$where $f^{(n)}$ is the $n$th derivative of $f\;(n=0,1,\dots).$ Clearly all polynomial functions are in $A$. Are any others?

Edit: Alfonso has answered this question well, but is there any characterization of $A$ in terms of familiar types of function?

答:
Consider an analytic function
$$\tag0f(x)=\sum_{n=0}^\infty \frac{a_n}{n!}x^n.$$
As $f^{(n)}(0)=a_n$, a necessary condition for such $f$ to be an example is that $a_n\to 0$.

But $a_n\to 0$ is also sufficient.
Indeed, with $b_n:=\sup_{k\ge n} |a_k|$ we have $b_n\to 0$ and hence
$$\begin{align}|f^{(n)}(x)|&= \left|\sum_{k=0}^\infty \frac{a_{n+k}}{k!}x^k\right|\\
&\le\sum_{k=0}^\infty\frac{|a_{n+k}|}{k!}|x|^k\\
&\le b_n\sum_{k=0}^\infty\frac{1}{k!}|x|^k\\&=b_ne^{|x|}\to 0. \end{align} $$
(Note that the calculation for $f^{(0)}(x)$ shows that $f$ is entire to begin with).


问:
Code that corrects a length $s$ burst error
Show that exists an $(n,k)=(155,135)$ binary code that corrects burst errors of length $6$.

P.S. A  Code with length $n$  fixes burst error (pattern) of length $s$ if the error set of this code consists of sets of length $n$ whose non-zero components are located in $s$  consecutive bits, that is, in each such set, the locations of the first and last non-zero components differ by no more than $s - 1$.

答:
The field of 32 elements, $GF(32)$ or $\Bbb{F}_{32}$, whichever name you prefer, has $31$ non-zero elements. Each element can be represented by a block of five bits in several ways. As $155=5\cdot31$, this is just what the doctor ordered.

A burst of length $6$ cannot disturb more than two (consecutive) $5$-bit symbols. Therefore a code correcting two symbol errors can handle bursts like this.

The Reed-Solomon codes are MDS, so a $32$-ary RS code with $(n,k)=(31,27)$ has minimum distance $d=n-k+1=5$. Therefore it can correct any two symbol errors, and the problem is solved as $135=5\cdot27$.

That RS-code does not care whether the two errorneous symbols are consecutive or not. Making this a bit of a cheat solution. But, hey, it works :-)


问:
Taylor series development with integral
So I've done this to a series
$$ \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!} \int_0^xt^{n-2} = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!}x^{n-2} $$
But looking at the solution it says $$ \sum_{n=2}^{\infty} \frac{1}{3^{n-1}n!} \int_0^xt^{n-2} = \sum_{n=2}^{\infty} \frac{1}{3^{n-1}(n-1)n!}x^{n-1} = \sum_{n=1}^{\infty} \frac{1}{3^{n}(n+1)!n}x^{n}$$

So my question is how can I simplify my solution to obtain it in that way, because I don't understand where does the $ (n-1)n! $ comes from

Title: Taylor series development with integral

Body: Note that $$\int_0^x t^{n-2} dt = \frac{t^{n-1}}{n-1} \bigg | _0^x = \frac{x^{n-1}}{n-1} \ne x^{n-2} $$


问:
Not sure how to go about solving this integral
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$

I've let $u=x^2+3$ but can't seem to get the right answer.

Really not sure what to do.

答:
Or please try putting $x = \sqrt{3} \tan\theta$. Then you have $dx = \sqrt{3}\cdot\sec^{2}\theta \ d\theta$. So you have your integral as:

\begin{align*}
\int\frac{1}{x^{2}+3} \ \text{dx} &= \int \frac{\sqrt{3} \cdot \sec^{2}\theta}{3 (1+\tan^{2}\theta)} \ \text{d}\theta \\ &=\frac{1}{\sqrt{3}} \int \text{d}\theta
\end{align*}


问:
Derive the Rodrigues' formula for Chebyshev Polynomials

Rodrigues' formula for Chebyshev Polynomials is stated as $$T_n(x)=(-1)^n2^n\frac{n!}{(2n)!}\sqrt{1-x^2}\frac{d^n}{dx^n}(1-x^2)^{n-1/2}$$

I understand how the Rodrigues formula for all other special functions can be derived. One that for Laguerre polynomials is asked at Derive Rodrigues’ formula for Laguerre polynomials
, but that for Chebyshev Polynomials is nowhere to be found.

The generating function for the Chebyshev polynomials is

$$
g(x,z)=\frac{1-zx}{1-2zx+z^2}=\sum_{n=0}^\infty T_n(x)z^n
$$

Can I derive it starting from the generating function or is there an easier approach ?

答: An alternative method is to check that the $T_n$, as you've defined them via the Rodrigues formula, have $\deg T_n=n$, $T_n(1)=1$ and $y=T_n$ solves the Chebyshev DE $$(1-x^2)y''-xy'+n^2y=0.$$


问:
Explanation of proof of transcendence of $e$
I'm following the proof for the transcendence of $e$ from the book "Transcendental Numbers" by M. Ram Murty and Purusottam Rath. I am struggling to understand the final few lines.

As far as I understand, by the triangle inequality we have

$$
|J|\leq \sum_{k=0}^n |a_k| |I(k,f)|.
$$
Now $I(0,f)=0$, so the sum starts from $k=1$. So then

$$
\sum_{k=1}^n |a_k| |I(k,f)|\leq nA\max_{1\leq k\leq n}|I(k,f)| \leq nA\max_{1\leq k\leq n}\{ke^kF(k)\} = n^2Ae^n\max_{1\leq k\leq n}F(k)
$$
where $A:=\max\limits_{1\leq k\leq n}|a_k|$.

This is where I'm stumped because I don't know how to bound $F(k)$. I've read some other proofs for help, but they all slightly differ at this step.

Also, where does the "elementary" observation that $e^p\geq p^{p-1}/(p-1)!$ come from? And why is this needed? Can't we just say that the factorial function grows faster than the exponential function so that eventually the inequality $(p-1)!\leq (2n)!^p$ would be false?

Any help would be greatly appreciated.

答:
As regards the bound on $F$, we have that
$$F(x)=x^{p-1}(x+1)^p\cdots(x+n)^p$$
and therefore, for $k=1,\dots,n$,
$$F(k)k=k^{p}(k+1)^p\cdots(k+n)^p\leq ((2n)!)^p.$$
Hence
$$\sum_{k=0}^n|a_k|e^kF(k)k\leq Ae^n\sum_{k=1}^nF(k)k\leq Ae^n\sum_{k=1}^n((2n)!)^p=Ane^n((2n)!)^p.$$

Note that
$$e^p=\sum_{k=0}^{\infty}\frac{p^k}{k!}>\sum_{k=0}^{p-1}\frac{p^k}{k!}=1+p+\frac{p^2}{2!}+\dots +\frac{p^{p-1}}{(p-1)!}>\frac{p^{p-1}}{(p-1)!}.$$
The contradiction follows from the fact that
$$p^{p-1}e^{-p}\leq Ane^n((2n)!)^p$$
does not eventually hold. Indeed, when $p$ is large enough, the LHS becomes greater than the RHS because
$$\lim_{p\to +\infty}\frac{p^{p-1}}{Ane^{p+n}((2n)!)^p}=+\infty.$$
You are right, it is not needed and your argument works as well.


问:
Find probability given mutual independence (of 3 events)
A bus operator runs two bus routes from station
to station
. The first route is a direct bus, bus
. The second route involves changing buses at an intermediate station: passengers take bus
from station
to station
and then bus
from station
to station
. The events of the three buses running are mutually independent and each bus runs with probability 1/3
. Which of the following gives the probability that a person can travel from station
to station
?

My working out
$P(AuBuC)=(1/3)+ (1/3)+ (1/3)-(1/3(1/3)-(1/3)(1/3)-(1/3)(1/3)+(1/3)(1/3)(1/3)$

Which gives me 19/27 , however this is incorrect, any hints to help me?

答: You actually need to calculate $$P( A \cup (B \cap C)) = P(A) + P(B\cap C) -P(A\cap (B\cap C)) $$ as to go from $X$ to $Y$, you either need bus $A$ to run, or both buses $B$ and $C$ to run.


问:
Order of centralizer and cyclic group over $(G,\ast)$
I'm struggling to answer this question,

Let $(G,\ast)$ be a group with $n$ elements and $a\in G$ with the following property
$ab=ba \iff b\in \{ a^k\mid k\in \mathbb{Z} \}$,
if $m$ is the order of $a$ in $G$, show that at least $\frac{n}{m}$ elements of order $m$ exist in $G$.

To me, it initially looks like something related to the centralizer and cyclic group, maybe $Z(x)=\langle x \rangle$? Some research landed me into thinking I might have to use something like if $\frac{G}{Z(G)}$ is cyclic $\implies G$ is abelian?

Any help would be appreciated.

答: The property in the claim means that $C_G(a)=\langle a\rangle$, where $C_G(a)$ is the centralizer of $a$ in $G$. Now, $C_G(a)=\operatorname{Stab}(a)$ with respect to the action by conjugation of $G$ on itself. Therefore (orbit-stabilizer theorem), all the $\frac{n}{m}$ elements of the orbit of $a$ are conjugate to $a$, and hence they have the same order of $a$ (see e.g. here), namely $m$.


问:
Using Square area finding quadrilateral area
Area of square ABCD is 169 and that of square EFGD is 49. Find area of quadrilateral FBCG

I am stuck and just thinking which way can be helpful for me finding this area of quadrilateral FBCG. Instead of hint can please suggest me perfect answer , so using this idea answer, i will solve other questions which are near about similar.

答:
By symmetry, the quadrilaterals $FBCG$ and $FBAE$ are congruent, so they have the same area. Their combined area is the area of the region which is in the big square but outside the little square. So their combined area is $169-49$, that is, $120$. It follows that quadrilateral $FBCG$ has area $60$.

Remark: The above is the quickest way to find the answer.

But we can do it in various other ways, that may be useful to you in other problems.   For example, the little square has area $49$, so it has sides $7$ each. The big square has area $169$, so has sides $13$.

It follows that the height $CG$ of the trapezoid $FBCG$ is $13-7=6$. The two bases are $7$ and $13$. So by a standard formula for the area of a trapezoid, the area is $(6)\left(\frac{7+13}{2}  \right)$.

Or else you can think of the region $FBCG$ as made up of a $6\times 7$ rectangle, with half of a $6\times 6$ square on top. That gives area $(6)(7)+\left(\frac{1}{2}\right)(6)(6)$.

Another computational approach is by using triangles. Because of your picture, we calculate the area of $ABFE$, which is the same as the area of our target region. This is made up, as your picture shows, of two triangles. Each has height $13-7=6$. Their bases are $7$ and $13$, Now huse the standard formula for the area of a triangle.


问:
Ring of rational functions for reducible variety
Let $X$ be some affine algebraic variety over $\mathbb{k}$ (i.e. some closed subset in $\mathbb{A}_\mathbb{k}^n$). First suppose $X$ to be irreducible. Then the algebra $\mathbb{k}[X]$ is a domain and we can consider the field of rational functions $\mathrm{Quot}_{\mathbb{k}[X]}=\mathbb{k}(X)$. Could you explain me how to build an analogue of this field in the case when $X$ is not necessarily irreducible? Then $\mathbb{k}[X]$ must not be a domain and we are to build some kind of localization?

Also, what is the destination of rational functions? Why we cannot be satisfied with only regular maps and regular functions?

答:
This question may have general interest. There is an exercise in Matsumura (exercise 9.11 page 70) saying the following:

Let $A$ be a noetherian ring with the propery that $A_{\mathfrak{p}}$ is a domain for all prime ideals $\mathfrak{p}$. Let $\mathfrak{p}_1,..,\mathfrak{p}_l$ be the minimal prime ideals of $A$. It follows there is an isomorphism of rings

I1. $p:A \cong A/\mathfrak{p}_1\oplus \cdots \oplus A/\mathfrak{p}_l:=A_1\oplus \cdots \oplus A_l$

defined by

$p(a):=(\overline{a}, ..., \overline{a})$. This gives at the level of schemes an isomorphism

$Spec(A) \cong Spec(A_1) \cup \cdots \cup Spec(A_l)$

where the union is the "disjoint union".

This gives a general formula for the total ring of fractions $Tot(A)$:

I2. $Tot(A)\cong K(A/\mathfrak{p}_1)\oplus \cdots \oplus K(A/\mathfrak{p}_l)$

where $K(A/\mathfrak{p}_i)$ is the quotient field of the integral domain $A/\mathfrak{p}_i$.

Question: "Then k[X] must not be a domain and we are to build some kind of localization?"

Answer: The "total ring of fractions" is defined on page 21 in Matsumura's book "Commutative Ring Theory". This is a construction generalizing the "quotient field" defined for integral domains. If the ring $k[X]$ satisfies formula I1, it follows I2 gives a general formula for $Tot(k[X])$.

PS: I have not done the exercise myself.