问:
Basis for Kernel and image of map
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Find the basis for the kernel and the image. Find their dimension.

my thoughts

I have shown it is not a isomorphism, that $V$ is a subspace and - I think the kernel would be $\begin{pmatrix}
1 &-1  &2 \\
 1&0  &-1 \\
 0&  1& -3
\end{pmatrix}$ and after reducing to reduced echelon form I get that the basis would be $(1,3,1)$ with dimension $1$ because of the free variable $x_3$. Now, for the image, I have no idea since I do not know how to find the image in the first place to use Gaussian elimination.

答: By definition$$\ker(f)=\left\{\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\in V\,\middle|\,f\left(\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\right)=\begin{bmatrix}0\\0\end{bmatrix}\right\}.$$So, you can compute it by solving the system$$\left\{\begin{array}{l}x_1-x_3=0\\x_2-3x_3=0\end{array}\right.$$and searching for solutions in $V$. You will get that$$\ker(f)=\left\{\begin{bmatrix}t\\3t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$It follows now from the rank-nullity theorem that $\dim\operatorname{Im}(f)=1$. So, and since$$f\left(\begin{bmatrix}1\\1\\0\end{bmatrix}\right)=\begin{bmatrix}1\\1\end{bmatrix},$$you have$$\operatorname{Im}(f)=\left\{\begin{bmatrix}t\\t\end{bmatrix}\,\middle|\,t\in K\right\}.$$Finally, it is clear that$$\left\{\begin{bmatrix}1\\3\\1\end{bmatrix}\right\}\quad\text{and}\quad\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$$are bases of $\ker(f)$ and of $\operatorname{Im}(f)$ respectively.

Title: Basis for Kernel and image of map


问:
Simulating Brownian motion in R. Is this correct?
I have a Weiner process $\{W(t)\}_{t\ge0}$ with $\sigma^2=\text{Var}(W(1))=1$. For a real constant $\epsilon>0$ consider the differential ratio process $\Delta_\epsilon=\{\Delta_\epsilon(t)\}_{r>0}$ given by
\begin{equation}
\Delta_\epsilon(t) = \frac{W(t+\epsilon)-W(t)}{\epsilon}, \quad \text{for} \ t>0.
\end{equation}

I want to simulate a sample path of $\{\Delta_\epsilon(t)\}_{t\in(0,1]}$ for small $\epsilon>0$ in Rstudio.

Simulating a basic Weinerprocess/Brownian motion is easy in R, one can do it by the function rweiner() or by plotting the cumulative sum of standard normally distributed variables. The part that confuses me is how to simulate the $W(t+\epsilon)$. What I've done is used rwiener() and simply passed the end parameter to $1+\epsilon$.

```<code>epsilon = 10e-1
weinerEps = rwiener(end = 1+epsilon, frequency = 1000)
weiner = rwiener(end = 1, frequency = 1000)

Delta = (wienerEps - wiener)/epsilon

plot(Delta , type="l")
</code>```

I'm not sure this is correct though.

答:
Well, if you have a difference of $W(t+\epsilon)$ and $W(t)$, don't you want the same realization of $W$ before you take the difference?

I think you should run $\texttt{rwiener}()$ once, store the result, and then pull out both $W(t+\epsilon)$ and $W(t)$, subtract and then divide by $\epsilon$.

```<code>library('e1071')
epsilon = 1e-1  # 1/10=1e-1 and not 10e-1=1.
freq    = 1000
wiener  = rwiener(end = 1+epsilon, frequency = freq)
w       = wiener[1:1000];
w_eps   = wiener[(1+freq*epsilon):(1000+freq*epsilon)];
Delta   = (w_eps - w)/epsilon
plot(Delta , type="l")
</code>```

ContentLicense: CC BY-SA 4.0

Body:
Well, if you have a difference of $W(t+\epsilon)$ and $W(t)$, don't you want the same realization of $W$ before you take the difference?

I think you should run $\texttt{rwiener}()$ once, store the result, and then pull out both $W(t+\epsilon)$ and $W(t)$, subtract and then divide by $\epsilon$.

library('e1071')
epsilon = 1e-1  # 1/10=1e-1 and not 10e-1=1.
freq    = 1000
wiener  = rwiener(end = 1+epsilon, frequency = freq)
w       = wiener[1:1000];
w_eps   = wiener[(1+freq*epsilon):(1000+freq*epsilon)];
Delta   = (w_eps - w)/epsilon
plot(Delta , type="l")


问:
Specific expectation of non-negative random variable
Let $T$ a random variable of non-negative integers. Suppose that there is $n \in \mathbb{N}$ and there is $\alpha>0$ such that for all $k \in \mathbb{N}$ we have
$$\mathbb{P}(T >kn) \le (1-\alpha)^k.$$
Prove that
$\mathbb{E}[T] \le \frac{n}{\alpha}$.

I tried to use the sum formula for expectation of nonnegative integer random variables
$$\mathbb{E}[T] = \sum_{i \ge 0}{\mathbb{P}(T > i)},$$
but I can't get a '$\le$' innequalty.

答:
Recall from real analysis that $$\sum_{k=1}^{\infty} (1-a)^k = \frac{1-a}{a}\quad \text{for}\quad 0<a\le1.$$
Consider the random variable $n^{-1} T$. We can split its expectation up as follows: $$E[n^{-1}T] = E[n^{-1} T\cdot \mathbf{1}_{\{T\ge n\}}] + E[n^{-1} T\cdot \mathbf{1}_{\{0\le T<n\}}].$$
Observe that $$E[n^{-1} T\cdot \mathbf{1}_{\{0\le T< n\}}]\le 1\cdot P(0\le T<n) = 1 - P(T\ge n).$$ On the other hand, $S:=\lceil n^{-1} T\rceil\cdot \mathbf{1}_{\{T\ge n\}}$ is a non-negative integer-valued random variable and $$n^{-1} T\cdot \mathbf{1}_{\{0\le T< n\}}\le S.$$ By its definition, note that the following hold:

$S>0$ if and only if $T\ge n$,

$S>k$ if and only if $T>kn$ for $k=1,2,\dots$.

Therefore, using the expectation identity for non-negative integer-valued random variables,
\begin{align*}
E[S]&= \sum_{k=0}^{\infty} P(S>k) = \sum_{k=0}^{\infty} P(S>k) = P(S>0) + \sum_{k=1}^{\infty} P(S>k)\\
&=P(T\ge n)+ \sum_{k=1}^{\infty} P(T>kn) \le P(T\ge n) + \sum_{k=1}^{\infty} (1-a)^k\\&= P(T\ge n) + \frac{1-a}{a}.
\end{align*}
Combining the above three inequalities, we have
$$E[n^{-1}T] \le [1-P(T\ge n)] + P(T\ge n) + \frac{1-a}{a} = \frac{1}{a}.$$
Complete the proof by multiplying both sides of the above inequality by $n$.


问:
Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$
$$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$

So I would just like to know how to begin to solve this limit, or what topic does this problem fall under so that I can search for examples online or in text.

I know the answer is 1/2 and I have tried plugging in numbers  smaller or greater than one but as I assumed that was the wrong way to go.

ContentLicense: CC BY-SA 3.0

Body:
with attention to hopital rule we have:
$$
\lim_{x\to1^+}\frac{x-1-\ln x}{(x-1)\ln x}=
\lim_{x\to 1^+}\frac{1-\frac{1}{x}}{\ln x+\frac{x-1}{x}}=
\lim_{x\to 1^+}\frac{\frac{1}{x^2}}{\frac{1}{x}+\frac{1}{x^2}}=\frac{1}{2}
$$


问:
Repeated application of L'Hopital's rule to evaluate $\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$
I came across an interesting limit while trying to draw the graph of the function $f(x)=\frac{3^{\ln x}}{x}$.

What is $$L=\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$

Since $\ln(0^+)\to -\infty$ and $3^{-\infty}=0$ we have the limit in indeterminate form of $\frac{0}{0}$.
So By L'Hopital's rule we get
$$L=\lim_{ x \to 0^+}\frac{3^{\ln x}\ln 3 \times  \frac{1}{x}}{1}=\ln 3\lim_{x \to 0^+}\frac{3^{\ln x}}{x}$$
So how to find this limit?

答:
ok, Here's another way:
$$
\overbrace{\frac{3^{\ln x}} x = \frac{x^{\ln 3}} x}^\text{by the change-of-base identity} \!\!\!\!\!\!\!\!\! = x^{(\ln3)-1} \quad \text{and } (\ln3)-1>0.
$$
And so on.


问:
The precise general expression for the sequence generated by my code
From the following code, I would like to extract the general formula in terms of $n$

```<code>   public static int second_loop(int n ){
        int count = 0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                for(int k=i; k<=2*j - 1; k++)
                    count++;
        return count;
    }
</code>```

I manage to get the following approximation: 0.5 * n^2 * (n + 1).

Comparing the values that I got from the approximation above with the count (i.e., expected value) from the code, I got the following:

```<code>       expected value | approximation
N=1 |         1       |    1
N=2 |         6       |    6
N=3 |        19       |    18
N=4 |        43       |    40
N=5 |        82       |    75
</code>```

However, the approximation is wrong. Someone has tried to help me answer it but it’s still wrong.

My question is how can I get the precise expression?

I am not working or study math at the moment so this question might be super simple to solve, but unfortunately not for me.

答:
The code can be translated thus

$$\sum_{j=1}^N\sum_{i=1}^{\max(2j-1,N)}\left(2j-i\right).
$$

Now let $N = 2m$ then $2j-1<N$ is equivalent to $j\le m$. So we split the sum:

$$\begin{align}
&\sum_{j=1}^m\sum_{i=1}^{2j-1}(2j-i) + \sum_{j=m+1}^{2m}\sum_{i=1}^{N}(2j-i) \\
&= \sum_{j=1}^m \left((2j)(2j-1)-\frac{(2j-1)(2j)}{2}\right) + \sum_{j=m+1}^{2m} \left(2jN-\frac{N(N+1)}{2}\right) \\
&= \sum_{j=1}^m\left( 2j^2 - j \right) + \left(2\cdot 2m\left(\sum_{j=1}^{2m}j-\sum_{j=1}^m j\right) - \frac{2m(2m+1)}{2}\cdot m\right)\\
&= 2 \left(\frac{m(m+1)(2m+1)}{6} \right) - \frac{m(m+1)}{2} + \left(2m(4m^2+2m-(m^2+m))-m(2m+1)(m)\right)\\
&= m(m+1)\left(\frac{2m+1}{3}-\frac12\right) + \left(2m(3m^2+m)-m^2(2m+1)\right)\\
&= \frac16(m)(m+1)(4m-1)+m^2(4m+1)\\
&=\frac16 m(28m^2+9m-1).
\end{align}$$

Now you should do the case of $N=2m+1$, and noting that 2j-1<N is also equivalent to $j\le m$. Then, the first sum is the same as above.


问:
Regularized sum of $1+(1+\frac1{2^2}) + (1+\frac1{2^2}+\frac1{3^2})+\cdots$, or $\sum_{k=1}^\infty \sum_{j=1}^k\frac1{j^2}$?
For my exercises on divergent summation I try to find a regularization of
$$   S^{^\star}_2 \underset{\mathcal R}= \sum_{k=1}^\infty \sum_{j=1}^k \frac1{j^2} \tag 1
$$
It is not difficult to derive the formula for the partial sums introducing the harmonic numbers of order $1$ and $2$
$$H_1(N)=\sum_{k=1}^N \frac1{k} \\
  H_2(N)=\sum_{k=1}^N \frac1{k^2} \tag {2.1}$$
and writing
$$ S_2(N) = (N+1) \cdot H_2(N) - H_1(N) \tag {2.2}$$
When going to the limit $N \to \infty$ and the serial summation according to eq 2.1 cannot be done, we can -using the $\psi$ (or "digamma", in Pari/GP  psi(1+N))-function-   approximate  $$H_1(N) \sim \gamma + \psi(1+N)  \tag {3.1}$$  and for $H_2(N)$ the derivative of the $\psi()$:
$$ H_2(N) \sim \zeta(2)-\psi^{(1)}(1+N) \tag {3.2}
$$

I don't see how I can proceed now. I'd assume for the regularization in the case for $N$ in the limit I can use $\gamma$ for the $H_1^{^\star}$.

Q:But how can I the do the $H_2$-part for the limit formula for eq (3.2) resp. (2.2)?

A bit background: In a treatize on the triangle of Eulerian numbers (pg.8) I hand-waving-generalize the triangle towards negative row numbers, and for the row-sum of the $-2$'nd row I get those formal sums.

update I just found my 2013-question about the same subject, and there was an answer but which I could not then believe it is a final answer, and still I'm unsure whether (and how) I can proceed to a "fixed statement" see earlierquestion The user @did gave the answer
$$
\lim_{n\to\infty}\left(s_2(n)-n\frac{\pi^2}6+\log n\right)=\frac{\pi^2}6-1-\gamma \qquad \text{introduced by me: } =\zeta(2)-1 - \gamma
$$

答:
Regularizing the sum when the analytically continued function has a pole at the value of interest for the parameter, does not seem to give a unique value. There are several options to regularize:

You could introduce a variable $x$ to make it a power-series s.t. the series converges for $|x|<1$ i.e.
$$\sum_{k=1}^\infty x^k \sum_{j=1}^k \frac{1}{j^2} = \sum_{j=1}^\infty \frac{1}{j^2} \sum_{k=j}^\infty x^k = \sum_{j=1}^\infty \frac{1}{j^2} \frac{x^j}{1-x} = \frac{{\rm Li_2}(x)}{1-x} \, .$$ Finally expand the polylogarithm about $x=1$ using the polylogarithm identity https://dlmf.nist.gov/25.12 mentioned in the comments, which gives $$\frac{{\rm Li_2}(x)}{1-x} = \frac{-{\rm Li_2}(1-x) + \zeta(2) - \log(x)\log(1-x)}{1-x} \\ = \frac{\zeta(2)}{1-x} + \log(1-x) - 1 + {\cal O}\left( (1-x)\log(1-x) \right)$$
for $x\rightarrow 1$. Hence, when the bare singularities are discarded you arrive at $-1$.

Cut-Off regularization introduces an upper limit $m$ into the sum and the result is then expanded in terms of the cut-off. This gives $$\sum_{k=1}^m H^{(2)}_k = (m+1) H_m^{(2)} - H_m = (m+1)\zeta(2) - \log (m) - \gamma - 1 + {\cal O}(1/m) \, .$$ The divergent parts in $m$ can be viewed as singularities at $\infty$ and are of the same type as in the first method for $x\rightarrow 1^{-}$. The constant term differs and reads $\zeta(2) - \gamma - 1$.

Zeta functional regularization is another commonly applied procedure to treat infinities. The generalized zeta-function which converges for $\Re(s)>1$ is sought to be continued to $s=0$. Since $s=1$ has convergence issues, the analytically continued version will turn out to have a pole at $s=1$. Therefore we first continue to $\Re(s)>0$ by simple manipulation
\begin{align} \sum_{k=1}^\infty \frac{H_k^{(2)}}{k^s} &= \sum_{k=1}^\infty k^{-s} \sum_{j=1}^k j^{-2} = \sum_{k=1}^\infty k^{-s} \left( \sum_{j=1}^\infty j^{-2} - \sum_{j=k+1}^\infty j^{-2} \right)  \\ &= \zeta(2) \zeta(s) + \zeta(s+2) - \sum_{k=1}^\infty k^{-s} \sum_{j=k}^\infty j^{-2} \\ &= \zeta(2) \zeta(s) + \zeta(s+2) - \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j k^{-s} \, . \end{align}
The Zeta-function of the first term has the aforementioned pole at $s=1$ and can be easily continued to $s=0$ and the rightmost sum is now convergent for $\Re(s)>0$ which implies there to be another pole at $s=0$. To shift the convergence further to the left in the complex plane we now introduce $$k^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty x^{s-1} e^{-kx} \, {\rm d}x$$
into the last term, which in turn becomes (after interchanging summation and integration)
\begin{align} \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j k^{-s} &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \sum_{j=1}^\infty j^{-2} \sum_{k=1}^j e^{-kx} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \sum_{j=1}^\infty j^{-2} \, \frac{1-e^{-jx}}{e^x-1} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{\zeta(2)-{\rm Li_2}\left(e^{-x}\right)}{e^x-1} \\ &= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{{\rm Li_2}\left(1-e^{-x}\right)-x \log\left( 1-e^{-x} \right)}{e^x-1}  \, , \end{align}
where in the last step the aforementioned polylogarithm identity was used. The integral converges like the sum for $\Re(s)>0$ and each term of the integral converges separately. The restriction follows from the behaviour at $x=0$. Because of the $\frac{1}{\Gamma(s)}$ it suffices to extract the singularities at $s=0$, since any finite contribution will vanish in the limit $s\rightarrow 0$. To do this, we subtract and add the leading behaviour  of the nominator for $x\rightarrow 0$ \begin{align} {\rm Li_2}\left(1-e^{-x}\right) &= x + {\cal O}(x^2)  \\ \log\left(1-e^{-x}\right) &= \log(x) + {\cal O}(x) \, . \end{align}
After the leading terms have been subtracted, the resulting integral converges for $\Re(s)>-1$, i.e. it is finite at $s=0$. Hence it suffices to calculate $$ \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}x \, x^{s-1} \, \frac{x-x \log\left( x \right)}{e^x-1} \\ = \frac{s}{\Gamma(s+1)} \int_0^\infty \frac{x^s}{e^x-1} \, {\rm d}x - \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^s}{e^x-1} \, \log(x) \, {\rm d}x \\ = s \, \zeta(s+1) - \frac{1}{\Gamma(s)} \frac{{\rm d}}{{\rm d}s} \Gamma(s+1)\zeta(s+1) \\ = 1 + \frac{1}{s} + \gamma + {\cal O}(s) $$
where the integral representation of the Zeta function was used from the second to the third line. Evidently there is no logarithmic singularity like in the other two cases, but they all share the leading pole singularity. The constant term in this case after using the well known result $\zeta(0)=-\frac{1}{2}$ finally reads $\frac{\zeta(2)}{2} - \gamma - 1$.

What's the moral of the story? When the analytically continued version of the to be regularized sum contains a pole at the value in question, the result can be more or less anything. Only when the analytically continued value is finite, it is also unique. You can also see the arbitrariness for instance when you substitute $m \rightarrow m+c$ in the second case, or $x \rightarrow \frac{x + c (1-x)}{1+c(1-x)}$ in the first case, for any constant $c$.


问:
Circumscribed circles of the triangles
I'm interested in the following problem

Let $\omega$ be a circle with O the center of the circle and I a straight line. Perpendicular from O on the line I cut $\omega$ into A and B. Let P and Q be two points on the $\omega$ and let $PA\cap I=X_1$,$PB\cap I=X_2$, $QA\cap I=Y_1$, $QB\cap I=Y_2$. Prove that the circumscribed circles of the triangles $AX_1 Y_1$ and $AX_2 Y_2$ intersect a second time at a point on $\omega$.

My attempt. I saw that the points $P$ and $Q$ are mobile so I tried finding a projective function and then applying the moving points method but I am not very good at this method. I also tried to use an inversion but I don't think that it would work.

Any ideas please!

答:
Lemma. Let $\ell$ be a line and $\mathcal C$ be a circle with center $\mathcal C_O$. Let the line passing through $\mathcal C_O$ perpendicular to $\ell$ intersect $\mathcal C$ at $\{\mathcal C_1, \mathcal C_2\}$. Let $(\ell_1,\ell_2)\in\ell^2$ be two points on $\ell$ such that $\ell_i\mathcal C_1\cap \mathcal C=\mathcal P_i(\neq \mathcal C_1)$ for $i\in\{1,2\}$. Then $\{\mathcal P_1,\mathcal P_2,\ell_1,\ell_2\}$ are concyclic.

Proof. Note that
$$\angle \ell_2\ell_1\mathcal C_1=90-\angle \mathcal P_1\mathcal C_1\mathcal C_2=\angle \mathcal C_1\mathcal C_2\mathcal P_1=\angle \mathcal C_1\mathcal P_2\mathcal P_1\implies \{\mathcal P_1,\mathcal P_2,\ell_1,\ell_2\}\text{ are concyclic.}$$
$$\tag*{$\square$}$$

Let $\{Y_1B,X_1B\}\cap \omega:=\{E,D\}$. Note that $X_1P\perp BX_2$ and $BA\perp \ell\implies A$ is orthocenter of $\triangle X_2BX_1$ and as $AD\perp BX_1$, we get $\{X_2-A-D\}$ are collinear. Similarly, $\{Y_2-A-E\}$ are collinear. Thus, by our lemma, $X_1Y_1ED$ and $Y_2X_2ED$ are cyclic.

We claim that $\{DE,X_2Y_2,PQ\}$ concur at a point $C$. This is obvious by pascal's theorem on $BPQADE$.

Now, note that by power of point, we get
$$CQ\cdot CP=CD\cdot CE=CY_2\cdot CX_2=CX_1\cdot CY_1$$
Thus, point $C$ has equal powers with respect to $\{\omega, \odot(AX_1Y_1),\odot(AX_2Y_2)\}$ and as $C\not\equiv A$, these three circles must be coaxial completing the proof.
$$\tag*{$\blacksquare$}$$

Title: Circumscribed circles of the triangles


问:
Linear/Integer Programming: Scheduling task with regular intervals
Given is the following problem: There are $N$ items that have to be processed in given intervals over a year. For example, item $i_1$ is processed about every 14 days, item $i_2$ is processed monthly. The actual processing intervals are flexible, but they should be as close to the given intervals as possible. During holidays, no items are processed. Processing item $i$ takes $t_i$ hours and every day, only 8 hours are available for processing. The task is to determine when each item $i$ should be processed.

I believe this task could be solved with integer programming. Hence, I created a model, but I failed to link the missing parts together (see below). My primary questions are:

Is integer programming suitable for this task?

Are there already known problems that have a similar setting (e.g., traveling sales person) that can serve as a basis for this task?

How would you define the model?

To show that I actually thought about this problem, I will describe my model and my issues with this model in the following. My model uses the following variables:

$x_{ijk} \in \lbrace 0, 1 \rbrace$ with $x_{ijk} = 1$ iff item $i$ is processed the $k$th time at day $j$

$y_{ik} \in \mathbb{N}_0$: day when item $i$ is processed the $k$th time

$z_{il} \in \mathbb{N}_0$: how much the $l$th interval diverges from the optimal interval

Here, I represent dates as numbers in $[0, 365)$, where 0 corresponds to January, 1st. In the following, $i$ refers to items, $j$ refers to days, $k$ refers to an index (the $k$th time processing), unless otherwise stated.

My objective is to minimize the total divergence from the optimal intervals $z$:

$$
\sum_i^N \sum_l^{K_i - 1} z_{ik} \rightarrow \min
$$
where $K_i$ is the number an item $i$ is processed throughout a year (e.g., 12, if item $i$ is processed monthly).

Then, I defined the following constraints:

An item is processed at most once a day: $\sum_k^{K_i} x_{ijk} \leq 1 \; \forall i, j$

An item must be processed the $k$th time at some day: $\sum_j^{365} x_{ijk} = 1 \; \forall i, k$

No processing on holidays: $\sum_i^N \sum_k^{K_i} x_{ijk} = 0 \; \forall j \in H$ (the set of holidays)

Maximum 8 hours per day: $\sum_i^N \sum_k^{K_i} x_{ijk} \cdot t_i \leq 8 \; \forall j$

Range of $y$: $0 \leq y_{ik} \leq 365 \; \forall i, k$

$k+1$th processing occurs after $k$th processing: $y_{i(k + 1)} \geq y_{ik} + 1 \; \forall i, k$

Bind $y$ and $z$: $z_{ik} \geq y_{i(k+1)} - y_{ik} - z_{ik} \; \wedge \; z_{ik} \geq -(y_{i(k+1)} - y_{ik} - z_{ik}) \; \forall i, k$

However, a constraint that binds $x$ and $y$ is still missing. I believe that this constraint should state that $x_{ijk} = 1$ iff $y_{ik} = j$ for all $i, j, k$. Something like $x_{ijk} \implies y_{ijk}$ is quite straightforward, but the other direction seems to be more difficult. Furthermore, I think that the numbers of variables might be too high, e.g., $x_{ij}$ could be sufficient, and lastly, I am not sure whether there are other issues with this model. Hence, I'd like to get some suggestions on this.

答:
Yes, integer programming is a natural choice.  You can link $x$ and $y$ via
$$\sum_j j x_{i,j,k} = y_{i,k}$$

An alternative formulation is to use network flow in a time-space network with a node $(i,j)$ for each item $i$ and day $j$ (just omit holidays), a directed arc from $(i,j)$ to $(i,j')$ with $j < j'$, source node $(s,0)$, and sink node $(t,366)$.  Let binary arc variable $w_{i,j,j'}$ indicate whether item $i$ is scheduled on days $j$ and $j'$ and not on any days in between.  Let $\ell_i$ be the desired length of the processing interval for item $i$.  The problem is to minimize
$$\sum_{i,j,j'} |j'-j-\ell_i| w_{i,j,j'}$$ subject to
\begin{align}
\sum_{i,j,j'} w_{i,j,j'} - \sum_{i,j',j} w_{i,j',j} &= \begin{cases} 1 &\text{if $(i,j)=(s,0)$} \\ -1 &\text{if $(i,j)=(t,366)$} \\ 0 &\text{otherwise} \end{cases} &&\text{for all $(i,j)$} \tag1\\
\sum_{j,j'} w_{i,j,j'} &= K_i+1 &&\text{for all $i$} \tag2\\
\sum_{i,j,j'} t_i w_{i,j,j'} &\le 8 &&\text{for all $j>0$} \tag3\\
\end{align}
Constraint $(1)$ is the flow balance constraint for each node.
Constraint $(2)$ enforces processing item $i$ a total of $K_i$ times.
Constraint $(3)$ enforces the daily processing hours limit.


问:
Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ$

Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ.$

We know that $\tan45^\circ=1$ because a right triangle with angle equal to $45^\circ$ is isosceles.

I don't know the exact values of $\tan80^\circ$ and $\tan 100^\circ.$ Can you give me a hint?

答:
Hint:

$\tan x \lt 0$ for $90^\circ\lt x \lt 180^\circ$ and $\tan x$ is an increasing function on $[0,90^\circ)$.

$\tan \theta$ is defined to be the orange line in the diagram. What happens to this line as $\theta$ goes from $0^\circ$ to $90^\circ$?


问:
Need help to check alternating series criterion
I know this is simple calculation based question but I got stuck:

I want to check convergence of the following series:
$$1-\frac{1}{2}(1+\frac{1}{3})+\frac{1}{3}(1+\frac{1}{3}+\frac{1}{5})-\cdots$$
Clearly this is alternating series. So I need satisfy alternating series criterion. That means I need to show that its terms are decreasing monotonically.

Let $a_n=\frac{1}{n}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1})$ and  $a_{n+1}=\frac{1}{n+1}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1})$.

I just have to show  $a_n-a_{n+1}>0$  but right here I got stuck because of the presence of the first factor. I think i am missing a simple trick here.

Anyway, I got $$a_n-a_{n+1}=(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{3}(\frac{1}{n}-\frac{1}{n+1})+\cdots+\frac{1}{2n-1}(\frac{1}{n}-\frac{1}{n+1}) {-\color{red}{\frac{1}{(2n+1)(n+1)}}}.$$

The last red color term making problem.

May be ratio test help as well ?

Any help please

答: $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}\gt\frac{1}{(2n+1)(n+1)}$.  Therefore that extra term is taken care of.


问:
Is the the number of generators of a group the number of different generators that one finds if one counts over every generating set of the group?
Consider the additive group of integers as an example as mentioned at the bottom of the Wikipedia article. There are two generating sets that are mentioned; The set consisting of the number 1, {1}, and the two-element set, {3,5}. So when we talk about the number of generators of a group do we mean the number of different elements within every generating set, all added together. So above we've counted 3 generators so far, 1,3, and 5 - although there is clearly more than this in total since we haven't mentioned all the generating sets of the group.

As another example, in SU(N) we have $N^{2}-1$ generators. Would this be the total number of generating matrices that we'd get if we counted every single generating set?

On a similar note, would we only count a generator once if it appeared in more than one generating set? So for example if some group had generating sets {1,5}, {5,9}, {3,4,12}, would we say this group had 6 generators, rather than 7?

答:
A group $G$ typically have many different generating sets, that is subsets $S\subseteq G$ with $\langle S\rangle = G$, and among these are sets of different cardinality (for example, trivially $S=G$ is a generating set).
Therefore it makes little sense to speak of the number of generators of a group (or even of the set of generators).

We do speak of the free group $F_S$ generated by the set $S$ and for such a free group, the set $S$ is a canonical choice of generators. Nevertheless, the free group over $S=\{a,b\}$ is also generated by $\{a,b,1,a^{-2},bab\}$.
In general, when we speak of a group generated by $n$ elements, we mean a group $G$ that allows an epimorphism $f\colon F_S\to G$ where $|S|=n$, typically given by a presentation $G=\langle a_1,\ldots,a_n\mid\ldots\rangle$. Depending on the author, it may additionally be understood that $f|_S$ is injective, that is that the generators of $G$ obtained this may are actually an $n$-element set. But I guess it is often not the case that this distinction is made.

In summary, "a group generated by $n$ elements" should usually be more precisely called "a group that has at least one generating set of at most $n$ elements".
For example, a cyclic group is a group generated by a single element. We do however also count the trivial group as cyclic, even though it can in fact be generated by zero elements.


答:
Write $\tan=\dfrac{\sin}{\cos}$

$$\tan80^\circ-\tan45^\circ=\cdots=\dfrac{\sin(80-45)^\circ }{\cos80^\circ\cos45^\circ}>0$$

We know $\tan45^\circ=?$

and $\tan100^\circ=\tan(180^\circ-80^\circ)=-\tan80^\circ<0$

Title: Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ$

Body:
Arrange in ascending order $\tan45^\circ,\tan80^\circ$ and $\tan100^\circ.$

We know that $\tan45^\circ=1$ because a right triangle with angle equal to $45^\circ$ is isosceles.

I don't know the exact values of $\tan80^\circ$ and $\tan 100^\circ.$ Can you give me a hint?


问:
Find the Area of the ellipse
Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$

where $a>0$, $b>0$

I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$

I can't get the answer $\pi ab$

答:
from the equation of the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
$$  y = \sqrt{\frac{a^2b^2 - x^2b^2}{a^2}}  = \frac{b}{a}\sqrt{a^2-x^2}$$

since proceding with change of variable in integration is, using x = a*cos(Θ)::
$$
\\
\sqrt{a^2 - x^2} = \sqrt{a^2-a^2cos^2\theta} =asin\theta
\\
\int_{}^{}\sqrt{a^2 - x^2} = a\int_{}^{}sin\theta \
dx \
= a^2\int_{}^{}sin^2\theta d\theta = a^2\int_{}^{}\frac{1 - cos 2\theta}{2} d\theta = \frac{a^2\theta}{2}-\frac{a^2sin^22\theta}{2}+C = \frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x\sqrt{1-\frac{x^2}a^{}}}{a})+C
$$
so the area in the first quadrant is:

$$ \frac{b}{a} \int_{0}^{a} \sqrt{a^2-x^2} dx$$
that is:
$$ \frac{b}{a}((\frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x \sqrt{1-\frac{x^2}{a^2}}}{a}))|^a_0) = \frac{ab \pi}{4}$$

so for the area of the entire ellipse:
$$ \frac{ab\pi}{4} 4  = ab\pi$$

Title: Find the Area of the ellipse


问:
Simulation of reflection model using a sphere and a light source
This is regarding a simulation of a reflection model using a sphere(in 2D) from the convex surface. I'm not being able to work out the geometry to calculate the angle between incident ray and the normal at the point of incidence. The center of the sphere and the light source are at the same height(y coordinate is same). The coordinates of the light source and center of sphere are known, and the initial angle is also known which the incident ray makes with the initial line joining the source and the center. Also, how should I set the limits such that the ray gets reflected off the surface itself(the x and y coordinates of the circle)? Can someone help me out in these? By the way, I'm using turtle graphics for the simulation in c++.

答:
Reflection angle $\theta$  can be found from the sine rule:

$$
{r\over\sin\alpha}={d\over\sin\theta},
$$
where $d$ is the distance from the light source to the centre of the sphere, while $\alpha$ is the angle the incident ray makes with the line joining the source and the centre.

You can obtain the limiting value of $\alpha$ by setting $\sin\theta=1$ in the above formula.


问:
Showing that $(\frac{\sin3A}{\sin A})^2 - (\frac{\cos3A}{\cos A})^2 = 8\cos2A$

Show that $$\left(\frac{\sin3A}{\sin A}\right)^2-\left(\frac{\cos3A}{cosA}\right)^2=8\cos2A$$

Solution

$$\begin{align}
\left(\frac{\sin3A}{\sin A}\right)^2-\left(\frac{\cos3A}{\cos A}\right)^2
&=\left(\frac{\sin3A}{\sin A}+\frac{\cos3A}{\cos A}\right)\left(\frac{\sin3A}{\sin A}-\frac{\cos3A}{\cos A}\right) \\[4pt]
&=\left(\frac{\sin3A\cos A+\sin A\cos3A}{\sin A\cos A}\right)\left(\frac{\sin3A\cos A-\sin A\cos3A}{\sin A\cos A}\right) \\[4pt]
&=\left(\frac{\sin(3A+A)}{\sin A\cos A}\right)\left(\frac{\sin(3A-A)}{\sin A\cos A}\right) \\[4pt]
&=\left(\frac{\sin4A}{\sin A\cos A}\right)\left(\frac{\sin2A}{\sin A\cos A}\right)
\end{align}$$

What do I do next?

The textbook from which the above question is taken hasn't introduced the double angle formulas yet. So I am trying to find a way to do this question without using the double angle formulas.

答:
Alternatively, using triple angle formula  assuming $\sin A\cos A\ne0$ we get

$$(3-4\sin^2A)^2-(4\cos^2A-3)^2$$

$$=24(\cos^2A-\sin^2A)-16(\cos^4A-\sin^4A)$$

$$=8(\cos^2A-\sin^2A)(3-2(\cos^2A+\sin^2A))=?$$


答:
$\tan(x)$ is defined on $\bigcup_{n\in \Bbb Z} (90(2n-1), 90(2n+1))$ (when working in degrees) and increasing inside each such set.

Moreover, recall that $\tan(x)\equiv\tan(x+180k)$ for each $k\in \Bbb Z$

So, we may say that $\tan(x)$ is increasing on the range $(-90, 90)$, as seen when $n=0$, and $\tan(100)=\tan(-80)<\tan(45)<\tan(80)$


问:
Struggling with intuition about this probability question. Symmetry argument of two balls drawn from an urn.
So the question is as follows:

An urn contains m red balls and n blue balls. Two balls are drawn uniformly at random
from the urn, without replacement.

(a) What is the probability that the first ball drawn is red?

(b) What is the probability that the second ball drawn is red?*

The answer to a) quite clearly works out to be $\frac{m}{(m+n)}$, but the answer to b turns out to be the same, and my tutor said this is intuitive by a symmetry argument.

i.e. that $P(A_1)$ = $P(A_2)$ where $A_i$ is the event that a red ball is drawn on the ith turn. However I am struggling to see how this is evident, can anyone explain this?

答:
The second probability is

$$\frac{m-1}{m+n-1},$$

which is not the same as the first, so there's nothing to be worried about.


答:
Part (b) doesn't give any information about the first ball, it is just asking for the probability that the second ball in the line is red.

Now red balls (or those of any other color!) don't have any preference for positions in the line, hence if you randomly pick up any ball from the line, its probability of being red will be the same.

Title: Struggling with intuition about this probability question. Symmetry argument of two balls drawn from an urn.

Body:
So the question is as follows:

An urn contains m red balls and n blue balls. Two balls are drawn uniformly at random
from the urn, without replacement.

(a) What is the probability that the first ball drawn is red?

(b) What is the probability that the second ball drawn is red?*

The answer to a) quite clearly works out to be $\frac{m}{(m+n)}$, but the answer to b turns out to be the same, and my tutor said this is intuitive by a symmetry argument.

i.e. that $P(A_1)$ = $P(A_2)$ where $A_i$ is the event that a red ball is drawn on the ith turn. However I am struggling to see how this is evident, can anyone explain this?


问:
How to prove uniform continuity of $e^\frac{1}{z^2}$ on $0\lt\vert z\vert\leq R$, $\vert\arg(z)\vert\leq\pi/6$?
I have to prove that the complex-valued function $f(z)=e^\frac{1}{z^2}$ is uniformly continuous on the region of the complex plane: $0\lt\vert z\vert\leq R$ , $\vert\arg(z)\vert\leq\pi/6$.

I have tried using parametric equations for the given domain but I don't know how to relate them to any continuity definitions. Any help is appreciated.

Body:
It appears that the book is wrong.

Let $\displaystyle z_n = \frac{1}{2n}e^{i\pi/6}$ and $\displaystyle w_n = \frac{1}{n}e^{i\pi/6}$.  We have $\displaystyle |z_n - w_n| = \frac{1}{2n}|e^{i\pi/6}| = \frac{1}{2n} \to 0$ as $n \to \infty$.

However, $z_n^{-2} = 4n^2e^{-i\pi/3}= 2n^2(1 - i\sqrt{3})$ and $w_n^{-2} = \frac{n^2}{2}e^{-i\pi/3}= \frac{n^2}{2}(1 - i\sqrt{3})$, and, hence, as $n \to \infty$,

$$|f(z_n) - f(w_n)| = \left|e^{2n^2}e^{-i2\sqrt{3}n^2} -  e^{n^2/2}e^{-i\sqrt{3}n^2/2}\right|\geqslant e^{2n^2}\left|e^{-i2\sqrt{3}n^2}\right| -  e^{n^2/2}\left|e^{-i\sqrt{3}n^2/2}\right|\\ = e^{2n^2}-e^{n^2/2}\to +\infty$$


问:
When can I change a summation to an integral
Often in my engineering courses, professors just switch out a summation with an integral and change some extensive property to an infinitesimal quantity. This is something I have to accept, but I feel like I've never received rigorous justification for this.

For example, I'm working with products of inertia, defined as:
$$I_{xy} = \sum m_{i} (x_{i} y_{i}) ,$$

and I'm presented with the substitution
$$\sum m_{i} (x_{i} y_{i}) = \int_{\text{body}}^{} (yx)dm$$

which isn't defined anywhere in my textbook or online. I think I'm ready for an explanation of why this is a mathematically sound process.

EDIT: I did a quick derivation myself, tell me if I did anything hand-wavey
given the definition of the integral

! $$\int_{a}^{b} F(x) dx = \lim_{n \to \infty}\sum_{i=1}^{n} f(x_{i}) \Delta x_{i}$$
and
$$I_{xy} = \sum m_{i} (x_{i} y_{i}) $$
where yi can be expressed explitily as some
$$y_{i} = f(x_{i})$$
under constant linear density define
$$\Delta x  = \frac{b-a}{n},\,\ and\,\ m_{i} = m/n $$
$$\lambda = \frac{m}{b-a} = \frac{\frac{m}{n}}{\frac{b-a}{n}}  =  \frac{m_{i}}{\Delta x} $$
$$m_{i} = \frac{m}{b-a} \Delta x $$
we get
$$I_{xy} = \sum m_{i} (x_{i} y_{i}) = \frac{m}{b-a} \sum x_{i} f(x_{i}) \Delta x $$
let
$$x_i = a + i\cdot \Delta x$$
then we can finally take the limit
$$I_{xy} = \lim_{n \to \infty} \sum_{i=1}^{n} (x_{i} y_{i}) = \frac{m}{b-a} \sum x_{i} f(x_{i}) \Delta x \\
I_{xy} = \int_{a}^{b} x\cdot f(x) dx$$

答:
It is not mathematically sound.

What they can claim to be doing is passing to the limit, but that's not just how to do it.


问:
Apply complex residue theory to non even function improper integration
I am considering the integral

$$\int_{-\infty}^{\infty}\frac{t-5}{(t - 3+i)(t - 3-i)(t-2i)(t+2i)} dt.$$

The integrand, say $f(t)$, is readily not even, right? How do we use residues to solve this? The common technique uses the interior of large $[-R,R]$ upper-half plane semicircle and line segment combo. Only two singularities of the integrand are inside that interior. The rest is as usual.

Similar example is Complex analysis: Improper integral of a non-even function.

However, in this question and in that link as well, we will only get the value of Cauchy principle value $\lim_{R\rightarrow\infty} \int_{-R}^{R} f(t)\text{ }dt$, right?

If $f(t)$ is even, then we finally can get the value we want. Since here it is not, how do we continue? Similarly, why in that link, the answer can conclude the value too?

Thanks for any suggestion! Much appreciated!

Body: $$\int_{-\infty}^{\infty}\frac{t-5}{(t - 3+i)(t - 3-i)(t-2i)(t+2i)} dt=2 \pi  i \left[\text{res}(f(t),(t,2 i))+\text{res}(f(t),(t,3+i))\right]=2\pi i\left(-\frac{1}{15}+\frac{3 i}{40}+\frac{1}{15}+\frac{i}{20}\right)=2\pi i\left(\frac{i}{8}\right)=-\frac{\pi}{4}$$


问:
About span and subspaces
I run into the theorem that states
"The span of any subset S of a vector space V is a subspace of V"
in Friedberg Linear Algebra(p30).
I know a lot are written about this theorem but I want to take a closer look at another part of this theorem, which, I at least, haven't seen in any websites thus far.
To prove that the span of any subset S is a vector space we need a couple of things:

span(S) is closed under addition and multiplication

Zero vector is in span(S)

span(S) ⊂ V
I see a lot of proofs on the net that prove this theorem by showing the first two conditions.

Here is the definition of a subspace according to Freidberg(p16):
A subset W of a vector space V over a field F is called a subspace of V if W is a vector space over F with the operations of addition and scalar multiplication defined on V.

It states W must be a subset of V. (at least this is what I got). Why do the proofs on the net don't consider to prove "span(S) ⊂ V" ? Is it so obvious to see? Or... Where am I missing. It's been two days and it's really buzzing me.

Thanks in advance

答: It is obvious. Since $S\subset V$,  vector operations on $S$ (which is what you do when you take the span: you take linear combinations) will stay in $V$.

Title: About span and subspaces


答: The secon ball probability is$\frac{m}{m+n}\frac{m-1}{m+n-1}+\frac{n}{n+m}\frac{m}{m+n-1}=\frac{m}{n+m}$ This is the same as the first, which, by symmetry, is what it should be.


答: The reason is that by definition, the vector space $V$ that you're working in is closed under addition and scalar multiplication, i.e. any linear combination of vectors in $V$ remains in $V$. Restricting your vectors to a subset $S\subseteq V$ leaves that property in place so to speak, so any element in the span is certainly in the underlying vector space. That is to say any linear combination of vectors in $S$ can also be viewed as a linear combination of vector in $V$ because $S \subseteq V$.

Title: About span and subspaces

Body:
I run into the theorem that states
"The span of any subset S of a vector space V is a subspace of V"
in Friedberg Linear Algebra(p30).
I know a lot are written about this theorem but I want to take a closer look at another part of this theorem, which, I at least, haven't seen in any websites thus far.
To prove that the span of any subset S is a vector space we need a couple of things:

span(S) is closed under addition and multiplication

Zero vector is in span(S)

span(S) ⊂ V
I see a lot of proofs on the net that prove this theorem by showing the first two conditions.

Here is the definition of a subspace according to Freidberg(p16):
A subset W of a vector space V over a field F is called a subspace of V if W is a vector space over F with the operations of addition and scalar multiplication defined on V.

It states W must be a subset of V. (at least this is what I got). Why do the proofs on the net don't consider to prove "span(S) ⊂ V" ? Is it so obvious to see? Or... Where am I missing. It's been two days and it's really buzzing me.

Thanks in advance


问:
Cone and space homeomorphism question
I am wondering:

Under what conditions is a space homeomorphic to its cone?
Moreover, if $M$ is a compact manifold, is it always homeomorphic to its cone? What about if for manifolds of different dimensions? That is, does there exists a dimension where $M^n$ is homeomorphic to its cone?

答:
There are certainly spaces where $CX \approx X$, but manifolds do never have this property.

Let $M$ be  manifold of dimension $n$. Then $CM$ contains an open subspace homeomorphic to $M \times (0,1)$ (simply remove the tip and the base $X \times \{0\}$ from $CX$). If $CM \approx M$, then $M$ would contain an open subspace homeomorphic to $M \times (0,1)$  i.e. a manifold of dimension $n+1$. This is impossible.

A positive example is the Hilbert cube $Q$  (the countable infinite product of closed unit intervals). We have $CQ \approx Q$; see here.

Title: Cone and space homeomorphism question


问:
Flow of $rot \overrightarrow{F}$
We've got vector field: $\overrightarrow{F} = \begin{bmatrix} yz\\x^3z\\e^z\end{bmatrix}$. I want to compute flow of $rot\overrightarrow{F} $($=curl \overrightarrow{F}$) through the area of the side of the cylinder $x^2+y^2=1$, between surfaces: $z=5$ and $2x+y+z=2$, in the direction from (not towards) coordinate $z$.
So, how to compute it? Let $C$ be the curve {$x^2 +y^2=1; z<5, 2x+y+z>2$} and $D$ be the domain bounded by curve C (so $D =${$x^2+y^2<1, 2-2x-y<z<5$}). But I guess I need to give a parametric representation of $D$ (in order to find a normal vector $\overrightarrow{n}$) and then check if my parametric representation is correct. How to do it? Cause then I will write $rot \overrightarrow{F}$ which is easy to compute and all I will be looking for is $\int\limits_D  rot \overrightarrow{F} \cdot \overrightarrow{n} d S$. So my major diificulty is how to find parametric representation of $D$ (which will give me $\overrightarrow{n}$ and check if it is correct. Can you help me?</p>

答: Hint: As @Ted noted you will have $$C: \mathbf{r(t)}=\big(\cos(\theta),~\sin(\theta),~~2-2\cos(\theta)-\sin(\theta)\big),~~~~0\le\theta\le2\pi$$ Now evaluate $$\oint_C F\cdot dr$$

Title: Flow of $rot \overrightarrow{F}$


问:
What is the basic idea behind calculation of area?
The system of calculating area in terms of square units is pretty philosophical and not very intuitive. It must have taken a great amount of time for humanity to arrive at such a convention and to spread it across different societies.
My question is about finding basics of such a convention and should the person who first thought of calculating areas in square units be regarded a great philosopher equivalent to Newton and Einstein?
Moreover, could we have evolved a different method of calculating areas?

答:
Area is a property of a planar region. The formulation in terms of "unit area" comes from the fact that many of the properties of planar regions that human beings care about are translation-invariant. For example, how much grain or how many sheep a region of land can support, or how long it will take one person to till it.

All of these quantities have two important properties:

Translation invariance. If $A$ and $B$ are the same shape and size but in different places, then $f(A)=f(B)$.

Additivity. If $A$ and $B$ are disjoint regions, then $f(A\cup B)=f(A)+f(B)$.

These two properties were presumably intuitively obvious to ancient humans, without them having to formulate them precisely or even consciously. From these properties it follows that if you take any given shape $U$, then for any region $A$, $f(A)$ is determined by $f(U)$ and by how many disjoint copies of $U$ it takes to tile $A$. Therefore, how many copies of $U$ it takes to tile a region is a good measure of the "size" of land. It's surely visually very obvious that any region can be reasonably tiled with small squares.

Of course, all this is just a very complicated, modern way of saying something very simple: since any given 1x1 meter square can contain the same amount of wheat, well, obviously if I know how many square meters cover a given region of land, I know how much wheat I can put on it.


问:
Can it be that the only way to prove a theorem is based on generation of new tautologies?
We can use a language of a given logic (for example propositional logic or predicate logic) to formulate statements. We can also use rules of inference to reformulate single statements (derive "new" statements from "old" ones) or to combine two existing statements to generate a new statement.

We can also say that we might have a minimal complete set of inference rules (we do not need more rules and we cannot remove any rules).

Can it be the case for a given statement (or a set of statements) none of the inference rule is applicable but a combination of them is applicable?

Here is what I mean in a bit extended form. Inference rules are tautologies likee: $A \land B \iff B \land A$, $A \land (B \lor C) \iff (A \land B) \lor (A \land C)$. We can combine these tautologies to create more complex (larger) tautologies. Now, I can imagine that one of these "larger" tautologies can be applied to one of my statements (or a combination of two statements) to generate a new statements that follows (in a logical sense) from the "old" statement(s).

Can it be the case that it would be impossible to go from the "old" statement(s) to the "new" one just by applying the original (basic, atomic) tautologies but it is possible to get the "new" statement if we combine the original (basic, atomic) tautologies to get a more complex one and then we apply it to derive a "new" statement from the old one?

ADDED

I feel like I need to clarify my question. In a logic (for example propositional logic) we have a small and limited set of inference rules (let's call the "basic rules") that we can use to derive "new" statements (theorems) from "old" statements (old theorems and axioms). The whole process of derivations can be imagined as applying "basic rules" to existing statements to generate new statements until we reach the statement we need.

However, alternatively, one can combine "basic rules" with each other to derive new rules (let's call them "derived rules"). For example, if the first basic rules is "a + b = b + a" and the second basic rule is "a * (b + c) = (a * b) + (a * c)", then we can combine these two rules to generate a new "derived" rule: "a * (b + c) = (a * c) + (a * b)". And then, as soon as we have a new rule, we can apply it to a statement, to transform it. For example if we have a "statement" "2 * (3 + 4)" we can transform it using our new "derived rule" to get "(2 * 4) + (2 *3)".

So, my question was, if it might be the case that some statements ("theorems") are unreachable by "basic rules" only. So, in that case, instead of applying our "basic rules" to the statements, we need first to combine our "basic rules" to get new rules ("derived") and then apply the derived rules to our statements.

答:
If I interpret your question correctly, it seems to me you are asking:

If you can prove a theorem using "derived" rules, is it possible to prove it using only "atomic" rules?

This is a very interesting question, and is actually the content of Gentzen's cut elimination theorem, which he needed to prove consistency of arithmetic.

Cut elimination concerns a special rule called the "cut" rule, which is written

$$\frac{\Gamma\vdash A\quad \Delta,A\vdash B}{\Gamma,\Delta\vdash B}$$

and is read "If $A$ can be proved from hypotheses $\Gamma$, and if $B$ can be proved from hypotheses $\Delta,A$, then $B$ can be proved from hypotheses $\Gamma,\Delta$".

The content of this rule is that $A$ can be viewed as a lemma (or "derived" rule, as you put it), and this rule allows you to use this lemma in your proof of $B$.

The cut elimination theorem tells us that this rule is actually not needed. This theorem actually gives us an algorithm for converting a proof that uses the cut rule into a proof that does not use it (that is only uses "atomic" rules). The caveat is that the resulting proof may be exponentially larger than the original proof using cut.


问:
simplifying the trigonometric expression $\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta} + \frac{1}{\cos\theta}$
I'm stuck on this expression for which I don't have any answer.
$$\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta}+\frac{1}{\cos\theta}$$
Need some help on the simplifying steps.

Here is what I get from a calculator

答:
Here is a detailed answer:
$$
\require{cancel}
\begin{align}
\frac{\cot\theta-\tan\theta}{\sin\theta+\cos\theta}+\frac{1}{\cos\theta}&=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{1}\label{ko-eq1}\\
&=\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{2}\label{ko-eq2}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{3}\label{ko-eq3}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}\tag{4}\label{ko-eq4}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta+\sin\theta\left(\sin \theta+\cos \theta\right)}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{5}\label{ko-eq5}\\
&=\frac{\cos^{2}\theta\cancel{-\sin^{2}\theta}\cancel{+\sin^{2}\theta}+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{6}\label{ko-eq6}\\
&=\frac{\cos^{2}\theta+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{7}\label{ko-eq7}\\
&=\frac{\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}{\sin\theta\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}\tag{8}\label{ko-eq8}\\
&=\frac{1}{\sin\theta}\\
&=\csc\theta
\end{align}\\
$$
\eqref{ko-eq1}: Expand $\cot\theta$ and $\tan\theta$ as $\frac{\cos\theta}{\sin\theta}$ and $\frac{\sin\theta}{\cos\theta}$, respectively.

\eqref{ko-eq2}: Find common denominator of $\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$.

\eqref{ko-eq3}: Rewrite $\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}$ as $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}$ and multiply.

\eqref{ko-eq4}: Multiplication resut.

\eqref{ko-eq5}: Find common denominator of $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}$ and continue.

\eqref{ko-eq6}: Cancel the terms.

\eqref{ko-eq7}: After cancellation.

\eqref{ko-eq8}: Final cancellation and finish.

I hope this helps.


问:
Is there a logic where variables can have more than two values but "truth values" are still binary (either "false" or "true")?
In the standard propositional logic we have a certain number of variables (let's say A, B, C, D, E, and F) which can take one of two values (0 or 1, False or True). If we know the values of each variable (let's say A = 0, B = 1, C = 1, D = 0, and so on), then we know everything about the "world" and we do not need logic. However, instead of a complete knowledge about the world (value of each variable) we might have some restrictions like: We do not know what values A, B and C have, but we know that either A and B are both equal to 1 or C is not equal to 1: $(A \land B) \lor (\lnot C)$. In addition to that we usualy have some other "restrictions" (for example $\lnot A \land (\lnot D \land B)$, and so on). The goal of the propositional logic is to combine those "restrictions" to say what the world can / can't be.

This picture naturally suggest a generalisation. What if our variables (A, B, C, D, and so on) can have more than two values? Let's say instead of allowing them to be "black" and "white" (as before), we allow them be "red", "green" and "blue". Is there a special logic for that? Within this logic we could make statements like this: If B is blue or green, then A is red, except C is black (in this case A is blue or yellow).

Is there a special logic for statements like this? Many-valued logic is the first what comes in mind, but I guess it is about allowing truth-values to have more than two values, which is not the same as the case I have described.

答:
The first thing to note is that there is not actually a widely accepted definition of what counts as a "Logic". The closest thing to that is Gurevich's definition, which is what I will be using to answer your question. I paraphrase it as follows. A logic $L$ is given by a pair of functions (Sen,Sat) satisfying the following conditions. Sen associates with every vocabulary $\sigma$ a recursive set Sen($\sigma$) whose elements are called $L$-sentences of vocabulary $\sigma$. Sat associates with every vocabulary $\sigma$ a recursive relation Sat$_\sigma(A,\varphi)$ where $A$ is a $\sigma$-structure and $\varphi$  an $L$-sentence of vocabulary $\sigma$. We say that $A$ satisfies $\varphi$ if Sat$_\sigma(A,\varphi)$ holds.

Unpacking this, Sen is meant to be the set of all sentences of the logic, and Sat determines whether a structure satisfies that sentence. As an example, there is an easy definition for Sen and Sat that give propositional logic, setting Sen($\tau$) to be the set of all propositional sentences with up to $k$ distinct propositions, where $\tau$ is some vocabulary of $k$ distinct propositions, and setting Sat($A,\varphi$) to be the function that verifies whether the assignment of the propositions given by $A$ satisfies $\varphi$. For all other vocabularies $\sigma$, Sen($\sigma$) gets the empty set.

Notice here that the binary-ness of propositional calculus comes from the fact that we take as convention that propositions are nullary relations. So these vocabularies $b_k$ are merely sets of $k$ nullary relations. Thus, there is no analagous version for what you're looking for under the classical definition of relations, and hence under this definition of logics. You can of course generalize the definition of vocabulary so that instead of a vocabulary having only a set of relations and functions within its universe, it also has a set of functions with co-domain some fixed boolean algebra, that's how you get many-valued logics.

There is really no good way to enforce what you are looking for, the best you can do is something similar to what follows. Define the vocabulary $\sigma$ to have unary relations $Red(x), Blue(x), Green(x),$etc. and $k$ constants $P_1,P_2,\ldots,P_k$ to represent the propositions, for some $k$. Then you can define Sen($\sigma$) to be the boolean formulas over $\sigma$ (where base terms are in the form $R(P_i)$, where $R$ is some relation in $\sigma$). For any $\sigma$-structure $A$ and any $L$-sentence $\varphi$, Sat($A,\varphi$) will be true if and only if $A$ satisfies $\varphi$ in the typical way and moreover, $A$ interprets the relations of $\sigma$ to form a partition over the constants $P_i$. That is, for any two relations $R_\ell,R_j \in \sigma$, $R_\ell \cap R_j = \emptyset$ and for some $j$, $P_i \in R_j$. This no longer satisfies some of the basic tautologies you find in boolean logic however, for example the sentence $Red(P)\vee\neg Red(P)$ will no longer be a tautology. To make that a tautology you would have to restrict the class of structures over which we consider the logic, but this does not count as a logic of its own.


问:
Is the group of (real) regular symmetric matrices the representation of any other group? What if we use finite fields?
This question came up in a discussion between me and a friend. We recently learned a bit about representation theory. We then realized that the regular symmetric matrices over any field form a group. I now wonder what kind of group this is and wether it is the representation of some otherwise known group.

I already realized that when using $\mathbb{F}_2$ symmetric matrices are always boring, if regular, as regular and symmetric is only satisfied by the identity matrix. Is this maybe true for all finite (prime) fields? I realized (from some experiments) it might be true for $\mathbb{F}_3$ as well. If so then what are interesting cases of symmetric matrix groups and what groups do those correspond to?

I am not only happy for some answers, but would also love to get some reading advice on this topic. Thank you everyone in advance.

Edit: To clarify two things that were asked. A matrix group is the representation of another group if it is isomorphic to this group. By regular I mean that the determinant is not 0.

答:
The invertible symmetric matrices don't form a group (in the obvious way, that is, they're not closed under multiplication). If $A$ and $B$ are symmetric then $(AB)^T = B^T A^T = BA$ so $AB$ is symmetric iff $A, B$ commute.

You can try to fix this by symmetrizing the product to $A \circ B = \frac{AB + BA}{2}$ (assuming $2$ is invertible) but 1) this does not preserve invertible matrices and 2) it is also not associative. However, it does endow the vector space of symmetric matrices with the structure of a Jordan algebra.


问:
Why doesn't Euler's $e$ in the continuous interest formula show the exponent $n$?
The continuous compounding formula is
$$A = Pe^{rt}$$

However, $e = (1+\frac{1}{n})^n$ provided that $x\rightarrow0$

Then is it correct to rewrite $A = Pe^{rt}$ as $A = P((1+\frac{1}{n})^n)^{rt}$ ?

答:
Yes, you can write it showing $n$ in the exponent
$$ Pe^{rt}=P\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{nrt} $$


问:
Can we say that $\tau_1 = \tau_2\ $?

Let $X$ be a non-empty set. Let $\tau_1$ and $\tau_2$ be two topologies on $X.$ If there exists a homeomorphism $f : (X, \tau_1) \longrightarrow (X,\tau_2)$ then can we say that $\tau_1 = \tau_2\ $?

Any help will be appreciated. Thanks for reading.

Title: Can we say that $\tau_1 = \tau_2\ $?

Body: Take $X = \{1,2\}$, $\tau_1 = \{\varnothing,\{1\},X\}$, $\tau_2 = \{\varnothing,\{2\},X\}$ and the map $f : X \to X$ such that $f(1)=2$ and $f(2) = 1$.


问:
Establish identitiy $(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2 = 2$
Hey guys this is the question:

$$(\cos\theta - \sin\theta)^2 +  (\cos\theta + \sin\theta)^2 = 2$$

I did this before but I forgot how I did it.. I tried to do this but don't know how to continue.. btw I can only use identity formulas, so please don't answer with other formulas I don't know about...

$$\cos^2\theta  - 2\cos\theta\sin\theta + \sin^2\theta + 1 = 2$$

$$1 - 2\cos\theta\sin\theta = 2$$

then idk.. please help!

答:
$$
\require{cancel}
\begin{align}
\left(\cos\theta - \sin\theta\right)^2 + \left(\cos\theta + \sin\theta\right)^2&=\cos^2\theta\cancel{-2\sin\theta\cos\theta}+\sin^2\theta+\cos^2\theta\cancel{+2\sin\theta\cos\theta}+\sin^2\theta\\
&=\cos^2\theta+\sin^2\theta+\cos^2\theta+\sin^2\theta\\
&=1+1\\
&=2
\end{align}
$$


问:
Trying to prove $\{[−\infty, a)\}_{a\in\mathbb{R}}$ generates The borel sigma algebra of the extended reals. Any hints?
I have already proved several similar questions but for the Borel sigma algebra on the reals and I am now a bit stuck when working with the extended reals.</p>

答: Taking complements, we see each $[a,\infty]$ is obtained. Since $(a,\infty]=\cup [a+1/n,\infty],$ we obtain each $(a,\infty].$ Given reals $a<b,$ we obtain $(a,\infty]\cap [-\infty,b) =(a,b).$ That's a good start ...


问:
Basis for Kernel and image of map
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Find the basis for the kernel and the image. Find their dimension.

my thoughts

I have shown it is not a isomorphism, that $V$ is a subspace and - I think the kernel would be $\begin{pmatrix}
1 &-1  &2 \\
 1&0  &-1 \\
 0&  1& -3
\end{pmatrix}$ and after reducing to reduced echelon form I get that the basis would be $(1,3,1)$ with dimension $1$ because of the free variable $x_3$. Now, for the image, I have no idea since I do not know how to find the image in the first place to use Gaussian elimination.

答:
The equation defining the subspace $V$ tells us that $x_1=x_2-2x_3$, where $x_2$ and $x_3$ are free variables. (In other words, $V$ is the subspace that is the solution set of this linear equation.) So we can rewrite the definition of $V$ as
$$\begin{aligned}
V &= \left\{\begin{bmatrix} x_2-2x_3 \\ x_2 \\ x_3 \end{bmatrix} \colon x_2,x_3\in K\right\} = \\
&= \left\{x_2\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \colon x_2,x_3\in K\right\} = \\
&= \operatorname{Span}\left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}\right\}\subseteq K.\end{aligned}$$

Then plugging in a generic vector from $V$ into the map $f$, and using $x_1=x_2-2x_3$, we get
$$f\left(\begin{bmatrix} x_2-2x_3 \\ x_2 \\ x_3 \end{bmatrix}\right) = \begin{bmatrix} (x_2-2x_3)-x_3 \\ x_2-3x_3 \end{bmatrix} = \begin{bmatrix} x_2-3x_3 \\ x_2-3x_3 \end{bmatrix} = (x_2-3x_3) \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

This last line shows everything we wanted to know. We can see that the image is one-dimensional, the span of $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. And we can see that a vector in $V$ belongs to the kernel of $f$ iff $x_2-3x_3=0$, from which it's easy enough to find a description of this kernel.

Title: Basis for Kernel and image of map


答:
Rather than directly defining $e$ as
$$
e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \, ,
$$
it is better to first define the exponential function $\exp$ or $e^x$ as
$$
\exp(x)=e^x=\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n \, ,
$$
in which case $e=\exp(1)$. It is not $e$ which is fundamental so much as the exponential function. If we define the exponential function in this way, then we can write $A=Pe^{rt}$ as
$$
P \cdot \lim_{n \to \infty}\left(1+\frac{rt}{n}\right)^n \, .
$$

Title: Why doesn't Euler's $e$ in the continuous interest formula show the exponent $n$?

Body:
The continuous compounding formula is
$$A = Pe^{rt}$$

However, $e = (1+\frac{1}{n})^n$ provided that $x\rightarrow0$

Then is it correct to rewrite $A = Pe^{rt}$ as $A = P((1+\frac{1}{n})^n)^{rt}$ ?


问:
Equal Sized Voronoi Cells minimize mean error/distance
Suppose I have $N$ objects $x_1, x_2, ..., x_N$, where $N$ is a large integer number.

Suppose I have $M$ representation points: $y_1, ..., y_M$ where $M < N$.

Consider the quantity $$S = \frac{1}{N} \sum_{i=1}^N d(x_i, \psi(x_i))$$

where $\psi(x_i) $ is equal to $y \in \{y_1, ..., y_M \}$ such that $d(x_i,y) \leq d(x_i, y_k)$ for all $k \in \{1,..., M \}$.

The set of $x_i$'s mapped to a particular $y$ is called its Voronoi cell. Suppose I am trying to minimize $S$ by choosing $y_1, ..., y_M$. Is it true that equal sized, disjoint Voronoi cells is like conducive to minimization?

In a research problem I am working on, I encounter an object like $S$ which I have to lower bound. I find that I can easily give the desired lower bound if I assume equal sized, disjoint Voronoi cells condition. But I want to make sure there's no loss of generality?

答:
There are a number of variations of this problem depending on metrics used and precise details of the objective function. The most common and clearly related formulation is k-means clustering which involves the using the squared Euclidean distance in the objective function you have given (unlike the facility location problems which tend to optimize the maximum distance of any of the points).

For the analogous continuous problem (where rather than summing over $N$ discrete points, the objective integrates over a continuous set), optimal solutions are characterized by centroidal Voronoi tessellations and as the number $M$ of representation points increases, the optimum tend to equal sized hexagons [1].

However, getting a lower bound on your error quantity that you describe is going to require more information about the the $N$ objects you describe. If these points are naturally grouped into $M$ tight clusters, the optimal solution can be arbitrarily small and the sizes of the Voronoi cells are just based on the locations of these clusters which may involve a very non-uniform distribution.

[1] Newman, Donald J., The hexagon theorem, IEEE Trans. Inf. Theory 28, 137-139 (1982). ZBL0476.94006.


问:
what is the diffrence between a term , constant and variable in first order logic languages ?
in the text , the author says that the language contains parenthises , sentintial connectives and n-place functions , n-place predicates , equality sign = , terms , constans and variables

i have two question ,

1- what is the definition of a term , constant and variable ?!
and what is the diffrence between them ?

2- what is the definition of n-place predicate ?

thanx !

答:
A constant is a symbol. A variable is also a symbol. The difference comes from how they can be used (syntactically). For example, a symbol $v$ can occur with a quantor (as in $\forall v$) only if $v$ is a vraiable, not if it is a constant.
Terms are words (symbol sequences) in the language, again obeying certain rules; for example constants and variables are the simplest terms and new terms are obtained from simpler terms by connectives.
As a matter of fact, the details of these distinctions should be apparent with what I assume the author writes in the nest few paragraphs after these remarks.</p>


问:
Is there simple proof that no finite polynomial with rational coefficients can generate only primes?
There are many proofs, suitable for beginners and new students, that there is no polynomial that generates only primes - with the constraint that the polynomial has finite terms, is non-constant, and has positive integer coefficients.

However, searching the internet and this maths stack exchange - there does not appear to be a simple proof that loosens the requirements to positive rational coefficients.

The question has been discussed in comments to other questions, but no proof set out, or linked to.

I would appreciate a short simple proof, suitable for beginners (not trained at university maths).

Also can the proofs be extended to negative coefficients, integer or rational?

答:
Let the polynomial be

$f(x) = a_0+a_1x+a_2x^2+...+a_nx^n$ where $a_i$ is rational for  $0 \le i \le n$

$a_i = \frac{p_i}{q_1}$ where $p_i,q_i$ are coprime

Assume $f(x)$ always produces primes.

Let $f(k)$ be integer for some $k$. Therefore $f(k)$ is prime.Let $f(k) = q$ where $q$ is prime.

$f(k) \equiv 0\pmod q$

$f(k+nq)\equiv 0\pmod q$. Here $f(k+nq)$ is an integer and n is any natural numer divisible by the LCM of all the $q_i$ for $0 \le i \le n$.

Since $f(n+kq)$ is prime, it has to be $q$.

But $f(x)$ is an increasing function(if it nonconstant)

So $f(k+nq)>f(k)$

but $f(k+nq)=f(k)=q$

There is a contradiction.

You can create a very simillar proof even if the coefficients are negative.

Hope this helps


问:
If $E(X_n) \to E(X)$, does this imply convergence in $L^1$?
My question is basically if there exists any counterexample where we have that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$, but $X_n$ does not converge to $X$ in $L^1$. If not, then it should be true that $\mathbb{E}(X_n)\to \mathbb{E}(X)$ as $n\to \infty$ implies that $X_n \overset{L^1}{\rightarrow}X$ but I cannot really get to prove that one.

Any tips on how to approach this problem? Or any counterexample you could maybe think of?

答: Suppose that all $X_n$ and $X$ are defined on different probability spaces and have standard normal distribution. In this case $E X_n = 0 \to 0 = EX$ and there's no convergence in $L_1$.


问:
The series $1 + v^2 + v^4 + v^6 + v^8 + ...$ is gamma squared
Can anyone tell me anything about the series
$ 1 + v^2 + v^4 + v^6 + v^8 ... $

Does it have a name? Does it have any special properties?

I ask because $ \gamma = \sqrt{1 + v^2 + v^4 + v^6 + v^8 ...} $

Obviously it only works for v < 1

(i.e. less than the speed of light)

edit: hmmm...subtract one and divide by v^2 and the result is itself

edit: oh and $ \gamma - 1 \approx v^2/2 $ for small values hence

kinetic energy = $m (\gamma - 1) c^2  \approx ( E = m v^2/2$)

答:
Taking $v$ to be the speed of an object relative to some inertial frame $O$, $\gamma = (1-v^2/c^2)^{-1/2}$, then by the geometric series formula:

$$1+\frac{v^2}{c^2} + \frac{v^4}{c^4} + \ldots = \frac{1}{1-\frac{v^2}{c^2}}=\gamma^2$$

Therefore if $v$ in your question is in natural units, you have $1+v^2+v^4 + \ldots = \gamma^2$.


答: This is a question about special relativity, in which $v$ (often denoted $\beta$) satisfies $\gamma=1/\sqrt{1-v^2}$. As @ziprirovich notes, it's a geometric series, which converges to $\frac{1}{1-v^2}=\gamma^2$ because $|v|<1$.

Title: The series $1 + v^2 + v^4 + v^6 + v^8 + ...$ is gamma squared

Body:
Can anyone tell me anything about the series
$ 1 + v^2 + v^4 + v^6 + v^8 ... $

Does it have a name? Does it have any special properties?

I ask because $ \gamma = \sqrt{1 + v^2 + v^4 + v^6 + v^8 ...} $

Obviously it only works for v < 1

(i.e. less than the speed of light)

edit: hmmm...subtract one and divide by v^2 and the result is itself

edit: oh and $ \gamma - 1 \approx v^2/2 $ for small values hence

kinetic energy = $m (\gamma - 1) c^2  \approx ( E = m v^2/2$)


问:
The right way of proving LHS = RHS
I'm currently brushing up on my math before heading back to school for my bachelor's and am practicing on some questions.

Given the following equation :

$$\frac{x^2 - 3x}{x^2 - 9} = 1-\frac{3x-9}{x^2-9}.$$

and that I have to prove that the LHS = RHS; How would I go about doing so?

What I've done:

I've simply brought $(3x-9)/(x^2-9)$ over to the LHS, resulting in the equation $(x^2 - 3x)/(x^2-9) + (3x-9)/(x^2-9) = 1$ , which gives me $(x^2-9)/(x^2-9) = 1$.(LHS = RHS)

However, I do not think that what I've done is the correct method of proving. Should I have instead transformed the LHS into RHS without bringing anything from LHS to RHS and vice versa?

Furthermore , after which I'm integrating $(x^2-3x)/(x^2-9)$ by doing the following:

What I first did was factorise the denominator into $(x+3)(x-3)$.
I then equated $(x^2 - 3x) = A(x + 3) + B(x - 3)$.

Given that $x = 3 \implies A = 0$ and given that $x = -3 \implies B = -3$.

This would then give me the following equation to integrate:

$[ (-3)/(x+3) + 0(x-3) ]dx$

which led me to the answer $-3ln|x+3| + C$.

However, when I checked my answer, the correct answer should be

$$x -3ln|x+3| + C.$$

Where did the missing $x$ come from?

Any assistance would be greatly appreciated

Thanks!

答:
I've simply brought $(3x−9)/(x^2−9)$ over to the LHS,...

By so doing you have already assumed the equality of the two sides. When proving an alleged equality, one has access to only one side, not both - the other side should pop out from the proof.

As for the rest, its integration via partial fractions, which I think you were on the right track.

Title: The right way of proving LHS = RHS


问:
Show that $\sin(45^\circ+\alpha)=\cos(45^\circ-\alpha)$ and $\cos(45^\circ+\alpha)=\sin(45^\circ-\alpha)$
Show that if $\alpha\in[0^\circ;45^\circ]:\sin(45^\circ+\alpha)=\cos(45^\circ-\alpha)$ and $\cos(45^\circ+\alpha)=\sin(45^\circ-\alpha).$

I tried to use the unit circle, but I am not sure how to draw the angles $\alpha, 45^\circ+\alpha$ and $45^\circ-\alpha.$ I have noticed $45^\circ+\alpha\le90^\circ$ and $45^\circ-\alpha\ge0^\circ.$
Can I use any other approaches? Why do we have restrictions on $\alpha$? Thank you in advance!

答: Hint: $$\sin \theta = \cos(90^\circ -\theta) \\ \cos\theta =\sin(90^\circ -\theta)$$ The range of $\alpha$ is irrelevant, as this identity holds for all $\theta\in\mathbb R$.


问:
$ (X,d)$ is a metric space and topology induced by a metric $\rho$ is topology induced by d then..
This question is number 3 of exercise 4.1 of Wayne patty's topology.

Let $(X,d) $ be a compact metric space and let $\rho $ be any metric on $ X$ such that the topology induced by $\rho$ is topology induced by d. Prove that $(X, \rho)$ is bounded.

I have problem understanding what it means by  "topology induced by $\rho$ is the topology induced by d". $d$ will induce topology $d(x,y) = |x-y |$ and $\rho (x,y) = |x-y|$ and open sets can be given by $d(x,y)< \epsilon$ and $\rho (x,y) < \epsilon$ .I think it means that open sets  generated by them will be same.

Am I right?

In the case I am right I don't get any breakthrough on how it can be proved as using compactness ie open cover having finite subcover that $(X, \rho)$ is bounded?

Please help

答:
The topology induced by a metric $\mu$ on $X$ is the topology whose base is the family

$$\{B_\mu(x,r):x\in X\text{ and }r>0\}$$

of open balls with respect to $\mu$. Metrics $d$ and $\rho$ on $X$ induce the same topology if the bases

$$\{B_d(x,r):x\in X\text{ and }r>0\}$$

and

$$\{B_\rho(x,r):x\in X\text{ and }r>0\}$$

generate the same topology, i.e., if the metric spaces $\langle X,d\rangle$ and $\langle X,\rho\rangle$ have the same open sets. Note, however, that in general this has nothing to do with $|x-y|$: there may well be no notion of subtraction of of absolute value in the space $X$.

HINT: Fix a point $p\in X$ and consider the family $$\mathscr{U}=\{B_\rho(p,n):n\in\Bbb Z^+\}\,.$$ Verify that $\mathscr{U}$ is an open cover of $X$ and apply the compactness of $\langle X,d\rangle$.


问:
Finding example of a compact metric space a non hausdorff Topological space and a continuous function that maps X onto Y
This question is from my topology assignment and I am struck on it.

Give an example of a compact metric space $(X,d)$ , a topological space $(Y,\mathcal{U})$ that is not Hausdorff and a continuous function $f$ that maps $X$ onto $Y$.

Topological space can be taken $Y=\mathbb{N}$ under usual topology. But I am not able to decide $(X,d)$ as compact so that a continuous function exists.

If I take Euclidean metric $d$ then I have to take a closed and bounded subset but I am not able to then find a continuous function then.

Kindly guide!

答: HINT: You know that the continuous image of a compact space is compact, so $Y$ must be a non-Hausdorff compact space. An easy way to make sure that $Y$ is compact is to make it finite. We need at least two points in $Y$ if it is to be non-Hausdorff, so let’s let $Y=\{0,1\}$ and $U=\{\varnothing,\{0\},Y\}$. $[0,1]$ is a nice compact metric space; we can try to find a continuous function $f$ mapping $[0,1]$ onto $Y$. What has to happen in order for $f$ to be continuous? We need $f^{-1}[\varnothing]$, $f^{-1}[\{0\}]$, and $f^{-1}[Y]$ to be open in $[0,1]$. Two of those are automatically true; which two? Now find a way to define $f$ so that the third is true as well and $f$ is surjective (onto), and you’ll be done.


问:
binary tree and one proof condition, why n should relate to three?
I see this proof in binary tree on Question $1$. I see an example that tells a fact:

in a binary tree whit $n$ elements (n divisible by three) there is a node $u$ that number of nodes in a sub-tree with root $u$ at least $\frac{n}{3}$ and at most $\frac{2n}{3}$.

I couldn't get the point why this is true for every binary tree but this facts tell for $n$ divisible by three.

I need a simple and most simple proof like the one includes in mentioned link to get the idea and relation to divisibility by three.

答:
Too long for comments....

Claim 1: For general binary trees, each vertex $y$ that is not a leaf has a child $y_1$ with size$(y_1)$ at least $\lfloor$size$(y)/2 \rfloor$; namely in general $y_1$ is the child of $y$ with the larger size. [If $y$ has 2 children with equal size then $y_1$ can be any child of $y$.]

This I trust you can see already.

This Claim 1 suffices to show that there is a separator w one edge $e$ where the smaller side has at least $n/3$ vertices if $n$ is divisible by 3.  Indeed, let $y$ be a vertex, and then let $y_1$ be $y$'s child, where both (a) and (b) are satisfied: (a) size$(y_1) \ge$ $\lfloor($size$(y)/2 \rfloor$, and (b) size$(y) > 2n/3$ but size$(y_1) \le 2n/3$. [Note that such vertices exist, indeed march down the tree from the root at each step going to the vertex with the larger size; clearly there are $y$, $y_1$ so that (b) will be satisfied, by Claim 1 so will (a).] Then as $2n/3$ is an integer if $n$ divides 3, it follows that size$(y) \ge 2n/3+1$ and so Claim 1 implies that size$(y_1) \ge n/3$. So then $n/3 \le$ size$(y_1)$ as well, thus the string of inequalities $n/3 \le$ size$(y_1) \le 2n/3$ holds, and cutting the edge $e$ between $y$ and $y_1$ will leave size$(y_1)$ vertices on one side and $n-$size$(y_1)$ vertices on the other side.

This Claim 1 suffices also to show that there is a separator w one edge $e$ where the smaller side has at least $n/3$ vertices if $n$ is 2 mod 3. Indeed, let $y$ be a vertex, and then let $y_1$ be $y$'s child, where both (a) and (b) are satisfied: (a) size$(y_1) \ge$ $\lfloor($size$(y)/2 \rfloor$, and (b) size$(y_1) \le 2n/3$ but size$(y_1)> 2n/3$. Now, suppose
size$(y_1) \le \lfloor \frac{n}{3} \rfloor$. Then by Claim 1,
size$(y) \le 2$ size$(y_1)+1$. However, for $n$ that is 2 mod 3, the inequality
$2\lfloor n/3 \rfloor +1$ $\le 2n/3$ holds. So this would imply that size$(y) \le 2n/3$ after all which is a contradiction. So it follows that size$(y_1)$ must be
at least $\lfloor n/3 \rfloor +1$ $\ge n/3$, so it follows that
$n/3 \le$ size$(y_1) \le 2n/3$. Finish as in Case 1.

Thus, the only case left is $n$ is 1 mod 3. And here there are some instances for which the best we can do is where the smaller side has $\lfloor n/3 \rfloor$ vertices, see the example given in your other question. If however we impose the additional condition that every interior vertex of $T$ with, the exception of exactly 1 vertex the root, has degree 3, then in fact a separator where the smaller side has at least ceiling of $n/3$ vertices even when $n$ is 1 mod 3. The proof in the link you provided establishes that.


答:
$HINT$:You should know that $\sin(x)\times \sin(y)=\frac{\cos(x+y)-\cos(x-y)}{-2}$ and  $\sin(x)\times \cos(y)=\frac{\sin(x+y)+\sin(x-y)}{2}$

so, try to multiply and divide the former by $\sin(45+a)$ and the latter by $\cos(45+a)$

for the former:

$\frac{\sin(45+a)\times \sin(45+a)}{\sin(45+a)}=[\frac{\cos(90+2a)-\cos(0)}{-2}]=\sin(45+a)\times \cos(45-a)=[\frac{\sin(90)+\sin(2a)}{2}]$

so,$\sin(2a)=\sin(2a)$

The second for yours :))

Title: Show that $\sin(45^\circ+\alpha)=\cos(45^\circ-\alpha)$ and $\cos(45^\circ+\alpha)=\sin(45^\circ-\alpha)$

Body:
Show that if $\alpha\in[0^\circ;45^\circ]:\sin(45^\circ+\alpha)=\cos(45^\circ-\alpha)$ and $\cos(45^\circ+\alpha)=\sin(45^\circ-\alpha).$

I tried to use the unit circle, but I am not sure how to draw the angles $\alpha, 45^\circ+\alpha$ and $45^\circ-\alpha.$ I have noticed $45^\circ+\alpha\le90^\circ$ and $45^\circ-\alpha\ge0^\circ.$
Can I use any other approaches? Why do we have restrictions on $\alpha$? Thank you in advance!


问:
A little doubt on the usage of the $(=\; elim)$ rule?
I'm reading Language, Proof and Logic. I'm trying to understand this proof:

"= Intro" and "= Elim" are:

I understand the use of "= Intro" in the proof, but how exactly are we using "= Elim"? What is $p(n)$? I suspect $p(n) := [n=b]$. So we have in the fourth line:

$p(b)$ which means $b=b$

$b=a$

$p(a)$ which means $a=b$

Is that it?

答:
Almost!  The $P(n)$ is a formula that has some number of instances of the variable-free term $n$, and the $P(m)$ is the formula that is the result of replacing zero or more of those $n$'s with $m$'s given that $n = m$

So ... the '$n=m$' claim corresponds to the $b = a$ claim on line 2, meaning that the '$n$' is $b$, and the '$m$' is $a$. The '$P(n)$' is $b = b$ on line 3, and the $P(m)$ is $a=b$ on line 4.

So yes, you correctly identified that the $b=a$ claim was the 'substitution' claim that you applied to the claim $b=b$ ... but it is not quite true that the '$P(n)$' claim is $n = b$.... it is simply $b=b$.

However, I think you were trying to think about it this way:

Let the formula $P(x)$ be the expression $x = b$

Then $P(b)$ is the result of taking formula $P(x)$, and instantiating the free variable $x$ with constant $b$.

So, $P(b)$ is $b = b$

Likewise, $P(a)$ is the result of taking formula $P(x)$, and instantiating the free variable $x$ with constant $a$.

So, $P(a)$ is $ a = b$

Identifying a formula $P(x)$ of which both the 'before' expression $P(n)$  and the 'after' expression $P(m)$ are instances (or in our case $P(a)$ and $P(b)$) is indeed a helpful way to think about $= Elim$. It may also make it a little more clear as to why you don't have to replace all occurrences of some constant in an expression with some other constant.

That is, had we defined $P(x)$ as $x = x$, then $P(b)$ would still have been $b = b$, but $P(a)$ would then have been $a = a$.  And if $P(x)$ would have been $b = x$, then $P(a)$ would have been $b = a$.

Perversely, we could even have defined $P(x)$ as $b = b$. Then $P(a)$ would still have been $b = b$. So, if you infer $b = b$ from $b = b$ and $b = a$ using $= Elim$ in the proof-checker software, it should still check out. Try it!


问:
Subgroup decomposition
Let $K_1,K_2$ be strict normal subgroups of group $G$, $G =K_1K_2$ and $H \leq G$. Is it true that $H = (H \cap K_1)(H \cap K_2)$?
Can we relax the normal condition by considering conditions on $H$ or $G$?

答:
This isn't true. Consider $G = \mathbb Z \times \mathbb Z$. Then $G = K_1 K_2$, where
$$
K_1 = \mathbb Z \times \{0\}$$ and $$K_2 = \{0\} \times \mathbb Z.
$$
Then consider the subgroup
$$
H := \{(n, n)|n \in \mathbb Z\}.
$$
Its intersection with both $K_1$ and $K_2$ is zero, yet it is nontrivial.

Title: Subgroup decomposition


问:
Need help to check alternating series criterion
I know this is simple calculation based question but I got stuck:

I want to check convergence of the following series:
$$1-\frac{1}{2}(1+\frac{1}{3})+\frac{1}{3}(1+\frac{1}{3}+\frac{1}{5})-\cdots$$
Clearly this is alternating series. So I need satisfy alternating series criterion. That means I need to show that its terms are decreasing monotonically.

Let $a_n=\frac{1}{n}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1})$ and  $a_{n+1}=\frac{1}{n+1}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1})$.

I just have to show  $a_n-a_{n+1}>0$  but right here I got stuck because of the presence of the first factor. I think i am missing a simple trick here.

Anyway, I got $$a_n-a_{n+1}=(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{3}(\frac{1}{n}-\frac{1}{n+1})+\cdots+\frac{1}{2n-1}(\frac{1}{n}-\frac{1}{n+1}) {-\color{red}{\frac{1}{(2n+1)(n+1)}}}.$$

The last red color term making problem.

May be ratio test help as well ?

Any help please

答:
Doesn't answer your question, since I think you've already got the answer to it. It ads something to it though.

Let's evaluate the sum of the series $ \sum\limits_{n\geq 0}{\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}}} $.

Denoting $\left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}} $, and $ b_{n}=\frac{\left(-1\right)^{n}}{2n+1} $.

Let $ n\in\mathbb{N} $, we have the following : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\frac{1}{2}\left(\sum_{k=0}^{n}{\frac{1}{2k+1}}+\sum_{k=0}^{n}{\frac{1}{2\left(n-k\right)+1}}\right) $$

We added the sum to itsefl, but by shifting the terms, then we divided by $ 2 $.

Thus : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\left(n+1\right)\sum_{k=0}^{n}{\frac{1}{\left(2k+1\right)\left(2\left(n-k\right)+1\right)}} $$

Hence : $$ a_{n}=\sum_{k=0}^{n}{\left(\frac{\left(-1\right)^{k}}{2k+1}\times\frac{\left(-1\right)^{n-k}}{2\left(n-k\right)+1}\right)} $$

Now, using a Cauchy product, we have : \begin{aligned}\sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\sum_{n=0}^{+\infty}{a_{n}}\\&=\sum_{n=0}^{+\infty}{\sum_{k=0}^{n}{b_{k}b_{n-k}}}\\ &=\left(\sum_{n=0}^{+\infty}{b_{n}}\right)^{2}\\ &=\left(\sum_{k=0}^{n}{\frac{\left(-1\right)^{n}}{2n+1}}\right)^{2}\\ \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\frac{\pi^{2}}{4}\end{aligned}


问:
Splitting Field, irreducibility and degree n!
Let $f(x) ∈ F[x]$ have degree $n > 0$, and let $L$ be the splitting field of $f(x)$over $F$. I wish to show if $[L : F] = n!$ then f(x) is irreducible over F.
Also, the converse of this is false with some counterexample? I know $[L:F]$ divides $n!$ from this result but i can't make the same method work to prove this~
Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.</p>

答: Let $f = gh$ with $\deg(g) = m$ and $\deg(h) = k$ be a reducible polynomial. Let $F(a) = K$ and $K(b) = L$ where $a$ is a root of $g$ and $b$ is a root of $h$. We know that $[L :F] = [L : K][K : F] \mid m! k!$, but also that $m!k! < (m + k)!$ (if $m, k > 0$). Hence, $[L : K] < (m + k)! = n!$; which is a contradiction, therefore our original assumption must have been false, meaning that $f$ is irreducible.


问:
What natural integers satisfy the relation $\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
I would like to find all natural triplets $(a, b, c)$ that satisfy the  relation
$\frac{3}{7}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\,.$

Body:
Wlog let $a\leq b\leq c$ then $\frac{1}{a}\geq \frac{1}{b}\geq\frac{1}{c}$

Hence

$$\frac{3}{7}=\frac{1}{a} + \frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$$

$$a\leq 7$$

$a=1$  NA

$a=2$  NA

$a=3$

$$\frac{2}{21}=\frac{1}{b}+\frac{1}{c}\leq\frac{2}{b}$$

$b\leq 21$

$b=11, c=231$

$b=12, c=84$

$b=14, c=42$

$b=15, c=35$

$b=21, c=21$

I hope you can find other solutions.


问:
How do I apply an angular velocity vector[3] to a unit quaternion orientation?
I have an angular velocity vector[3] in three dimensions and a unit quaternion (magnitude of 1) representing an orientation in three dimensions.

I need to apply the angular velocity to the quaternion orientation, such that the rotation is correct in a Newtonian universe.

I'm sure it was in the books somewhere, but now I can't find it.

Thanks.

Alright, in reply to help; would this be the equations?

Where:

C is q-1 (well, the conjugate, but my books tell me a unit quaterinion conjugate is the same as the negate)

b is the angular velocity, already scaled by time.

a is q and the quaternion to rotate.

So: qbq-1, or abc for quaternion rotation. It's rotating, that's fine.

However, it's applying the angular velocity as if it were a rotation from (1,0,0,0).

To explain what I'm seeing: Let's say we apply an angular velocity, pre-multiplied by dt (delta time). It gets multiplied by the conjugate of the current rotation (which, currently, is (1,0,0,0) and the current rotation, to result in the next rotation.

So the next rotation is initially equal to the angular velocity.

Then we multiply the angular velocity, as a quaternion, by the conjugate of the current rotation.

Which is the conjugate of the angular velocity as a quaternion.

So essentially, we end up with a rotation from (1, 0, 0, 0) each time step.

...And now it's swinging back and forth along the Y axis.

I suspect the problem is in how I'm turning angular momentum into angular velocity; should I make a new topic? Not entirely sure of informal etiquette here. Makes sense too, but figured I'd ask.

答:
It's not entirely clear (to me) what you mean by "applying an angular velocity". Angular velocity is an instantaneous property; if you only have a single angular velocity vector, you only have the first derivative of the motion at one point in time, and thus can't say anything about the orientation at other times.

I'll assume that you're assuming the angular velocity to be constant, and that by "applying" it you mean calculating the orientation at later times of a body that has the given initial orientation and undergoes a motion with the given constant angular velocity.

To do this, you need to know a) how to find the unit quaternion corresponding to a rotation vector and b) how to compose two unit quaternions in order get the unit quaternion corresponding to the composition of the corresponding rotations.

For a), let's denote the angular velocity vector by $\vec{\omega}$. Over a time $t$, a motion with this (constant) angular velocity results in a rotation that can be described by the vector $\vec{w}=\vec{\omega}t$, where the direction and magnitude of $\vec{w}$ specify the axis and angle of the rotation, respectively. This rotation corresponds to the unit quaternion

$$q=\left(\cos\frac{|\vec{w}|}{2},\frac{\vec{w}}{|\vec{w}|}\sin\frac{|\vec{w}|}{2}\right)\;.$$

For b), since applying the rotation corresponding to a unit quaternion $q$ to a three-dimensional vector $\vec{v}$ consists in conjugating a quaternion $\nu$ corresponding to $\vec{v}$ by $q$, composing rotations just corresponds to multiplying the corresponding quaternions:

$$q(\vec{v})=q\nu q^{-1}\;,$$
$$q_1(q_2(\vec{v}))=q_1(q_2\nu q_2^{-1})=q_1q_2\nu q_2^{-1} q_1^{-1}=(q_1q_2)(\vec{v})\;.$$

(This is part of what makes quaternions so useful for representing three-dimensional rotations). So now you just multiply $q$ and your initial unit quaternion, say, $q_0$. Since the initial rotation is applied first, the corresponding unit quaternion is on the inside of the nested conjugations, so it has to be the right-hand factor: $q'=qq_0$.

For more on all this, see this Wikipedia article.


答:
A variable is a symbol like $x$ standing for a variable, so these are the basic bricks. Terms are built up from variables and operation symbols, recursively:

Each variable itself is also a term,

and whenever $\tau_1,..,\tau_n$ are terms, and $f$ is an $n$-ary operation symbol ($n$ place function, as you called), then the string $f(\tau_1,..,\tau_n)$ is again a term.

A constant is basically a nullary operation, i.e. it inputs $0$ pieces of elements and outputs one element, so in practice, a constant will be a particular element in any structure ('incarnation') of the given language. Constants themselves are also terms.

An $n$-ary predicate symbol $R$ is then realized in an $n$-ary relation in the structrues. If $a_1,..,a_n$ are concrete elements in a concrete structure, then $R(a_1,..,a_n)$ is either true or false.

For example consider the language with one binary operation symbol $+$ and one binary relation symbol $\le$ and a constant $0$. A structure of this is e.g. $\Bbb N$ (or $\Bbb Z$ or $\Bbb R$) with the usual interpretations of these symbols. A quantifier-free formula is for example
$$(x+y)+x\le 0$$

Title: what is the diffrence between a term , constant and variable in first order logic languages ?

Body:
in the text , the author says that the language contains parenthises , sentintial connectives and n-place functions , n-place predicates , equality sign = , terms , constans and variables

i have two question ,

1- what is the definition of a term , constant and variable ?!
and what is the diffrence between them ?

2- what is the definition of n-place predicate ?

thanx !


问:
A point about integrals in Residue formula proof
I am reading the following proof in Complex Analysis by Stein and Shakarchi, page 76.

They are using the concept of a keyhole contour, with which you are probably familiar. Let's call this keyhole contour $K$ and follow the book's notation for other curves.

Letting the width of the keyhole's corridor tend to zero, we are left with the following (after the corridor walls cancel out):

$$\int_K f(z)dz \; = \int_Cf(z)dz \; + \int_{C_{\epsilon}} f(z)dz \; = \; 0$$

Now, this seemingly contradicts the second equality presented in the image. But I suppose this is just a notation issue, because the book uses $\int_{C_{\epsilon}} f(z)dz$ to denote integral over positively oriented curve $C_\epsilon$, while I use it to denote integral over negatively oriented curve. It makes more sense to me to so because $C_\epsilon$ is a part of a bigger $K$ contour for which the orientation was already chosen.

My only question is whether this is correct?

EDIT:

Here are diagrams of what I mean by $K$, $C$ and $C_\epsilon$:

I took $z_0 = 0$ for simplicity and $\delta$ is the width of the corridor.
First I let $\delta \to 0$ which cancels the corridor and in my understanding leaves me with positively oriented circle $C$ and negatively oriented circle $C_\epsilon$. Also integral over the whole keyhole $K$ shown in the first image must be $0$ because $f(z)$ is holomorphic everywhere except at the point $z_0$.

答:
I think part of the difficulty here is that the direction of the paths is being changed from that in the book, so let's assume both $C$ and $C_\epsilon$ are counter-clockwise. You have the correct idea: Since $f$ is holomorphic in the shaded region, the integral of $f$ along the entire contour $K$ below is $0$.

As the gap vanishes, the integral along the black segments cancel, and we get
$$
\int_Kf(z)\,\mathrm{d}z=\int_Cf(z)\,\mathrm{d}z-\int_{C_\epsilon}f(z)\,\mathrm{d}z=0
$$
with a minus instead of a plus as in the first equation in the question (after the image from the book).


问:
Derivative of an even function is odd and vice versa
This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even.
(Write the equation that says $f$ is even, and differentiate both sides, using the
chain rule.)"

I already read numerous solutions online. The following block shows the official solution, but I didn't quite understand it. (Particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$.)

(1F-6). Following the hint, let $z=-x$. If $f$ is even, then $f(x)=f(z).$ Differentiating and using the chain rule:
$$f'(x)=f'(z) \frac{dz}{dx}=-f'(z),$$ because $\frac{dz}{dx}=-1.$ But this means that $f'$ is odd.

Similarly, if $g$ is odd, then $g$($x=-g(z)$. Differentiating and using the chain rule: $$g'(x)=-g'(z) \frac{dz}{dx}=g'(z),$$ because $\frac{dz}{dx}=-1.$

Thanks in advance =]

答:
If $f(x)$ is an even function,

Then: $f(−x)=f(x)$

Now, differentiate above equation both side:
$f'(−x)(−1)=f'(x)$

$⇒f'(−x)=−f' (x)$

Hence $f'(x)$ is an odd function.

Rest part can be proved similarly.


问:
Is this a relation of equivalence? $R := \{(x,y) \in \mathbb{R} \times \mathbb{R} ; \sin(x)=\sin(y)\}$
We have to prove reflexivity, symmetry, and transitivity.

$1.)$ Reflexivity: If our relation is reflexive we know that if $(x,y)\in R$ then $(x,x)\in R$

$(x,x)\in R \to \sin(x)=\sin(x)$ which is true. Out relation is reflexive.

$2.)$ Symmetry: If our relation is symmetric we know that if $(x,y)\in R$ then $(y,x)\in R$

$(y,x)\in R \to \sin(y)=\sin(x)$ which is the same as our original relation: $\sin(x)=\sin(y)$

$3.)$ Transitivity: If our relation is symmetric we know that if $(x,y) \in R \land (y,z) \in R$ then $(x,z) \in R$

$(x,y) \in R \to \sin(x)=\sin(y)$

$(y,z) \in R \to \sin(y)=\sin(z)$

$\sin(x)=\sin(y) \land \sin(y)=\sin(z) \to \sin(x)=\sin(z)$ Which ios what we wanted to prove.

So $R$ is an equivalent relation on $\mathbb{R} \times \mathbb{R}$

答:
It's correct.

Actually it holds for an arbitrary function in place of $\sin$, moreover, using the quotient construction one can prove that a relation $R\subseteq X\times X$ is an equivalence relation if and only if there is a function $f:X\to Y$ such that
$$R=\{(x,x'):f(x)=f(x')\}\,.$$


问:
How to solve an exponential differential equation?
My electrical engineering class just moved into differential equations from linear algebra, which is a topic I've never touched on before. The professor doesn't work hardly any examples on the board, so I've found myself a bit confused on how to solve differential equations that aren't linear. Below I've attached a homework problem that I'm having quite a bit of trouble solving. Is there any way I could get some help with the steps on how to solve an exponential differential equation (in simplistic terms too, please!) from an example in this homework problem? I don't want the entire problem answered, just one or two with an explanation so I can wrap my head around how to solve these things. Thanks!

Problem:

答:
The algebraic equation associated to your ODE is
$$z^2+16z+63=0.$$
It has positive discriminant $\Delta=4$ and (real) roots:
$${z_1=\frac{-16-2}{2}=-9,}\quad{z_2=\frac{-16+2}{2}=-7.}$$
Hence, the general integral is the one in answer

B.$\qquad C_1e^{-9t}+C_2e^{-7t}$


问:
Way to have inutition about the shape of a curve?
Past the usual memorized curves like $y=\sin(x), y=|x|, y=1/x, y=\ln(x), \ldots,$ is there a way to have an intuition about the shape of a curve from looking at an arbitrary function/term?  (that is, other than for transformations of the usual memorized functions and asymptotes that result from prevention of division by zero)

I can’t think of any good examples right now, but a few times in classes the teachers have commented something like “you can see that this function’s/term’s/expression’s  curve will be … (he/she draws on board)…” despite the fact that the function/term/expression is a conglomeration of the basic ones we have memorized the curves for.

答:
The way I try to imagine curves is try to identify zeros of the function. And then try to see how the slope is changing by finding its first derivative. And if I want more accurate information I ll try to look for other properties like second derivative to see how the curvature of the graph looks like....

But these kind of things will be learnt only on practice...


问:
Roots of the Cubic equation

The Cubic formula:

$ax^3+bx^2+cx+d=0$

With under the following conditions:

$a \neq 0 $

$a,b,c,d \in \Bbb{R} $

We can derive the following formula as a the root of $x$:

$u= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}+\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}  $

$v= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}-\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}  $

$x_1= u + v-{b\over 3a}$

But the proof of derivation of the formula mentioned above was only limited to 1 root. In addition, that formula had no complex conjugates. However as we know, there must be 2 other roots which includes complex conjugates in their formulas.

In reference, Wikipedia: Cubic equation also says that there should be 2 other roots at maximum.

So, altogether the 3 roots are:

$i= \sqrt {-1} $

$ \omega = -{1\over 2} + {{\sqrt 3}i\over 2} $

$x_1= u + v-{b\over 3a}$

$x_2= {\omega }u + {\omega}^2v -{b\over 3a} $

$x_3= {\omega }^2 u + {\omega}v -{b\over 3a} $

And unfortunately, I didn't find or know know the proof of any other 2 roots, i.e $x_2$ & $x_3$.

So would you please show me the proof of the other 2 roots of Cubic formula?

```<code>Note: Please, no synthetic division. I need proof by formula
</code>```

答:
By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;

and replacing $a, b, c$ by $x, -u, -v$ respectively,

$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.

$u, v$ are known as resolvents.

Title: Roots of the Cubic equation