问:
Is it the case that there is some string, S, such that every string ending in S ends a perfect square? If so, is the same true for every power?
I am not a mathematician, but I gather from this paper that any string of digits ending in '1', '3', '7', or '9', no matter what it is, is the ending sequence of the decimal representation of infinitely many cubes. This other paper extends this result to other exponents that are relatively prime to 10. Each paper holds out the case of squares for further study.

My question is: does this hold for squares? Is it the case that there is some string of digits, S, such that every string ending in S is itself the ending of infinitely many squares? If so, does this result hold for every other exponent too, so that fourth powers, fifth powers etc. will also have some string, S, such that every string ending in S is itself the ending of infinitely many fourth powers (or fifth powers, etc. as the case may be)?

(I am told that this German paper should answer my question, but, alas, I don't read German.)

I am not a mathematician, so I am afraid that any answers will have to be spelt out very clearly for me.

Thanks very much for any help that anyone can give -- much appreciated.

答:
The answer to your question is yes, there do exist strings $S$ for which, given any string $T$ that finishes with $S$, there exist infinitely many square numbers whose final digits are the string $T$.

(S = 001 is such a sequence, and in fact we will show below how to construct 49 other suitable sequences, all of length 3.)

The proof can be given in a single paragraph, but this will require me to use certain mathematical notions (from group theory) that I suspect will not be familiar to you as a non-mathematician.

We know from the Chinese remainder theorem that the group of units $(\mathbb{Z}/10^k \mathbb{Z})^{\times} \cong (\mathbb{Z}/2^k \mathbb{Z})^{\times} \times (\mathbb{Z}/5^k \mathbb{Z})^{\times}$. For all $k \geqslant 1$ we have $(\mathbb{Z}/5^k \mathbb{Z})^{\times} \cong C_{\varphi(5^k)} \cong C_{4 \cdot 5^{k-1}}$ by the theory of primitive roots. Similarly, for all $k \geqslant 3$, we have $(\mathbb{Z}/2^k \mathbb{Z})^{\times} \cong C_2 \times C_{2^{k-2}}$. Putting this all together we have $$(\mathbb{Z}/10^k \mathbb{Z})^{\times} \cong C_2 \times C_{2^{k-2}} \times C_{4 \cdot 5^{k-1}}$$ and so there are $\frac{1}{8}\vert (\mathbb{Z}/10^k \mathbb{Z})^{\times}\vert = 2^{k-2} \cdot 5^{k-1}$ squares modulo $10^{k}$ that are comprime to $10$. This increases by a factor $10$ when $k$ goes to $k+1$, so it follows that all possible extensions of the squares in $(\mathbb{Z}/10^k \mathbb{Z})^{\times}$ must be squares in $(\mathbb{Z}/10^{k+1} \mathbb{Z})^{\times}$. Taking $S$ to be the set of quadratic residues modulo $10^3$ that are comprime to $10$ (chosen to lie between $1$ and $999$), we see that all strings in $S$ are suitable.

In thinking about how I would explain this argument in non-technical terms, I've realised that it would be almost impossible to do without knowing more details about your particular mathematical background. Do get in touch if you need to discuss this proof further!


问:
Proving Young's inequality for n numbers whitout using AM-GM inequality

Let $\alpha_1, ..., \alpha_n \geq 0$, and $p_1,...,p_n \geq 0$ such
that $\sum_{i=1}^{n} p_i^{-1} =1$. Show that $\alpha_1...\alpha_n \leq p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n}$.

It is Young's inequality for n numbers. I know the proof that uses AM-GM, but in this case, I can't use it.

My attempt: I tried to solve it by induction, I suppose that it was true for $n-1$ numbers and summed all the possibilities to obtain $\sum_{k=1}^{n} \prod_{i \in ([k]-\{k\})} \alpha_i \leq (n-1)(p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n})$, but don't seem to be that the left right hand is greater or equal to $\alpha_1...\alpha_n$.

答:
We note that Jensen's Inequality states that for a convex function $f:I\to\mathbb{R}$ defined on an interval $I$, given $x_1,x_2,\dots,x_n\in I$ and $\omega_1,\omega_2,\dots,\omega_n\geq 0$, then we have $$f\left(\frac{\omega_1x_1+\omega_2x_2+\dots+\omega_nx_n}{\omega_1+\omega_2+\dots+\omega_n}\right)\leq\frac{\omega_1f(x_1)+\omega_2f(x_2)+\dots+\omega_nf(x_n)}{\omega_1+\omega_2+\dots+\omega_n} $$

or more compactly,
$$f\left(\frac{\sum_{i=1}^{n}\omega_ix_i}{\sum_{i=1}^{n}\omega_i}\right)\leq\frac{\sum_{i=1}^{n}\omega_if(x_i)}{\sum_{i=1}^{n}\omega_i} $$

Let $\gamma_1,\gamma_2,\dots,\gamma_n$ be positive real numbers and let  $p_1,p_2,\dots,p_n$ be positive real numbers such that $\sum_{i=1}^{n}p_i^{-1}=1$.

Recall that the function $\exp:\mathbb{R}\to(0,\infty)$ is convex, since its second derivative is always positive.

Now, we observe that we can write our weights $\omega_i$ as $p_i^{-1}$ and our $x_i$ as $\ln(\gamma_i^{p_i})$. Then we have

$$\begin{align*}\prod_{i=1}^{n}\gamma_i&=\exp\left(\ln\left(\prod_{i=1}^{n}\gamma_i\right)\right)\\
&=\exp\left(\sum_{i=1}^{n}\ln(\gamma_i)\right)\\
&=\exp\left(\sum_{i=1}^{n}p_i^{-1}\ln(\gamma_i^{p_i})\right)\tag{Mult. by $1=p_i\cdot p_i^{-1}$}\\
&=\exp\left(\frac{\sum_{i=1}^{n}p_i^{-1}\ln(\gamma_i^{p_i})}{\sum_{i=1}^{n}p_i^{-1}}\right)\tag{Divide by $1$}\\
&\leq\frac{\sum_{i=1}^{n}p_i^{-1}\exp(\ln(\gamma_i^{p_i}))}{\sum_{i=1}^{n}p_i^{-1}}\tag{Applying Jensen's}\\
&=\sum_{i=1}^{n}p_i^{-1}\gamma_i^{p_i}\tag{$\sum_{i=1}^{n}p_i^{-1}=1$}\end{align*}$$

ContentLicense: CC BY-SA 4.0


问:
Why Can't These Linear Equations Be Solved?
I have the following three linear equations:

$$s_1 = k^{-1}(h_1 + rx_1)$$
$$s_2 = k^{-1}(h_2 + rx_2)$$
$$s_3 = k^{-1}(h_3 + rx_3)$$

I have the values of following parameters:

$s_n$ (different for each equation but known)

$h_n$ (different for each equation but known)

$r$ (the same across all equations)

I do not have the value of the following parameters:

$k$ (the same across all equations)

$x_n$ (different for each equation)

I know that $k$ and $r$ are equal across all of the equations, but I do not know the value of $k$.

Is it possible to calculate $k$ or $x_n$?  If not, why not? Can the equality of $k$ across the equations provide help in solving the equations?

Note: I asked a related question the other day about using Gaussian elimination to solve three linear equations but did not manage to obtain an answer.  I've posted this separately to attempt to understand why it can't be solved.

答: You can't determine the four unknown $k,\,x_j$ from these three equations. But if $r\ne0$ then since $x_j=(ks_j-h_j)/r$, if we knew $k$ we could determine the $x_j$. Construed as the vector equation $x=(ks-h)/r$, the solution set is a line in $x$-space, on which each point has a different value of $k$.


问:
What is a counterexample to the converse of the Absolute Value Theorem for sequences?
So I just recently learned this about explicit sequences:

$\displaystyle\text{If }\lim_{n\to\infty}\vert a_n\vert=0\text{, }\text{then }\lim_{n\to\infty}a_n=0$

I'm cool with this being true, looks good to me.

So I thought, "what about the converse?"

$\displaystyle\text{If }\lim_{n\to\infty}a_n=0\text{, }\text{then }\lim_{n\to\infty}\vert a_n\vert=0$

I was told this is false, but I'm having trouble thinking of a counterexample that proves that. Anybody know one?

答:
The converese is true, this follows immediatelly from $||a_n|-0|=|a_n-0|$.

In particular, $\lim_n a_n=0 \Leftrightarrow \lim_n|a_n|=0$.

What it was probably told to you is the following:

Lemma If $\lim_n a_n=L$ then $\lim_n |a_n|=|L|$.

Note that if $L \neq 0$ the converse of this statement is false, and a counterexample is
$$
a_n=(-1)^n L
$$


问:
Solve the recurrence relation $T(n)=2T(n/2)+\log(n)$ given that $T(1) = 1$
Question:

Assume that a recurrence relation is given as below:

$T(n)=2T(n/2)+\log(n)$

and we know that $T(1) = 1$

We want to solve the relation (find an explicit definition of $T(n)$
which does not rely on itself).

My solving:

I wanted to solve it using substitution method but when i tried I get the series
$$\log(n)+2\log(n/2)+4\log(n/4)+8\log(n/8) ... 2^k\log(n/2^k)$$
which I am not able to solve.

答:
Assuming $\log(n) = \log_2 n$ we have

$$
T\left(2^{\log_2 n}\right) = 2T\left(2^{\log_2 \frac n2}\right)+\log_2 n
$$

now calling $\mathcal{T}(\cdot) = T\left(2^{(\cdot)}\right)$ and $z = \log_2 n$ we follow with the linear recurrence

$$
\mathcal{T}(n) = 2\mathcal{T}(z-1) + z
$$

with solution

$$
\mathcal{T}(n) =2^{z-1}c_0 + 2^{z+1}-(z+2)
$$

and now going backwards with $z = \log_2 n$ we arrive at

$$
T(n) = \frac n2 c_0 +2(n-1)-\log_2 n
$$

Finally with $T(1) = 1$ follows $c_0 = 2$


问:
Correctly serving drinks in a cafe problem
Five friends go into a cafe, sit at a table and order 3 coffees and 2 teas.  The waiter brings the drinks to the table and randomly serves the drinks to the five friends.

How do I compute the following?

The probability that the first served drink was a correct match

The probability that the first two served drinks were correct matches

Attempt:

Coffee is given to someone who orders coffee, or tea is given to someone who orders tea.

$\frac{3}{5}$ $\cdot$ $\frac{3}{5}$ + $\frac{2}{5}$ $\cdot$ $\frac{2}{5}$ = $\frac{13}{25}$. Am I doing it right?

I don't know how to appraoch this 2nd one. Any help is much apprecaited.

答:
The first two drinks served were coffee, and these matched, has a probability of:

$$\frac35\frac24\frac35\frac24=\frac{36}{400}$$

Similarly for CT:

$$\frac35\frac24\frac35\frac24=\frac{36}{400}$$

TC:

$$\frac25\frac34\frac25\frac34=\frac{36}{400}$$

TT:

$$\frac25\frac14\frac25\frac14=\frac{4}{400}$$

Adding gives the answer as $\frac{112}{400}=\frac{7}{25}$.


问:
Two continuous functions on closed interval and smallest element
Assume $f,g\colon [0,1]\to [0,1]$ are continuous functions and $$ g(0)<f(0) \ \ \text{and}\ \ g(1)>f(1)$$Is it true that there exists smallest $t\in(0,1)$ such that $$ g(t)\geq  f(t)$$
I know continuous will have max and min since it defined on closed interval. But I know If we could ensure such smallest element.

答:
Let $h(x) = f(x) - g(x)$, $h$ is continuous.

$$h(0)= f(0) -g(0) \gt 0$$
$$h(1)=f(1)-g(1) \lt 0$$

By the intermediate value theorem there is some $x_0$ in $(0,1)$ such that $h(x_0)=0$.
Also, since $h(x)=0$ is a level curve then its preimage is closed, since it is also bounded in $(0,1)$ then it is compact; so by the minimum value theorem there is such a least element.


问:
What is the Hadamard product of matrix $A$ with $BC$, for $B, C$ being matrices?
Say we have three matrices $A,B,C$. Define $\circ$ to be the Hadamard product. The usual matrix product of two matrices $B,C$ is denoted simply as $BC$. Is there an easy expression for
$$A\circ(BC)$$?

答:
$\def\D{\operatorname{Diag}}$Let $I$ be the identity matrix. Use the column vectors $(a_k,e_k)$ denote the $k^{th}$ columns of the matrices $(A,I),\,$ respectively. Then one can write the matrix $A$ as a sum of dyadics
$$A = \sum_{k=1}^n a_ke_k^T$$
The Hadamard product of a dyad and a matrix obeys
$$xy^T\circ M \;=\; \D(x)\cdot M\cdot\D(y)$$
Substituting into the given equation yields
$$\eqalign{
A\circ(BC) &= \sum_{k=1}^n a_ke_k\circ(BC) \\
 &= \sum_{k=1}^n \D(a_k)\;BC\,\D(e_k) \\
 &= \sum_{k=1}^n A_kBCE_k \\
}$$
where $A_k$ is a diagonal matrix constructed from the vector $a_k\,$ (and $E_k$ from $e_k$).


问:
Markov Chain transitional probability query.
Say I have the transitional probability matrix P= $\begin{bmatrix}.8 & .2\\.6 & .4\end{bmatrix}$ And the entry (1,1) denotes the probability that I stay in state 0, (1,2) I move from state 0 to state 1, (2,1) I move from state 1 to state 0 and (2,2) stay in state 1. Now suppose I am on day 1 and in state 0, and I want to find the probability that I will remain in state 0 over day 2, 3 and 4. Am I right in thinking that I compute $P^2 \ entry (1,1) + P^3 \ entry (1,1)+P^4 \ entry (1,1). $
And for when I need to find the probability of being in state 0 on day 4, I only compute $P^4 \ entry (1,1)$.</p>

答:
Revising my opinion, based on comment of @Did. I would like explain his answer in detail to make sure you avoid my mistake.

Let $X_t$ be the position of your process at time $t$. You start with $X_1 = 0$.

You would like to end up with $X_2 = 0$. The probability of that event is $P_{11}$. Since your Markov chain is time homogeneous,
$$\mathbb{P}[X_3 = 0|X_2 = 0] = \mathbb{P}[X_2 = 0|X_1 = 0] = P_{11}.$$
So, to go from $X_1 = 0$ to $X_2 = 0$, it's also a transition with probability $P_{11}$.

Now, what is the probability of tracing that path entirely?
$$
\begin{split}
\mathbb{P}[X_3 = 0,X_2=0|X_1 = 0]
  &= \mathbb{P}[X_3 = 0|X_2 = 0,X_1 = 0] \cdot \mathbb{P}[X_2 = 0|X_1 = 0] \\
  &= \mathbb{P}[X_3 = 0|X_2 = 0] \cdot \mathbb{P}[X_2 = 0|X_1 = 0] \\
  &= \left(P_{11}\right)^2.
\end{split}
$$

Along similar lines, you can show
$$
\mathbb{P}[X_n = 0,X_{n-1}=0, \ldots, X_2=0|X_1 = 0] = \left(P_{11}\right)^{n-1}.
$$

ContentLicense: CC BY-SA 3.0


问:
Compute homology groups of space $\Bbb RP^2$ attached with Mobius band using Mayer Vietories
(This is exercise 2.2.28 from Hatcher) Consider the space $X$ (say) obtained from a $\Bbb RP^2$, by attaching a Mobius band $M$ via a homeomorphism from the boundary circle of the Mobius band to the circle $\Bbb RP^1 \subset \Bbb RP^2$. Compute the homology of the space

We set $A=N_{\epsilon}(\Bbb RP^2)$, $B=N_{\epsilon}(M)$, $X:=A^o\cup B^o $, s.t $A$, $B$ and  $A\cap B$ deformation retracts to $\Bbb RP^2$, $M$ and $ S^1$, so the (reduced) Mayer-Vietoris sequence yields

$$ \tilde H_2(S^1)\to \tilde H_2(\Bbb RP^2)\oplus\tilde H_2(M)\to \tilde H_2(X)\to \tilde H_1(S^1)\to \tilde H_1(\Bbb RP^2)\oplus \tilde H_1(M)\to \tilde H_1(X) \to 0$$
and replacing the homology groups we already know we have
$$0\to 0 \to \tilde H_2(X) \overset{\partial}{\to} \mathbb Z \overset{\phi}{\to} \mathbb Z_2\oplus \mathbb Z\overset{\psi}{\to} \tilde H_1(X)\to 0$$

Now, I have to figure out what the maps $\partial,\phi,\psi$ do, and this is where I'm stuck. I'd appreciate a detailed explanation.

What I am thinking that $\partial$ is injective and $\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi$. Again $\phi=(j_*,i_*)$ where $i: S^1 \hookrightarrow M , j: S^1 \hookrightarrow \Bbb RP^2$. Then $i_*(1)=2$, but little bit confused about $j_*$ is $j_*(1)=[0]$ because in calculating homology of $\Bbb RP^2$ the map from $e^1 \to e^0$ is zero map again I am thinking that the map might be $1 \mapsto [1]$. Now for $j_*(1)=[0]$.

$\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi= (\mathbb Z_2\oplus \mathbb Z)/<(2,[0])>=\Bbb Z_2 \oplus \Bbb Z_2$. Then $ker \phi=0=Im \partial$. So, $\tilde H_2(X)=0$. Can anyone please tell me whether I am right or wrong or how to rectify it, moreover what is $ H_0(X)$?

答:
We have that $j$ takes $S^1$ onto $S^1\cong\mathbb{R}P^1\subset\mathbb{R}P^2$, a cycle that represents the generator of $H_1(\mathbb{R}P^2)$, so $j_*(1)=[1]$ and $\text{im}(\phi)=<([1],2)>.$ The kernel of $\phi$ is still zero, so $H_2(X)=0$.

As commented, $H_0(X)=\mathbb{Z}$ since $X$ is path connected.

We will argue $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$. We have the following partition $$\{([k],2k): k\in\mathbb{Z}\}\sqcup\{([k+1],2k): k\in\mathbb{Z}\}\sqcup\{([k],2k+1): k\in\mathbb{Z}\}\sqcup\{([k+1],2k+1): k\in\mathbb{Z}\}\\ = \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}.$$

Under the quotient map $\pi: \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z} \to  (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})/<([1],2)> \cong(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}) / \text{im}(\phi)$,

$$\pi([k],2k)=[([0],0)] $$ $$\pi([k+1],2k)=[([1],0)] $$ $$\pi([k],2k+1)=[([0],1])] $$ $$\pi([k+1],2k+1)=[([1],1)]. $$

With $H_1(X)\cong (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})\ /\ \text{im}(\phi)$, the quotient map shows $|H_1(X)|=4$. Since $2[([0],1)]=[([0],2)]$ and $([0],2)=([k+1],2k)$ for $k=1$, we have $2[([0],1)]=[([1],0)]\neq[([0],0)]$ implying $H_1(X)\neq \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ Thus, $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$, the only other group of order 4.


问:
Simulating the HP memristor in Python
I am trying to simulate the non-linear electrical resistive device called memristor in python. The device is characterized by a change in conductivity as an electrical field is applied, and so the relationship between applied voltage and current through the device is hysteretic and non-linear. The system is described by the equations

\begin{equation}
    M(x) = R_\text{ON} \cdot x + R_\text{off} \cdot (1-x)
\end{equation}
\begin{equation}
    \frac{dx}{dt} = \frac{\mu_x \cdot R_\text{ON}}{D^2}\cdot i_\text{tran}(t)
\end{equation}
When a voltage is applied, the region where there is charge change in thickness, so x is used here as the internal state that describes how the region of charge changes in relation to the total thickness of the layer.
\begin{equation}
    x = \frac{w}{D}, \hspace{10pt} 0 \le x \le 1
\end{equation}
I have tried to set this up in Python, but lack of understanding of differential equations and little practical experience with simulating in Python makes it difficult to spot my mistakes. I am pretty sure I have some of the basics wrong. The code looks like this:

```<code>from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt

R_ON = 100      #Resistance maximum conducting state [ohm]
R_off = 16e3    #Resistance minimum conducting state [ohm]
W = 2*np.pi     #Angular frequancy
V_0 = 5         #Volts
t0 = 0          #Time array start
tf = 1          #Time array stop
N = 50          #Array length
mu_V = 1e-14    #Oxygen vacancy mobility
D = 10e-9       #Titanium layer thickness
X0 = 0.76       #Normed charge region (initial value)

def M_tran(x):
    return R_ON*x+R_off*(1-x)

def i_tran(t, x):
    return (V_0*np.sin(W*t))/M_tran(x)

def odefun2(t, x):
    c = (mu_V*R_ON)/(D**2)
    return c*i_tran(t, x)

t = np.linspace(t0,tf,N+1)
T = odeint(odefun2, X0, t, tfirst=True)

U = np.multiply(i_tran(t, X0), M_tran(T[:,0]))

plt.plot(i_tran(t, X0), U)
</code>```

Do I need a window function for x perhaps?

I'll attach the plot I get up if it can be of any help, although it makes no sense to me.
Plot generated by this code

答:
For completeness, the analytic solution to this question is as follows. Your ODE can be written as

$$\frac{dx}{dt}=\frac{f(t)}{x+b(1-x)}$$

where $f(t)$ is a given forcing (which has constants as well as the actual input voltage in it); note that I have absorbed your $R_{on}$ into $f$ as well. This equation can be solved as

$$\int_{x_0}^x y+b(1-y) dy = \int_0^t f(s) ds.$$

Thus

$$\frac{(1-b)x(t)^2}{2} + b x(t) - \left ( \frac{(1-b) x_0^2}{2} + bx_0 \right ) = \int_0^t f(s) ds.$$

This is just a quadratic equation that you can solve for $x(t)$ for each $t$; the coefficients are actually structured such that a solution is guaranteed to exist at least for a short time no matter what $b$ or $x_0$ is, which I find quite neat.

Numerically it looks like the only real problem with your implementation is that in the call to i_tran inside the plotter, you need to use the time-dependent values of $x$ not the initial value.

Title: Simulating the HP memristor in Python


问:
What is the limit of $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ { k }^{ \left( \frac { 1 }{ 2k } \right) } } } $
When I was trying to solve this question

I came up to this: $$\lim _{ n\rightarrow \infty  }{ \sum _{ k=1 }^{ n }{ { k }^{ \left( \frac { 1 }{ 2k }  \right)  } }  } ={ 1 }^{ 1/2 }+{ 2 }^{ 1/4 }+{ 3 }^{ 1/6 }+...+{ n }^{ 1/(2n) }.$$Does it converge? What is the limit of the sum?

答:
This has no limit since $k^{\frac{1}{2k}}\to 1$ and $k\to\infty$.

You don't really need that to see that it can't converge, since $k^{\frac{1}{2k}}>1$ for all $k$.


问:
Poisson equation, exact solution
I have to prove that the solution $u$ of the following Poisson equation
$$-\Delta u(x,y)=\sin(\pi x)\sin(\pi y)$$
for $(x,y)\in[0,1]^2$ with boundary conditions
$$u(0,y)=u(1,y)=u(x,0)=u(x,1)=0$$
can be written as $v(x)\sin(\pi y)$.

So far I've written $u(x,y)=v(x,y)\sin(\pi y)$ and tried to get $\partial_y v(x,y)=0$, but I am stuck at
$$\left((-\partial_x^2-\partial_y^2+\pi^2)v(x,y)-\sin(\pi x)\right)\sin(\pi y)=2\pi\partial_y v(x,y)\cos(\pi y)$$

Any idea?

答:
Solution can be founded in a form
$$
u(x, y) = \int\limits_0^1 dx_0 \int\limits_0^1 G(x_0, y_0, x, y) \sin(\pi x_0)\sin(\pi y_0) dy_0,
$$
where $G$ is Green's function. In your problem $G$ has a form

$$G(x_0, y_0, x, y) = 4 \sum\limits_{n,m=1}^{\infty} \frac{\sin(\pi n x) \sin(\pi n x_0) \sin(\pi m y) \sin(\pi m y_0)}{ \pi^2 (n^2 + m^2) }$$

Using orthogonality of system $\{\sin(\pi m y)\}_{m=1}^\infty$ on range $y \in [0,1]$, we can get $\int\limits_0^1 \sin(\pi y_0) \sin(\pi m y_0) dy_0 = \frac{\delta_{1,m}}2 $. Now, we can write

$$
u(x, y) = \int\limits_0^1 dx_0 \int\limits_0^1  4 \sum\limits_{n=1}^{\infty} \frac{\sin(\pi n x) \sin(\pi n x_0) \sin(\pi y) \sin(\pi y_0)}{ \pi^2 (n^2 + 1) } \sin(\pi x_0)\sin(\pi y_0) dy_0,
$$

or
$$
u(x, y) = \sin(\pi y) \left( 2\int\limits_0^1 \sum\limits_{n=1}^{\infty} \frac{\sin(\pi n x) \sin(\pi n x_0)}{ \pi^2 (n^2 + 1) } \sin(\pi x_0) dx_0  \right).
$$

Now, we define $v(x) = 2\int\limits_0^1 \sum\limits_{n=1}^{\infty} \frac{\sin(\pi n x) \sin(\pi n x_0)}{ \pi^2 (n^2 + 1) } \sin(\pi x_0) dx_0$, and we get $u(x, y) = v(x)\sin(\pi y)$.

Using the same technique, you can simplify summation and integral in expression of $v(x)$.

Title: Poisson equation, exact solution


问:
Why is this a smooth manifold?
I read that the space of quantum states, i.e. the space of density operators

$\mathcal{S}_n = \{ \rho \in H_n : \rho \geq 0, \, \, Tr[\rho]=1\}$

is a smooth manifold of dimension $n^2-1$, without further explanation. Could anyone help me understand why this is the case?

Here $H_n$ denotes the space of hermitian matrices: $H_n = \{ A \in \mathbb{C}^{n \times n} : \, A^* =A  \}$, and $\rho \geq 0$ means that the matrix $\rho$ is positive semi-definite.

答:
Preimage theorem:

Theorem. Let $f:X\to Y$ be a smooth map, and let $y\in Y$ be a regular value of $f$. Then $f^{-1}(y)$ is a submanifold of $X$. If $y\in {\text{im}}(f)$, then the codimension of $f^{-1}(y)$ is equal to the dimension of $Y$. Also, the tangent space of $f^{-1}(y)$ at $x$ is equal to $\ker(df_{x})$.

Title: Why is this a smooth manifold?


问:
Properties of a set of all isomorphisms $ f: G \to G $
I'm kinda stuck with this task.

Let $G$ be a group and $ S $ the set of all isomorphisms $ f: G \to G$. I first want to show that $ (S, \circ) $ is also a group.

I believe I've shown that all the properties of a group is fulfilled with $ (S, \circ) $:

i) Assume that $ x \in $ and $ f_1,f_2 \in S$. Then $f_2(x) = y \in G$, since $ f_1 $ is an ismorphism. $ f_2(y) = z \in G $ since $ f_2 $ is an ismorphism. Then $ f_1(f_2(x)) = f_1(y) = z = f_1 \circ f_2(x),$ hence $f_1, f_2 \in S \longrightarrow f_1 \circ f_2 \in S.$

ii) $id$ is an isomorphism $ \longrightarrow id \in S$.

iii) $ f $ is an isomorphism $ \longrightarrow \exists f^{-1}$, one can show that $ f^{-1} $ is also an isomorphism, $ \longrightarrow f^{-1} \in S$.

Now, assume that $ | S | = 1$, then I want to show that $ G $ is abelian and each element has an order of 1 or 2.

Im kinda lost with that $ | S | = 1 $. If $ f \in S $ then $f^{-1} \in S $ due to previous result, should not $ | S | = 1 $ imply that $ f = f^{-1} = id$? And how do I move forward from this? Any hints is muy appreciated!

答:
Since $Id_G\in \operatorname{Aut}(G)\subseteq\operatorname{Sym}(G)$, then (subgroup test) you have to prove that from $f,f_1,f_2\in \operatorname{Aut}(G)$ follow that $f_1f_2$ and $f^{-1}$ are operation-preserving. So:

$(f_1f_2)(gh)=f_1(f_2(gh))=f_1(f_2(g)f_2(h))=f_1(f_2(g))f_1(f_2(h))=(f_1f_2)(g)(f_1f_2)(h)\Longrightarrow f_1f_2\in\operatorname{Aut}(G)$

$f^{-1}(gh)=f^{-1}(g)f^{-1}(h)\iff f(f^{-1}(gh))=f(f^{-1}(g)f^{-1}(h))\iff gh=gh\Longrightarrow  f^{-1}\in\operatorname{Aut}(G)$

Now, $gh=g(hg)g^{-1}=\varphi_g(hg)$ for every $g,h\in G$, where $\varphi_g$ is the automorphism "conjugation by $g$". So, if $\mid\operatorname{Aut}(G)|=1$, then $\varphi_g=Id_G$ for every $g\in G$, whence $gh=hg$ for every $g,h\in G$, and $G$ is abelian.


问:
Lemma 7.8, Applied Analysis (Hunter)
Here's an image of the lemma and the relevant definition preceding it:

I'm trying to understand the approach to the proof. It's not clear to me why it's sufficient to prove that
$$
||S_M - S_N||_\infty \leq \frac{C_k}{N^{k - 1/2}} \left\vert\left\vert f^{(k)} \right\vert\right\vert.
$$
The way I understand it so far is that if this can be done, then this shows that $S_N$ is Cauchy in $C(\mathbb{T}),$ so it converges to a limit there in the uniform norm. Since $S_N$ is the $N$-th partial sum of the Fourier series of $f$, $S_N \to f$ in the $L^2$-norm. Since $S_N$ converges in the uniform norm, and $S_N \to f$ in the $L^2$-norm, then $S_N \to f$ in the uniform norm. This logic makes sense in the sort of naive "you can't converge to two different limits" intuition, but is that what is actually going on? If this is the case, how can I see this more precisely? And if not, what is this all about?

Thanks in advance.

答: That indeed is an "you can't converge to two different limits"-like argument but you have to be a little careful here. First of all by showing this inequality, as you said, it is proved that $(S_N)$ is Cauchy in $C(\mathbb{T})$. Thus, by completeness there exist a $\tilde{f}\in C(\mathbb{T})$ such that $S_N \to \tilde{f}$ with respect to the supremum norm. In particular $S_N \to \tilde{f}$ with respect to the $L^2(\mathbb{T})$-norm. But as we know the $L^2(\mathbb{T})$-limit of $(S_N)$ is $f$. Therefore $f=\tilde{f}$ a.e.. Then if you consider the essential supremum norm, which in many case has the same notation as the supremum norm, then $\| S_N -f \|_{\infty} = \| S_N - \tilde{f} \|_{\infty}$

Title: Lemma 7.8, Applied Analysis (Hunter)


问:
inequalities on Riemmanian manifolds in local coordiantes
Given a compact Riemmanian manifold $(X, g)$ and a smooth function $f$ on it. Recall that $\nabla f = g^{i j} \partial_{i} f \partial_{j}$, so $|\nabla f|^2 = g_{jq}g^{ij}g^{pq}\partial_{i} f \partial_{p} f$.

We can compute in normal coordinates that $\Delta_{g} |\nabla f |^2 = \partial_{k} \partial_{k}((g_{jq}g^{ij}g^{pq})) \partial_{i} f \partial_{p} f +$ a bunch of other terms.  Let me denote this first term on the right hand side by $R$. I have the following computation to get a lower bound for $R$:

$R \geq -C \sum |\partial_{i} f| |\partial_{p} f| \geq - C (\sum_{i} |\partial_{i} f|^2 + \sum_{p} |\partial_{p}f|^2) =  -C |\nabla f|^2$. This $C$ depends on some curvature of the metric.

I am not so sure about this computation and specifically my question is:

I found online that if the manifold is compact, then one can treat the norm of a tensor $T$ using the norm of their components in local coordinates. This is equivalent to the norm defined as $|\nabla^{k} T|_{g}$. However, if we use the norm in local coordinates, it is going to be dependent on the choice of coordinates. What does it mean for it to be equivalent to the second definition?

How do you treat inequalities in local coordinates? In the example I gave, can I just bound any random terms with constants like I did for the derivatives of the metric or does it have to be a tensor (globally defined objects) for me to throw it away using its bound.

1 is related to 2 because 1 seems to be suggesting that invariance does not matter much but that does not make sense to me.

答:
I found online that if the manifold is compact, then one can treat the norm of a tensor  using the norm of their components in local coordinates. This is equivalent to the norm defined as |∇|. However, if we use the norm in local coordinates, it is going to be dependent on the choice of coordinates. What does it mean for it to be equivalent to the second definition?

Potentially what is meant here is that there exists constants $C_0, C_1>0$ such that
$$C_0 \sum_{|I|=k} T_I^2\leq |\nabla^kT|^2_g \leq C_1\sum_{|I|=k} T_I^2$$

where $I$ denotes a multi-index of length $|I|=k$. (Notice that $C_0$ and $C_1$ here are independent of $x\in M$). On the left and right hand side, we have the Frobenius norm computed using the coordinate components of the tensor.

I believe this result is not independent of coordinates. However, it is independent of choice of normal coordinates. This should not be too surpising, since normal coordinates carry some information about the Riemannian metric on $M$, while general coordinates are not guaranteed to do so (although I believe the result might also hold for harmonic coordinates as well.)

This can be shown in the case $k=1$ using the two equivalent definitions of manifolds of bounded geometry.

A manifold of $C^k$-bounded geometry is a manifold $(M,g)$ with positive injectivity radius on which for each $i\leq k$, the $k$-th covariant derivative of the Riemannian curvature $|\nabla^i R|_g$ is bounded by a constant. Thus, a compact manifold is of $C^k$-bounded geometry for any choice of $k$.

Suppose that $(M,g)$ is of $C^k$-bounded geometry, then for $i\leq k$, there exists a $C_0, C_1>0$ such that in any normal coordinate chart of size less than injectivity radius,
$$
C_0 \leq |D^\alpha g_{ij}| \leq C_1 \text{ and }C_0 \leq |D^\alpha g^{ij}|\leq C_1
$$

where $D^\alpha$ indicates the ordinary derivative with multi-index $\alpha$ such that $|\alpha|\leq k$ in $\mathbb{R}^d$ taken in coordinates.

A classical result (I believe of Eichorn, but I need to track down the original paper) shows that the above property is actually equivalent to the bounded curvature condition above. Thus, if $(M,g)$ is a compact manifold, one has bounds on the components $g_{ij}$ and $g^{ij}$ as well as their derivatives in any normal coordinate chart.

Applying this to your original question, observe the norm of the gradient in a normal coordinate chart:
$$
|\nabla f|_g= g_{pq}g^{iq}g^{jp}\partial_i f \partial_j.
$$

Observe that for each $x$ in the domain of the coordinate chart, $g_{ij}(x)$ is a symmetric positive definite matrix. Since we have bounds on the entrees of these matrices which are independent of $x$, we have an upper bound on the maximum eigenvalue of $g_{ij}$ and a lower positive bound on the minimum eigenvalue by taking the maximum eigenvalue of $g^{ij}$. In other words, there is a $K_0,K_1>0$ such that for any vector $s= (s^1,...,s^d)$

$$
K_0 \delta_{ij}s^is^j \leq g_{ij}(x) s^is^j\leq K_1 \delta_{ij}s^is^j.$$

Using this together with the definition of the norm you gave, we get some $K'_0,K'_1>0$ only dependent on the bounded geometry constants and the dimension of $M$ such that

$$K'_0 \sum_i (\partial_i f(x))^2\leq |\nabla f(x)| \leq K'_1 \sum_i (\partial_i f(x))^2.$$

Where the derivatives on the left and rightmost sides are taken in any choice of normal coordinate chart on $M$ and the constants are independent of this choice. This is the Frobenius norm of the coordinate Jacobian.

A note that for higher-order covariant derivatives this becomes even more messy, since the covariant derivative of order $k$ in coordinates contains derivatives of  $f$ of all orders from $1$ to $k$. This means that a very messy inductive argument must be used. I am not sure if this has been written down, but I intend to write up some notes in the future on this.


问:
Can a set equipped with two different topology makes into topological manifold with different dimension?
Let $M$ equipped with two different topology $\tau_1$ and $\tau_2$ .Is it possible that $\tau_1$  and $\tau_2$ makes $M$ into topological manifold with different dimension ?

There is a bit difference between this question and topological invariance of dimension (which states that dimension of topological manifold is invariance under homeomorphism).

答:
All nonempty manifolds of any positive dimension have the cardinality $\mathfrak c$ of the continuum because they are covered by countably many open sets homeomorphic to $\mathbb R^n$ for some $n > 0$. Since all $\mathbb R^n$ have the cardinality $\mathfrak c$ (you do not need topology for this fact!), our claim follows.

Thus for any two nonempty manifolds $(M,\tau_M), (N,\tau_N)$ of positive dimension there exists a bijection $f : M \to N$. Now consider the topology $f^{-1}(\tau_N) = \{f^{-1}(U) \mid U \in \tau_N\}$. Then $(M,f^{-1}(\tau_N))$ is homeomorphic to $(N,\tau_N)$.


问:
Does the limit of gradient flow from a given initial value depend on the choice of inner product?
I will start with an example to motivate my question, and then ask it more generally.

Example. Let us consider the Hilbert spaces $H^1 \subset L^2$, and a Frechet-differentiable functional $E:H^1\to\mathbb{R}$, e.g. $E(f)=\int (f')^2$.
The gradient flow equation for this functional has the form
$$\frac{d}{dt}f=-\mathrm{grad}_V E(f),$$
where $\mathrm{grad}_V E$ is such that $dE(f)(u)=\langle\mathrm{grad}_V E(f), u\rangle_V$. Choosing $V=L^2$ and $V=H^1$ gives different gradient-flow equations.
Let us choose an initial value $f|_{t=0}\in H^1$ and look at the solutions for both choices.
For the Dirichlet energy, they will naturally converge to the same steady-state solution, because there is only one.

Question. How far can this behavior be generalized? Assume I have two Hilbert spaces $U\subset V$, a functional $E:U\to\mathbb{R}$, and I prescribe $f|_{t=0}\in U$. Assume furthermore that the gradient flows for $\mathrm{grad}_U E$ and $\mathrm{grad}_V E$, starting from the same initial value, both converge in $U$. Does it follow that they converge to the same solution? To frame it in more intuitive terms: If the functional has more than one local minimum, can the choice of inner product "redirect" the flow enough, so it converges to different local minima from the same initial point?

I would assume that at least in finite-dimensional spaces, the choice of inner product does not matter. Does the same hold in infinite-dimensional spaces? If not, can we get there with additional assumptions?

答:
Yes, as a minimal example consider $H=\Bbb R^2$. Let $\langle, \rangle_1$ be the usual euclidean scalar product and consider an additional product
$$\langle (x_1, y_1) , (x_2,y_2)\rangle_2 :=2x_1x_2+x_1y_2+x_2y_1+2y_1y_2 $$
Now look at the functional
$$E:\Bbb R^2\to\Bbb R, \quad (x,y) \mapsto \sin(x)^2$$
then the respective gradients are (exercise):
$$\nabla_1E = 2\sin(x)\cos(x)\cdot (1,0),\qquad \nabla_2 E = 2\sin(x)\cos(x)\cdot (2/3, -1/3)$$
Now start at $f=(1,0)$. The first gradient will move you to $(0,0)$, the second will however have you end up in $(0,1/2)$. For both cases the value of the functional itself doesnt change but you can smoothly modify the function close to $(0,1/2)$ if you want so that the second gradient falls into a shallower (or deeper) hole.


问:
Confirming correct distribution assumption
Suppose that the chance of making one typo on a page is $p_1$, two typos is $p_2$, and making none is $p_3$. what distribution has the random variable "sum of typos on $n$ pages"?

sine the number of typos is categorically distributed (3 outcomes), the sum of $n$ iid copied is multinomial, right?

答: Not quite.  A multinomial distribution is a multivariate distribution when the number of categories is more than $2$.  So what it is going to count is the number of times each of the three categories occurs across $n$ pages.  For instance, you might observe an outcome where there are $12$ pages with one typo, $3$ pages with two typos, and $57$ pages with zero typos, out of a total of $n = 12 + 3 + 57 = 72$ pages, but the total number of typos is a function of this vector-valued outcome:  it is $$12(1) + 3(2) + 57(0) = 18.$$  The same thing applies to the case where there is only one page--the random variable for which typo category is categorical, but the number of typos is a function of that category.


问:
Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$
Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$

From Eves theorem we have as expectation is non-negative,

$V(Y) \geq V(E(Y|X))$

It means after knowing information X, my prior variance V(Y) has reduced

Is posterior variance  $V(Y|X)$ or $V(E(Y|X))$. Both look similar to me. Please elaborate intuitive difference /relation between these 2. I am a beginner. Request your patience.

答:
$V(Y|X)$ is the posterior variance of $Y$ given $X$. It is a random variable which is a function of $X$; for each possible value $x$ of $X$, $V(Y|X=x)$ is the variance of the random variable $Y$ conditioned on $X=x$.

$V(E(Y|X))$ is harder to give an intuitive explanation for. Just as $V(Y|X)$ is a random variable, $E(Y|X)$ is a random variable, where $E(Y|X=x)$ is the expected value of $Y$ given $X=x$. Since $E(Y|X)$ is a random variable, it has a variance, and $V(E(Y|X))$ is that variance. But I do not know what that quantity is really useful for. As mentioned in the other answer, $$V(E(Y|X))=V(Y)-E(V(Y|X))=E[V(Y)-V(Y|X)],$$ so you can interpret $V(E(Y|X))$ as the "average reduction in variance in $Y$ you get by learning the value of $X$".


问:
Can a number have a prime factor that isn't a part of the number's prime factorization?
Is it possible to have a number, $x$, divisible by some prime, such that that prime does not appear in the unique prime factorization of $x$?</p>

答: If $p|x$, then write the prime factorization of $\frac{x}{p} = p_1p_2....p_n$.  But then $x=pp_1p_2...p_n$ is a prime factorization of $x$, and unique factorization shows it is the only prime factorization of $x$.</p>


答:
$$\lim_n \sqrt[2n]{n} =1 \neq 0$$

thus the series diverges.

Title: What is the limit of $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ { k }^{ \left( \frac { 1 }{ 2k } \right) } } } $

Body:
When I was trying to solve this question

I came up to this: $$\lim _{ n\rightarrow \infty  }{ \sum _{ k=1 }^{ n }{ { k }^{ \left( \frac { 1 }{ 2k }  \right)  } }  } ={ 1 }^{ 1/2 }+{ 2 }^{ 1/4 }+{ 3 }^{ 1/6 }+...+{ n }^{ 1/(2n) }.$$Does it converge? What is the limit of the sum?


问:
Incidence geometry in 3 dimensions from Euclidean and non-Euclidean geometries by Marvin Greenberg
The Question

Prove: If each of two intersecting lines is parallel to a third line, then the three lines are coplanar

My Attempt

This exercise appears in Euclidean and non-Euclidean geometries by Marvin Greenberg. I tried combining between

Theorem: if there is a line and a point not on the line then there is a unique plane containing them

and

Corollary: two parallel lines have a unique plane containing them.

but yet couldn't figure a way to prove this completely.

答:
If two lines $a,b$ intersect in a point, let's say $P$ and are parallel to the same line $c$ then we must have $a\equiv b$ for the Fifth Postulate, since the parallel is unique.

Plus for the first axiom, the line $c$ contains two distinct points $Q,R$.

Thus they are coplanar since three not collinear points $P,Q,R$ lie in the same plane (axiom).


问:
In a velocity-time graph, if the velocity is negative and the slope is negative, is it still accelerating?
So in a velocity-time graph, a negative velocity means the speed in the opposite direction. So from -10 m/s to -30 m/s, it is accelerating in the opposite direction. However the slope is negative, so is it still accelerating?</p>

答: The issue here is with the word 'accelerating', which might not align with how you think about it intuitively. Suppose we define the positive direction as moving to the right. Then, moving to the left means that you have a negative velocity. Hence, saying that we are going from $-10 \mathrm{m/s}$ from $-30 \mathrm{m/s}$ is a shorthand for 'at first, we were moving at $10$ metres per second to the left; now, we are moving at $30$ metres per second to the left'. Hence, the speed of the car has increased, since $30>10$. However, the velocity, which tells us not just the speed, but the direction in which the speed is acting, has decreased from $-10\mathrm{m/s}$ to $-30\mathrm{m/s}$. Since acceleration is measure of the change in velocity, not speed, over this time period the car has experienced negative acceleration. This is despite the fact that the car is now moving faster.


问:
Something similar to Kronecker's Theorem
I would like to prove (or have a reference to) the following:

Given $n$ real numbers $a_1,\ldots,a_n>0$ and $\varepsilon>0$, there exist $k,k_1,\ldots,k_n\in\mathbb N\setminus\{0\}$ such that $|ka_j-k_j|<\varepsilon$ for $j=1,\ldots,n-1$.

This is an easy consequence of Kronecker's Theorem if $a_1,\ldots,a_n,1$ are rationally independent. In this case, Kronecker says that the set $\{ka\mod\mathbb Z^n : k\in\mathbb Z\}$ is dense in $[0,1)^n$, where $a = (a_1,\ldots,a_n)\in\mathbb R^n$. In particular, there exist points close to zero in this set, which is what I am looking for.

However, if there are rational dependencies, I can't seem to find a way. Any suggestions anyone?

答:
[This answer was refering to a earlier version of the question.]

A partial solution for $n=2$. Let $\varepsilon>0$ be fixed. For simplicity denote $a:=a_1$ and $b:=a_2$. In fact, without loss of generality, one can even assume that $b=1$ by applying the injective scaling $x\mapsto x/b$.

So now we have the set $A=a\mathbb{N}+I_{\varepsilon}$ and we want to make sure that this intersects the naturals.
Case 1: If $a=p/q$ is rational, this is immediate as $qa=p\in\mathbb{N}$.
Case 2: Assume $a$ is irrational. Then by the Weil equidistribution theorem, the fractional parts $\{ka\}$ with $k\in\mathbb{N}$ are uniformly distributed on $[0,1]$. Since the measure of the interval $[0,\varepsilon]$ is strictly positive it means that one can find some infinite subset $M\subset\mathbb{N}$ such that $0\leq\{ma\}\leq\varepsilon$ for all $m\in M$, in other words it follows that $M\subset\mathbb{N}\cap A$. Clearly $M$ is unbounded.

Now for $n\geq3$, it would suffice by an inductive argument to know that $M$ contains an arithmetic progression. Indeed, if that is the case, one can just restrict the intersection of the sets for $a_1$ and $a_2$ to a set of the form $c\mathbb{N}$ and apply the same argument again.

So can an infinite set of naturals not contain an arithmetic progression of the form $c\mathbb{N}$ for some $c\in\mathbb{N}$? In general yes, for example a lacunary sequence like $1,2,4,8,16,\dots$. This occurs if and only if the gap between consecutive numbers in our set can be arbitrarily large (otherwise I think a pigeon-hole argument guarantees the existence of an arithmetic progression).

But what about our infinte set $M$? If it were to have arbitrarily large gaps between consecutive elements, that would imply that one can find arbitrarily many consecutive naturals such that $\{na\}>\varepsilon$. I suspect that this contradicts the Weil equidistribution theorem, but i am not sure.


问:
What does invalid input mean for the diff function?
Currently I'm learning about maple. I'm studying structural mechanics and we need to use maple to evaluate expressions. The following code is for an exercise to use castigliano's second theorem to evalute the moment T. This is probably carelessness/inexperience from my side, but due to Covid the university is closed and I can't get help from instructors. If somebody could give me feedback, that would be great!

```<code>M1 := 1/2*q*x*(l - x);
M2 := -T + T*x/l;

Ec := int((M1 + M2)^2/(2*EI), x = 0 .. l);

eq1 := diff(Ec, T) = 0;

Error, invalid input: diff received (1/8)*l^2*q, which is not valid for its 2nd argument

sol := solve(eq1, T);
assign(sol);

Warning, solving for expressions other than names or functions is not recommended.
Error, invalid left hand side in assignment

evalf(T);
</code>```

Although it throws errors and warnings, it still evaluates the correct answer. But these warnings are bothering me.

Question:

What did I do incorrect to cause these warnings/errors?

How can I
avoid errors/warnings?

答:
Look at the error message coming from the call to diff.

It indicates that the second argument being passed to diff is actually (1/8)*l^2*q and not actually just the name T.

It appears that the name T evaluates to (1/8)*l^2*q, ie. T has a previously assigned value.

My guess is that you attempted to run your code twice, but forgot that the final command is a call to assign that will assign a value to T using the prior equation returned by solve.

It's unclear what is your intention. (You didn't show us whether M2 contains the plain name T or something involving that value that seems assigned to it.)

You could run the code in a fresh session, or after restart, or after unassigning T, ie.,

```<code>T := 'T':

M1 := 1/2*q*x*(l - x);

            1
      M1 := - q x (l - x)
            2

M2 := -T + T*x/l;

                 T x
      M2 := -T + ---
                  l

Ec := int((M1 + M2)^2/(2*EI), x = 0 .. l):
Ec := simplify(Ec);

        / 4  2         2         2\
      l \l  q  - 10 T l  q + 40 T /
Ec := -----------------------------
                 240 EI

eq1 := diff(Ec, T) = 0;

            /     2         \
          l \-10 l  q + 80 T/
   eq1 := ------------------- = 0
                 240 EI

sol := solve(eq1, T);

                   1    2
           sol := - q l
                   8
</code>```

Note that your don't have to assign the result from solve in order to utilize that value as a replacement for T later on. For example, after the above code, your can use the eval command to substitute that value for T into arbitrary formulas:

```<code>soleq := solve(eq1, {T});

                /    1    2\
      soleq := { T = - q l  }
                \    8     /

eval( T^2 + T, soleq );

         1   4  2   1    2
         -- l  q  + - q l
         64         8
</code>```

My tip is: read the error messages and try to understand their meaning.

By the way, this Question is off-topic in this forum because it about Maple programming rather than mathematics per se. More appropriate would be either stackoverflow.com or www.mapleprimes.com  (Maplesoft user forum).

Body:
Look at the error message coming from the call to diff.

It indicates that the second argument being passed to diff is actually (1/8)*l^2*q and not actually just the name T.

It appears that the name T evaluates to (1/8)*l^2*q, ie. T has a previously assigned value.

My guess is that you attempted to run your code twice, but forgot that the final command is a call to assign that will assign a value to T using the prior equation returned by solve.

It's unclear what is your intention. (You didn't show us whether M2 contains the plain name T or something involving that value that seems assigned to it.)

You could run the code in a fresh session, or after restart, or after unassigning T, ie.,

T := 'T':

M1 := 1/2*q*x*(l - x);

            1
      M1 := - q x (l - x)
            2

M2 := -T + T*x/l;

                 T x
      M2 := -T + ---
                  l

Ec := int((M1 + M2)^2/(2*EI), x = 0 .. l):
Ec := simplify(Ec);

        / 4  2         2         2\
      l \l  q  - 10 T l  q + 40 T /
Ec := -----------------------------
                 240 EI

eq1 := diff(Ec, T) = 0;

            /     2         \
          l \-10 l  q + 80 T/
   eq1 := ------------------- = 0
                 240 EI

sol := solve(eq1, T);

                   1    2
           sol := - q l
                   8

Note that your don't have to assign the result from solve in order to utilize that value as a replacement for T later on. For example, after the above code, your can use the eval command to substitute that value for T into arbitrary formulas:

soleq := solve(eq1, {T});

                /    1    2\
      soleq := { T = - q l  }
                \    8     /

eval( T^2 + T, soleq );

         1   4  2   1    2
         -- l  q  + - q l
         64         8

My tip is: read the error messages and try to understand their meaning.

By the way, this Question is off-topic in this forum because it about Maple programming rather than mathematics per se. More appropriate would be either stackoverflow.com or www.mapleprimes.com  (Maplesoft user forum).


问:
$\operatorname{\mathcal{Jac}}\left( \mathbb{Q}[x] / (x^8-1) \right)$
$\DeclareMathOperator{\Jac}{\mathcal{Jac}}$
Using the fact that $R := \mathbb{Q}[x]/(x^8-1)$ is a Jacobson ring and thus its Jacobson radical is equal to its Nilradical, I already computed that $\Jac \left( \mathbb{Q}[x] / (x^8-1) \right) = \{0\}$. But I'd like to calculate this in an elementary way, preferably using nothing beyond the very basic properties of the Jacobson radical, quotients of rings and maximal ideals. And here starts my struggle.</p>

答:
Hints:

$\Bbb Q[x]$ is a principal ideal domain

The ideals (including the maximal ideals) sitting above $(x^8-1)$ just depend on the factorization $x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$.

If $p$ and $q$ are coprime polynomials, then $(p)\cap (q)=(pq)$.

The nilradical of a commutative ring is certainly always contained in the Jacobson radical.


问:
Branch and bound method in case the non integer value is less than 1.
If I need to find the integer valued solution to Linear programming problem using branch and bound method, but the non integer x value at the last iteration is less than 1, let's say 3/4. Then I have one constrain $x_1 \le 0$ and $x_1 \ge 1$. How do I proceed with the first case, when $x_1 \le 0$? I cannot add another constrain with auxiliary variable $x_1 + s_1 \le 0$, because all $x_i \ge 0$. Can I simply assume that $x_1 = 0$ and do the linear programming problem ignoring the $x_1$  variable?

答:
If you started with nonnegative constraint and you branch on $x_1 \le 0$ and $x_1 \ge 1$, under that branch of $$x_1 \le 0$$ and $$x_1 \ge 0,$$ you can indeed conclude that $$x_1=0.$$

The value indeed has been fixed for that branch.


问:
Let $(X,d)$ be a metric space and let $x,y$ be two different points in it. Show that $B(x,r) \cap B(y,r) = \emptyset$

Let $(X,d)$ be a metric space and let $x,y$ be two different points in it. Show that $$B(x,r) \cap B(y,r) = \emptyset$$ when $r=\frac{d(x,y)}{2}$.

So the objective is to prove that the neighbourhoods of $x$ and $y$ share no elements if $r=\frac{d(x,y)}{2}$. If I pick $a \in B(x,r)$ then I have that $d(a,x) < \frac{d(x,y)}{2}$ and similarly if $b \in B(y,r)$ then $d(b,y) <\frac{d(x,y)}{2}$. I'm wondering how I can use these inequalities to show this? The fact that $d$ is a metric in $X$ would imply also that the triangle inequality would be applicable here, but it doesn't seem to help. What can I do to proceed?

答: Suppose that $a$ is in the intersection, $d(x,y)\leq d(x,a)+d(a,y)<2{{d(x,y)}\over 2}=d(x,y)$ contradiction


问:
Search for simpler (other) idea to show $(1.001)^{1000}>2$
The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$

Thanks in advance for any hint.

答:
Let's pretend we know nothing of mathematical induction, so we can't just say $(1+x)^n\ge1+nx$ for all $n\ge1$ (and $x\ge-1$), but let's assume we can do enough algebra to know that

$$(1+x)^2=1+2x+x^2\gt1+2x$$

and

$$(1+x)^5=1+5x+10x^2+10x^3+5x^4+x^5\gt1+5x$$

for all $x\gt0$.  Then we can show

$$(1+x)^{10}\gt(1+5x)^2\gt1+10x$$

from which it follows that

$$(1+x)^{100}\gt(1+10x)^{10}\gt1+100x$$

and, finally,

$$(1+x)^{1000}\gt(1+100x)^{10}\gt1+1000x$$

at which point we can plug in $x=1/1000$ to conclude

$$1.001^{1000}=\left(1+{1\over1000}\right)^{1000}\gt1+{1000\over1000}=2$$


问:
Find an example of its sequence
$\newcommand{\N}{\mathbb{N}}$
$\newcommand{\F}{\mathbb{F}}$
$\newcommand{\coloneqq}{:=}$

It is my $\limsup$ and $\liminf$ sor a field $\F$:

$$\limsup(\liminf)x_{n} \coloneqq \inf\{x \in \F : x_{n} \leq (\geq) x \text{ for almost all $x_{n}$ }\}$$

Find a sequence $(a_{n})_{n \in \N}$ such that:

$$\inf\left(\{ a_{n} \,:\, n \in \N\}\right) < \liminf\left(\{ a_{n} \,:\, n \in \N\}\right) < \limsup\left(\{ a_{n} \,:\, n \in \N\}\right) < \sup\left(\{ a_{n} \,:\, n \in \N\}\right)$$

In its case, must the set $\{ a_{n} \,:\, n \in \N\}$ have a maximum?

I think is true but I do not whatr can I do...

Suppse by contraction it do not have a maximum. Call $S =\sup\{a_{n} : n \in \N\}$So $a_{n}$. By definition for any $n \in \N$ $a_{n} \leq S$

答:
$a_1=-2$, $a_2=2$ and $a_n=(-1)^n$ for $n \ge 3$.

Regarding your second question. Denote $L=\limsup_n \{a_n\} \lt \sup_n \{a_n\}=S$. $L \lt S$ implies that it exists $N \in \mathbb N$ such that $a_n \le \frac{L+S}{2}$ for $n \ge N$. Therefore

$$\sup \{a_n \mid n \in \mathbb N\} = \sup \{a_n \mid n \lt N\}$$ and the $\sup$ of a finite set is attained.

Title: Find an example of its sequence


问:
Is vector space $\mathbb{R}^{n}$ the same that $\mathfrak{M}_{n\times 1}$?
Being $\mathfrak{M}_{n\times 1}$ a vector space of matrix of $n$ rows and $1$ columns.

I  can't found a substantial difference between $\mathfrak{M}_{n\times 1}$ and $\mathbb{R}^{n}$. Both have the same properties and I saw in Linear Algebra's books that are used interchangeably.

答: Yes . The map $$(x_1,x_2,\cdots x_n) \longmapsto \begin{pmatrix}x_1 &x_2& \cdots&x_n  \end{pmatrix}^T \in M_{n \times 1}$$ is an isomorphism. So we can identify $M_{n \times 1}$ by $\Bbb R^n$. Consequently, $M_{n \times n}$ can be identified with $\Bbb R^{n^2}$


问:
How to Taylor expand this function, which is a function of vectors?
Suppose that $a,b,c,d$ are vectors in $\mathbb{R}^2$, so $a_{\alpha}$ where $\alpha = 1,2$. We define a scalar product as $a \cdot b := a_1 b_2 - a_2 b_1$. Suppose also that we have a real-valued function $f(a,b,c)$ which is a function of some $3$ vectors.

What does the Taylor expansion of $f$ look like to first order? For example, I want to Taylor expand $f(a+b,c,d)$ around $(a, c, d)$, so $$f (a+b, c, d) = f(a,c,d) + \ldots $$  but I don't know how to write the remaining terms. Do I get a scalar product of $b$ with some term?

答:
There is something to confront here in your question, which has been pointed out already. The first thing is to see that your "scalar product" is not a scalar product: it's a skew symmetric bilinear form. More specifically, notice that $a \cdot a = 0  \hspace{4pt} \forall a$. What you've written is actually the magnitude of the vector-cross product of $a$ and $b$ (More precisely, given a linear map $T: \mathbb{R}^2 \to \mathbb{R}^3 $, $a \cdot b = \vert Ta \times Tb \vert$). This is not a scalar product.

We can relate your skew symmetric form to the usual Euclidean inner product $( \cdot, \cdot)_E $ by the formula
$$
a \cdot b = (a, Jb)_E
$$

Where $J$ is a matrix,
$$
J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$

So that's what we're dealing with. Now let's look at a Taylor expansion of your function. Any sensible Taylor expansion should look like
$$
F(a + tb) = F(a) + (b \cdot DF(a)) t + \mathcal{O}(t^2)
$$
Where $DF(a)$ is some suitable notion of derivative at $a$.
From this we can see that
$$
b \cdot DF(a) = \frac{d}{dt} F(a + tb) \vert_{t=0}
$$
I think that formula should get at the crux of your question. However, there's more to notice here. We can write out the right side as the usual inner product with the gradient $ \nabla F $:
$$
\frac{d}{dt} F(a + tb) \vert_{t=0} = (b, \nabla F(a))_E
$$
and using our relation of the skew symmetric form to the euclidean inner product, we have
$$
b \cdot DF(a) = (b, J DF(a))_E
$$
These two are equal, and if they are equal for all $b$, this implies
$$
J DF(a) = \nabla F(a)
$$
$J$ is invertible, so we see that $ DF(a) = J^{-1} \nabla F(a)$. Summarizing,
$$
F(a+b) = F(a) + b \cdot (J^{-1} \nabla F(a)) + \mathcal{O}(t^2)
$$


问:
Use washer and shell method to find $k$

Let $R$ denote the first quadrant region bounded by $y=k-kx^2$, and $y=kx^2$. $k> 0, k \in \mathbb{R}$. If $V_1$ is the volume obtained by rotating $R$ about the line $y=0$, and $V_2$ is the volume obtained by rotating R about the line $x=-2$, and $V_1=V_2$, then what is $k$?

My plan is to evaluate exact volume of $V_1$ and $V_2$, then make the equation $V_1=V_2$ and solve for $k$. The first one $V_1$ can use the washer method, and second one might be shell, since it is rotating around a vertical line $x = 2$.

答:
Note that the two curves meet at $x= \frac1{\sqrt2}$. Then, washer-integrate $V_1$ and shell  $V_2$ over $x\in(0,\frac1{\sqrt2})$

\begin{align}
&V_2-V_1\\
=&2\pi\int_0^{1/\sqrt2} (x+2)(k(1-x^2)-kx^2))dx - \pi\int_0^{1/\sqrt2}(k^2(1-x^2)^2-k^2 x^4)dx\\
= &\left(\frac{4\sqrt2}3\pi k +\frac\pi4k \right)- \left(\frac{\sqrt2}3\pi k^2  \right)= 0
\end{align}
which yields

$$k= 4+ \frac{3\sqrt2}{8}$$

Title: Use washer and shell method to find $k$


问:
Best polynomial approximation of a partially constant function
Suppose $\alpha \in [0;1]$. Let's define $f_\alpha$ as the function $[0;1] \to [0;1]$ with the following formula:

$$f_\alpha(x) = \begin{cases} 0 & \quad x < \alpha \\ 1 & \quad x \geq \alpha \end{cases}$$

What is the best $n$-degree polynomial approximation of $f_\alpha(x)$?

By best $n$-degree polynomial approximation I mean a polynomial $p_n(x)$ of degree $\leq n$ such that $\sup_{x \in [0;1]} |f_\alpha(x) - p_n(x)|$ is smallest possible.

The best polynomial approximation exists due to the facts that $L_\infty[0;1]$ is a Banach space, all polynomials of degree at most $n$ form its closed convex subspace and there is a projection theorem for Banach spaces.

答:
For all $n$, the constant polynomial
$$c(x) = \frac{1}{2}$$
is the best approximation for the function $f_{\alpha}$.

To see this, we show that for every polynomial $p : [0,1] \to \Bbb R$ the following holds:
$$\sup_{x \in [0,1]} |f_{\alpha} (x) - c(x)| \le \sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|$$

First: obviously
$$\sup_{x \in [0,1]} |f_{\alpha} (x) - c(x)|= \frac{1}{2}$$

So it is enough to prove that for any polynomial $p$ the inequality
$$\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|\ge \frac{1}{2}$$ holds. There are two cases:

First case: $p( \alpha ) \le 1/2$. In this case you have
$$|f_{\alpha} (\alpha) - p(\alpha)| = |1-p(\alpha)| = \left|\frac{1}{2} + \frac{1}{2} - p( \alpha) \right| = \frac{1}{2}  + \left( \frac{1}{2} - p( \alpha) \right) \ge \frac{1}{2}$$

In particular the supremum is larger or equal to $1/2$.

Second case: $p( \alpha ) > 1/2$. In this case consider arbitrary $x \in [0 ;\alpha )$. You have
$$\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)| \ge \sup_{x \in [0,\alpha)} |f_{\alpha} (x) - p(x)| \ge \lim_{x \to \alpha^-} |f_{\alpha} (x) - p(x)| = p(\alpha) > \frac{1}{2}$$

This completes the proof.

Remark: the polynomial minimizing the quantity $\sup_{x \in [0,1]} |f_{\alpha} (x) - p(x)|$ need not to be unique. There may be more than one, but I think that the constant polynomial is the simplest one.


问:
Find the limit of its fractions!
$\newcommand{\N}{\mathbb{N}}$
$\newcommand{\F}{\mathbb{F}}$
$\newcommand{\coloneqq}{:=}$

If it exists find the value of:
$$1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \ddots}}}$$
i.e. the limit of the sequence $(x_{n})_{n \in \N}$ with:
$$x_{1} \coloneqq 1 \hspace{0.5cm}\text{and}\hspace{0.5cm} x_{n+1} \coloneqq 1 + \frac{1}{x_{n}}$$

May you help with hints to begin?

答:
Hint: If the sequence has a limit, then it must hold that
$$
\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty}\left(1 + \frac 1{x_n} \right) = 1 + \frac{1}{\lim_{n \to \infty} x_n}.
$$

Title: Find the limit of its fractions!


问:
Find the limit. Given that the function is continuous on [0,1] how could we find it?
Find $$\lim_{n\to\infty} n\int_{0}^{1} x^{n} f(x)dx$$ could there be any generalisation for the powers of the form $$n^k$$ instead of n within the limit of the integral ? how could we find it? I tried to use the fact that f being continuous on the compact it is bounded but that would work if wr hadnt had that n in front of the integrand

答:
Let $\varepsilon>0$ be given.  For this $\varepsilon$, fix the number $\delta>0$ such that whenever $1-\delta<x\le 1$, $|f(x)-f(1)|<\varepsilon$.

Then, we have the estimates

$$\begin{align}
\left|\int_0^1 nx^n (f(x)-f(1))\,dx\right|&\le \int_0^{1-\delta}nx^n \left|f(x)-f(1)\right|\,dx+\int_{1-\delta}^1 nx^n \left|f(x)-f(1)\right|\,dx\\\\
&\le 2||f||_{\infty}\frac n{n+1}\left(1-\delta\right)^{n+1}+\varepsilon \frac n{n+1}\left(1-\delta^{n+1}\right)
\end{align}$$

Let $n\to \infty$ and conclude.


问:
Dimension of irreducible components of $f^{-1}(X)$, where $X$ is a hypersurface
I am trying to understand the following problem from Gathmann's algebraic geometry notes:If $f:\mathbb{P}^n\rightarrow\mathbb{P}^m$ is a morphism and $X\subset\mathbb{P}^m$ is a hypersurface then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.

$X$ a hypersurface means it has dimension $m-1$. And $f$ a morphism means it is given locally by polynomials. Further, $f$ is continuous, and so $f^{-1}(X)$ is closed in $\mathbb{P}^n$. My only idea is to start with a chain of irreducible closed subsets $X_0\subsetneq X_1\subsetneq\cdots\subsetneq X_{n-1}=X$ and take the preimage. But I'm not sure where to go from here. Any suggestions?

答:
If $X$ is a hypersurface, it is given by one homogeneous equation, i.e.
$$ X = \{\, [x_0: \dotsc: x_m] \in \mathbb P^m \, | \, p(x_0, \dotsc, x_m) = 0 \,\}$$
for some homogeneous polynomial $p \in \mathbb C[x_0, \dotsc, x_m]$. The preimage $f^{-1}(X)$ is then the vanishing locus of
$$ f^* p = p(f_0, \dotsc, f_m) \in \mathbb C[y_0, \dotsc, y_n],$$
where $f_i \in \mathbb C[y_0, \dotsc, y_n]$ are the homogeneous polynomial defining $f$.


问:
Show that a continuous function with a zero in between every zero is identically 0
Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$ and for every $0\le a,b\le 1$, there exists a $c\in(a,b)$ such that $f(c)=0.$ Show that $f$ is identically $0$.

Suppose there exists an $x_0\in(0,1)$ such that $f(x_0)\ne0.$ Thus, by continuity, either there exists a $\delta_1$ such that if $x\in(x_0-\delta_1,x_0+\delta_1),$ then $f(x)>0$ or, similarly there is a $\delta_2$ such that if $x\in(x_0-\delta_2,x_0+\delta_2),$ then $f(x)<0.$ Im a bit stuck as to where to go from here. I want to contradict this by showing there must be a zero in one of these regions but I can't figure out how. Does anyone have any hints

答:
Your idea is right! If there exist $x_0\in (0,1)$ such that $f(x_0)\neq 0$, we can suposse that $f(x_0)>0$. Then we can take $\delta>0$ such that

For all $x\in (x_0-\delta,x_0+\delta)$ we have that $f(x)>0$

$(x_0-\delta,x_0+\delta)\subseteq (0,1)$

By hyphotesis, there exist $c\in (x_0-\delta,x_0+\delta)$ such that $f(c)=0$. This is a contradiction!


问:
Usefulness of generating functions
I am able to come up with good generating functions to solve combinatorial problems, but unable to appreciate its usefulness. For example, I have following combinatorial problem.

How many ways to pay $50$ using unlimited $1, 2 , 5$ coins, (1) when order matters, (2) when order does not matter.

My generating functions:

(1) Number of ways = coefficient of $x^{50}$ in $(1+x+x^2+x^3+...+x^{50})(1+x^2+x^4+x^6+...+x^{50})(1+x^5+x^{10}+x^{15}+...+x^{50})$

Till now, it's fun. But now what? Calculating the coefficient of $x^{50}$ from above expression is as good as counting all individual combination of $1,2,5$ to make $50$, manually. So how is generating function useful in counting?

For case (2) I have following generating function.

Number of ways = coefficient of $x^{50}$ in $(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...+\frac{x^{50}}{50!})(1+x^{2}+\frac{x^{4}}{2!}+\frac{x^{6}}{3!}+...+\frac{x^{50}}{25!})(1+x^{5}+\frac{x^{10}}{2!}+...+\frac{x^{50}}{10!})$

However this permutes only same coins. Unable to come up how all types of coins will be permuted among themselves. Also, same question like above, how generating function is helping in count? I can very well count it manually, like $5$ can occur in $0$ to $10$ times, and for each of these occurrences of $5$, count occurrences of $2$ and $1$.

答:
I would say your generating function in (1) is actually the answer to the "order does not matter" question.  A simple function which also works is $\dfrac{1}{(1-x)(1-x^2)(1-x^5)}$.  This gives a sequence (starting at zero money) of $1,1,2,2,3,4,5,6,7,8,10,\ldots$ ways of paying coins.

For order mattering the numbers should be higher, you can use the generating function $\dfrac{1}{(1-x-x^2-x^5)}$. This gives a sequence (starting at zero money) of $1,1,2,3,5,9,15,26,44,75,128,\ldots$ ways of paying coins.

In both cases you can turn these into a recurrence for calculation, respectively

$a_n=a_{n-1}+a_{n-2}-a_{n-3}+a_{n-5}-a_{n-6}-a_{n-7}+a_{n-8}$ when order does not matter

$a_n=a_{n-1}+a_{n-2}+a_{n-5}$ when order matters

each starting $a_0=1$ and $a_n=0$ for $n<0$.

Title: Usefulness of generating functions


问:
Solving PDE with non constant coefficients when the unknown function depends on more than one variable
Need to solve this PDE for my research work. Tried substitution $y=x^m$ and also power series method but could not get the solution to $a_3$.
Equation: $$\left[-\frac{2m}{r}-2\left(1-\frac{2m}r\right)^{-1}\cdot \frac{m}{r^2}\right]a_3 + m\frac{\partial a_3}{\partial r}=0.$$
where $a_3$ depends on $4$ functions $t,r, \theta,\phi$.
If anybody can help!

答:
According to wolfram alpha for example (if I typed it right)
$$
a_3(t,r,\phi,\theta)=g(t,\phi,\theta)(r-2m)^{1/m}r^{2-\frac1m}.
$$
An equation with only one partial derivative means that you treat the other variables as constants. A PDE means several different partial derivatives at the same time: then things become more involved.


问:
Inner tangent between two circles formula
As a programmer I need to draw the inner tangents between two circles, but only the segments, not the whole line.

But the internet is surprisingly hostile to lazy programmers who don't know their math :).
I know the center points of both circles (R1x, R1y, R2x, R2y) and their radius (r1, r2). What are the points for the tangent segments? (Ax, Ay, Bx, By and Cx, Cy, Dx, Dy).

Thanks!

EDIT: I am not really that lazy, I was on wikipedia (actually I did lot of googling first) but the formula is not there and I can not derive it myself :(

答:
We can find both the inner and outer tangents between two circles with some simple mathematical observations.

Outer tangents

Let's look at the outer tangents first, because it's easier to intuit, after which the inner tangent image makes more sense:

Treating the bigger circle as C1 with radius $r1$, and the smaller circle C2 with radius $r2$, we can create a new circle C3 inside C1 with radius $r3 = r1 - r2$, and then construct the green triangle: we know two of its points, but we don't know its third point, and knowing that third point lets us calculate the two outer tangent points.

While we don't know the third point, we do know all the side lengths:

```<code>hypotenuse = dist(c1, c2)
short = r3
</code>```

And because we're dealing with a right-angled triangle, we also know:

```<code>long = sqrt(hypotenuse*hypotenuse - short*short)
</code>```

But we don't actually need to know how long this edge is, for the same reason: because we're dealing with a right-angled triangle, we can use trigonometry to find the angle we want using only the hypothenuse and short edge:

```<code>φ = acos(short/hypotenuse)
</code>```

However, one thing that's easy to overlook is that angles in computer graphics are universally relative to the x-axis, whereas this angle is relative to the hypotenuse, so we'll need to add it to the angle from the x-axis to the hypotenuse:

```<code>φ = atan2(c2.y-c1.y, c2.x-c1.x) + acos(short/hypotenuse)
</code>```

With that angle, we can now calculate our first circle's tangent point:

```<code>t1x = c1.x + c1.r * cos(φ)
t1y = c1.y + c1.r * sin(φ)
</code>```

And our second circle's tangent point:

```<code>t2x = c2.x + c2.r * cos(φ)
t2y = c2.y + c2.r * sin(φ)
</code>```

And we're done. To get the two tangent points on the other side, we simply pick "the other angle", being the angle from the x-axis to the hypotenuse, minus the angle from the hypotenuse to the short edge:

```<code>φ2 = atan2(c2.y-c1.y, c2.x-c1.x) - acos(short/hypotenuse)
s1x = c1.x + c1.r * cos(φ2)
s1y = c1.y + c1.r * sin(φ2)
s2x = c2.x + c2.r * cos(φ2)
s2y = c2.y + c2.r * sin(φ2)
</code>```

And that's the outer tangents sorted.

(Also note, of course, that if you're using a vector based programming language, all of these things become "run the relevant function(s) on your vector inputs" instead of computing x and y coordinates independently)

Inner tangents

With an understanding of the outer tangents, the graphic for the inner tangent should at least make sense: we have the same "green triangle" that we need to find a third point for, which is one of the two tangent points, and getting the other will then require using the opposite angle, rather than the same angle.

First, rather than a third circle C3 inside C1 with radius $r1 - r2$, we want a circle C3 around C2 instead, with radius $r1 + r2$. This gives us the two edges we need to compute the triangle's angles:

```<code>hypotenuse = dist(c1,c2)
short = r1 + r2
φ = asin(short/hypotenuse)
</code>```

Again though, we now have the angle between the hypotenuse and the long edge, when what we really want is "the angle from the x-axis to the line from our circle center to its tangent point" so: first off, that would be the angle in the red highlighted area, not the green highlighted area:

```<code>φ = asin(short/hypotenuse) - π/2
</code>```

but then we also need to remember to start at the angle from the x-axis to the hypotenuse:

```<code>φ = atan2(c2.y-c1.y, c2.x-c1.x) + asin(short/hypotenuse) - π/2
</code>```

And with that, the tangent point on C1 is calculated the same:

```<code>long = sqrt(hypotenuse*hypotenuse - short*short)
</code>```0

But the tangent on C2 requires the angle for the opposite direction, or, rotating it 180 degrees. For trig functions, that means just adding pi to the angle, so let's do that:

```<code>long = sqrt(hypotenuse*hypotenuse - short*short)
</code>```1

And to get the other two points, we flip the plus/minus signs in our angle formula:

```<code>long = sqrt(hypotenuse*hypotenuse - short*short)
</code>```2

And we're done. Fairly straight forward, really easy to program.

Additional links

While searching for inner and outer tangents on the web will get you a decent number of results, one result you're almost certainly not going to find is the wikipedia article on Powers of a Point, and specifically the Similarity Points, which give you formulae for computing some of the points related to inner and outer tangents without ever computing the tangents themselves.

Body:
We can find both the inner and outer tangents between two circles with some simple mathematical observations.

Outer tangents

Let's look at the outer tangents first, because it's easier to intuit, after which the inner tangent image makes more sense:

Treating the bigger circle as C1 with radius $r1$, and the smaller circle C2 with radius $r2$, we can create a new circle C3 inside C1 with radius $r3 = r1 - r2$, and then construct the green triangle: we know two of its points, but we don't know its third point, and knowing that third point lets us calculate the two outer tangent points.

While we don't know the third point, we do know all the side lengths:

hypotenuse = dist(c1, c2)
short = r3

And because we're dealing with a right-angled triangle, we also know:

long = sqrt(hypotenuse*hypotenuse - short*short)

But we don't actually need to know how long this edge is, for the same reason: because we're dealing with a right-angled triangle, we can use trigonometry to find the angle we want using only the hypothenuse and short edge:

φ = acos(short/hypotenuse)

However, one thing that's easy to overlook is that angles in computer graphics are universally relative to the x-axis, whereas this angle is relative to the hypotenuse, so we'll need to add it to the angle from the x-axis to the hypotenuse:

φ = atan2(c2.y-c1.y, c2.x-c1.x) + acos(short/hypotenuse)

With that angle, we can now calculate our first circle's tangent point:

t1x = c1.x + c1.r * cos(φ)
t1y = c1.y + c1.r * sin(φ)

And our second circle's tangent point:

t2x = c2.x + c2.r * cos(φ)
t2y = c2.y + c2.r * sin(φ)

And we're done. To get the two tangent points on the other side, we simply pick "the other angle", being the angle from the x-axis to the hypotenuse, minus the angle from the hypotenuse to the short edge:

φ2 = atan2(c2.y-c1.y, c2.x-c1.x) - acos(short/hypotenuse)
s1x = c1.x + c1.r * cos(φ2)
s1y = c1.y + c1.r * sin(φ2)
s2x = c2.x + c2.r * cos(φ2)
s2y = c2.y + c2.r * sin(φ2)

And that's the outer tangents sorted.

(Also note, of course, that if you're using a vector based programming language, all of these things become "run the relevant function(s) on your vector inputs" instead of computing x and y coordinates independently)

Inner tangents

With an understanding of the outer tangents, the graphic for the inner tangent should at least make sense: we have the same "green triangle" that we need to find a third point for, which is one of the two tangent points, and getting the other will then require using the opposite angle, rather than the same angle.

First, rather than a third circle C3 inside C1 with radius $r1 - r2$, we want a circle C3 around C2 instead, with radius $r1 + r2$. This gives us the two edges we need to compute the triangle's angles:

hypotenuse = dist(c1,c2)
short = r1 + r2
φ = asin(short/hypotenuse)

Again though, we now have the angle between the hypotenuse and the long edge, when what we really want is "the angle from the x-axis to the line from our circle center to its tangent point" so: first off, that would be the angle in the red highlighted area, not the green highlighted area:

φ = asin(short/hypotenuse) - π/2

but then we also need to remember to start at the angle from the x-axis to the hypotenuse:

φ = atan2(c2.y-c1.y, c2.x-c1.x) + asin(short/hypotenuse) - π/2

And with that, the tangent point on C1 is calculated the same:

t1x = c1.x + c1.r * cos(φ)
t1y = c1.y + c1.r * sin(φ)

But the tangent on C2 requires the angle for the opposite direction, or, rotating it 180 degrees. For trig functions, that means just adding pi to the angle, so let's do that:

t2x = c2.x + c2.r * cos(φ + π)
t2y = c2.y + c2.r * sin(φ + π)

And to get the other two points, we flip the plus/minus signs in our angle formula:

φ2 = atan2(c2.y-c1.y, c2.x-c1.x) - asin(short/hypotenuse) + π/2
s1x = c1.x + c1.r * cos(φ2)
s1y = c1.y + c1.r * sin(φ2)
s2x = c2.x + c2.r * cos(φ2 + π)
s2y = c2.y + c2.r * sin(φ2 + π)

And we're done. Fairly straight forward, really easy to program.

Additional links

While searching for inner and outer tangents on the web will get you a decent number of results, one result you're almost certainly not going to find is the wikipedia article on Powers of a Point, and specifically the Similarity Points, which give you formulae for computing some of the points related to inner and outer tangents without ever computing the tangents themselves.


问:
Non trivial Eigenvectors of the Laplacian in $H^1$
Let $N$ be natural and $\Omega \subset \mathbb R^N$ be open and bounded. I want to show that
$$f-\Delta f=0$$
in $D'(\Omega)$ has non trivial solutions in $H^1$. So far I've managed to show solutions to this PDE (for general open $\Omega$) are orthogonal to $H^1_0$ in $H^1$ and using Fourier transform I've found the solutions for different $\Omega \subseteq \mathbb R$. I'm still not sure how to prove existence for general bounded open $\Omega$.

答:
$H_0^1(\Omega)$ is a proper, closed subspace of $H^1(\Omega)$, therefore its orthogonal complement $H_0^1(\Omega)^{\perp}$ is non-trivial. Since this is the space of solutions, there exists at least a nontrivial solution.

In fact, we can show that for any $h\in H^1(\Omega)$ there exists a uniqe solution to
$$\begin{cases}\Delta u= u & \Omega \\
u=h & \partial \Omega.
 \end{cases} $$

Fix $h\in H^1(\Omega)\setminus H_0^1(\Omega)$.
Let $$H^1_h(\Omega):=\left\{u\in H^1(\Omega):u-h\in H_0^1(\Omega)\right\}$$
Thus $0\notin H^1_h(\Omega)$, and $H_h^1(\Omega)$ is closed in $H^1(\Omega)$. Since an element $u\in H^1(\Omega)$ solves the equation if and only if $u\in H_0^1(\Omega)^{\perp}$,  we just have to show that
$$H^1_h(\Omega)\cap H_0^1(\Omega)^{\perp}\neq \emptyset. $$
An element of this set is the element of $H_h^1(\Omega)$ with the smallest $H^1$ norm.  To show that it is the only element, notice that if $u_1,u_2\in H_h^1(\Omega)\cap H_0^1(\Omega)^{\perp}$, then  $u_1-u_2\in H_0^1(\Omega)\cap H_0^1(\Omega)^{\perp}=\left\{0\right\}$.


问:
Power set of the Universal Set
In set theories with a universal set $U$, can we take the powerset of $U$? In particular, can we look at $\mathcal{P}(U)$ and compare with $U$, i.e. can we say something like $U \in \mathcal{P}(U)$, $U \notin \mathcal{P}(U)$, $|U| < |\mathcal{P}(U)|$, or $U \neq \mathcal{P}(U)$?</p>

答:
Some nonwell-founded set theorizing (§4.4) has mappings $\wp(V) \rightarrow V$ and vice versa. I'm not sure if this is a theorem of the system's axioms or an axiom itself (it appears to be a conjunction or alternation of axioms; the mappings correspond to an initial algebra and a final coalgebra, but I won't pretend to understand that difference!).

In paraconsistent set theory, things are even weirder. It appears to be the case here that a cardinal for ORD is equivalent to its own powerset while sort of being smaller than itself as such. Or at any rate, the situation is so exotic that it's hard to tell how the mapping between $V$ and $\wp(V)$ works except to say that it's not the same as usual!

Title: Power set of the Universal Set


问:
How to express numbers between 1 and 1000 as a sum of consecutive integers in two different ways
How many integers 1≤N≤1000 can be written both as the sum of 26 consecutive integers and as the sum of 13 consecutive integers?</p>

答:
HINT:

If $a$ is the first term of the AP series with $13$ members, the sum will be $$\frac{13}2\cdot \{2\cdot a+(13-1)1\}=13(a+6)$$ as the common difference $=1$ here.

Similarly, for $26$ consecutive integers, the sum will be $$\frac{26}2\cdot \{2\cdot b+(26-1)1\}=13(2b+25)$$

We need  $$13(a+6)=13(2b+25)\iff a+6=2b+25\implies a=2b+19$$

and  to check $$1\le 13(a+6)\le 1000 $$

$$\implies 1\le 13(2b+25)\le 1000\implies 1\le 2b+25\le 76\implies -12\le b\le 25$$


答:
The region being revolved is bound by curves $y = k - kx^2$ and $y = kx^2$ for $k \gt 0$.

At intersection of both curves, $k - kx^2 = kx^2 \implies x =  \frac{1}{\sqrt2}$ as we are in first quadrant.

For revolution around $y = 0$, we apply washer method. Radius of revolution $y$ is between $kx^2 \leq y \leq k-kx^2$.

$V_1 = \displaystyle \int_{0}^{1/\sqrt2}\int_{kx^2}^{k-kx^2} 2  \pi \ y \ dy \ dx = \frac{\sqrt2}{3} \pi \ k^2$

For revolution around $x=-2$, we apply shell method. The size of shell is given by $k-kx^2 - kx^2 = k - 2kx^2$. Radius of revolution is $2+x$.

$V_2 = \displaystyle \int_{0}^{1/\sqrt2} 2  \pi \ (2+x) (k-2kx^2) \ dx = \frac{3 + 16\sqrt2}{12} \pi \ k$

When $V_1 = V_2, k = \frac{3 + 16\sqrt2}{4\sqrt2} \approx 4.53$.

Title: Use washer and shell method to find $k$

Body:
Let $R$ denote the first quadrant region bounded by $y=k-kx^2$, and $y=kx^2$. $k> 0, k \in \mathbb{R}$. If $V_1$ is the volume obtained by rotating $R$ about the line $y=0$, and $V_2$ is the volume obtained by rotating R about the line $x=-2$, and $V_1=V_2$, then what is $k$?

My plan is to evaluate exact volume of $V_1$ and $V_2$, then make the equation $V_1=V_2$ and solve for $k$. The first one $V_1$ can use the washer method, and second one might be shell, since it is rotating around a vertical line $x = 2$.


问:
Cayley graph with opposite action: is the group abelian?
Let $G$ be a group, let $S$ be a set of generators and let $\Gamma=\Gamma(G,S)$ be the Cayley graph, where there is an edge between $g$ and $h$ if and only if $h=gs$ for some $s\in S$.

We know that the action of $G$ on $\Gamma$ via left multiplication acts on $\Gamma$ via graph automorphisms: if $h=gs$, then for each $a\in G$,
$$
ah = a(gs) = (ag)s ,
$$
and so $ag$ ahd $ah$ are connected by an edge whenever $g$ and $h$ are.

Conversely, the action of $G$ on $\Gamma$ via right multiplication does not in general induce graph automorphisms:
$$
ha = gsa \ne ags ,
$$
unless $G$ is abelian.

Now, is the converse true? If the action of $G$ via right multiplication induces graph automorphisms, can we conclude that $G$ is abelian?

答:
No, this is wrong.
Let us take the union of several conjugacy classes in the non-Abelian group G as a generating set:
$S=a_1^G\cup\ldots\cup a_k^G$.
It is easy to see that the action of $G$ via right multiplication induces graph automorphisms. For instance, $G=S_3$, $S=\{(12),(12),(23)\}$.


问:
prove :nth root of the Product of arithmetic sequence is bigger than the sqrt of the first and last numbers
let $ \{a_n\} $ be an arithmetic sequence of positive numbers.

prove that for all $ n\ge2 $:

$ \frac{a_{1}+a_{n}}{2}\ge\sqrt[n]{a_{1}a_{2}a_{3}\dots a_{n}}\ge\sqrt{a_{1}a_{n}} $

I managed to solve the left part of the inequality by using the sum of the arithmetic sequence and the EM-GM inequality.

I'm having issues with the right side, tried using ln on everything and induction but the algebra was too tough. Also tried comparing with the harmonic mean
Any other methods to solve this with calculus 1 tools(haven't learned about $ \Gamma $ function yet) ?

答:
Let $a_n = a + (n-1)d$.

Then for $0 < m < n-1$ we have:

$$a_{1 + m}a_{n-m} = (a_1 + md)(a_n -md) \\= a_1a_n+(a_n-a_1)md-(md)^2 =a_1a_n +md((n-1-m)d)\ge a_1a_n$$

Now consider $(\sqrt[n]{a_1a_2\dots a_n})^2 = \sqrt[n]{a_1a_2\dots a_n}\sqrt[n]{a_na_{n-1}\dots a_1} = \sqrt[n]{(a_1a_n)(a_2a_{n-1})\dots(a_na_1)}$.

By our result above, that expression cannot be less than $\sqrt[n]{(a_1a_n)^n} = a_1a_n$.

Therefore $\sqrt[n]{a_1a_2\dots a_n} \ge \sqrt{a_1a_n}$.


问:
Finding domain of function and writing in interval notation
I think I have the answer but it always takes me a long time to find it and wanted to check it.  Equation is;$$ f(x) =\frac{\sqrt {x+4}}{x-4}$$ Interval notation then is $$ [-4, 4) \cup (4, \infty)$$

答: Note that $\sqrt{x+4}$ yields $x\ge -4$ and $\frac{1}{x-4}$ yields $x\ne 4$, so $x$ can be anything greater than $-4$ excluding $4$. Therefore we can write $x \in [-4,+\infty) \setminus \{ 4\}$ or equivalently $x \in [-4,4)\cup (4, +\infty)$.


问:
Evaluate the given limit by recognizing it as a Riemann sum.
Please note that this is homework. Please excuse my lack of $\LaTeX{}$ knowledge.

The Problem:

Evaluate the given limit by first recognizing the sum (possibly after taking the logarithm to transform the product into a sum) as a Riemann sum of appropriate function associated to a regular partition of $[0,1]$.

$\lim_{n\to \infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}$

My petty Attempt

Okay so taking the hint from the problem. I tried this:

$\Delta x = \dfrac{1-0}{n} = \frac{1}{n}$

Then $x^*_k = 0 + k\Delta x = \frac{k}{n}$

Then I took the logarithm and expanded the logarithm.

$e^{[\lim_{n\to \infty}ln{(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!})}]}$

$e^{[\lim_{n\to \infty}ln{(\frac{1}{n})} + ln{(\sqrt[n]\frac{(2n)!}{n!})}]}$

$e^{[\lim_{n\to \infty}-ln{(n)} + \frac{ln(\frac{(2n)!}{n!})}{n}]}$

$e^{[\lim_{n\to \infty}-ln{(n)} + \frac{ln(2n!) - ln(n!)}{n}]}$

Now after this I am totally stuck. I am not even sure if I am going on the right track. All I ask is for some guidance. I know that the Riemann Sum is defined as:

$\sum^n_{k=1}{f(x^*_k)\Delta x_k}$

I also understand that the stuff inside the limit is already in closed form. Somehow I have to use the properties of logarithm to transform it to a $\sum$ but how?

答:
We apply the $\log$ function and we use the Riemann sum and we integrate:
$$\log\left(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}\right)=-\log n+\frac{1}{n}\sum_{k=1}^n\log(k+n)=\frac{1}{n}\sum_{k=1}^n\log(k+n)-\log n\\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_0^1 \log(1+x)dx=2\log(2)-1$$

hence
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}=\frac{4}{e}$$


问:
Why is this sum equal to 1?
I'm sorry for opening a new question for a triviality, but I couldn't find this result online.

The following holds:

$$\sum_{j=0}^{n-1} {\dfrac{ {{K-1}\choose{j}} {{N-1-(K-1)}\choose{n-1-j}} } {{N-1}\choose{n-1} }}=1$$

It looks like an identity that I'm not aware of. I checked with easy numbers and all terms are different than 0. I only know of this identity $\sum_{k=0}^n {{n}\choose{k}}=2^n$ which I don't think is any helpful.

Both $-1$ in the second binomial in the numerators have been added from a previous step, I guess in order to make it easier to understand why the sum is actually $1$.

I'd be glad if you can help me. Thank you.

答:
Observe that by the 1 Chu-Vandemonde Identity

$$\sum_{j=0}^{n-1}\binom{k-1}j\binom{N-k}{n-1-j}= \binom{N-1}{n-1}$$

...and we're done

Title: Why is this sum equal to 1?


问:
About cross product, differentiation and perpendicular vectors
I have the following problem.

$\vec{r} (t)$ is a twice-differentiable vector function and $\vec{B}$ is a fixed vector with magnitude $4$. The cross product between $\vec{r} (t)$ and $\vec{B}$ is $4t \hat{i}$ for all t.

If vector $\vec{r}'(t)$ is perpendicular to vector $\vec{B}$. Determine if vectors $\vec{r}'(t)$ and $\vec{r}'' (t)$ are perpendicular.

I know that $4t \hat{i}$ is orthogonal to vectors $\vec{r}(t)$ and $B$, and that if a differentiable vector has constant length then the vector and its first derivative are orthogonal.

Could you please show me an example or give any tips on how to solve this?

Please excuse my english. Thanks.

答:
If you take the second derivative in your relation you get
$$
r''\times B+2r'\times B'+r\times B''=0
$$
Since $B$ is fixed, the last two terms disappear and hence $r''\times B=0$. Therefore $r''$ is parallel to $B$ and (given that $r'$ is perpendicular to $B$), it is perpendicular to $r'$.

I do not see any need for the condition $|B|=4$.


问:
Probability of A and not B
I'm studying Introduction to probability and currently, I'm stuck with the following problem. Given:

$P(A)=0.7$,
$P(B)=0.5$,
$P(A\cap B)=0.45$

What is the probability of A and not B?

I've checked this similar question but I don't understand the answers. Also I've asked my instructor and she told me that $1-P(A\cap B)(P(B^c))$ is the answer (As the answers suggested, this result is not correct). Why is that? She does not provided me a completely explanation.

Update 1: The original problem is the following

In a multiplex cinema, there are two different rooms, $A$ and $B$, working simultaneously. Let $SA$ be the event that, during a certain showing, room $A$ becomes full before the film begins, and let $SB$ be the event that, during the same showing, room $B$ becomes full before the beginning of the movie. We know that $P(SA)=0.7$; $P(SB)=0.5$ and $P(SA∩SB)=0.45$

Calculate the probability that room $A$ will become full and room $B$ will not.

Did I state the problem correctly?

Update 2: Add a Venn diagram. Following the advice of Ethan Bolker. Here is the Venn diagram that I made.

答: Hint: try drawing a Venn diagram.</p>

Title: Probability of A and not B

Body: Hint: try drawing a Venn diagram.


答:
Hint:
$$P(A \cap \overline{B}) = P(A\setminus(A\cap B)) = P(A) - P(A\cap B)$$

Title: Probability of A and not B

Body:
I'm studying Introduction to probability and currently, I'm stuck with the following problem. Given:

$P(A)=0.7$,
$P(B)=0.5$,
$P(A\cap B)=0.45$

What is the probability of A and not B?

I've checked this similar question but I don't understand the answers. Also I've asked my instructor and she told me that $1-P(A\cap B)(P(B^c))$ is the answer (As the answers suggested, this result is not correct). Why is that? She does not provided me a completely explanation.

Update 1: The original problem is the following

In a multiplex cinema, there are two different rooms, $A$ and $B$, working simultaneously. Let $SA$ be the event that, during a certain showing, room $A$ becomes full before the film begins, and let $SB$ be the event that, during the same showing, room $B$ becomes full before the beginning of the movie. We know that $P(SA)=0.7$; $P(SB)=0.5$ and $P(SA∩SB)=0.45$

Calculate the probability that room $A$ will become full and room $B$ will not.

Did I state the problem correctly?

Update 2: Add a Venn diagram. Following the advice of Ethan Bolker. Here is the Venn diagram that I made.


问:
Covariance matrix multiplied by its transpose
Suppose we have a covariance matrix $A$ (it is automatically symmetric).  So, $A^TA$ is also symmetric.
However, do the elements of the matrix $A^TA$ keep the same pair-values as that of the matrix $A$?

Thank you in advance.

答:
In the two by two case (similarly for higher dimensions), a covariance matrix is represented by:

$$A=\pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}$$  so that

$$A^T A = A^2 = \pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}\pmatrix{\sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2}=\pmatrix{\sigma_1^4+\sigma_{12}^2 & \sigma_{12}(\sigma_1^2 + \sigma_2^2) \\ \sigma_{12}(\sigma_1^2 + \sigma_2^2)& \sigma_2^4+ \sigma_{12}^2}.$$

We see that each term is a combination (sum of products) of elements of $A$.

Not sure exactly what you are asking, but each element of $A^TA=A^2$ is not exactly a function only of the corresponding element of $A$.


问:
Numerical Approximation: Integral of Exponential to an Exponential power
I am trying to solve a self-consistent equation for the variable $S$.

$$
\mathbf{S} = \int_0^t C_1 \mathbf{S} e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da \\
%
1 = \int_0^t C_1 e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da
$$

I would input values of $C_1, C_2, C_3$ and $C_4$ and I am interested in finding the $\mathbf{S}\neq 0$ that satisfies the above equation.

Things I have tried:

I am thinking numerical approximation but I am unfamiliar with the methods used and not sure what to read up on. Perhaps Runge-Kutta or Eulers method? Would appreciate any tips.

I attempted Bisection Method using a Matlab code, trying to solve for the zero of $\int_0^t C_1 e^{-C_2 a-C_3\cdot\mathbf{S}\cdot e^{-C_4 a}} da-1$ but it was giving me an error (I think because of the integral.)
Here is the bisection code:

```<code>% Clearing Screen
clc

% Setting x as symbolic variable
syms x;

% Input Section
y = input('Enter non-linear equations: ');
a = input('Enter first guess: ');
b = input('Enter second guess: ');
e = input('Tolerable error: ');

% Finding Functional Value
fa = eval(subs(y,x,a));
fb = eval(subs(y,x,b));

% Implementing Bisection Method
if fa*fb > 0
    disp('Given initial values do not bracket the root.');
else
    c = (a+b)/2;
    fc = eval(subs(y,x,c));
    fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');
    while abs(fc)>e
        fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);
        if fa*fc< 0
            b =c;
        else
            a =c;
        end
        c = (a+b)/2;
        fc = eval(subs(y,x,c));
    end
    fprintf('\nRoot is: %f\n', c);
end
</code>```

In case it is of interest:
I am vaguely familiar with python, Matlab, C++. More familiar with Mathematica.

答:
You have that $S(0)=0$ and, for $t>0$, your integral equation is of the form
$$
\int_0^t f(a) e^{S(t) g(a)} da = 1.
$$

This can be numerically solved using a numerical quadrature. Suppose that you restrict yourself to a uniform mesh $t_i = i h$, and use the same mesh for the numerical quadrature. You then have
$$
\sum_{j=0}^i \omega_j^{(i)}f(t_j)e^{S_j g(t_j)}=1
$$

solving with respect to $S_i$, you get
$$
\omega_i^{(i)}f(t_i)e^{S_i g(t_i)} =1-\sum_{j<i} \omega_j^{(i)}f(t_j)e^{S_j g(t_j)}
$$

$$
e^{S_i g(t_i)} = \dfrac{1-\sum_{j<i} \omega_j^{(i)}f(t_j)e^{S_j g(t_j)}}{\omega_i^{(i)}f(t_i)}
$$

$$
S_i = \frac{1}{g(t_i)}\log \left(\dfrac{1-\sum_{j<i} \omega_j^{(i)}f(t_j)e^{S_j g(t_j)}}{\omega_i^{(i)}f(t_i)}\right)
$$

This last recurrence equation approximates the values of $s(t)$ on the mesh. Now, you just need to choose a numerical quadrature, like the trapezoidal rule, and properly compute the weights $\omega_j^{(i)}$.


问:
How to Taylor expand this function, which is a function of vectors?
Suppose that $a,b,c,d$ are vectors in $\mathbb{R}^2$, so $a_{\alpha}$ where $\alpha = 1,2$. We define a scalar product as $a \cdot b := a_1 b_2 - a_2 b_1$. Suppose also that we have a real-valued function $f(a,b,c)$ which is a function of some $3$ vectors.

What does the Taylor expansion of $f$ look like to first order? For example, I want to Taylor expand $f(a+b,c,d)$ around $(a, c, d)$, so $$f (a+b, c, d) = f(a,c,d) + \ldots $$  but I don't know how to write the remaining terms. Do I get a scalar product of $b$ with some term?

答:
$\def\va{{\bf a}}
\def\vb{{\bf b}}
\def\vc{{\bf c}}
\def\vd{{\bf d}}
\def\e{\varepsilon}
\def\np{\odot}$We have
\begin{align*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \sum_i b_i\left.\frac{\partial f(\va+\vb,\vc,\vd)}{\partial b_i}\right|_{b_i=0} + \ldots \\
&= f(\va,\vc,\vd) + \sum_i b_i \frac{\partial f(\va,\vc,\vd)}{\partial a_i} + \ldots \\
&= f(\va,\vc,\vd) + (\vb,\nabla_\va f(\va,\vc,\vd)) + \ldots,
\end{align*}
where $(,)$ is the usual inner product and where
$\nabla_\va f = \langle \partial f/\partial a_1,\partial f/\partial a_2\rangle$ is the gradient.
But
$(\langle a_1,a_2\rangle,\langle b_1,b_2\rangle)
= \langle a_1,a_2\rangle\np\langle -b_2,b_1\rangle$.
(To avoid confusion with the dot product we will write $\va\np\vb=a_1b_2-a_2b_1$.)
Thus,
\begin{align*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \vb\np \widetilde\nabla_\va f(\va,\vc,\vd) + \ldots,
\end{align*}
where
$\widetilde\nabla_\va f = \langle -\partial f/\partial a_2,\partial f/\partial a_1\rangle$.

Addendum

As is discussed in detail in the other answers, the product $\va\np\vb$ is only a "scalar product" in the sense that it is a product of vectors returning a real value.


问:
Closed form for $\sum{\frac{\zeta(2n)-1}{(2n)^2}}$
I am interested in finding a closed form that involve some known constants to express following infinite sum:

$$\sum_{n=1}^{\infty}{\frac{\zeta(2n)-1}{(2n)^2}}$$

you can find out some similar results when denominator exponent is 0 or 1 in wikipedia zeta function entry, but I was not able to find anything related to $n^2$ denominator.

答:
This not-an-answer explains the integral representation given in my comment.

Using the infinite product for the sine, for $0<x<1$ we have

\begin{align*}
\log\frac{x\pi}{\sin x\pi}&=-\sum_{n=1}^\infty\log\left(1-\frac{x^2}{n^2}\right)=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1k\left(\frac{x^2}{n^2}\right)^k
\\&=\sum_{k=1}^\infty\frac{x^{2k}}{k}\sum_{n=1}^\infty\frac{1}{n^{2k}}=\sum_{k=1}^\infty\frac{\zeta(2k)}{k}x^{2k}.
\end{align*}

Dividing by $x$ and integrating, we get the desired: $$\sum_{k=1}^\infty\frac{\zeta(2k)}{2k^2}=\int_0^1\log\frac{x\pi}{\sin x\pi}\frac{dx}{x}\underset{\color{gray}{x\pi=t}}{\phantom{\big[}=\phantom{\big]}}\int_0^\pi\log\frac{t}{\sin t}\frac{dt}{t}.$$

(If we denote this by $S$, the sum in question is $S/2-\pi^2/24$.)


问:
Some function of order statistics of Exp(1) random variables
Let $X_1$, ..., $X_n$ be independent, $Exp(1)$ distributed random variables, where $X_{(0)}=c \in \mathbb{R}$ ist fixed. Define $Z_i := (n-i+1)(X_{(i)}-X_{(i-1)})$.

What is the distribution of $\sum_{i=1}^n Z_i$?

I tried to solve this problem in two steps. First, I looked at the distribution of $Z_1$, ..., $Z_n$ and via transformation could (making use of a hint from the lecture) show that they are exponentially distributed and independent again. Their density is $$f_{Z_1, ..., Z_n}(z_1, ..., z_n)=\exp(\sum_{i=1}^n-z_i).$$

I know that I should be able to conclude the distrubtion of $\sum_{i=1}^n Z_i$ from here. But I don't see how.

I have had problems with this general problem pattern before: Given the joint density of random variables $Y_1$, ..., $Y_n$, how do I calculate e.g. the density of some real valued mesaruable function $g(Y_1, ..., Y_n)$?

I was able to do it for the $Z_i$ because we've had it for the range $R=X_{(i)}-X_{(i-1)}$ in the uniform case, but I'm kind of missing the big picture here.

Can anyone help me with this?

答: The density of $(Z_1,...,Z_n)$ is the product of the densities of each $Z_i$, therefore they are independent. And a sum of independent exponential random variables with the same parameter is Gamma distributed.  </p>


问:
inequalities on Riemmanian manifolds in local coordiantes
Given a compact Riemmanian manifold $(X, g)$ and a smooth function $f$ on it. Recall that $\nabla f = g^{i j} \partial_{i} f \partial_{j}$, so $|\nabla f|^2 = g_{jq}g^{ij}g^{pq}\partial_{i} f \partial_{p} f$.

We can compute in normal coordinates that $\Delta_{g} |\nabla f |^2 = \partial_{k} \partial_{k}((g_{jq}g^{ij}g^{pq})) \partial_{i} f \partial_{p} f +$ a bunch of other terms.  Let me denote this first term on the right hand side by $R$. I have the following computation to get a lower bound for $R$:

$R \geq -C \sum |\partial_{i} f| |\partial_{p} f| \geq - C (\sum_{i} |\partial_{i} f|^2 + \sum_{p} |\partial_{p}f|^2) =  -C |\nabla f|^2$. This $C$ depends on some curvature of the metric.

I am not so sure about this computation and specifically my question is:

I found online that if the manifold is compact, then one can treat the norm of a tensor $T$ using the norm of their components in local coordinates. This is equivalent to the norm defined as $|\nabla^{k} T|_{g}$. However, if we use the norm in local coordinates, it is going to be dependent on the choice of coordinates. What does it mean for it to be equivalent to the second definition?

How do you treat inequalities in local coordinates? In the example I gave, can I just bound any random terms with constants like I did for the derivatives of the metric or does it have to be a tensor (globally defined objects) for me to throw it away using its bound.

1 is related to 2 because 1 seems to be suggesting that invariance does not matter much but that does not make sense to me.

答:
The first thing to note is that $g_{jq}g^{pq} = \delta_j^p$, so
$$ |\nabla f|^2 = g^{ij}\partial_if\partial_jf $$

Second, if you want a lower bound for $\Delta|\nabla f|^2$, it follows directly from the Bochner formula.

As for your questions about estimates for the norm of a tensor, it depends on what the situation is. If you're working on a fixed compact Riemannian manifold, and you need an estimate for the norm of a tensor that's independent of the tensor itself, then using the local coordinate version is fine. If, however, you want an estimate where the constants do not depend on the specific manifold but only on geometric properties of the manifold, then it's possible to still use local coordinates (but carefully) but it's usually better to find coordinate-free estimates of the norm of the tensor with respect to the metric. The Bochner formula is an example of this approach.