问:
Poisson Integral for half plane approaches argument
I'm looking at an exercise in Ahlfors' Complex Analysis. It states the following:

Assume $U(\zeta)$ is piecewise continuous and bounded for all real $\zeta$, and assume $U$ has a jump at $0$ (i.e. $U(0^-)=1$ and $U(0^+)=0$). Define $P_U(z)=\frac{1}{\pi}\int^\infty_{-\infty} \frac{y}{(x-\zeta)^2+y^2}U(\zeta)d\zeta$ (the Poisson Integral on the half plane). Prove that $P_U(z)-\frac{1}{\pi}arg(z)\rightarrow0$ as $z\rightarrow 0$.

From what I can tell, the problem seems to boil down to showing that the integral portion of $P_U(z)$ approaches $arg(z)$ as $z\rightarrow 0$. I've shown in a previous problem that $P_U(z)$ is harmonic and its boundary values are equal to $U(\zeta)$, but don't quite know how to use it here. Any help is appreaciated!

答:
Define $V: \Bbb R \to \Bbb R$ as $V(\zeta) = 1$ for $\zeta < 0$, $V(0)=U(0)$, and $V(\zeta) = 0$ for $\zeta > 0$. Show that

$U-V$ is continuous at $\zeta=0$.

$P_V(z) = \frac 1\pi \arg(z)$.

It follows that
$$
P_U(z)-\frac{1}{\pi}arg(z) = P_U(z)-P_V(z) = P_{U-V}(z) \to (U-V)(0) = 0
$$
for $z \to 0$.


问:
Simplex example to be applied on Zoutendijk method for constrained optimization
The Zoutendijk method of feasible directions for the optimization with non-linear constraints involves a step where a linear program is solved. In a particular example, the linear program is defined as follows:

$$
\hbox{Minimize}_{d_1,d_2,z}\quad z\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(*)\\ \hbox{subject to:}\\
\begin{align*}
\qquad\qquad\qquad-5.5d_1-3d_2-z&\leq 0\\
\qquad\qquad\qquad-d_1-z&\leq 0\\
\qquad\qquad\qquad-1\leq d_1&\leq 1\\
\qquad\qquad\qquad-1\leq d_2&\leq 1\\
\end{align*}
$$
where $z$ is $\hbox{Max}\{\nabla^tf(x)d,\, \nabla^tg_i(x)d\}$; $f$ is the function to be optimized and $g_i$ the restrictions that are active for the current parameter values.

According to the textbook, the problem above can be solved by means of the simplex method and the following solution is obtained: $d_1=1, d_2=-1, z=-1$.

I am stuck trying to solve this linear program. What I have done is the following. First, transform the variables $d_1$, $d_2$ and substitute them respectively by $y_1=(1-d_1)/2$ and $y_2=(1-d_2)/2$. As $d_1,d_2\in[-1,1]$, the variables $y_1,y_2\in[0,1]$. In particular $y_1,y_2$ are nonnegative and so, I understand they are suitable for the simplex method.

For the new variables, the program is (after including stack variables to transform the $\leq$ inequalities into equalities):

Edit:
I have realized that $z$ should be allowed to take positive, zero or negative values. By means of new variables $s_5,s_6\geq 0$ I define now $z=s_5-s_6$, so that now $z$ is a free variable.

$$
\hbox{Minimize}_{y_1,y_2,z,s_1,s_2,s_3,s_4,s_5,s_6}\quad z\color{red}{=s_5-s_6}\\ \hbox{subject to:}\\
\begin{align*}
\qquad\qquad\qquad 11y_1+6y_2-s_5+s_6+s_1&=8.5\\
\qquad\qquad\qquad 2y_1-s_5+s_6+s_2&= 1\\
\qquad\qquad\qquad y_1+s_3&= 1\\
\qquad\qquad\qquad y_2+s_4&= 1\\
\qquad\qquad\qquad y_1,\,y_2,s_1,s_2,s_3,s_4,s_5,s_6&\geq 0\\
\end{align*}
$$

Now, I proceed with the simplex method, starting as follows:

Edit: The initial Tableau becomes now:
$$
\begin{eqnarray*}
\begin{array}{cc|cccccccc|c}
&&s_5&s_6&y_1&y_2&s_1&s_2&s_3&s_4\\
&&1&-1&0&0&0&0&0&0\\
\hline
s_1&0&-1&1&11&6&1&0&0&0&8.5\\
s_2&0&-1&1&2&0&0&1&0&0&1\\
s_3&0&0&0&1&0&0&0&1&0&1\\
s_4&0&0&0&0&1&0&0&0&1&1\\
\hline
&&1&-1&0&0&0&0&0&0
\end{array}
\end{eqnarray*}
$$

Edit:
I choose the starting values proposed in the textbook, $d_1=0$, $d_2=0.75$, which imply $y_1=0.5$, $y_2=0.125$ and $z=\hbox{Max}\{\nabla^tf(x)d,\, \nabla^tg_i(x)d\}=0$. Thus, the values for the stack variables compatible with the constraints are the values marked in red.Those starting values are actually proposed in the book for the original non linear program (not for the search of the direction sought in this linear program). I choose initial values setting the slack variables equal zero. I'm not sure if this is the right approach.

Applying the steps of the simplex, I obtain the following Tableau:

$$
\begin{eqnarray*}
\begin{array}{c|cccccccc|c}
&s_5&s_6&y_1&y_2&s_1&s_2&s_3&s_4\\
\hline
\color{red}{s_1}&\color{red}{0}&0&9&6&1&-1&0&0&7.5\\
\color{red}{s_6}&\color{red}{-1}&1&2&0&0&1&0&0&1\\
\color{red}{s_3}&\color{red}{0}&0&1&0&0&0&1&0&1\\
\color{red}{s_4}&\color{red}{0}&0&0&1&0&0&0&1&1\\
\hline
&0&0&0&0&0&0&0&0&
\end{array}
\end{eqnarray*}
$$

This is the final Tableau, as there are no negative elements in the last row. The values for the solutions indicated in red imply: $z=s_5-s_6=0-1=-1$  and $y_1=y_2=0$, which imply in turn $d_1=d_2=1-2\times 0=1$. This is not the solution reported in the textbook, but according to the answer by RobPratt it is an admissible solution. It can be checked that the constraints are met.

Yet, I'm not completely sure whether the above calculations are correct. I thought that the slack variables should be initialized to the elements in the right-hand-side of the constraint they belong to, i.e.: $s_1=8.5$, $s_2=1$, $s_3=1$, $s_4=1$. I tried this option, it generates many more Tableaus until no more negative elements show up in the last row. In that case I obtained the other solution mentioned in the answer by RobPratt and in the textbook, $d_1=1$, $d_2=-1$, but the value for $z$ turned out to be $5/2>-1$, so something was not correct. Maybe I made some mistake in the calculations since several Tableaus were generated, but I'm not sure if it could bee some other type of mistake.

Edit: My questions are more concrete now:

1. Are the above calculations and solution obtained correctly?

2. How should I adjust the initial Tableu in order to get the other solution?

It seems I made a mistake in the calculations, but I am not sure either about the approach I took for solving program (*) above. I am not familiar with software tools to solve this program. It would be helpful to me if someone could check the solution of the program described above. If it yields the solution of the textbook $d_1=1$, $d_2=-1$, then I would at least know that my interpretation of the program is most likely correct and could focus on checking the calculations.

答: Your transformation is correct.  Both $(d_1,d_2,y_1,y_2,z)=(1,-1,0,1,-1)$ and $(d_1,d_2,y_1,y_2,z)=(1,1,0,0,-1)$ are optimal, and so is every point on the line segment that joins them.

ContentLicense: CC BY-SA 4.0


问:
A basic question on the derivative of a continuous function
Is the following condition necessary for the existence of derivative of a continuous function at point $x$: </p>

<p>$$\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^-}\frac{f(x)-f(x+h)}{h}$$</p>

答:
No. The readily available counterexample is $f(x)=x$ which though differentiable at $0,$ </p>

<p>$$\displaystyle\lim_{h\to0+}\frac{f(h)}{h}=1\\\displaystyle\lim_{h\to0-}\frac{f(h)}{-h}=-1$$</p>

ContentLicense: CC BY-SA 3.0

Body:
No. The readily available counterexample is $f(x)=x$ which though differentiable at $0,$

$$\displaystyle\lim_{h\to0+}\frac{f(h)}{h}=1\\\displaystyle\lim_{h\to0-}\frac{f(h)}{-h}=-1$$


问:
Question about the Limit of a Sum of a Series
I am a Calculus II student and we have been learning about series, summation, factorials, etc. I thought about this while I was in the shower and I cannot seem to get it out of my head.

What would this ever be equal to?

$\lim_{p\to \infty}\lim_{t\to p^-}\sum_{n=0}^t \frac{1}{(p-n)!}$

For the first real terms of $n$, it is clearly converged on 0. However, as $t$ approaches $p$ from the left side, the denominator begins to approach $0$. This would indicate that we are now adding terms that approach 1. For instance...

When $t=(p-1)$, the term is $=\frac{1}{p-(p-1)}=\frac{1}{p-p+1}=\frac{1}{1}=1$

Does it converge? Does it diverge? What would this series even look like as $n$ approaches $p^-$?

Body:
$$\lim_{p\to \infty}\lim_{t\to p^-}\sum_{n=0}^t \frac{1}{(p-n)!}
 = \lim_{p\to \infty}\sum_{n=0}^p \frac{1}{(p-n)!} = \lim_{p\to \infty}\sum_{n=0}^p \frac{1}{n!} = e$$


问:
If $A$ is a non empty subset of the metric space ($\mathbb{R}$,d) and $A$ = $A'$, then $A$ is uncountable.
I have to prove this proposition "If $A$ is a non empty subset of the metric space ($\mathbb{R}$,d) and $A$ = $A'$, then $A$ is uncountable. " but I dont know how. Can anybody help me ?

答:
The set A cannot be finte as every element in that  finte set is an isolated point.Now if A is countably infinite then A can be written as a countable union of closed sets (since every singleton set is closed).Now A being a closed set of a complete space  is complete.But by Baire category th then at least one closed set must have non -empty interior but singleton set has empty interior .
So A must be uncountable.</p>


问:
$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x\log \pi + (n-x)\log(1-\pi)\;\;?$
$$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x \log \pi + (n-x)\log(1-\pi)$$  this is what i have.

i dont understand how $\binom{n}{x}$ disappears, but the rest is fine.

I tried this, but it is getting complicated. Is there some way I can do things like this quickly and at the first glance?

$$\begin{align}\log\left(\binom{n}{x}\right) & = \log\left(\dfrac{n!}{x!(n-x)!}\right)\\ \\
& =\log(n!)-\log\left[x!(n-x)!\right] \\ \\
& = \log(n!)-\left[\log(x!)+\log(n-x)!\right] \\ \\
& = \;\; \cdots
\end{align}$$

答:
so we use the property

$\log abc=\log a+\log b+\log c$

$\log(\left(\begin{matrix}n \\ x\end{matrix}\right)\pi^x(1-\pi)^{n-x})$

$=\log\left(\begin{matrix}n \\ x\end{matrix}\right)+\log \pi^x+\log(1-\pi)^{n-x}$

$=\log\left(\begin{matrix}n \\ x\end{matrix}\right)+x\log \pi+(n-x)\log(1-\pi)$

$\log\left(\begin{matrix}n \\ x\end{matrix}\right)=\log \Gamma(n+1)-\log\Gamma(x+1)-\log\Gamma(n-x+1)$


问:
On Kunen's Set Theory, Exercise II.9.3
The exercise is as follows:

Suppose that $F : \mathbf{V} \to \mathbf{V}$ is a permutation of $V$ and $E = E_F$. Define $i : \mathbf{V} \to \mathbf{V}$ by recursion on $\in$ so that $i(x)$ is the $y$ such that $F(y) = \{i(w) : w \in x\}$. Prove that $i$ is $1$-$1$, and is an isomorphism between $(\mathbf{V},\in)$ and $(\mathbf{WF}^{(V,E)},E)$.

Also, $\mathsf{AC} \leftrightarrow \mathsf{AC}^{(\mathbf{V},E)}$.

Also, the model $(\mathbf{V},E)$ satisfies the axiom that every relation $R$ on a set $A$ is isomorphic to a relation on a well-founded set.

Note that here, $x \, E_F \, y$ iff $x \in F(y)$.

I'm unable to comprehend how does $i$ work or "look like". Intuitively, I try to think of $i$ as an embedding (as hinted in the previous paragraph), so $i(x)$ should be a set that "contains" (in the sense of $E$) all $i(w)$'s, where $w \in x$. But this does not help me in proving the claims in the exercise.

Thanks in advance.

答:
The idea is that for any set in $(V,\in)$ we can construct an isomorphic copy of it in $(V,E)$ using $\in$-recursion. The isomorphic copy of $x$ is $i(x).$ First we construct the empty set, which is the set with no $E$-elements. This is just $F^{-1}(\emptyset),$ so $i(\emptyset) = F^{-1}(\emptyset).$ Then, if we have constructed all the elements of a given set and want to construct the set itself, we just take $F^{-1}(\{i(w): w\in x\}),$ which is the set whose $E-$elements are the isomorphic copies of all the elements of $x.$ So the first part is just to prove that this works as advertised, and that furthermore every well-founded set in $(V,E)$ is in the range of $i$ (think about the Mostowski collapse of the relation $E$ on the given set).

As for the other parts, it is good to think about what a relation in $(V,E)$ looks like. A relation $R$ on $X$ should be a set whose elements are ordered pairs of elements of $X,$ but what are ordered pairs? Write $$ (x,y)^E = F^{-1}\{F^{-1}(\{x\}), F^{-1}(\{x,y\})\},$$ i.e. (according to $(V,E)$) the set whose elements are the set whose only element is $x$ and the set whose elements are $x$ and $y$. So $R$ is a relation on $X$ means that $F(R)$ is a set consisting of elements of the form $(x,y)^E$ where $x,y\in F(X).$

For any relation $R$ on $X$ (according to $(V,E)$), we can find a corresponding relation (according to $(V,\in)$) on $F(X)$ $$ \tilde R = \{(x,y): (x,y)^E \in F(R)\},$$ and we know from the first part that $(V,E)$ has an isomorphic copy of this, $i(\tilde R),$ which according to $(V,E)$ will be a relation on a well-founded set that is isomorphic to $R.$

Title: On Kunen's Set Theory, Exercise II.9.3


问:
Proving $\lim_{n \to \infty} \frac{1-e^{-n}}{2} = \frac{1}{2}$
I attempted to prove this using the definition of convergence to a real number. I am new to proofs and would like to know whether this is valid?

Also, is there a general rule on how much information needs to be specified in proofs? Thanks

I am new to posting questions on this site so if I have posted incorrectly please let me know.

https://i.stack.imgur.com/wgxl3.jpg

--

Update: added Mathjax text

$
\text { Prove } \lim _{n \rightarrow \infty} \frac{n-1}{n} \rightarrow 1 \\
\text { To show: } \forall \varepsilon>0 . \exists N \in \mathbb{N} \text { .st. } \forall n>N . \\
\left|\frac{n-1}{n}-1\right|<\varepsilon \\
\left|\frac{n-1}{n}-\frac{n}{n}\right|<\varepsilon \\
\left|-\frac{1}{n}\right|<\varepsilon \\
\frac{1}{n}<\varepsilon \\
 n>\frac{1}{\varepsilon} \\
\text {take } N=\left\lceil\frac{1}{\varepsilon}\right\rceil
$

答:
By the definition of a limit,

$$
\lim\limits_{n\rightarrow +\infty}a_n = a, \text{ if }\forall \varepsilon > 0, \exists N\in \mathbb{N}\text{ s.t. }\forall n > N, |a_n - a| < \varepsilon.
$$

Let's take $\varepsilon > 0$.

Let's consider

$$
\left|\frac{1-e^{-n}}{2}-\frac{1}{2}\right| = \left|\frac{-e^{-n}}{2}\right| = \frac{e^{-n}}{2} < \varepsilon \Rightarrow n > \ln\left(\frac{1}{2\varepsilon}\right),
$$

which means that $N$ can be taken as an integer part of $\ln\left(\frac{1}{2\varepsilon}\right)$, i.e. $N = \left[\ln\left(\frac{1}{2\varepsilon}\right)\right]$.

Now, we can conclude that

$$
\lim\limits_{n\rightarrow +\infty}\frac{1-e^{-n}}{2}=\frac{1}{2},
$$

because $\forall \varepsilon > 0, \exists N = \left[\ln\left(\frac{1}{2\varepsilon}\right)\right]\text{ s.t. }\forall n > N, \left|\frac{1-e^{-n}}{2}-\frac{1}{2}\right| < \varepsilon.$


问:
Show that any convex function is locally bounded
Show that a convex function $f:\mathbb{R}^n \rightarrow \overline{\mathbb{R}}$ is bounded in a neighborhood of $x\in \text{ri}(\text{dom}(f))$. Showing that it has an upper bound is not difficult using Jensen's inequality. To show that it has a lower bound I showed that if it wasn't bound in $B(x,\delta)$, then the epigraph in this region would be $B(x,\delta)\times \langle-\infty,+\infty \rangle$, but to do this I made a weird assumption which might or might not be true. If $f$ didn't had a lower bound in  $B(x,\delta)$, then for every $M\in \mathbb{R}$ it exists an $x_M$ such that $f(x_M)<M$. Naturally $(x_M,M)\in \text{epi}(f_{|B(x_0,\delta)})$ and since $\text{epi}(f_{|B(x_0,\delta)})$ is convex for any $(x,\lambda) \in \text{epi}(f_{|B(x_0,\delta)})$ it would follow that $(tx+(1-t)x_M,t\lambda+(1-t)M)\in \text{epi}(f_{|B(x_0,\delta)})$ for any $t\in ]0,1[$. My assumption is that with this convex combination I can reach any point in $B(x,\delta)\times \langle-\infty,+\infty \rangle$. Any help regarding my assumption or another way to show  that there's a real lower bound would be welcome, thanks in advance.

For this problem $\overline{\mathbb{R}}=\mathbb{R}\cup\{+\infty\}$,$\text{dom}(f)=\{x\in \mathbb{R}^n/ f(x)<+\infty\}$ and $\text{ri}(C)$ is the relative interior of $C$.

After reading some books I found that not only is the function bound but it's uniformly continuous, the proof wasn't that simple. I feel there must be an easy way to find the lower bound maybe showing that it's lower semicontinuous at a neighborhood of $x$?

答:
Here is one way to prove that $f$ is lower semicontinuous at every point $x$ in the relative interior of its effective domain.

Get rid of "relative" by restricting attention to an appropriate affine subspace.

You already know that $f$ has an upper bound $M$ in some closed neighborhood $\overline B(x,\delta)$.

Suppose, to the contrary, that there is $\epsilon>0$ and a sequence $x_n\to x$ such that $f(x_n)\le f(x)-\epsilon$ for all $n$.

Let $y_n=x-\delta\dfrac{x_n-x}{\|x_n-x\|}$. Observe that $x=(1-\lambda_n)x_n+\lambda_n y_n$ where $\lambda_n=\dfrac{\|x_n-x\|}{\|x_n-x\|+\delta}\to 0$ as $n\to\infty$.

By virtue of convexity, $f(x)\le (1-\lambda_n)( f(x)-\epsilon)+\lambda_n M$. Let $n\to\infty$ to reach a contradiction.


问:
combinatorial argument that $ [\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}]^{2} = \sum_{k=0}^{2n}\binom{2n}{k}$

Give a combinatorial argument with double counting showing that
$$ \Bigg[\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}\Bigg]^{2} = \sum_{k=0}^{2n}\binom{2n}{k}$$

I am unsure on how to approach this problem from a combinatorial argument perspective. I have found an analytical proof using the binomial theorem, but can't formulate an explanation in the former way.

From my understanding, $\displaystyle \sum_{k=0}^{n} {n \choose k}$ represents the number of ways we can chose a set of $k$ objects from a group of $n$ objects, for $k \le n$. So this quantity squared should be equivalent the number of ways to pick a set of $k$ objects from a group of $2n$ objects.

Could anyone provide me any hints on how to approach this problem combinatorically ?

答:
Both sides describe the number of ways to select any number of elements from $2n$ elements (and are both equal to $2^{2n}$)

Let's write $A=\{1,\dots,n\}$ and $B=\{n+1,\dots,2n\}$.

Now the subsets of $A\cup B$ are in bijection to the pairs of a subset of $A$ and a subset of $B$.

$$\mathcal{P}(A\cup B) \to \mathcal{P}(A)\times \mathcal{P}(B)$$
$$S \mapsto (S\cap A, S \cap B)$$

Also
$$|\mathcal{P}(A\cup B)| = \sum_{k=0}^{2n} |\{S\subseteq A\cup B \mid |S|=k\}| = \sum_{k=0}^{2n} \binom{2n}{k}$$
$$|\mathcal{P}(A)| = \sum_{k=0}^{n} |\{S\subseteq A \mid |S|=k\}| = \sum_{k=0}^{n} \binom{n}{k}$$
$$|\mathcal{P}(B)| = \sum_{k=0}^{n} |\{S\subseteq B \mid |S|=k\}| = \sum_{k=0}^{n} \binom{n}{k}$$

So
$$RHS = |\mathcal{P}(A\cup B)| = |\mathcal{P}(A)||\mathcal{P}(B)| = LHS$$


问:
Integral of $\cos(\theta_1-\theta_2)/\sqrt{1-\cos(\theta_1-\theta_2)}$
I am trying to see if the following integral converges, and if it has a simple solution

$\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{\cos(\theta_1-\theta_2)}{\sqrt{1-\cos(\theta_1-\theta_2)}}d\theta_1d\theta_2$.

So far, the integral is eluding me. At first I thought about writing it as a Fourier series, but this does not help me solve the integral.

Any help or guidance would be appreciated.

答: Mathematica says the integral converges to $(2\sqrt{2}\pi^2)i$. So it probably involves a complex substitution. I'm working on it. I will let you know if I figure it out.


问:
If two functions f,g are both discontinuous at some point a in the domain, does this mean that the sum f+g is also discontinuous at a?
Real analysis question. Having trouble either coming up with a counterexample where the sum would be continuous at a, or proving that the sum would be discontinuous at a for all cases.</p>

答:
Pick $$f(x)=
\begin{cases}1 & \text{if } x \neq 0, \\
0 & \text{if } x =0.
\end{cases}$$

and

$$g(x)=
\begin{cases}0 & \text{if } x \neq 0, \\
1 & \text{if } x =0.
\end{cases}$$


问:
Minimum formation sequence length for a logic formula equals number of of subformulas
I am trying to prove the following statement:

Let $SF(\varphi)$ be the group of subformulas of formula $\varphi$, defined as follows:

If $\varphi=p$ is atomic variable, $SF(\varphi)=\{p\}$

If $\varphi=\neg\alpha$ then, $SF(\varphi)=\{\neg\alpha\} \cup SF(\alpha)$

If $\varphi=\alpha \circ \beta$, then $SF(\varphi)=\{\alpha \circ \beta\} \cup SF(\alpha)\cup SF(\beta)$

Theorem:

Let $\varphi$ be a formula, then $|SF(\varphi)|$ = minimum formation sequence length for $\varphi$.

(a formation sequence is defined in this post for example: Formation sequence for a logic formula)

I tried to prove it using structural induction on a formula $\varphi$, but I failed to show the induction step for binary connective. I managed only to achieve the following bound:
$|SF(\varphi)|=|\{\alpha \circ \beta\} \cup SF(\alpha)\cup SF(\beta)|\leq1+| SF(\alpha)| +  |SF(\beta)|$.

Please advise.

Thank you in advance.

答:
There is indeed some trickiness in the case of a binary connective.  The issue is that $SF(\alpha)$ and $SF(\beta)$ might contain some elements in common, and so the obvious formation sequence of length $1+|SF(\alpha)|+|SF(\beta)|$ you can get by concatenating formation sequences for $\alpha$ and $\beta$ might actually be longer than $|SF(\varphi)|$.  The idea to fix this is that any common subformula of $\alpha$ and $\beta$ will appear both in the formation sequence for $\alpha$ and the formation sequence for $\beta$, and so when you concatenate the formation sequences you can delete the duplicate entries (since they are redundant) and end up with a sequence of length just $|SF(\varphi)|$.

There are various ways you can make this idea formal.  For instance, you could first prove that any formation sequence for $\varphi$ must contain all subformulas of $\varphi$.  Then, you can prove by induction on $\varphi$ that there exists a formation sequence for $\varphi$ in which every term of the sequence is a subformula of $\varphi$.  Then, you can take such a sequence and remove each term which is a duplicate of a previous term, and this will give you a formation sequence of length $|SF(\varphi)|$ since every subformula will appear exactly once.


问:
Rewrite statement using symbols
I have a statement: Every real number has a unique multiplicative inverse.

My answer is ($\forall x \in \mathbb{R}$)($\exists! y \in \mathbb{R}$)($xy = 1$). Is this correct? Thank you! Please do not care about the truth value of the statement, I know it is wrong because x must be nonzero.
I was not sure about the unique part.

答:
It looks good, although I'm not sure about the way you use parentheses.  I tend to use commas:$$\forall x\in\mathbb{R},\exists!y\in\mathbb{R},xy=1.$$
The "exists unique" quantifier could also be expanded out, depending on what you're going to do with this formalized statement:
$$\forall x\in\mathbb{R},\exists y\in\mathbb{R},xy=1\wedge(\forall y'\in\mathbb{R},xy'=1\to y'=y).$$

Title: Rewrite statement using symbols


答: Take any fonction $f$ that is discontinuous at a point $x$. Then so is $g = -f$. But $f+g = 0$ is continuous everywhere.

Title: If two functions f,g are both discontinuous at some point a in the domain, does this mean that the sum f+g is also discontinuous at a?

Body: Real analysis question. Having trouble either coming up with a counterexample where the sum would be continuous at a, or proving that the sum would be discontinuous at a for all cases.</p>


问:
Limit of floor function and sqrt
Is anyone able to help me with the following limit question
$$\lim_{x \to +\infty} \frac{ \left\lfloor \sqrt{x}  \right\rfloor  }{ \sqrt{x} } $$

答: Hint: use the fact that for any $y$, $y - 1 < \lfloor y \rfloor \le y$.

Title: Limit of floor function and sqrt


问:
Show that the polynomial $p$ that is defined as a determinant is linear in $x$
We have a matrix:

We have a polynomial defined as:

We need to show that this polynomial is linear in $x$:

What I tried:
1.) $(p+q)(x) = p(x)+q(x)$ .... here I do not know  how to separate the sum of the determinant, I refreshed my memory of properties of the determinants, and there is no property, where I could just split $x$ and $a_{ij}$.

2.) $(\alpha p)(x) = \alpha p(x)$ ... I think we also need to show this as the second property of linearity.

答:
Simply subtract the first row from every other row. This operation doesn't change the determinant.

After that, there is only $x$ in the first row. You then expand the determinant with respect to the first row.


问:
How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$
With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?

$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$

Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count different cases:

first: having all $1,2,3,4$ digits and choose another digit from $1,3,4$:
$${3\choose1}\times\frac{5!}{2!}$$
second: having two pair of same digit and choose another digit:
$${3\choose2}\times2\times\frac{5!}{2!2!}$$
Summing them we have $3\times\dfrac{5!}{2}+3\times\dfrac{5!}{2}=5!\times3$ but I don't have this in the options.

答:
Either you don't select $2$ or you select $2$. In the first case you have 3 choices: $(2, 2, 1); (2, 1, 2); (1, 2, 2)$. For each such choice you have $\frac{5!}{2!2!}$ allocations. If you select $2$, you have two paths:$(2, 2, 0), (2, 0, 2), (0,2,2)$, in which case you have $\frac{5!}{2!2!}$ allocations and $(1, 1, 2), (1,2,1),(1,1,2)$, in which case you have $\frac{5!}{2!}$ allocations. Putting it together,
$$
\frac{5!3}{2!2!} + \frac{5!3}{2!2!} +\frac{5!3}{2!} = 5!3
$$


问:
Discrete Bracket and convergence of a martingale
Let's assume that we have a discrete martingale $(M_n)_{n\in\mathbb{N}}$ with respect to a filtration $(\mathcal{F}_n)_{n\in\mathbb{N}}$. It takes only real values.
We define the discrete bracket $\langle M\rangle_n=\sum\limits_{i=1}^{n-1} \mathbb{E}(M_{k+1}^2-M_k^2|\mathcal{F}_k)$.

Of course, if $\langle M\rangle_n$ is bounded in $L^2$ then $M_n$ converges in $L^2$.
However if we only assume that $\langle M\rangle_n$ converges almost surely toward a finite random variable, is it true that $M_n$ converges almost surely?

If it is true in the discrete case, is it true in the continuous case?

答: Yes, this is true in both the discrete and continuous setting, assuming $M$ is continuous.  Let $\tau_m := \inf \{t : \langle M \rangle_t > m\}$ and $\lim_{t \rightarrow \infty} \langle M \rangle_t = \langle M \rangle_\infty$.  Then $\langle M \rangle^{\tau_m}$ is bounded and in particular in $L^2$ so $\lim_{t \rightarrow \infty}M_t^{\tau_m}$ exists almost surely.  But since for almost every $\omega$ we have $\langle M \rangle_\infty(\omega) < \infty$, we know $\tau_m(\omega) =\infty$ for all $m$ sufficiently large, so $\lim_{t \rightarrow \infty} M_t$ exists as well.


答:
As $e^{-n}$ decreases to $0$, this term vanishes and the limit is obviously $\dfrac12$. Now you have to show that the deviation from the limit, $\left|\dfrac{1-e^{-n}}2-\dfrac12\right|=\dfrac{e^{-n}}2$ can be made as small as you want ($<\epsilon$) for all $n$ sufficiently large.

If you find $N$ such that $\dfrac{e^{-N}}2<\epsilon$, it is clear that $n>N\implies\dfrac{e^{-n}}2$ because the negative exponential is a decreasing function. Remains to show that $N$ can always be found, for any $\epsilon>0$.

Using logarithms, we find that $N\ge-\log(2\epsilon)$ will work, and is always possible.

You can also reason without logarithms. Take a number above $\dfrac1{2\epsilon}$, which has $d$ digits. Now if you divide $d$ times by $e^3>10$, the result will certainly be less than $1$. Hence $N=3d$ will do.

Title: Proving $\lim_{n \to \infty} \frac{1-e^{-n}}{2} = \frac{1}{2}$

Body:
I attempted to prove this using the definition of convergence to a real number. I am new to proofs and would like to know whether this is valid?

Also, is there a general rule on how much information needs to be specified in proofs? Thanks

I am new to posting questions on this site so if I have posted incorrectly please let me know.

https://i.stack.imgur.com/wgxl3.jpg

--

Update: added Mathjax text

$
\text { Prove } \lim _{n \rightarrow \infty} \frac{n-1}{n} \rightarrow 1 \\
\text { To show: } \forall \varepsilon>0 . \exists N \in \mathbb{N} \text { .st. } \forall n>N . \\
\left|\frac{n-1}{n}-1\right|<\varepsilon \\
\left|\frac{n-1}{n}-\frac{n}{n}\right|<\varepsilon \\
\left|-\frac{1}{n}\right|<\varepsilon \\
\frac{1}{n}<\varepsilon \\
 n>\frac{1}{\varepsilon} \\
\text {take } N=\left\lceil\frac{1}{\varepsilon}\right\rceil
$


问:
Could it be generally proved that some positive integer is not of the form $x+y+2xy$?
I believe that the answer to the question is negative, because if it were possible, we would be generally able to tell if some positive integer is prime or not using the sieve of Sundaram, but I would like to know additional arguments (or counterarguments!) supporting a negative answer.

To be more especific, with "generally" I mean using some general method to show that some given positive integer is not of the form mentioned. The question could be rephrased as how to find the positive integers which are not solutions to the equation $x+y+2xy=z$, being $x,y,z$ positive integers.

Thanks in advance!

答:
If $n=x+y+2xy$, then we have that

$$2n+1=1+2x+2y+4xy$$

$$2n+1=(2x+1)(2y+1)$$

It's kind of like completing the square. Now we want $2n+1$ to be the product of two integers greater than one, otherwise x or y is zero. That is possible if and only if $2n+1$ is not prime. Luckily since $2n+1$ is odd all its factors are odd so you can always select $x,y$ for any factor. Therefore $n$ is of the form $x+y+2xy$ if and only if it is not of the form $\frac{p-1}2$ where $p$ is a prime.


答:
Let's say we have $n$ pizzas and they either have cheese or don't have cheese. Each pizza can have cheese or no cheese, $2$ choices for $n$ pizzas, so $2^n$ ways. However, you can also calculate it this way. First, find the number of ways to have $0$ no cheese and $n$ cheese. We can choose $0$ pizza to be no cheese, so $n \choose 0$. Also, we can have $1$ no cheese and $n-1$ cheese, so choose $1$ pizza to have $1$ no cheese out of $n$ pizzas, thus we get $n \choose 1$. See where we are going? We keep on calculating through and eventually get the sum ${n \choose 0} + {n \choose 1} + \dots + {n \choose {n-1}} +{n \choose n}$, and we know this is equal to $2^n$, so this sum squared should be $2^{2n}$

Now look at the right side. Each pizza can have cheese or no cheese, $2$ choices for $2n$ pizzas, so $2^{2n}$ ways. However, you can also calculate it this way. First, find the number of ways to have $0$ no cheese and $2n$ cheese. We can choose $0$ pizza to be no cheese, so $2n \choose 0$. Also, we can have $1$ no cheese and $2n-1$ cheese, so choose $1$ pizza to have $1$ no cheese out of $2n$ pizzas, thus we get $2n \choose 1$. See where we are going? We keep on calculating through and eventually get the sum ${2n \choose 0} + {2n \choose 1} + \dots + {2n \choose {n-1}} +{2n \choose n}$, and we know this is equal to $2^{2n}$.

So, we have shown both sides equal to $2^{2n}$, thus we are done.

Title: combinatorial argument that $ [\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}]^{2} = \sum_{k=0}^{2n}\binom{2n}{k}$

Body:
Give a combinatorial argument with double counting showing that
$$ \Bigg[\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}\Bigg]^{2} = \sum_{k=0}^{2n}\binom{2n}{k}$$

I am unsure on how to approach this problem from a combinatorial argument perspective. I have found an analytical proof using the binomial theorem, but can't formulate an explanation in the former way.

From my understanding, $\displaystyle \sum_{k=0}^{n} {n \choose k}$ represents the number of ways we can chose a set of $k$ objects from a group of $n$ objects, for $k \le n$. So this quantity squared should be equivalent the number of ways to pick a set of $k$ objects from a group of $2n$ objects.

Could anyone provide me any hints on how to approach this problem combinatorically ?


问:
Mean Value Theorem related problem
Let $f(x)\leq g(x)$ for all $x \in I$, where $I$ is an interval $\subseteq$ R. Also, let $f(c) = g(c)$ for some $c \in I$ but not an endpoint.
Prove that $f'(c) = g'(c)$ (assume differentiablity)

I have tried the Mean Value Theorem, let I = [a,b]. So for $f'(c) = g'(c)$ I will need to prove that $f(a) - f(b) = g(a) - g(b)$, but I am stuck at this. I have also tried the definition of derivative but I am still unable to go ahead.

Please give me some hints on this. Thank you!

答:
If we take the function $h(x) = g(x) - f(x)$

we see that $h(x) \ge 0 = h(c)$ for all $x \in I$

So $c$ is a point of local minimum for $h$ is in the interval $I$.

Since $c$ is internal point to $I$, and $h$ is differentiable at $c$

(and that's so because both $f$ and $g$ are differentiable at $c$),

it follows that $h'(c) = 0$ (there is a very basic theorem stating this).

Fermat's theorem (stationary points)

This theorem is more basic than the mean value theorem (MVT),

meaning it's usually proven before the MVT is proven (in real analysis courses).

So now we have $0 = h'(c) = f'(c) - g'(c)$

So it follows that $f'(c) = g'(c)$

So all in all, your problem is just a simple exercise of this theorem.

I don't think the problem is MVT related.

If you're interested in a proof, it's very simple

Proof of Fermat's theorem

Title: Mean Value Theorem related problem


答: Both sides count the number of subsets chosen from among $n$ men and $n$ women.  The LHS does so by counting the number of men and women independently, conditioning on the number $k$ of each.  The RHS does the count by conditioning on the number $k$ of people out of $2n$.


问:
If A and B are n×n matrices, then |A+B^T |=|A^T+B|
If $A$ and $B$ are $n×n$ matrices, then $|A+B^T|=|A^T+B|$

can I prove that it's t/f without giving a value to the matrices?

答:
Nice two-liner solves this for you:
$$\mathrm{det}(A) = \mathrm{det}(A^T)$$
$$\mathrm{det}(A+B^T) = \mathrm{det}([A+B^T])^T) = \mathrm{det}(A^T+B)$$

You can show $(A+B)^T = A^T + B^T$ in the middle there for a more complete proof.


问:
What are flimsy numbers?
I was reading the binary weight sequence information on OEIS (its number is A000120), when I found this sequence too. Their called the flimsy numbers, but I can't seem to find anything on google about them (minus one random email that is in reference to the OEIS). I don't see anything particularly interesting about these numbers. Is there anything of note about the flimsy numbers, or their counterpart the sturdy numbers?</p>

答:
It's been several years, but now that I've had some real math education I've been scrolling through my old questions and feel I'm equipped to actually read the OEIS article. According to the page, flimsy numbers are natural numbers $n$ such that

some multiple of n has fewer ones in its binary expansion than does n.

What makes these numbers interesting is noted in this quote also from the OEIS,

Stolarsky says "at least half the primes are flimsy."

The list of flimsy prime numbers is given by A330696.

Title: What are flimsy numbers?


问:
Prove the existence (or well-definedness) of the induced connection in tensor bundle
Given a connection $\nabla$ on a vector bundle $E$ over a smooth manifold $M$, we know there is a unique extension of $\nabla$ to all tensor bundles of $E$ that satisfies Leibniz rule and contraction. I am going to prove this.

We can first define the connection on the dual space of $E$ using the formula forced by the axioms above. Then the rest is essentially just proving one lemma:

Given two bundles $E$, $F$ on $M$ and connections on $E$ and $F$, both denoted by $\nabla$, then there is a unique connection $\nabla$ on $E\otimes F$ such that
$$\nabla_X(s_E\otimes s_F)=\nabla_X s_E\otimes s_F+s_E\otimes \nabla_X s_F$$

I tried to invoke the universal property of tensor product as usual, to use bilinearity to prove well-definedness. However this tensor product of sections is not the strict tensor product in linear algebra tense (it takes tensor product pointwise, and it cannot be understood as a tensor product of two $\mathbb R$-vector spaces. It can be tensor product of $C^\infty(M)$ modules though, but $\nabla_X$ is not $C^\infty$ linear). So I get in problem here, and I ask for a conceptual way to show why $\nabla_X$ is a well defined map from the space of sections of $E\otimes F$ to itself.

答:
It seems to me that people usually include the $\mathbb{R}$-linearity as part of the definition (correct me if I'm wrong); i.e. define $\nabla$ on the simple tensors by the product rule above, and then extend it linearly over $\mathbb{R}$ to the whole space $C^{\infty}(E\otimes F)$.

Actually, due to the fact that $C^{\infty}(E\otimes F)\simeq C^{\infty}(E)\otimes_{C^{\infty}(M)}C^{\infty}(F)$ which OP has mentioned in the comment above, it follows that additivity (i.e.
\begin{align}
\nabla_X(s\otimes t+s'\otimes t')=\nabla_X(s\otimes t)+\nabla_X(s'\otimes t')
\end{align}
which is weaker than $\mathbb{R}$-linearity) is sufficient.

With this in mind, here is an alternative proof, which is essentially a direct computation, though perhaps it may look less elegant.

Let $A\in C^{\infty}(E\otimes F)$. Such $A$ can be expressed as a finite sum of simple tensors, though not uniquely in general. So we want to show that if it can be written in the following two ways
\begin{align}
A=\sum_is_i\otimes t_i=\sum_j\tilde{s}_j\otimes\tilde{t}_j
\end{align}
(the number of summands in these two expressions may be different in general), then
\begin{align}
\nabla_X\left(\sum_is_i\otimes t_i\right)
=\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) & & (*)
\end{align}
Let $(e_{\alpha})$ and $(\epsilon_{\beta})$ be smooth local frames for $E$ and $F$ respectively. Then one can write
\begin{align}
s_i=s_i^{\alpha}e_{\alpha},\qquad
t_j=t_j^{\beta}\epsilon_{\beta},\qquad
\tilde{s}_i=\tilde{s}_i^{\alpha}e_{\alpha},\qquad
\tilde{t}_j=\tilde{t}_j^{\beta}\epsilon_{\beta}
\end{align}
(Einstein summation convention is assumed.) Then we will have
\begin{align}
\sum_is_i\otimes t_i
=\left(\sum_is_i^{\alpha}t_i^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta},
\qquad
\sum_j\tilde{s}_j\otimes\tilde{t}_j
=\left(\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Since $\{e_{\alpha}\otimes\epsilon_{\beta}\}$ is a smooth local frame for $E\otimes F$, by uniqueness of local components we must have
\begin{align}
\sum_is_i^{\alpha}t_i^{\beta}=\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta} & & (1)
\end{align}
Taking exterior derivative we then have
\begin{align}
\sum_i\left(t_i^{\beta}ds_i^{\alpha}+s_i^{\alpha}dt_i^{\beta}\right)
=\sum_j\left(\tilde{t}_j^{\beta}d\tilde{s}_j^{\alpha}
+\tilde{s}_j^{\alpha}d\tilde{t}_j^{\beta}\right) & & (2)
\end{align}
Now let $\omega_{\alpha}^{\beta}$ and $\theta_{\alpha}^{\beta}$ be the connection 1-forms of $\nabla^E$ and $\nabla^F$ respectively, so that e.g.
\begin{align}
\nabla_X s_i=\left[Xs_i^{\alpha}+s_i^{\beta}\omega_{\beta}^{\alpha} (X)\right]e_{\alpha} & & (3)
\end{align}
and similar identities hold for $\nabla_Xt_i$, $\nabla_X\tilde{s}_j$ and $\nabla_X\tilde{t}_j$. Then we can compute
\begin{align}
&\nabla_X\left(\sum_is_i\otimes t_i\right) \\
%%%
&=\sum_i\left(\nabla_Xs_i\otimes t_i+s_i\otimes\nabla_Xt_i\right) \\
%%%
&=\underbrace{\left(\sum_i\left((Xs_i^{\alpha})t_i^{\beta}
+s_i^{\alpha}(Xt_i^{\beta})\right)\right)}_{=:I}e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\gamma}t_i^{\beta}\right)}_{=:II}
\omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\alpha}t_i^{\gamma}\right)}_{=:III}
\theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
where the first step follows by additivity and product rule, while the second step is obtained by substitution of (3) and rearranging the terms.

Now $\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)$ will have the same expression, except that all of the $s_i,t_i$ are replaced by $\tilde{s}_j,\tilde{t}_j$; says
\begin{align}
\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)
=\tilde{I}\cdot e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{II}\cdot \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{III}\cdot \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Then:

By (2), we have $I=\tilde{I}$.

By (1), we have $II=\tilde{II}$ and $III=\tilde{III}$.

Hence, (*) holds as desired.


问:
Increasing and subadditivity implies a convex inverse exists
If $f$ is an increasing and subadditive function on the nonnegative reals such that $f(0) =0$, then $f$ has an inverse and furthermore $f^{-1}$ is convex.

How would I go about proving existence? My intuition is once existence has been proved I can use the fact that subadditivity implies concave and so the inverse must be convex. Is this correct?

答: For existence, you note that any strictly monotone function has an inverse function from its range to its domain. In this case, you know that the infimum of its range is $0.$ You do not know that the range is all of $[0, \infty]$ The rest of your argument is fine.


答:
You can combine both combinatorial and analytical reasoning as follows.
Given a set $\,A = \{a_1,a_2,\dots,a_n\}\,$ with $\,n\,$ elements, and
a subset $\,S\subseteq A,\,$ for each $\,i\in\{1,2,\dots,n\}\,$
the binomial $\,x_i+y_i\,$ represents the condition that $\,a_i\in S.\,$
That is, the choice $\,x_i\,$ indicates that $\,a_i\in S\,$ and $\,y_i\,$
indicates that $\,a_i\notin S.\,$ Using the multiplicative principle of
independent choices, the product $\,P_A:=(x_1+y_1)(x_2+y_2)\cdots(x_n+y_n)\,$
expands into a sum of monomials which represents all subsets of $\,A.\,$
If $\,B=\{b_1,b_2,\dots,b_m\}\,$ is another set disjoint from $\,A,\,$
and the disjoint union $\,C=A\cup B,\,$ then all the subsets of $\,C\,$
is represented by the product $\,P_C = P_AP_B\,$ where each subset
$\,S\subseteq C\,$ is partitioned uniquely as $\,S=S_A\cup S_B.\,$
Here $\,x_{i+n}\,$ indicates that $\,b_i\in S_B\,$ and $\,y_{i+n}\,$
indicates that $\,b_i\notin S_B.\,$

Given this result, consider the special case where $\,x_i=x,\, y_i=y\,$
for all $\,i\,$. Then
$$ P_A = (x+y)^n = \sum_{j=0}^n x^jy^{n-j}{n\choose j},$$
$$ P_B = (x+y)^m = \sum_{k=0}^m x^ky^{m-k}{m\choose k},$$
$$ P_C = (x+y)^{n+m} = \sum_{i=0}^{n+m} x^iy^{n+m-i}{n+m\choose i}. $$
When $\,n=m\,$ and $\,x=y=1\,$ your result follows.


问:
$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x\log \pi + (n-x)\log(1-\pi)\;\;?$
$$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x \log \pi + (n-x)\log(1-\pi)$$  this is what i have.

i dont understand how $\binom{n}{x}$ disappears, but the rest is fine.

I tried this, but it is getting complicated. Is there some way I can do things like this quickly and at the first glance?

$$\begin{align}\log\left(\binom{n}{x}\right) & = \log\left(\dfrac{n!}{x!(n-x)!}\right)\\ \\
& =\log(n!)-\log\left[x!(n-x)!\right] \\ \\
& = \log(n!)-\left[\log(x!)+\log(n-x)!\right] \\ \\
& = \;\; \cdots
\end{align}$$

答: This is only true when $x=0$ or $x=n$. Otherwise the final term does not disappear.</p>


问:
How can I prove that every word has a unique representation?
Let $\Sigma = \{0,1,*\}$ be a finite alphabet and let $\mathcal{L}\subseteq \Sigma^*$ be a formal language defined as follows

$\{0,1\}\subseteq \mathcal{L}$.

If $w_1,\dots,w_n\in\mathcal{L}$ then $*w_1w_2\dots w_n*\in\mathcal{L}$

Nothing more is a an element of $\mathcal{L}$.

I want to prove that

If $*w_1w_2\dots w_n*=*v_1v_2\dots v_m*$ where $w_i, v_j \in \mathcal{L}$ then $n=m$ and $w_i=v_i$ for
all $i$.

I was trying to use the fact that every element of $\mathcal{L}$ has an even number of $*$ but maybe I am missing something.

答:
Hint: if you're at all familiar with logic, this proof is (perhaps superficially) very similar to a standard proof of the unique readability theorem in propositional calculus.

So, the first step is going to be to refine our definitions slightly. Your guess about using the even parity of the number of $*$ in each word of $\mathcal{L}$ is correct, and there is a pattern behind it. As a matter of fact, since the stars come in pairs, we can distinguish between left and right stars, so from now on instead of, for example, $*0*$ I am going to write $*_l 0 *_r$ (this is ungainly notation, but is highly illustrative for the proof). So, given a word $w\in \mathcal{L}$ we are going to write $l(w)$ for the number of left stars in $w$, and $r(w)$ for the number of right stars, accordingly. In particular, it should be easy to see that if $w$ is a word, then $r(w)=l(w)$.

Thus begin with a lemma: let $w\in \mathcal{L}$, and let $x$ be a non-empty proper left subword of $w$. Then $l(x)>r(x)$.

Proof: we prove this by induction on the length of $w$. The base case is trivial. So assume the claim holds for all $w$ up to length $n$. Consider a word $w$ with length $n+1$. If $w$ is of the form $*_l\ldots *_l\{0,1\}*_r\ldots *_r$, then we're clearly done as for any proper left subword $x$ we have $l(x)>r(x)$, and hence it's not a word in $\mathcal{L}$. So suppose more generally $w=*_lv_1v_2\ldots v_m*_r$, where $v_i\in \mathcal{L}$, and let $x$ be a proper left subword of $x$. Clearly, depending on how many $v_i$ are included in $x$ (and not forgetting the very first $*_l$!), we have $l(x) = 1 + l(v_1)+l(v_2)+\ldots + l(y_k)$, where $y_k$ is a left subword of $v_k$. By the induction hypothesis we have $$\begin{align} l(x) & \geq 1 + r(v_1) + r(v_2) + \ldots + r(y_k) \\ & > r(v_1) + r(v_2) + \ldots + r(y_k) \\ &= r(x).\end{align}$$ This concludes the proof. I hope you see where this is going. We can now tackle the main question.

Let $w:= *_lw_1w_2\ldots w_n*_r$ and $v:=*_lv_1v_2\ldots v_m*_r$ be equal words in the language $\mathcal{L}$. We will try to derive a contradiction. Suppose there exists $i$ such that $w_i\neq v_i$. Pick the smallest such $i$ (since there might has to be more than one), and assume W.L.O.G. that $w_i$ is a proper left subword of $v_i$ (again, this assumption is tenable because $w$ and $v$ are identical symbol-wise). But since $v_i\in \mathcal{L}$, it follows by the lemma I had just proved that $l(w_i)>r(w_i)$ and hence $w_i\notin \mathcal{L}$. This provides the required contradiction, and so we are done.


问:
Suppose the 10% of apples in a supermarket are rotten. The manager checks the stock of apples, by inspecting them one at a time.
What is the probability that the 7th apple he inspects is the rotten one he finds? What distribution does the variable "number of inspected apples until the first rotten one" follow?</p>

答:
Did you mean what are the odds that the 7th apple he inspects is the first rotten one?

If you have unlimited apples,

$$P(R=7) = \left(\frac{9}{10}\right)^6 * \frac{1}{10}$$
Which is around 0.05.

Check out a geometric distribution - defined as the distribution of the number of Bernoulli trials to get one success. Generally for the first rotten apple to be the kth apple:

$$P(X=k) = (1-p)^{k-1}p$$


问:
Proving $\lim_{n \to \infty} \frac{1-e^{-n}}{2} = \frac{1}{2}$
I attempted to prove this using the definition of convergence to a real number. I am new to proofs and would like to know whether this is valid?

Also, is there a general rule on how much information needs to be specified in proofs? Thanks

I am new to posting questions on this site so if I have posted incorrectly please let me know.

https://i.stack.imgur.com/wgxl3.jpg

--

Update: added Mathjax text

$
\text { Prove } \lim _{n \rightarrow \infty} \frac{n-1}{n} \rightarrow 1 \\
\text { To show: } \forall \varepsilon>0 . \exists N \in \mathbb{N} \text { .st. } \forall n>N . \\
\left|\frac{n-1}{n}-1\right|<\varepsilon \\
\left|\frac{n-1}{n}-\frac{n}{n}\right|<\varepsilon \\
\left|-\frac{1}{n}\right|<\varepsilon \\
\frac{1}{n}<\varepsilon \\
 n>\frac{1}{\varepsilon} \\
\text {take } N=\left\lceil\frac{1}{\varepsilon}\right\rceil
$

答:
$n \ge 1:$

Recall: $e^n=1+n+n^2/2! +.... >n$

$f(n):=|\dfrac{1-e^{-n}}{2}-1/2|=$

$\dfrac{e^{-n}} {2}= \dfrac{1}{2e^n} <\dfrac{1}{n}.$

Let $\epsilon >0$ be given.

Archimedean Principle:

There is a $N \in \mathbb{N}$ s.t.

$N >1/\epsilon.$

For $n \ge N$

$f(n) <\dfrac{1}{n} \le \dfrac{1}{N} <\epsilon. $


问:
About nilpotent groups of orders $1$ to $59$ up to isomorphism

Find the number of all nilpotent groups of order $<60$, up to isomorphism - i.e. for every $n \in \{1,2,\ldots,59\}$, find the number of nilpotent groups up to isomorphism.

We know that

Result 1: Every abelian group is nilpotent.

Result 2: Every finite $p$-group is nilpotent.

Hence, all abelian groups of orders $1$ to $59$ are nilpotent. Additionally, all groups of orders $2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41,43,47,49,53,59$ are nilpotent, due to Result 2.

So the only groups left to worry about, are non-abelian groups of non-prime (or prime power) order. This feels really random, and cumbersome. Is there a systematic approach to listing all the desired nilpotent groups? Thanks!

答:
For a prime $p$ there is one group of order $p$, and two (cyclic and $p\times p$) of order $p^2$.
Also (and this is the part that is hard in that hand calculation will require days or more) there are $5$ groups each of orders $8$ and $27$, $14$ groups of order $16$ and $51$ groups of order $32$. With this information you can calculate the numbers of nilpotent groups of these orders as products of the numbers for the dividing prime powers.  For example, the number of nilpotent groups of order $12=4\cdot 3$ is the number of groups of order $4$ (which is $2$ because of the $p^2$ rule) times the number of groups of order $3$ (which is one) for a total of $2\cdot 1=2$.


问:
Number of ways of spelling Abracadabra in this grid
This is embarrasingly, the first problem in notes on introductory combinatorics by Polya and Tarjan. (Solved, but I havent looked).

Problem statement: Find the number of ways of spelling "abracadabra" always going from one letter to the adjacent.

$$A$$
$$B \quad B$$
$$R \quad R \quad R $$
$$A\quad A \quad A \quad A $$
$$C\quad C\quad C\quad C\quad C$$
$$A\quad A \quad A \quad A \quad A\quad A$$
$$D\quad D\quad D\quad D\quad D$$
$$A\quad A \quad A \quad A $$
$$B \quad B \quad B $$
$$R \quad R$$
$$A$$

I got a very improbable answer of $2^{25}$ so I tried a simpler case to understand it.

Starting at the northmost A there are two routes. At each of the two B's on the second row there are $2$ routes, so uptil this point

$$A$$
$$B \quad B$$
$$R \quad R \quad R $$

there should be $2^3$  ways to get the three letter word "ABR" but on manually counting the number of ways is just 4 (LL, LR, RR, RL where R=right/L=left). What is wrong with my approach? More precisely, where have I overcounted?

Edit: I understood my problem. I used the product rule instead of the sum rule. I think I will stop paying attention to these "rules" as they are hindering my problem solving anyway. (I have asked a previous questions on the subject)

答:
This is essentially the same as this problem. Start by rewriting your array as a square, with the top now at the bottom left, and the bottom now at the top right:
$$\begin{array}{cccccc}
A & D & A & B & R & A\\
C & A & D & A & B & R\\
A & C & A & D & A & B\\
R & A & C & A & D & A\\
B & R & A & C & A & D\\
A & B & R & A & C & A
\end{array}$$
You want to get from the bottom left to the top right, using only moves right or up. A simple thing is to note that there is only one way to get to the bottom left, and at any point, the number of ways to get to a point is the sum of the number of ways to get to the point directly below and the number of ways to get to the point directly left. This gives:
$$\begin{array}{cccccc}
1 & 6 & 21 & 56 & 126 & 252\\
1 & 5 & 15 & 35 & 70 & 126\\
1 & 4 & 10 & 20 & 35 & 56\\
1 & 3 & 6 & 10 & 15 & 21\\
1 & 2 & 3 & 4 & 5 & 6\\
1 & 1 & 1 & 1 & 1 & 1
\end{array}$$
giving $252$ ways.

Alternatively, in the end, you must make five moves right and five moves up. Your only decisions to make is when you make the moves up. You have six locations (before all the moves right, between any two moves right, and after all moves right), and you have to choose them, allowing repetitions and without caring about the order. That is, you want to count combinations with repetitions, selecting six from seven possibililties. The way to make $r$ choices from $n$ possibilities, allowing repetitions and where order does not matter, is $\binom{n+r-1}{r}$, so in this case we have
$$\binom{6+5-1}{5} = \binom{10}{5} = \frac{10!}{5!5!} = 252.$$


答: $\binom{n}{x}$ does not disappear and the identity does not hold. Try $n=4, x=3, \pi=0.5$, say, to get $\log(1/4)=2\log(1/2)$ on the left and $4\log(1/2)$ on the right.</p>

Title: $\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x\log \pi + (n-x)\log(1-\pi)\;\;?$

Body:
$$\log\left(\binom{n}{x} \pi^x (1-\pi)^{n-x}\right)=x \log \pi + (n-x)\log(1-\pi)$$  this is what i have.

i dont understand how $\binom{n}{x}$ disappears, but the rest is fine.

I tried this, but it is getting complicated. Is there some way I can do things like this quickly and at the first glance?

$$\begin{align}\log\left(\binom{n}{x}\right) & = \log\left(\dfrac{n!}{x!(n-x)!}\right)\\ \\
& =\log(n!)-\log\left[x!(n-x)!\right] \\ \\
& = \log(n!)-\left[\log(x!)+\log(n-x)!\right] \\ \\
& = \;\; \cdots
\end{align}$$


问:
Approximation of a function with this condition
I'm wondering if I have the expression

$$f = y(ln(x) +1)$$

and I'm given $x \gg 1$. Is it valid to approximate

$$f \approx yln(x)$$

答:
There is no strict rule for "approximating". So your question should perhaps be something like "How good or bad is this approximation?" I also cannot answer this question with certainty, since you did not provide any further context. It really depends on what your definition of "good" is and what you want to do with the approximation.

Generally speaking, this is a good approximation (again - it depends on what your understanding of "valid" and "good" is):

Define $f(x)=y(\log(x) +1)$ and $g(x)=y\log(x)$, where $y$ is a real constant. The functions $f$ and $g$ have the same time complexity, so $f=O(g)$ and $g=O(f)$. If you're unfamiliar with the big $O$ notation, here are two equivalent definitions of it for $x\to\infty$:

Def 1.
$$\exists\ C > 0\ \exists\ x_0 > 0\ \forall\ x > x_0: |f(x)| \le C\cdot|g(x)|$$

Def 2.
$$\limsup_{x \to \infty} \left|\frac{f(x)}{g(x)}\right| < \infty$$

I will leave the proof for $f=O(g)$ and $g=O(f)$ to you. In this case I would use the first definition but try to prove it with both. If you still have question ask in the comments.


问:
Recostruct a function from the coefficients of its Chebyshev expansion
The Chebyshev polynomials of the first kind are obtained from the recurrence relation
$$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$
I know the numerical result of the integral:
$$c_{k}=\int_{-1}^{1} \frac{f(x) T_{k}(x)}{\sqrt{1-x^{2}}} d x$$
for $k=0,1,\dots,n$. From these values is it possible to reconstruct a possible trend of the function?

答:
Sure. The projection coefficient is $\frac{c_k}{(T_k,T_k)}=\frac{2}{\pi} c_k$ and so a Fourier-like approximation of $f$ is

$$\frac{2}{\pi} \sum_{k=0}^n c_k T_k(x)$$

which will converge to $f$ in the $L^2((1-x^2)^{-1/2} \, dx)$ norm as $n \to \infty$.


问:
How to add two bi-vectors geometrically?
We know that how to add two vectors in the plan. i.e. The resultant of two vectors can be found using either the parallelogram method or the triangle method. For this we should place both vectors  at the same initial point (one dim less than vector dim).

I want to know how to add two bivectors in $\Bbb R^n$ geometrically? for example $v\wedge w+x\wedge y$. The first things come to my mind is that we should place these two bivectors with one edge in common,  then construct a parallelepiped with them and the plane through the diagonal of this parallelepiped is the wanted sum.

but I think one cannot always place two bivectors with one edge in common unless they have one common vector in their span. e.g. $v\wedge w$ and $w\wedge z$. Or perhaps my question is meaningless?

答:
Only decomposable $2$-vectors (i.e., ones that can be written as $v\wedge w$ for some vectors $v,w$) represent $2$-dimensional parallelograms/planes. The sum is an algebraic construction, not — sad to say for me — a geometric one. I mean, abstractly, in the vector space $\Lambda^2\Bbb R^n \cong \Bbb R^{n(n-1)/2}$, you are still adding vectors in a geometric way, but in general this says nothing about $2$-planes. Only when you have a sum of the form $v\wedge w + cv\wedge u = v\wedge (w+cu)$ will you get another decomposable $2$-vector as the sum.

(Here's a dimension count for you. As I indicated, the dimension of the vector space of bivectors is $n(n-1)/2$. The dimension of the Grassmannian of (oriented) $2$-planes in $\Bbb R^n$ is $2(n-2)$, and so if we add in one dimension for scalar multiples, we get $2n-3$, which is less than $n(n-1)/2$, unless $n=2$ or $3$.)


答:
The symbol used is $\sim$ not $\approx$ when taking about asymptotic behaviour of functions.

Assuming $g(x)\neq 0$ then $f(x)\sim g(x)$ is a notation for $\lim\limits_{x\to x_0}\dfrac{f(x)}{g(x)}=1$

Most of the time the context (i.e. $x_0$ which is often $0$ or $\infty$) is implicit and is omitted, though you can use $f(x)\sim_{x_0} g(x)$ to specify it explicitly.

This notation goes along with Landau $o(\ )$ notation to denote something negligible.

Where we use $x\ll 1$ for numbers, or in the context of $x\to 0$, we can use $f(x)=o(g(x))$ to signify $f(x)\ll g(x)$ with definition $\lim\limits_{x\to x_0} \dfrac{f(x)}{g(x)}=0$

It is immediate that $$f(x)=g(x)+o(g(x))\iff f(x)\sim g(x)$$

Rem: you noticed we ask for $g(x)\neq 0$, but we can also rewrite this in epsilon-delta as  $|f(x)|\le \epsilon|g(x)|$ equivalently to get rid of the division, and keep the same meaning.

In the present case $f(x)=y\ln(x)+y$

Since $y$ is constant $\frac y{\ln(x)}\to 0$ at infinity since $\ln(x)\to+\infty$

It means $y=o(\ln(x))$ ($y$ is small/negligible compared to $\ln(x)$) and we have $f(x)\sim y\ln(x)$.

Title: Approximation of a function with this condition

Body:
I'm wondering if I have the expression

$$f = y(ln(x) +1)$$

and I'm given $x \gg 1$. Is it valid to approximate

$$f \approx yln(x)$$


问:
Find the roots of the cubic below
$$3x^3 + 2x^2 +75x -50 = 0$$
Solve for $x$

I've tried to solve this cubic by factoring, but it doesn't factor. Factoring by grouping doesn't work, and using the factor theorem also doesn't work.
How would one go about finding the root/s in a situation like this? I've plugged the equation into a couple of online calculators, they give the roots, but don't explain how they got their results. I have little experience with cubic equations, and I would like to know how you can solve this particular one.

Title: Find the roots of the cubic below

Body:
Our trick is to find one real root, the other two can be found by complex number or polynomial division
$$3x^3+2x^2+75x-50 = 0$$
$$x^3+\frac{2}{3}x^2+\frac{75}{3}x-\frac{50}{3} = 0$$
$$x = y -\frac{1}{3}\frac{2}{3}$$
$$(y-\frac{2}{9})^3+\frac{2}{3}(y-\frac{2}{9})^2+25(y-\frac{2}{9})-\frac{50}{3} = 0$$
$$y^3+\frac{671}{27}y-\frac{16184}{729} = 0$$
We would use the Tartaglia's formula here
$$y^3+py+q = 0$$
$$y = \sqrt[3]{ -\frac{q}{2} +\sqrt{ (\frac{p}{3})^3+(\frac{q}{2})^2 } } + \sqrt[3]{ -\frac{q}{2} -\sqrt{ (\frac{p}{3})^3+(\frac{q}{2})^2 } } $$
Applying to get
$$y^3+\frac{671}{27}y-\frac{16184}{729} = 0$$
$$ y = \sqrt[3]{ -\frac{-16184}{2\cdot 729} +\sqrt{ (\frac{671}{3\cdot 27})^3+(\frac{-16184}{2\cdot 729})^2 } } + \sqrt[3]{ -\frac{-16184}{2\cdot 729} -\sqrt{ (\frac{671}{3\cdot 27})^3+(\frac{-16184}{2\cdot 729})^2 } } $$
$$ y = \sqrt[3]{ \frac{8092}{729} +\sqrt{ \frac{1512725}{2187} } } + \sqrt[3]{ \frac{8092}{729} -\sqrt{ \frac{1512725}{2187} } } $$
So the solution to your cubic $3x^3+2x^2+75x-50 = 0$ is
$$ x = -\frac{2}{9} + \sqrt[3]{ \frac{8092}{729} +\sqrt{ \frac{1512725}{2187} } } + \sqrt[3]{ \frac{8092}{729} -\sqrt{ \frac{1512725}{2187} } } $$
For other methods to solve the cubic like trigonometry and tschirnhaus see here


问:
If the operation $\&$ has the following properties: $x\&0=0\forall x\in \mathbb{R}$ and $x\&(y+1)=x\&y+(x-y)\forall x,y\in \mathbb{R}$
If the operation $\&$ has the following properties: $x\&0=0\forall x\in \mathbb{R}$ and $x\&(y+1)=x\&y+(x-y)\forall x,y\in \mathbb{R}$ Find the value of $2014\& 10$

I tried to solve this question as follows:

$x\&-1+x+1=0$

$x\&-1=-(x+1)$

Using $x=0$, gives you nothing. I know that the answer is $20095$, but I don't know how to work it out. Could you please explain to me how to solve this question?

答:
Hint: Show by induction on $k$ that $$x\& (y+k)=x\& y +kx-\sum_{j=0}^{k-1}(y+j)$$

and then use that with $y=0$.


问:
What is the use of, and intuition behind, writing $\frac{d^2}{dx^2}$ for the second derivative?
Is it possible to take a second derivative without taking the first derivative before?

 Why do we multiply the $d$ and $dx$ operators? Like, does $\dfrac{d^2}{dx^2}$ really mean $\dfrac{d}{dx} \cdot \dfrac{d}{dx}$?

What's the intuitive understanding about this? Can it be represented in a graph? Like... 'Little change squared in $y$ over little change squared in $x$'?

答:
It depends what you mean by take a second derivative without taking the first derivative before. You can use formulae to calculate a second derivative without calculating the first derivative but this formula would have relied on first taking the first derivative.

The second derivative is the little change in the first derivative over the little change in x.


问:
When continuous random variable is in both the conditional and the conditioned
Let $A,B$ be continuous random variables. Let $E,G,K$ be events. Let $t \in image(B)$.

(I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf. If not, then assume joint pdf is well-defined whenever necessary.)

Question 1: Can we say $P(B \le A|B=t) = P(t \le A|B=t)$?

Question 2: If so, then how? If not then why?

What I've tried: I think we're doing something like $P(E|K)=P(E \cap K|K)$, and I get it when $P(G|K)$ is defined for $P(K)>0$. What is being done here when technically $P(K)=0$? I mean, of course, in the 1st place when we say like '$P(\cdot|B=t)$', this is notational, we're not really conditioning on the $P$-null event $\{B=t\}$. But I still don't get exactly what's being done here. Well...

As I understand, $$RHS = P(t \le A|B=t) := \frac{\int_{t \le A} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da} = \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da}$$

$$= \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{f_{B}(t)}= \int_{t}^{\infty} \frac{f_{A,B}(a,t)}{f_{B}(t)} da,$$ where/whence $f_{A|B=t}(a) := \frac{f_{A,B}(a,t)}{f_{B}(t)}$.

That's all I got.

Not really sure how to evaluate LHS. Perhaps LHS := RHS, i.e. LHS is defined as RHS?

Update: Wait I think I have an idea how to do LHS

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db$$

Here,

'$f_{(A,B)|B=t}(a,b)$' is I guess meant like $f_{(A,B)|T=t}(a,b)$ where $T$ is a random variable that is (not just almost surely but really) surely equal to $B$.

'$f_{(A,B)|T=t}(a,b)$' is meant like 2 random variables conditioned on a 3rd random variable. As I understand multivariate conditional joint distributions, we have this is $f_{(A,B)|T=t}(a,b) := \frac{f_{A,B,T}(a,b,t)}{f_T(t)}$ and then...

$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db = \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{f_T(t)} da db$$

$$= \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} da db$$

$$= \frac{\int \int_{b \le a} f_{A,B,T}(a,b,t) da db}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} $$

And then I don't know. Apparently we might not be able to have a joint pdf if 2 of the random variables are surely equal (maybe even for almost surely):

4.1. joint distribution of x with..itself

4.2. https://stats.stackexchange.com/questions/47162/what-is-the-joint-probability-distribution-of-two-same-variables

答:
Here is a third approach that assumes the joint PDF $f_{A,B}(a, b)$ exists.

Define $Z=B-A$.   We have the PDF transformation:
$$ f_{Z,B}(z,b) = f_{A,B}(b-z,b)$$
Then, assuming $f_B(t)>0$:
\begin{align}
P[B\leq A | B=t] &= P[Z\leq 0|B=t]\\
&=\int_{-\infty}^0 f_{Z|B}(z|t)dz\\
&=\int_{-\infty}^0 \frac{f_{Z,B}(z,t)}{f_B(t)} dz\\
&=\int_{-\infty}^0 \frac{f_{A,B}(t-z,t)}{f_B(t)}dz\\
&\overset{(a)}{=}\int_t^{\infty} \frac{f_{A,B}(u,t)}{f_B(t)}du\\
&=\int_t^{\infty} f_{A|B}(u|t)du\\
&=P[A\geq t|B=t]
\end{align}
where (a) uses the change of variable $u=t-z$ and $du=-dz$.


问:
what is the value of a correct to the nearest integer

This is an A-Level past paper question statistics and probability I managed to solve the first bit I  cant seem to figure this out because i have no teacher I don't know who to ask I would really appreciate any help. Thank you
 This is the mark scheme but I don't understand the steps and whats going on.

答:
The first line is the upper limit of a confidence interval for proportions  $$p\pm z\sqrt{\frac{p(1-p)}{n}}$$  where p is the sample proportion and n the sample size. For your example $p=\frac{6}{60} = 0.1$ for the sample size n = 60.

For a 95% confidence interval 5% of values are outside the interval -z to z, which means 2.5% of values are greater than z i.e. $P(Z>z) = 0.025$. In the Normal Table we look up $P(Z<z) = 0.975$ and see z is 1.96.  Solving the first equation for z in the sample answer you get z = 1.291. Going to the normal table, this goes with a probability value of 0.9015, near enough. This then corresponds to P(-1.291<Z<1.291) = 0.8 or $\alpha = 80%$


问:
If $n$ distinct balls are distributed to $r$ distinct boxes in such a way that each n ball can go any one of the r boxes.
If $n$ distinct balls are distributed to $r$ distinct boxes in such a way that each n ball can go any one of the r boxes. Then find the probability that,

A) All boxes are occupied.

B) Exactly K boxes are occupied.

My attempt: Firstly I choose r many balls from $n$ and then distribute these to $r$ boxes one by one. Then the possible cases to occupy the r boxes is ( $n_{C_r}\times r!$), so the probability will be $\frac{n_{C_r}\times r!\times r^{n-r}}{r^n}$ but in this way I am not getting the correct answer. I think it can be done using Starling Number. But I am not getting what is the problem in my attempt. Can anyone please provide the explicite solution of this problem.

答:
There are $r^n$ possible ways to distribute the balls.
There are ${(r-1)}^n$ ways to distribute the balls that avoid one particular box.
Of those, there are ${(r-2)}^n$ ways to distribute that avoid two particular boxes, and so on.  We have to deal with overlap, however.

For $K = 1$, there are only $r$ ways to have all the balls in the same box.  So $P(K=1) = \frac{r}{r^n}$.

For $K = 2$, there are ${r \choose 2}$ ways to choose which two boxes, but we care about the order, so it is $r(r-1)$ instead, and then given that choice, there are $2^n$ ways to arrange the balls in those two boxes, including the possibility of none in one of the two boxes.  So $$P(K\leq 2) = \frac{r(r-1) \cdot 2^n}{r^n}$$
Continuing in this way, the probability of all boxes being occupied is $1-P(K \leq (r-1))
$.


问:
What is the dimension of the space of traceless $n × n$ matrices?
I am trying to figure out what would be the dimenstion of the space of the traceless $n \times n$ matrices. Please help me  with this question.</p>

答:
Let $M_{n\times n}$ be the vector space of n by n square matrices. And so
$\forall A\in M_{n\times n}, \ A = \left \{ a_{ij}  | (i,j)\in [1,n] \right \}$.

Now every vector space has a basis(not going to prove this here but you can find proofs of it elsewhere on this site) so let's construct one:

Let $E_{ij}$ be the collection of n by n matrices with only 1 nonzero entry.
That is, for some $i,j\in [1,n]$,

$E_{ij}= \left \{  e_{kl}\  | e_{kl}=1 \leftrightarrow kl=ij,\ e_{kl}=0\leftrightarrow kl\neq ij
\ (1\leq k,l\leq n)  \right \}$

This is the standard basis for $M_{n\times n}$. We now can write

$A =  \sum_{i,j=1}^{n} a_{ij}E_{ij} \ | \ \forall A\in M_{n\times n}$  and this vector space has dimension $n^{2}$. Now let $K\subset M_{n\times n}$ be subspace of n by n matrices whose trace is equal to zero. Bear in mind that $ (\forall A \in M_{n\times n} ) \  tr(A)=0 \ \leftrightarrow  \  \sum_{i=1}^{n}a_{ii}=0$.

So for some $1\leq i\leq n$,
$trace(A)-a_{ii}$ = $0-a_{ii} =-a_{ii} = \sum_{j=1}^{i-1}a_{jj} +\sum_{j=i+1}^{n}a_{jj}$. And so you have a linear dependence relation.  We can see this by expanding our result for the i^th entry and writing it out as $a_{ii} = -(\sum_{j=1}^{i-1}a_{jj} +\sum_{j=i+1}^{n}a_{jj})$.

So if the trace of a square matrix A = 0, then one of the diagonal entries is constrained by the values of the other (n-1) diagonal entries as it is the negative sum of the other values. Now there are $n^{2}-n$ non-diagonal entries that do not affect the trace so they are independent of tr(A). So the Dim(K)= $(n^{2}-n)+(n-1)=n^{2}-n+n-1= n^{2}-1$.