问:
Why was I wrong on this permutation / combination / multinomial problem?
Question: We want to make a string of Christmas light to string across a room. The string
contain 36 lights. There are 10 red bulbs, 12 blue bulbs, 10 green bulbs, and
4 yellow bulbs. How many visually distinguishable light strings can we create?
Similar question here.

My Attempt:

Red = R, Blue = B, Green = G, Yellow = 4, then the string of lights can be represented:

RRRRRRRRRRBBBBBBBBBBBBGGGGGGGGGGYYYY

The number of permutations of this string is $${36!}\over{10!12!10!4!}$$
But this is apparently wrong.

Textbook solution:

There are $C(36, 10)$ ways to choose the positions for the red bulbs. There are
$C(36 − 10, 12) = C(26, 12)$ ways to choose the positions for the blue bulbs.
There are $C(14, 10)$ ways to choose the positions for the green bulbs. There
are $C(4, 4)$ ways to choose the positions for the yellow bulbs.

Hence,

$C(36, 10)C(26, 12)C(14, 10)C(4, 4) = {{36!}\over{10!12!14!4!}}$

Now $${{36!}\over{10!12!10!4!}} > {{36!}\over{10!12!14!4!}}$$

However, I'm failing to see where the extra cases are coming from. Where is the discrepency in my approach to this problem?

Body: C(36,10)C(26,12)C(14,10)C(4,4) = $$\frac{36!}{26!10!} * \frac{26!}{14!12!} * \frac{14!}{4!10!} * \frac{4!}{0!4!} = \frac{36!}{10!12!10!4!}$$


问:
Can a sum of first consecutive $n$th numbers ever equal a power of three?
Define $S(n)=1+2+3+\dots+n$

Note: $S(2)=3$

Is the following claim true?

For all $n>2$, there is no such $S(n)=3^t$ where $t$ positive integer?

Small attempt, let $S(n)=\frac{n(n+1)}2=3^t$ then $\frac{S(n)}{S(2)}=3^{t-1}=\frac{n(n+1)}6$ consider $n=3k+1$ then $\frac{n(n+1)}6$ never been an integer.
So we can conclude $n$ at least not equal to $3k+1$

Source code Pari/GP

```<code>for(n=1,3000,for(t=1,10,if(sum(i=1,n,i)==3^t,print([n,t]))))

[2, 1]
</code>```

Related post :Do the squares of an arithmetic progression ever sum to a power of three?

答: $\gcd(2,3^t)=1$ so if $n(n+1)/2=3^t$ then $2\cdot 3^t=n(n+1)$ and can see one of $n,n+1$ must be $2$ which means either $n=2$ [OK gives $n+1=3=3^1$] or $n+1=2$ so $n=1$ giving $n(n+1)=2$ which isn't a power of $3.$

ContentLicense: CC BY-SA 4.0


问:
All ring homomorphisms from $\mathbb{Z}[X]$ to $\mathbb{Z}$
I understand that the homomorhpism $\phi: \mathbb{Z}[X] \rightarrow \mathbb{Z}$ has to have a property that $\phi(1) = 1$. But how do I proceed with that to count all homomorphisms?

答:
Note that every such homomorphism is completely determined by the value of $\phi(x)$. Indeed, since $\phi(1)=1$ it easily follows by induction that $\phi(n)=n$ for all $n\in\mathbb{Z}$. And then:

$\phi(\sum_{i=0}^m a_ix^i)=\sum_{i=0}^m \phi(a_i)\phi(x^i)=\sum_{i=0}^m a_i[\phi(x)]^i$

So this shows that every homomorphism must have the form $\phi(\sum_{i=0}^m a_ix^i)=\sum_{i=0}^m a_ib^i$ for some $b\in\mathbb{Z}$. What is left to show is that for every $b$ this is indeed a homomorphism. This is really easy, so I'll leave it to you. So this gives you all the homomorphisms.


问:
For the following S and T find the number of ways of expressing T as the sum offour elements of S.

My Approach:

okay my first approach to this problem was calculating all the sets of quadruples and then subtracting it with the ones that would not fit, this worked for the first a sub-question but later when I checked it was wrong for the example. I found a lot of programming answers for this but none of them without them so please help me out.

答:
(a): Let $x_1$ be the number of $1$s used and $x_2$ the number of $2$s. The total number of addends is $4$, while the sum is $6$, therefore:

$$\begin{cases}
x_1+x_2=4 \\
x_1+2x_2=6
\end{cases}
$$

which has only one solution: $x_2=2$ and $x_1=2$. There are ${5 \choose 2}$ ways to choose two $1$s from $S$ and ${4 \choose 2}$ ways to choose two $2$s from $S$. Then the number of solutions is:

$${5 \choose 2}{4 \choose 2}=60$$

(b): Let $x_1$ be the number of $1$s used, $x_2$ the number of $2$s and $x_3$ the number of $3$s. The total number of addends is $4$, while the sum is $9$, therefore:

$$\begin{cases}
x_1+x_2+x_3=4 \\
x_1+2x_2+3x_3=9
\end{cases}
$$

If we subtract the first equation from the second, we get $x_2=5-2x_3$, therefore $x_2$ is odd and then $x_2=1$ or $x_2=3$, then

(b.1): $x_2=1$, the system reduces to:

$$\begin{cases}
x_1+x_3=3 \\
x_1+3x_3=7
\end{cases}
$$

and then $x_3=2$ and $x_1=1$, with a total number of choosing $x_1$ $1$s, $x_2$ $2$s, $x_3$ $3$s equal to:

$${4 \choose 1}{3 \choose 1}{2 \choose 2} = 12$$

(b.2): $x_2=3$, the system reduces to:

$$\begin{cases}
x_1+x_3=1 \\
x_1+3x_3=3
\end{cases}
$$

and then $x_3=1$ and $x_1=0$, with a total number of choosing $x_2$ $2$s, $x_3$ $3$s equal to:

$${3 \choose 3}{2 \choose 1} = 2$$

If you like, you can write that as ${4 \choose 0}{3 \choose 3}{2 \choose 1}$.

Finally, we sum the two cases (b.1) and (b.2) to get $12+2=14$.

(c) is left to you.


问:
Generalising the Chinese Remainder Theorem
We have that for $I,J$ ideals of some ring $R$ with $R=I+J$, $$\frac{R}{I\cap J} \cong \frac{R}{I} \times \frac{R}{J}$$
My question is whether the analogous expression for three ideals $I,J,K$ where $R=I+J+K$ is true?

I think I have found a counterexample with $R=\mathbb{Z}$, $I=2\mathbb{Z}$, $J=2\mathbb{Z}$, $K=3\mathbb{Z}$.

Here, $R=I+J+K$ and $\frac{R}{I\cap J\cap K}=\frac{R}{I \cap K}=\frac{\mathbb{Z}}{6\mathbb{Z}} \cong \frac{R}{I} \times \frac{R}{K}$

We have that $\frac{R}{I} \times \frac{R}{K}$ is not isomorphic to $\frac{R}{I} \times \frac{R}{J}\times \frac{R}{K}$ and so this is our counterexample.

Can someone please check this counterexample is sound? Thanks

答:
Your counterexample is correct. The correct generalization of the Chinese Remainder Theorem, obtained by inducting on the statement you gave is:

Suppose $I_1,\ldots,I_n$ are ideals of a ring $R$ which are pairwise comaximal, i.e. for any $1\leq i\ne j\leq n$ we have $I_i+I_j=R$. Then
  $$\frac{R}{I_1\cap\cdots\cap I_n}\cong \frac{R}{I_1}\times\cdots\times\frac{R}{I_n}$$

ContentLicense: CC BY-SA 3.0


问:
Basis transformation matrix and eigenvectors
we generally know that if we have some matrix $A$ and we diagonalize it and get a matrix $D$, there is some matrix $P$ such that:

$$ D = P^{-1}AP$$

My question is then, does the matrix P generally contain the eigenvectors corresponding to the eigenvalues on the diagonal of D? If yes, why? I fail to see how that comes out explicitly, but it seems it is that way (is it true only for orthogonal matrices?).

Thank you!

答:
That is indeed true. If

$$A = PDP^{-1}$$

is a diagonalization of $A$, $P$ is a matrix whose column vectors are eiegenvectors corresponding to the diagonal entries in $D$, which would then be the eigenvalues. How can you see this? First let $P = [v_1 \quad v_2 \quad \cdots \quad v_n]$ and assume that the diagonal elements of $D$ are $\lambda_1, \lambda_2, \ldots, \lambda_n$. Observe that

$$PD = [v_1 \quad v_2 \quad \cdots \quad v_n]D  = [\lambda_1 v_1 \quad \lambda v_2 \quad \cdots  \quad \lambda_n v_n] $$

$$AP = A[v_1 \quad v_2 \quad \cdots \quad v_n] = [Av_1 \quad Av_2 \quad \cdots \quad Av_n]$$

Since $AP = DP$ by assumption, we easily see that the $v_i$-s are eigenvectors of $A$ with corresponding eigenvalues $\lambda_i$-s.


问:
Integration of $e^{-\langle Ax , x \rangle}$ over $\mathbb{R}^n$

Problem :

If $A_{n \times n}$ is a symmetric, positive-definite matrix, show that : $$\int_{\mathbb{R}^n} e^{-\langle Ax , x \rangle}~ dx = \sqrt{\dfrac{\pi^n}{\det(A)}}$$ where $\langle a , b\rangle$ denotes the inner product of $a$ and $b$.

Approach :

I was approaching the problem using the Change of Variable Formula, using the function $\varphi(x) = A^{-1}x$ . Since $A$ is p.d., I can show that it is invertible. But I can't proceed anymore.

I found a similar-looking problem here, but couldn't understand anything.

答:
Since $A$ is symmetric, there exists some orthogonal $S \in \mathbb{R}^{n \times n}$ (i.e. $S^{-1} = S^\top$) such that $A = S^{-1}DS$ where $D := \mathrm{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix containing all the eigenvalues of $A$. Note that they are positive because of the assumption of $A$ being positive definite. So, because of $S^{-1} = S^\top$:
$$
\int_{\mathbb{R}^n} e^{-\langle Ax, x\rangle} ~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\langle Sx, DSx \rangle}~\mathrm{d}x
$$
Now introduce an operator $\Phi: \mathbb{R}^n\rightarrow \mathbb{R}^n$, $\Phi(x):= S^{-1}x$. $\Phi$ is bijective because of $S$ being invertible. One furthermore easily finds $D\Phi(x) = S^{-1}$ for all $x \in \mathbb{R}^n$. We also know that $\lvert \det(S^{-1}) \rvert = 1$ because $S$ is orthogonal. So the transformation formula yields:
$$
\int_{\mathbb{R}^n} e^{-\langle Sx, DSx \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{- \langle S \Phi(x), DS \Phi(x) \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\langle x, Dx \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\sum_{j = 1}^n \lambda_jx_j^2}~\mathrm{d}x
$$
Use that $e^{x+y} = e^x e^y$ for all $x, y \in \mathbb{R}$ and Fubini to conclude:
$$
\int_{\mathbb{R}^n} e^{-\sum_{j = 1}^n \lambda_jx_j^2}~\mathrm{d}x  = \prod_{j = 1}^n \int_{-\infty}^\infty e^{-\lambda_j x_j^2}~\mathrm{d}x_j
$$
Now consider
$$
I_j := \int_{-\infty}^\infty e^{-\lambda_j x_j^2}~\mathrm{d}x_j.
$$
Introduce a substitution $y := \sqrt{\lambda_j}x_j$. Then:
$$
I_j  = \frac{1}{\sqrt{\lambda_j}} \int_{-\infty}^\infty e^{-y^2}~\mathrm{d}y = \sqrt{\frac{\pi}{\lambda_j}}
$$
Putting everything together:
$$
\int_{\mathbb{R}^n} e^{-\langle Ax, x\rangle} = \prod_{j = 1}^n I_j = \frac{\sqrt{\pi}^n}{\sqrt{\prod_{j = 1}^n \lambda_j}} = \frac{\sqrt{\pi}^n}{\sqrt{\det(A)}} = \sqrt{\frac{\pi^n}{\det (A)}}
$$
In the last step we used that the product of eigenvalues is the determinant.


问:
Convolution & DFTs: How much zero padding is necessary to avoid circular convolution?
When performing discrete [spatial] convolutions in the frequency domain, how much zero-padding is necessary to avoid the effects of circular convolution?

I have a book that almost certainly answers the question, but I've loaned it out and don't have easy access to it for now.  So far, my google searches have turned up published papers with an abstract that vaguely mentions the topic, but I'd need access to that journal to see if the paper has the info I need -- info I figured would be easy to find.

答: The convolution of two functions with non-zero values at indices from $0$ to $N-1$, i.e. $N$ values, will have non-zero values at indices $0$ to $2(N-1)$, i.e. $2N-1$ values. Thus, it suffices to use a DFT of length $2N-1$ to avoid cyclical overlap. However, if $N$ is chosen so as to factor into small primes, this property is lost if a transform with length $2N-1$ is performed; in this case, it is far more efficient to pad to $2N$ instead.</p>


答:
Now

$n(n+1)=2×3^t$

So now one of $n,n+1$ must be divisible be 3, and so because they are consecutive other will not be divisible by three
Let $3|n$ and not $n+1$ (the other case follows the same proof)

$3^t|n(n+1)$ and as $gcd(n+1,3)=1$

$3^t|n$ and $n=3^t×k$

Substituting in the original...

$2 = k(n+1)$ and now it's obvious...

Title: Can a sum of first consecutive $n$th numbers ever equal a power of three?

Body:
Define $S(n)=1+2+3+\dots+n$

Note: $S(2)=3$

Is the following claim true?

For all $n>2$, there is no such $S(n)=3^t$ where $t$ positive integer?

Small attempt, let $S(n)=\frac{n(n+1)}2=3^t$ then $\frac{S(n)}{S(2)}=3^{t-1}=\frac{n(n+1)}6$ consider $n=3k+1$ then $\frac{n(n+1)}6$ never been an integer.
So we can conclude $n$ at least not equal to $3k+1$

Source code Pari/GP

```<code>for(n=1,3000,for(t=1,10,if(sum(i=1,n,i)==3^t,print([n,t]))))

[2, 1]
</code>```

Related post :Do the squares of an arithmetic progression ever sum to a power of three?


问:
The existence of continuous and concave/convex extension of discrete function
Given a finite $A\subset R$ and a nondecreasing $f:A\rightarrow R$ such that if $c\geq \frac{a+b}{2}$ then $f(c)\geq \frac{f(a)+f(b)}{2}$, is there always exist a continous, nondecreasing, twice differentiable, and concave extension $g$ of $f$ such that $g_{|A}=f$? If the intuition is correct, what formal results can I cite?

答:
You can't always extend to a differentiable function. Intuitively, if you plot the points $(a,f(a))$ in the plane and any three of them are collinear, then any concave (or convex) extension must be coincide with the line segments connecting them.

For example, suppose $\{(a,f(a)) \ | a \in A \} = \{(0,0),(1,1),(2,2),(3,2),(4,2)$}.
Then any concave extension must satisfy:
$$
g(x) = \begin{cases}
x & 0 \le x \le  2 \\
2 & 2 \le x \le 4
\end{cases}
$$
This is not differentiable at $x=2$.

So why is it true that a concave extension must satisfy the above? It's pretty clear if you draw a picture. For example, for $x \in (0,1)$, the concavity between the points $0,x,1$ implies $g(x) \ge x$, but concavity on the points $x,1,2$ implies $g(x) \le x$. So you must have $g(x)=x$ for all  $x \in (0,1)$. Likewise for the other segments, $g$ must be the affine function connecting the graph of $f$.


问:
Prove $Z=1+X$ has a geometric distribution
I have $(=)=\frac{1}{2^{i+1}}$ where i is a natural number

I need to prove $Z=1+X$ has a geometric distribution and find the $E(Z),V(Z),V(X),V(E)$

My attempt:

$X$ is a geometric distribution with $p=1/2$ so $E(X)=V(X)=2$

$Z=\frac{1}{2^{i+1}}+1=\frac{1+2^{i+1}}{2^{i+1}}$ and I got stuck here, how can I show this is geometric?

答:
If $Z=X+1$, then $$P(Z=i)=P(X+1=i)=P(X=i-1)=\frac1{2^{i+1-1}}=\frac1{2^i}$$

Hence we have shown that $Z$ is Geometric distribution with success probability $\frac12$.


问:
Find $x$ if $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$

The relation $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$ holds true for all $1.$ $ x\in \mathbb R$, $2.$ $ x\in (-1,\infty) $, $3.$ $ x\in (-\infty,1) $, $4.$ $ x\in (-\infty,2)$

I took RHS=$\frac{\pi}{4}+\tan^{-1}x=\tan^{-1}1+\tan^{-1}x=\tan^{-1}(\frac{1+x}{1-x})$,if $x\gt0, x\lt1$

(using the property $\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy})$, if $x\gt0,y\gt0,xy\lt1$)

Therefore, my answer is $x\in(0,1)$, which is not an option.

The answer in the key is $2. $ $x\in(-\infty,1)$.

Looks like, they just used $x\lt1$ condition and not $x\gt0$. Why?

答:
Note that the function
$$f(x)= \tan^{-1}\frac{1+x}{1-x}-(\frac{\pi}{4}+\tan^{-1}x)$$
is continuous everywhere except at the break point $x=1$. Then, evaluate
$$\lim_{x\to 1^-} f(x)=0,\>\>\>\>\>\>\>
\lim_{x\to 1^+}f(x)=-\pi\ne 0
$$
Thus, the equality holds over $(-\infty,1)$.


问:
What does a proof in an internal logic actually look like?
The nLab has a lot of nice things to say about how you can use the internal logic of various kinds of categories to prove interesting statements using more or less ordinary mathematical reasoning. However, I can't find a single example on the nLab of what such a proof actually looks like. (The nLab has a frustrating lack of examples in general.)

Can anyone supply me with some examples? I'd be particularly interested in the following kinds of examples:

I've heard in a topos one can internalize the real numbers and, in the topos $\text{Sh}(X)$ of sheaves on a topological space, this reproduces the sheaf of continuous real-valued functions $X \to \mathbb{R}$. Moreover, one can internalize "finitely generated projective $\mathbb{R}$-module" and in $\text{Sh}(X)$ this reproduces real vector bundles on $X$. What can you prove about vector bundles this way?

I'd also like to see examples of what you can do in the internal logic of Cartesian closed categories.

This MO question is related but it doesn't really satisfy my curiosity.

答:
Here is an arbitrary example from algebraic geometry. We'll prove the following
well-known statement about $\mathcal{O}_X$-modules on reduced schemes $X$ by
reducing to constructive linear algebra interpreted in the topos
$\mathrm{Sh}(X)$ of sheaves on $X$:

Let $\mathcal{F}$ be an $\mathcal{O}_X$-module locally of finite type.
  Then $\mathcal{F}$ is locally free iff its rank is constant.

We can translate this statement into the internal language of $\mathrm{Sh}(X)$
by the following dictionary:

In the internal language, the sheaf of rings $\mathcal{O}_X$ looks like an ordinary ring.

Accordingly, $\mathcal{F}$ looks like an ordinary module on that ring.

$\mathcal{F}$ is locally of finite type iff it is finitely generated from
the internal point of view.

$\mathcal{F}$ is locally free iff it is a free module from the internal
point of view.

Internally, we can define the rank of $\mathcal{F}$ as the minimal number
of elements needed to generate $\mathcal{F}$. But constructively,
the natural numbers may fail to have minima of arbitrary inhabited sets (see
this enlightening blog post by Andrej Bauer), so this minimal number
might not actually be an (internal) natural number, but be an element of a
suitable completion. Externally, the rank defined this way induces a upper
semicontinuous function on $X$ (see nLab and the Mulvey reference therein); it is constant iff internally, the minimal number of
generators is an actual natural number.

Finally, the scheme $X$ is reduced iff $\mathcal{O}_X$ looks like an
ordinary reduced ring from the internal perspective. This in turn is
equivalent to $\mathcal{O}_X$ being a so-called residue field from the
internal point of view (i.e. a non-trivial ring with every non-unit being
zero).

So the statement follows if we can give a constructive proof of the following
linear algebra fact:

Let $A$ be a residue field and let $M$ be a finitely generated $A$-module.
  Then $A$ is free iff the minimal number of elements needed to generate $M$
  as an $A$-module is an actual natural number.

The direction "$\Rightarrow$" is clear. For the direction "$\Leftarrow$",
consider a minimal generating family $x_1,\ldots,x_n$ of $M$ (which exists by
assumption). This family is linearly independent (and therefore a basis): Let $\sum_i \lambda_i x_i =
0$. If any $\lambda_i$ were invertible, the family
$x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n$ would too generate $M$, contradicting
the minimality. So each $\lambda_i$ is not invertible and thus zero (by assumption on $A$).


答:
By virtue of the universal property of polynomial algebras, given a commutative ring $A$ and a associative & unitary $A$-algebra $E$, there is a canonical bijection between the set of (unitary) $A$-algebra morphisms from $A[X]$ to $E$ and the (support) set $E$ itself.

More generally, given a family $X$ of indeterminates indexed by $I$, there is a canonical bijection:
$$\mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Alg_1}}}(A[X], E) \to \mathrm{Hom}_{\mathbf{Ens}}(I, E)$$
between the set of all (unitary) $A$-algebra morphisms from $A[X]\colon=A[X_i]_{i \in I}$ to $E$ and the set of all maps from $I$ to $E$.

It thus follows that in terms of cardinalities you have $\left|\mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Alg_1}}}(A[X], E)\right|=|E|^{|I|}$. In the case of just one indeterminate - when $I$ is a singleton - this relation takes the simpler form $\left|\mathrm{Hom}_{\operatorname{\mathit{A}-\mathbf{Alg_1}}}(A[X], E)\right|=|E|$.

In your particular case $A=E=\mathbb{Z}$, hence $\left|\mathrm{Hom}_{\operatorname{\mathbb{Z}-\mathbf{Alg_1}}}(\mathbb{Z}[X], \mathbb{Z})\right|=|\mathbb{Z}|=\aleph_0$.

Title: All ring homomorphisms from $\mathbb{Z}[X]$ to $\mathbb{Z}$

Body: I understand that the homomorhpism $\phi: \mathbb{Z}[X] \rightarrow \mathbb{Z}$ has to have a property that $\phi(1) = 1$. But how do I proceed with that to count all homomorphisms?


问:
Show that $ f $ is infinitely differentiable function on $ \mathbb{R} \text{ and } f^{(n)}(0) = 0$
Hello can you help me with that...

Let $ f \colon \mathbb{R} \rightarrow \mathbb{R} $ be the function defined as follows: $ f(x) = 0 \text{ if } x \leq 0 \text{ and } f(x) = e^{- \frac{1}{x}} \text{ if } x > 0 $. Show that $ f $ is infinitely differentiable function on $ \mathbb{R} \text{ and } f^{(n)}(0) = 0 \text{ for every } n \in \mathbb{N} $

I tried to find the derivative of $ f(0) $

$ f^{'} (0) $ $= \lim_{h\to 0} $ $ \frac{e^{\frac{-1}{h}}}{h} $

$ \lim_{h \to 0^\mathtt{\text{-}}} \frac{e^{- \frac{1}{h}}}{h}=-\infty  $

$ \lim_{h \to 0^\mathtt{\text{+}}} \frac{e^{- \frac{1}{h}}}{h}=0 $

So this limit doesn't exists and this function doesn't have derivative at zero
so please tell me what's my mistake

答: @DavidMitra has pointed out that $f(x) = 0$ on $(-\infty,0)$ by definition, so your left-handed limit should be zero.  Your right-handed limit is correct, so the limit is zero.


问:
Approximation Algorithm question
Define an independent set of a graph $G = (V, E)$ to be a subset $S$ of vertices such that $V-S$ is a vertex cover of $G$. Is every $2$-approximation algorithm for finding a minimum vertex cover also a $2$-approximation algorithm for finding a maximum independent set?

I have read about approximation algorithms, but could relate vertex cover problem to independent set. Do you have any suggestions on why such algorithm can be found or not?

答: Could you please suggest any ideas? I really need your answers. I will delete these comments immediately if anyone just comments and revives this thread.</p>

Title: Approximation Algorithm question


答:
Observe that if

$$\mathbb{P}[X=x]=\Bigg(\frac{1}{2}\Bigg)^{x}\frac{1}{2}\cdot \mathbb{1}_{\{0;1;2;...\}}(x)$$

$$Z=X+1$$

Is the same Geometric distribution with parameter $0.5$ with the only difference that $X$ counts the failures before the first success while $Z$ counts the trials.

Thus

$$\mathbb{E}[Z]=\frac{1}{p}=2$$

and

$$\mathbb{E}[X]=\frac{1}{p}-1=\frac{1-p}{p}=1$$

It is self evident that

$$\mathbb{V}[X]=\mathbb{V}[Z]$$

Title: Prove $Z=1+X$ has a geometric distribution

Body:
I have $(=)=\frac{1}{2^{i+1}}$ where i is a natural number

I need to prove $Z=1+X$ has a geometric distribution and find the $E(Z),V(Z),V(X),V(E)$

My attempt:

$X$ is a geometric distribution with $p=1/2$ so $E(X)=V(X)=2$

$Z=\frac{1}{2^{i+1}}+1=\frac{1+2^{i+1}}{2^{i+1}}$ and I got stuck here, how can I show this is geometric?


问:
Deductive proofs of basic "laws" of arithmetic
I recently started studying math, one of the things that really interested me about it was the idea that math, unlike science, depended solely on deductive proof. With this in mind, I've went back to arithmetic in search of deductive proof for "laws" (hope that's the right word) such as the commutative law of multiplication. However, I'm having trouble locating a deductive proof of even this most basic "law". Can anyone point me in the right direction?

Thanks!

答:
Using the Peano axioms, you can prove that all of the "laws" for addition and multiplication hold in the natural numbers (i.e. the non-negative integers).  From there, we normally define the integers, rational numbers, and real numbers as incremental extensions of the natural numbers, and part of that development is showing that the corresponding definitions of addition and multiplication in those sets still preserve the associative, commutative, distributive, and any other laws that you're familiar with.

There are any number of elementary math courses and textbooks that cover this "program", like introductory set theory, real analysis, discrete math, or a first proof course.


问:
Simplify the Boolean expression A'BC + A'BD' + AB' + AC + ABC + ACD
I'm trying to simplify the expression: A'BC + A'BD' + AB' + AC + ABC + ACD

I got to the expression: BC + AC + A'BD' + AB'

I know this expression is equvalant to the one above because I verified using truth tables.

The simplified expression, according to online calculators is: BC + A'BD' + AB'

How do I continue with my simplification?

答:
This is an application of the Consensus Theorem:

$xy + y'z = xy + y'z + xz$

Applied to your case, we thus have that:

$AB' + BC + AC = AB' +BC$

So, if you have the Consensus Theorem in your arsenal, you can just do in 1 step:

$BC + AC + A'BD' + AB' \overset{Consensus}{=}$

$BC + A'BD'+AB'$

If you can't use Consensus, but you do have Adjacency ($xy +xy'=x$) and Absorption ($p + pq =p$), then you can do:

$BC + AC + A'BD' + AB' \overset{Adjacency}{=}$

$BC + ABC + AB'C + A'BD' + AB' \overset{Absorption \ \times \ 2}{=}$

$BC+ A'BD' + AB'$

And if you don't have those either, do what tranceloction does in the other answer.


答:
Taking the tangent of the two members, you check that this equation holds to a multiple of $\pi$.

The function $\arctan$ maps the real axis to the range $\left(-\dfrac\pi4,\dfrac\pi4\right)$ and is odd. Then $\dfrac\pi4+\arctan x\in\left(0,\dfrac\pi2\right)$ and this excludes the case of $x\ge1$.

Title: Find $x$ if $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$

Body:
The relation $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$ holds true for all $1.$ $ x\in \mathbb R$, $2.$ $ x\in (-1,\infty) $, $3.$ $ x\in (-\infty,1) $, $4.$ $ x\in (-\infty,2)$

I took RHS=$\frac{\pi}{4}+\tan^{-1}x=\tan^{-1}1+\tan^{-1}x=\tan^{-1}(\frac{1+x}{1-x})$,if $x\gt0, x\lt1$

(using the property $\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy})$, if $x\gt0,y\gt0,xy\lt1$)

Therefore, my answer is $x\in(0,1)$, which is not an option.

The answer in the key is $2. $ $x\in(-\infty,1)$.

Looks like, they just used $x\lt1$ condition and not $x\gt0$. Why?


问:
Find the intersection point and verifying its direction
The tangent to the curve $y = e^x$  drawn at the point $(c,e^c)$ intersects the line joining the points $(c – 1, e^{c-1})$ and $(c+1, e^{c+1})$

(A) on the left of x = c

(B) on the right of x = c

(C) at no point

(D) at all point

My approach is as follow

$y = {e^x},y' = {e^x} \Rightarrow y' = {e^c}$

$y = {e^c}x + {c_{in}}$

Where $c_{in}$ represent the y intercept

${c_{in}} = {e^c}\left( {1 - c} \right)y = {e^c}x + {e^c}\left( {1 - c} \right)$

The points are $\left( {c - 1,{e^{c - 1}}} \right)\& \left( {c + 1,{e^{c + 1}}} \right) \Rightarrow m = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}$

$y = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}x + c{'_{in}}$...eq(i)

${e^{c + 1}} = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)\left( {c + 1} \right)}}{2} + c{'_{in}} \Rightarrow {e^c}\left( {e - \frac{{\left( {e - {e^{ - 1}}} \right)\left( {c + 1} \right)}}{2}} \right) = c{'_{in}} \Rightarrow c{'_{in}} = {e^c}\left( {\frac{{ - e\left( {c - 1} \right) + {e^{ - 1}}\left( {c + 1} \right)}}{2}} \right)$

$y = \frac{{{e^c}\left( {e - {e^{ - 1}}} \right)}}{2}x + {e^c}\left( {\frac{{ - e\left( {c - 1} \right) + {e^{ - 1}}\left( {c + 1} \right)}}{2}} \right)$...eq(ii)

How we will proceed

Definitely we need to find the value of intersection x and see whether it is left or right of c

答:
There is more conceptual approach. Since the function $y=e^x$ is convex (concave upwards), tangent at a given point lies below the segment joining two points on either side of given point.

One can see geometrically, if upper line has more (positive) slope, the two lines intersect on the left and if the lower line has more (positive) slope, the two lines intersect on the right.

Slope of tangent is $m_T=e^c$

Slope of line is $$m_L=\dfrac{e^{c+1}-e^{c-1}}{2}$$

All one has to do is compare $m_T$ and $m_L$. Can you finish?


问:
Help with a solution about entropy!
I'm triying to solve this question question, but I don't understand why
$$ -\log_2 (2\pi n pq)/(2n) $$ transforms into  $$O(\ln n/ n)$$
can you help me please :( !! </p>

答: Use that $\log ab=\log a+\log b$ and $\log_b a=\dfrac{\log a}{\log b}$. Here $\log$ is the natural logarithm. Remember you want to show there exists $C$ such that your expression in absolute value is $$\leq C\frac{\log n}n$$</p>

Title: Help with a solution about entropy!


问:
Proof check on reduced residue system proof
Claim: if $a_1,a_2,\dots,a_k$ is a reduced residue system modulo $n$ then $aa_1,aa_2,\dots,aa_k$ is also a reduced residue system modulo $n$.

proof:

Given $\gcd(a,n)=1$. We can find an $x$ such that $ax\equiv 1\pmod{n}$. Also for $a_i$ in the range $1\leq i\leq k$ we have $\gcd(a_i,n)=1$ by definition. So we can find a $y_i$ such that $a_iy_i\equiv 1\pmod{n}$. Then $(xy_i)(aa_i)\equiv(ax)(y_ia_i)\equiv 1\pmod{n}$. This proves that $aa_i$ is invertible modulo $n$. Hence $\gcd(aa_i,n)=1$, that is the $aa_i$'s are coprime to $n$. If $aa_i\equiv aa_j\pmod{n}$, then $xaa_i\equiv xaa_j\pmod{n}$, or $a_i\equiv a_j\pmod{n}$. Since the $a_i$'s are distinct modulo $n$, this is only true if $i=j$. Hence the $aa_i$'s are also distinct modulo $n$. Therefore, $aa_1,aa_2,\dots,aa_k$ is also a reduced residue system modulo $n$.

Is my proof correct? I am unsure about the step where I regroup the product, that is if that actually shows $aa_i$ is invertible or not.

答:
You just need to show $|\{aa_1,aa_2,\dots,aa_k\}|=k$. Suppose it is less than $k$. Then exist $a_i,a_j$, $i\ne j$, such that $$aa_i\equiv aa_j\pmod n$$

But then $n\mid a(a_i-a_j)$ so, by Gauss lemma ($n$ and $a$ are relatively prime) we have $n\mid a_i-a_j$, but this means $a_i\equiv a_j\pmod n$. A contradiction.


问:
Proving $\frac{\cos(3A)-\sin(3A)}{1-2\sin(2A)}=\cos(A)+\sin(A)$

Can someone please show me step-by-step how to prove this trigonometric identity?
$$\frac{\cos(3A)-\sin(3A)}{1-2\sin(2A)}=\cos(A)+\sin(A)$$

This is what I've done so far:

Using the LHS,
$$\frac{cos(2A+A)-sin(2A+A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(2A)cos(A)-sin(2A)sin(A)-sin(2A)cos(A)-cos(2A)sin(A)}{1-4sin(A)cos(A)}$$
$$=\frac{cos(A)(cos(2A)-sin(2A))-sin(A)(sin(2A)+cos(2A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)-2cos^2(A)sin(A)-sin^2(A)cos(A)-cos^2(A)sin(A)-2sin^2(A)cos(A+sin^3(A))}{1-4sin(A)cos(A)}$$
$$=\frac{cos^3(A)+sin^3(A)-3cos^2(A)sin(A)-3sin^2(A)cos(A)}{1-4sin(A)cos(A)}$$

After this I have no idea how to continue to prove the identity.

If anyone could help I would be extremely grateful.

答:
Let $\cos A+\sin A=\sqrt2\cos\left(A-\dfrac\pi4\right)$

$\cos3A-\sin3A$

$=\sqrt2\cos\left(3A+\dfrac\pi4\right)$

$=\sqrt2\cos\left(3A-\dfrac{3\pi}4+\pi\right)$

$=-\sqrt2\cos3\left(A-\dfrac\pi4\right)$ as $\cos(\pi+C)=-\cos C$

Finally $\cos3B=\cos B(4\cos^2B-1)=\cos B(2(1+\cos2B)-1)=?$

Here $B=A-\dfrac\pi4$


问:
Understanding the differences between $\omega + 1$ and $1 + \omega$ using Kenneth Kunen's definition of Ordinal Addition
In Kenneth Kunen's The Foundations of Mathematics, the following statement is proven:

If $R$ well-orders $A$, then there is a unique $\alpha \in ON$ such that $(A;R) \cong (\alpha;\in)$.

In case there is any confusion (because I've seen that sometimes there are slight differences in convention), Kunen establishes the following definitions:

$R$ well-orders $A$ iff $R$ totally orders $A$ strictly and $R$ is well-founded on $A$

$R$ is well-founded on $A$ iff for all non-empty sets $X \subseteq A$, there is a $y \in X$ that is $R$-minimal in $X$

$y \in X$ is $R$-minimal in $X$ iff $\neg \exists z (z \in X \land zRy)$. Alternatively, $y \in X$ is $R$-maximal in $X$ iff $\neg \exists z (z \in X \land yRz))$

$R$ totally orders $A$ strictly iff $R$ is transitive and irreflexive on $A$ and satisfies trichotomy on $A$

Informally we define $ON = \{x: x \text{ is an ordinal} \}$, and the author has proven that $ON$ is well-ordered by $\in$

If $<$ and$\prec$ are relations, then their lexicographic product on $S \times T$ is the relation $\triangleleft$ on $S \times T$ defined by:

$$\langle s,t \rangle \triangleleft \langle s',t' \rangle \leftrightarrow [s < s' \lor [s=s' \land t \prec t']]$$

Finally:

$F$ is an isomorphism from $(A;<)$ onto $(B; \triangleleft)$ iff $F$ is a bijection $(F: A \to B$) and $\forall x,y \in A [ x < y \leftrightarrow F(x) \triangleleft F(y)]$. Then, $(A;<)$ and $(B;\triangleleft)$ are isomorphic (in symbols, $(A;<) \cong (B;\triangleleft)$ ) iff there exists an isomorphism from $(A;<)$ onto $(B;\triangleleft)$.

After proving the above statement, Kunen provides the following two definitions (which are what my question relates to):

Definition 1: If $R$ well-orders $A$ then $\text{type}(A;R)$ is the unique $\alpha \in ON$ such that $(A;R) \cong (\alpha;\in)$. We also write $\text{type}(A)$ (when $R$ is clear from context). or $\text{type}(R)$ (when $A$ is clear from context).

Definition 2: $\alpha \cdot \beta = \text{ type}(\beta \times \alpha) $ and $\alpha + \beta = \text{ type}(\{0\} \times \alpha \ \ \cup \ \ \{1\} \times \beta)$.

Kunen adds: "In both cases, we're using lexicographic order to compare ordered pairs of ordinals"

Using Kunen's definitions of ordinal multiplication and addition, I want to make sure that I am correctly understanding the major features. The two examples I would like to consider are:

$\omega + 1 $ versus $1 + \omega$

$\omega \cdot 2$ versus $2 \cdot \omega$ (Edit: Because this post is running long, I'll just focus on the addition)

Because of Kunen's definitions of ordinal arithmetic, a useful lemma is the following:

If $<$ and $\prec$ are well-orders of $S,T$, respectively, then their lexicographic product $\triangleleft$ on $S \times T$ is a well-order of $S \times T$.

$\color{blue}{\text{ Also, although I only did an informal proof, I am fairly certain that } \{0\} \times \alpha \ \ \cup \ \ \{1\} \times \beta}$ $\color{blue}{\text{is well-ordered, too}}.$

Consider $\omega + 1 $

$\omega$ is an ordinal and $1$ is an ordinal, which we will denote as $\{0\}$. Presumably, the implicit relation associated with these sets is $\in$. Thus:

$\omega + \{0\} = \text{type}(\{0\} \times \omega \ \cup \  \{1\} \times \{0\})=\text{type}(\{ \langle 0,0 \rangle, \langle 0,1 \rangle, \langle 0,2 \rangle,...,\langle 1,0 \rangle\})$

Importantly, note that $\langle 1,0 \rangle$ is a maximal element. Let $A$ be the set $\{ \langle 0,0 \rangle, \langle 0,1 \rangle, \langle 0,2 \rangle,...,\langle 1,0 \rangle\}$.

By the initial statement this post began with: there is a unique $\alpha \in ON$ such that $(A;\in) \cong (\alpha;\in)$

Consider $1 + \omega$

$\{0\} + \omega = \text{type}(\{0\} \times \{0\} \  \cup \  \{1\} \times \omega)=\text{type}(\{ \langle 0,0 \rangle, \langle 1,0 \rangle, \langle 1,1 \rangle \langle 1,2 \rangle, ...\}) $

Importantly, note that there is no maximal element. Let $A'$ be the set $\{ \langle 0,0 \rangle, \langle 1,0 \rangle, \langle 1,1 \rangle \langle 1,2 \rangle, ...\}$

By the initial statement this post began with: there is a unique $\alpha' \in ON$ such that $(A';\in) \cong (\alpha';\in)$

Now, the trouble I am having in both cases is determining which function I should use to describe the isomorphisms...and then, using these functions, demonstrate that $\alpha = S(\omega)$ and $\alpha'=\omega$

For $\alpha \in ON$ such that $(A;\in) \cong (\alpha;\in)$, a good candidate seems to be something along the lines of:

$$f: \begin{cases}
\langle 0,n \rangle \mapsto n \\
\langle 1, 0 \rangle \mapsto \omega
\end{cases}$$

However, the fact that I am mapping $\langle 1, 0 \rangle$ to $\omega$ seems arbitrary. Couldn't I have mapped it to any ordinal that lays "above" the natural numbers?

For $\alpha' \in ON$ such that $(A';\in) \cong (\alpha';\in)$, a good candidate seems to be:

$$f': \begin{cases}
\langle 0,0 \rangle \mapsto 0 \\
\langle 1, n \rangle \mapsto n+1
\end{cases}$$

Once again, however, I'm not quite sure I understand why my mappings are justified. I feel like I am creating these functions rather blindly, and without the proper motivation/justification.

I apologize if this question does not make much sense (still getting used to working with ordinals).

答:
Ordinal addition is "stacking up orders". So $\alpha+\beta$ means that we put a copy of $\beta$ on top of a copy of $\alpha$. Now, what does it mean for $1$ and $\omega$? Well, putting a copy of $\omega$ on top of $1$ is just $\{-1\}\cup\Bbb N$, really. And quite easily, this is again isomorphic to $\Bbb N$ by the obvious function. On the other hand, $\omega+1$ means adding a copy of $1$, i.e. a point, on top of $\omega$. So $\omega+1$ is $\Bbb N$ with a "cap", a maximum. Clearly, not the same order as $\omega$.

Since you bring up order multiplication, the idea is that $\alpha\cdot\beta$ is replacing each point in $\beta$ with a copy of $\alpha$. So $\omega\cdot 2$ means consider $\{0,1\}$ and then replacing each of those with a copy of $\omega$: so we really get $\omega+\omega$ (in fact, the very same set, by definition!). On the other hand, $2\cdot\omega$ means replacing each natural number with a copy of $\{0,1\}$. But this is the same as considering $\{0,1\}$ and then $\{2,3\}$, and then $\{4,5\}$, etc. So $\{2n,2n+1\}$ is the $n$th copy. Easily, again, this is the same as $\omega$.


问:
How to solve 3d inequality or optimisation problems?
I want to find a value of $x$ for which these equations hold for the maximum number of values of $y$ and $z$.

$$ 0<x<1 $$
$$ 0<y<1 $$
$$ 0<z<1 $$
$$ 0< x < 2z $$
$$ y<(x+y)/2 < z $$

I am a high schooler who has never come across such '3d inequality/optimisation'. I tried graphing these but could not figure out how.
Can someone please guide me on how to solve these kind of problems, it would provide me great satifaction. Further is there a book or a course I could take which would help me be able to solve these kind of problems? Thanks!

答:
A linear equation $ax+by+cz=d$ determines a plane $$\pi:=\bigl\{(x,y,z\in{\mathbb R}^3\bigm|ax+by+cz=d\bigr\}$$ in our $(x,y,z)$-space, and the linear inequality $ax+by+cz<d$ one of the two half-spaces defined by this plane $\pi$. You are given $10$ such inequalities, and are told to draw the set $S$ of points that fulfill all of them simultaneously. This set $S$ then is the intersection of ten half-spaces. This intersection is a certain convex polytope, it could also be empty.

The inequalities $0<x<1$, $\>0<y<1$, $\>0<z<1$ obviously define the open unit cube $C:=\>]0,1[\>\times\>]0,1[\>\times]0,1[\>$. Hence we know that $S\subset C$, and a nice drawing should be possible. We then are given the additional inequalities
$$y<x,\qquad z>{x+y\over2},\qquad z>{x\over2}\ ,$$
the first of which is a rewriting of $y<{x+y\over2}$. We already know that $y>0$ in $S$. It follows that the inequality $z>{x\over2}$ is automatically fulfilled when $z>{x+y\over2}$.

In all, we have to draw the part of $C$ where $y<x$ and $z>{x+y\over2}$. The equality $y=x$ splits $C$ diagonally in two halves, and we are left with a triangular prism $P$ standing on the $(x,y)$-plane. This prism has three vertical edges of length $1$. The last plane $\pi:\ z={x+y\over2}$ intersects these edges in the points $(0,0,0)$, $\bigl(1,0,{1\over2}\bigr)$, and $(1,1,1)$. The set $S$ then consists of all points in $P\subset C$ lying above $\pi$.


问:
Does negative third derivative imply negative first derivative?
Does negative third derivative imply negative first derivative? For a system, the negative of the derivative of the Lyapunov function means the system is stable. How about the negative third derivative?</p>

答:
No.  Consider $f(x) = x^4 - 4x^3 + 5x^2 + 1$.  This function is everywhere positive: $f > 0$.

Also, $f' > 0$ on $(0,\infty)$ and $f''' < 0$ on $(-\infty, 1)$.

This means that on the interval $(0,1)$ we have both $f''' < 0$ and $f' > 0$.


问:
Valid proof technique for proving compactness of a topological space?
My goal is to prove that if given that the product space $X \times Y$ of two topological spaces is compact, then $X$ and $Y$ are compact. My idea for proof technique is as follows.

Prove that the projection mappings $\operatorname{\pi}: X \times Y \to X$ and $\operatorname{\pi}: X \times Y \to Y$ are continuous.

Prove that the image of the projection mapping is the whole space $X$ and $Y$.

Suppose that either space $X$ or $Y$ is not compact. Use Theorem 26.5 from Munkres ("the image of a compact space under a continuous map is compact") to show that this results in a contradiction.

I am fairly confident in step one and step three is a gimme. My questioning lies in step 2. I am wondering:

Is step 2 even true

If it is true, is this something that is generally accepted in topology or if it needs proving.

I also am looking for feedback to see if as a whole this proof would be sound and valid. If anyone has alternate proof technique ideas, I would gladly be open to listening.

答:
This community wiki solution is intended to clear the question from the unanswered queue.

If exactly one of the spaces $X , Y$ is empty, then the projection to the other space is not onto because $X \times Y = \emptyset$.

If $X, Y = \emptyset$ or $X, Y \ne \emptyset$, then 2. is true by definition of the product.


问:
If the suspensions of two simply connected spaces are homotopy equivalent, are the spaces homotopy equivalent?
If $X$ and $Y$ are two CW complexes such that $\Sigma X$ and $\Sigma Y$ are homotopy equivalent, are the spaces $X$ and $Y$ homotopy equivalent? I have seen a counterexample when $\pi_1(X)$ is not zero, so I am really looking for counterexamples where $X$ and $Y$ are $1$ connected.</p>

答: The spaces $\Sigma(S^n\times S^m)$ and $\Sigma\left(S^n\vee S^m\vee S^{m+n}\right) \cong S^{n+1}\vee S^{m+1}\vee S^{m+n+1}$ are homotopy eqiuvalent, while $S^n\times S^m$ and $S^n\vee S^m\vee S^{m+n}$ are not.


问:
Are the number of maximal opens sets in a Noetherian topological space finite?
I know that a Noetherian topological space (such as Zariski topological space on Affine space) has ascending chain condition on open sets (descending chain condition on closed sets). This means that every ascending chain of open set in such a space has a maximal open set. Now, my question is: are the number of chains in such Noetherian space finite? in other words is the number of maximal open sets finite?</p>

答: No. For instance, $\Bbb A^n$ and $\Bbb P^n$ are T1, and therefore the maximal proper open sets are the complements of points.


问:
Finding the dimension of these ${\mathbb{R}}^4$ subspaces
Consider the vector subspaces $W_1 = \{(x, y, z, t) \in\mathbb R^4 \mid −y + t = 0,\ z + x = 0\}$ and $W_2 = \{(x, y, z, t) \in\mathbb R^4 \mid −2x + 2y − 2t + 2z = 0\}$ from $\mathbb R^4$.

Find their dimensions: $\dim_{\mathbb{R}} (W_1), \dim_{\mathbb{R}} (W_2)$.

I found that $\dim_{\mathbb{R}}W_1 = 2$

(Row Reduction)

$$[1,0,1, 0]$$
$$[0,1,0, -1]$$

But how do I find the dimension of $W_2$?

答:
We have $-2x+2y-2t+2z=0\implies x-y-z+t=0$. Now see one variable $t$ (say) depends upon others.

It's easy to see the set $\{(1,0,0,-1),(0,1,0,1),(0,0,1,1)\}$ is a set of LI vectors.

You can prove that this is the basis as any vector in $W_2$ is of the form $(x,y,z,t)=(x,y,z,y+z-x)=x(1,0,0,-1)+y(0,1,0,1)+z(0,0,1,1)$.


问:
Applying Menger's theorem to graph with 2k-1 non-disjoint paths
Let $G$ be a graph, $s,t$ be distinct non-adjacent vertices of $G$, and let $k\ge 1$.

If there exist $(2k-1)$ paths with ends $s$ and $t$. other vertices except for $s$ and $t$ belong to at most $2$ paths.

Show that there exists $k$ pairwise disjoint paths with ends $s$ and $t$.

My attempt is trying to prove that there is no seperation with order $<k$.

答:
Theorem: (Menger) If $s,t$ are vertices of a graph $G$ and $st \not\in E(G)$, then the minimum size of an $s,t$-cut equals the maximum number of pairwise internally disjoint $s,t$-paths.

Hint: You are on the right track. Why is it that any set of $k-1$ vertices cannot possibly disconnect $s$ and $t$? What is the most number of $s,t$ paths that a set of $k-1$ vertices can get rid of? On the flip side, what contradiction can you reach if a $k-1$ $s,t$-cut exists?

Obtaining a lower bound on the size of a $s,t$-cut allows you to conclude the desired result using Menger's Theorem.


问:
$0^\infty$ and indeterminate forms
I have actually two related questions:

1. Should indeterminate forms be able to attain an infinite number of values to be considered indeterminate?

I’m asking because Wikipedia says:

The expression
1/0 is not commonly regarded as an indeterminate form, because there is not an infinite range of values that
f/g could approach.

2. I know that $0^\infty$ is not indeterminate.

But i want to know why this is wrong:

$\lim_{(x,y)\to(0,\infty)}x^y=e^{y\cdot \ln(x)}$ and if we take the path $y=\frac{a}{\ln(x)}$ it becomes $e^a$ and could attain infinite number of values and therefore is indeterminate.

答:
The issue is that as $x\to 0$ then $y:=a/\ln(x)\to 0$, so what you have written is a justification of $0^0$ being indeterminate.

Whatever path we take, $0^\infty$ will give $0$: this is because if $y\geq 1$ and $0<x<1$ then $0<x^y\leq x$.

Title: $0^\infty$ and indeterminate forms


问:
Show that space with inner product $\sum_{f(x)g(x)\ne 0} f(x)\overline{g(x)}$ is not separable
Consider the space of all functions $f$ on $\mathbb{R}$ for which $f(x)=0$ everywhere, except in a countable number of points in $\mathbb{R}$ and $$ \sum_{f(x)\ne 0} |f(x)|^2<\infty.$$ Define the inner product $$ \langle f,g\rangle = \sum_{f(x)g(x)\ne 0}f(x)\overline{g(x)}.$$

My notes proceed by saying that this space is not separable. The given reasoning: for each sequence of function $f_n$ in this space, there exist functions $f\ne 0$, s.t. $\langle f,f_n\rangle=0,\forall n\in \mathbb{N}$.

I'm trying to find such functions. It's fairly easy to construct an example for specific sequences (f.ex. $f_n(x) = 1$ for $x=n$, $f_n(x)=0$ otherwise, $\forall n\in \mathbb{N}$). My goal is to do this for general sequences. Is following strategy ok?

Consider $(f_n)_n$ in this space. For $n\in\mathbb{N}$, define $I_n=\{x\in\mathbb{R}:f_n(x)\ne 0\}$. Let $I = \bigcup_{n\in\mathbb{N}} I_n$. Then $I$ is countable as the countable union of countable sets. We can now construct a function $f$ such that $f(x)=0,\forall x\in I$. Indeed, let $y\not\in I$, then we can set $f(x)=1$ for $x=y$ and $f(x)=0$ otherwise. Then $$ |\langle f,f_n\rangle|\le \sum_{x\in I} f(x)\overline{f_n(x)}=0, \forall n\in\mathbb{N},$$ but clearly, $f\ne 0$.

Is this ok? Are there other constructions or, more generally, other reasonings to show that the given space is not separable?

Thanks!

答:
Yes, this is a good construction.

Another (related) one goes like this.  For each $t \in \mathbb R$, define $e_t$ by $e_t(t) = 1$ and $e_t(x) = 0$ otherwise.  Then $\{e_t : t \in \mathbb R\}$ is an uncountable orthonormal system in this inner product.


问:
We are to evaluate the problem at the given limit using pi and redicals in our answer as needed.
The Problem:
$$
\int\!\sin^5(4x)\,dx
$$
The formula that I used from the integration tables is:</p>

<p>$$
\int\!\sin^n(u)\,du
$$
My final answer is
$$
(-1/5)\left(\sin^4(4x)\cos(4x)-(4/5)\cos(4x)+(4/15)\right)\cos^3(4x)+C
$$
This answer is wrong.  If anyone can tell me if I am using the wrong formula or shine some light on what I am doing wrong, I would appreciate it.  </p>

答:
The formula is probably a recurrence down to $\sin^{n-2}$.  But for odd $n$ we can do this, after substituting $u=4x, dx=\frac 14 du$: $$\begin {align}\int \sin^5 (4x)\; dx&=\frac 14 \int \sin^5 u\; du\\&=\frac 14\int(1-\cos^2u)^2\sin u \;du\\&=\frac 14\int(1-2\cos^2 u +\cos^4 u)\sin u \; du\\&=\frac 14(-\cos u+\frac 23\cos^3 u-\frac 15 \cos^5 u)
\\ &=\frac 14(-\cos (4x)+\frac 23\cos^3 (4x)-\frac 15 \cos^5 (4x)) \end {align}$$</p>


问:
Dirichlet series for square root of Riemann Zeta function
Can we obtain Dirichlet series for the function $\sqrt{\zeta(s)}$? Is it possible via Euler product for $\zeta(s)$?

答:
If there is a solution then it will be the product of the Dirichlet series of   terms $(a_n)_{n\ge 1}$ with itself and thus have to verify the Dirichlet convolution $\;\displaystyle\sum_{d|n}\;a_d\cdot a_{n/d}=1$.

From this we obtain $\,a_1=1\,$ and $\ \ \displaystyle 2\cdot a_n=1-\sum_{d|n,d>1,d<n}\;a_d\cdot a_{n/d},\ \ n>1$ and thus :

$$\sqrt{\zeta(s)}=1+\frac 1{2\cdot 2^s}+\frac 1{2\cdot 3^s}+\frac 3{8\cdot 4^s}+\frac 1{2\cdot 5^s}+\frac 1{4\cdot 6^s}+\frac 1{2\cdot 7^s}+\frac 5{16\cdot 8^s}+\frac 3{8\cdot 9^s}+\frac 1{4\cdot 10^s}\\+\frac 1{2\cdot 11^s}+\frac 3{16\cdot 12^s}+\frac 1{2\cdot 13^s}+\frac 1{4\cdot 14^s}+\frac 1{4\cdot 15^s}+\frac {35}{128\cdot 16^s}+\frac 1{2\cdot 17^s}+\frac 3{16\cdot 18^s}+\cdots$$

Which is OEIS A046643  and A046644 for the numerators and denominators with more references.

For fun let's add the expansion of the cube root :
$$\sqrt[3]{\zeta(s)}=1+\frac 1{3\cdot 2^s}+\frac 1{3\cdot 3^s}+\frac 2{9\cdot 4^s}+\frac 1{3\cdot 5^s}+\frac 1{9\cdot 6^s}+\frac 1{3\cdot 7^s}+\frac {14}{81\cdot 8^s}+\frac 2{9\cdot 9^s}+\frac 1{9\cdot 10^s}+\frac 1{3\cdot 11^s}+\frac 2{27\cdot 12^s}+\frac 1{3\cdot 13^s}+\frac 1{9\cdot 14^s}+\frac 1{9\cdot 15^s}+\frac {35}{243\cdot 16^s}+\frac 1{3\cdot 17^s}+\frac 2{27\cdot 18^s}+\cdots$$
quartic root :
$$\sqrt[4]{\zeta(s)}=1+\frac 1{4\cdot 2^s}+\frac 1{4\cdot 3^s}+\frac 5{32\cdot 4^s}+\frac 1{4\cdot 5^s}+\frac 1{16\cdot 6^s}+\frac 1{4\cdot 7^s}+\frac {15}{128\cdot 8^s}+\frac 5{32\cdot 9^s}\\+\frac 1{16\cdot 10^s}+\frac 1{4\cdot 11^s}+\frac 5{128\cdot 12^s}+\frac 1{4\cdot 13^s}+\frac 1{16\cdot 14^s}+\frac 1{16\cdot 15^s}+\frac{195}{2048\cdot 16^s}+\frac 1{4\cdot 17^s}+..$$
and quintic root :
$$\sqrt[5]{\zeta(s)}=1+\frac 1{5\cdot 2^s}+\frac 1{5\cdot 3^s}+\frac 3{25\cdot 4^s}+\frac 1{5\cdot 5^s}+\frac 1{25\cdot 6^s}+\frac 1{5\cdot 7^s}+\frac {11}{125\cdot 8^s}+\frac 3{25\cdot 9^s}+\frac 1{25\cdot 10^s}+\frac 1{5\cdot 11^s}+\frac 3{125\cdot 12^s}+\frac 1{5\cdot 13^s}+\frac 1{25\cdot 14^s}+\frac 1{25\cdot 15^s}+\frac {44}{625\cdot 16^s}+\frac 1{5\cdot 17^s}+.$$
$$ $$
Interesting! In fact the coefficient of the $k$-term of the Dirichet series for $\,\sqrt[N]{\zeta(s)}\,$ appears to be :

$\displaystyle\dfrac 1{N^d}\quad$ if $k$ is the product of $d$ distinct primes.

$\displaystyle\dfrac {\prod_{i=1}^{m-1} N+\frac 1i}{m\,N^m}\quad$ if $k$ is the $m$-th power of a prime ($m>1$).

a product of the previous ones for products of powers of distinct primes !?


问:
Changing the order of an integral
$\int_{0}^{1} \int_{\sqrt{x}}^{1}  \int_{0}^{1-y} f(x,y,z) \ dz \ dy \ dx $

To change the order of the integral in the form

$\int \int  \int f(x,y,z) \ dy \ dz \ dx $

My work:

$z = 1-y$

$y = 1-z$

$0< y < 1-z$

$\sqrt{x} < z< 1$

$0<x<1$

That is

$\int_{0}^{1} \int_{ \sqrt{x}}^{1}  \int_{0}^{1-z} f(x,y,z) \ dy \ dz \ dx $

Is this correct. Please advise.

答:
No that is not correct. You have changed the region itself. You only need to change the order of integration and hence the bounds of variables.

For example, you changed the parabolic cylinder from $y = \sqrt x \,$ to $ \,z = \sqrt x \,$.

Please note this is the region is in the first octant bound below by the parabolic cylinder $y = \sqrt x$ and above by the plane $y + z = 1 \,$.

So for order of integration $dy \, dz \, dx$,

$\sqrt x \leq y \leq 1 - z, 0 \leq z \leq 1 - \sqrt x, \,0 \leq x \leq 1 $

Title: Changing the order of an integral


答: Short answer:$$0^\infty=(\exp-\infty)^\infty=\exp(-\infty\times\infty)=\exp(-\infty)=0.$$Long answer: suppose $f(x) \to0^+,\,g(x)\to\infty$ as $x\to c$. Then$$\ln f\to-\infty\implies g\ln f\to-\infty\implies f^g\to0.$$

Title: $0^\infty$ and indeterminate forms

Body:
I have actually two related questions:

1. Should indeterminate forms be able to attain an infinite number of values to be considered indeterminate?

I’m asking because Wikipedia says:

The expression
1/0 is not commonly regarded as an indeterminate form, because there is not an infinite range of values that
f/g could approach.

2. I know that $0^\infty$ is not indeterminate.

But i want to know why this is wrong:

$\lim_{(x,y)\to(0,\infty)}x^y=e^{y\cdot \ln(x)}$ and if we take the path $y=\frac{a}{\ln(x)}$ it becomes $e^a$ and could attain infinite number of values and therefore is indeterminate.


答:
An alternative way is to use Taylor series:
$$
\sqrt {\zeta (s)}  = \sqrt {1 + (\zeta (s) - 1)}  = 1 + \frac{1}{2}(\zeta (s) - 1) - \frac{1}{8}(\zeta (s) - 1)^2  +  \cdots .
$$
Now just substitute the Dirichlet series of the zeta function into this expansion, re-expand it and collect the terms involving $\frac{1}{n^s}$.

Title: Dirichlet series for square root of Riemann Zeta function

Body: Can we obtain Dirichlet series for the function $\sqrt{\zeta(s)}$? Is it possible via Euler product for $\zeta(s)$?


问:
Dice probability when a number is disallowed in the first round
In the game Settlers of Catan a player starts each turn rolling 2 six sided dice. There's a variation of the game where if a 7 is rolled in the first round of a game (a 'round' is when each player has taken a turn) it doesn't count an needs to be re-rolled. This must lower the overall probability of a 7 being rolled for the rest of the game (and I'd imagine increase the odds for all the other numbers). How would this probability be calculated?</p>

答:
There are exactly 6 ways to roll a seven with two dice:

\begin{array}{|c|c|}
\hline
\text{First die} & \text{Second die} \\
\hline
1 & 6 \\ \hline
2 & 5 \\ \hline
3 & 4 \\ \hline
4 & 3 \\ \hline
5 & 2 \\ \hline
6 & 1 \\ \hline
\end{array}

The total number of ways of rolling two dice is $6 \times 6 = 36$, since there are $6$ ways of rolling each die individually.  Therefore, if you don't reroll sevens, you will get a seven with probability $\frac{6}{36} = \frac16$.

If you do reroll sevens, during the time you reroll them
you will have $0$ probability of getting a seven.  You will keep rerolling until you get some other case, and each other case is still equally likely, so now there are only $36 - 7 = 29$ cases left, instead of the usual $36$.

The effect of the first round of the game has no bearing on the probability of a seven being rolled for the rest of the game.  This is the nature of probability; the two parts of the game are entirely independent.  However, one other thing we can compute that maybe you wanted to know is how much rerolling sevens increases the probability of rolling other things.  Well, each individual case (such as 1, 1 or 3, 1) now has probability $\frac1{29}$ of being rolled instead of the original $\frac{1}{36}$, so the probability has increased by
$$
\frac{1/29}{1/36} = \frac{39}{29} \approx 1.24137
$$
when sevens are rerolled.
So each number (2,3,4,5,6,8,9,10,11,12) appears about $24$ percent more often than normal during the first round.


问:
Let $f(x)$ be differentiable on for all $x\in\mathbb{R}$, and $|f'(x)|\le2$, for every $x$, and $f(9)=33$, $f(1)=17$. what is $f(5)=?$
I can't get how I can know stuff about $f(5)$ from the information that I got.
I tried to apply Lagrange: $\exists  c\space such\space  that\space 1<c<9 \space \space :f'(c)=\frac{f(9)-f(1)}{9-1} =2 $
 and I have no idea how to reach $f(5)$ from here.

any hints or tips or solutions will be appreciated!

答:
$g(x):=f(x)-2x$;

$g(9)=g(1)=15$

$g'(x)=f'(x)-2 \le 0$ meaning $g(x)$ is monotone non-increasing.

Therefore $g$ is constant between $1$ and $9$, $f(5)=g(5)+10=25$.


问:
How to find the limit $\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$
Find the limit
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx
$$

By intuition, I can guess the answer is $0$, but I have no idea how to start to prove it.

What I have tried is using the mean value theorem. There exists $c \in (0, 1)$:
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx = \lim_{n \to \infty}\frac{c^ne^c}{1 +e^c}
$$
and I don't see how I should continue from here.

答:
From
$$
0\le \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \int_0^1{\frac{x^n}{1}}dx=\int_0^1{x^n}dx
$$
From the squeeze theorem, take limits on both sides and we get
$$
0\le \lim_{n\rightarrow \infty} \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\lim_{n\rightarrow \infty} \int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \lim_{n\rightarrow \infty} \int_0^1{x^n}dx=0
$$
and thus
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx=0
$$


问:
Question about matrix with orthonormal columns.
I met a question which says, if I have two matrix $A\in M_{n\times k}$ and $B\in M_{n\times l}$, both with orthonormal columns, with $n>k>l$. The question is, how to prove that $(AA^T - BB^T)x = x$ has a solution for some $x\neq0$?

答:
Since the columns of $A$ are orthonormal, then $AA^T$ is the identity on any column on $A$. So if $V$ is the space generated by the columns of $A$, then $AA^T$ is the identity when restricted to $V$.

Let $W$ be the space generated by the columns of $B$.  Since $k>l$, the dimension of $W^\perp$, wich is $n-l$ is bigger than $n-k$. So $V$ and $W^\perp$ have a nontrivial intesection.

Let $x\in V\cap W^\perp$. For such $x$ we have $AA^Tx=x$ and $BB^Tx=B(B^Tx)=B0=0$

So $(AA^T-BB^T)x=x$.


问:
Use the quotient rule of logs to expand $\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$
I am to expand $\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$ using the quotient rule of logs. I arrived at:

$$\log_3(7)+\log_3(4)+\log_3(x)-\log_3(x-1)-\log_3(x-2)$$

Whereas my textbook solution is:

$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$

It looks like I got it right for the denominator part but not the numerator. I don't see how to arrive at $\log_3(x+3)$ from $7x^2+21x$?

My train of thought was to divide $7x^2$ in the numerator by the $7x$ in the denominator leaving me with $7x+21x = 28x$ in the numerator and just $(x-1)(x-2)$ in the denominator.

With $28x$ in the numerator I expanded out to be $$\log_3(7)+\log_3(4)+\log_3(x).$$

Where did I go wrong, and how can I arrive at $$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$

答: First isolate $7x$ from the numerator. You then have $\log_3\left[\frac{7x(x+3)}{7x(x-1)(x-2)}\right] = \log_3\left[\frac{(x+3)}{(x-1)(x-2)}\right] _{(x\ne0)} = \log_3(x+3)-\log_3(x-1)-\log_3(x-2)$


答:
We have
$$
0 < \frac{{e^x }}{{1 + e^x }} < 1
$$
for $0<x<1$, thus
$$
0 < \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx}  < \int_0^1 {x^n dx}  = \frac{1}{{n + 1}}.
$$
Consequently, by the squeeze theorem,
$$
\mathop {\lim }\limits_{n \to  + \infty } \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx}  = 0.
$$

Title: How to find the limit $\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$

Body:
Find the limit
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx
$$

By intuition, I can guess the answer is $0$, but I have no idea how to start to prove it.

What I have tried is using the mean value theorem. There exists $c \in (0, 1)$:
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx = \lim_{n \to \infty}\frac{c^ne^c}{1 +e^c}
$$
and I don't see how I should continue from here.


答:
"My train of thought was to divide 7$x^2$ in the numerator by the 7 in the denominator leaving me with 7+21=28 in the numerator and just (−1)(−2) in the denominator."

Once re-read your statement!!
when you divide 7$x^2$ with 7x you end up with x not 7x;

you will be left with (x+3) not (7x+21x) when you divide (7$x^2$ + 21x)/7x .

Title: Use the quotient rule of logs to expand $\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$

Body:
I am to expand $\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$ using the quotient rule of logs. I arrived at:

$$\log_3(7)+\log_3(4)+\log_3(x)-\log_3(x-1)-\log_3(x-2)$$

Whereas my textbook solution is:

$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$

It looks like I got it right for the denominator part but not the numerator. I don't see how to arrive at $\log_3(x+3)$ from $7x^2+21x$?

My train of thought was to divide $7x^2$ in the numerator by the $7x$ in the denominator leaving me with $7x+21x = 28x$ in the numerator and just $(x-1)(x-2)$ in the denominator.

With $28x$ in the numerator I expanded out to be $$\log_3(7)+\log_3(4)+\log_3(x).$$

Where did I go wrong, and how can I arrive at $$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$


答:
You can factor the numerator and denominator, both have a common factor $7x$, so you have
$$\frac{7x^2+21x}{7x(x-1)(x-2)}=\frac{x+3}{(x-1)(x-2)},$$ now you can use the $\log$ rules and get the answer that the book gives


问:
Using recurrence relation to solve the given equation for $a_{n+2}= 7a_{n+1} -12a_n$.
Find the general solution of the resurrence relation $a_{n+2}-7a_{n+1}+12a_n=0$.

so what I did is as follows:
I am using recurrence to solve this.

So the equation for the above will be $r^2-7r+12=0$. Solving the same we get roots as $r=4, 3$

so we can write $a_n = C_1 (4^n) + C_2(3^n)$

Is this okay or do I have to do some more steps after this?

答: Unless you are given some initial conditions, you are done.  If you have values at two points you can set up a pair of simultaneous equations to evaluate $C_1,C_2$


问:
What is the basis of a subspace/plane
Given the subspace $L=\{(a,b,c)\in \mathbb{R}^3 : 2a-4b-c=0 \}$ I need to find the projection of a vector onto this subspace. I therefore need to find the span of this subspace/plane. How do I do that?

Edit: the span of a subspace is probably not what I am looking for. What I believe I actually wanted to ask is what a basis of this subspace is such than $span=\{ \vec{V_1}, \vec{V_2}... \}$ spans the subspace $L$ and $V_1$ and $V_2$... is a basis for $L$

答:
$L$ has codimension $1$, as it is given by one linear equation (hyperplane) and so dimension $2$.
A basis is e.g. $\{(2,1,0), (0,1,-4)\}$.

Title: What is the basis of a subspace/plane


问:
Tackling a slightly modified sum identity involving binomials
Summing binomials with alternating signs is known in certain cases, for instance:
$$
\sum_{j=0}^{n} (-1)^j\binom nj=0 \tag{1}
$$
I am trying to show that when the prefactor of the binomial is divided by a first order polynomial of type $j+1$, the sum becomes $\frac{1}{1+n}$. In other words, that

$$
\sum_{j=0}^{n} \frac{(-1)^j\binom nj}{j+1}=\frac{1}{1+n} \tag{2}
$$
I've tried first tackling the case of $j$ instead of $j+1$ in the denom., but it didn't lead to anything simpler than the current problem. Any ideas for how to make progress showing such identity would be much appreciated, or referring to similar problems that are solved.

答:
This class of techniques - which I love - is called generating functions.
The idea is that some algebraic manipulations on series and polynomials translate some non-intuitive operations on a sequence. In this case you know the summation of a sequence, and you want to know what's the summation of the sequence with a $1/(j+1)$ in front of. You will see this principle in action in a while.

Consider the polynomial

$$p(x) = \sum_{j=0}^n \frac{x^{j+1}}{j+1} \binom{n}{j} $$

Note that $x^{j+1}/(j+1)$ is the antiderivative of $x^j$, so that

$$ p(x) = \sum_{j=0}^n \int x^j \binom{n}{j} = \int \sum_{j=0}^n x^j \binom{n}{j} = \int (1+x)^n = \frac{(1+x)^{n+1} -1}{n+1} $$

The $-1$ pops up to match the constant term ($p(0) = 0$). Now note that when we evaluate $p(x)$ in $-1$ we almost get our result; indeed

$$ -p(-1) = -\sum_{j=0}^n \frac{(-1)^{j+1}}{j+1} \binom{n}{j} = \sum_{j=0}^n \frac{(-1)^j}{j+1}\binom{n}{j} $$

is what we are looking for. Using the antiderivative like manipulation we get that

$$ -p(-1) = -\frac{ (1-1)^{n+1}-1}{n+1} = \frac{1}{n+1}$$

As desired.


问:
Derivative question
I need to find the derivative of:

$$
h(x) = \int_{0}^{x^2} (1-t^2)^{1/3} \, dt
$$

Would the answer to that just be:

$$
(1-x^4)^{1/3}?
$$

答:
If $F(x)=\int_{a}^{x}f(t)\;\mathrm{d}t$, then $F^{\prime }(x)=f(x)$. By the
chain rule if $F(x)=\int_{a}^{u(x)}f(t)\;\mathrm{d}t$, then

$$F^{\prime }(x)=F^{\prime }(u)u^{\prime }(x)=f(u(x))u^{\prime }(x).$$

In the present case $F(x)=h(x)$, $u(x)=x^{2}$ and $f(t)=(1-t^{2})^{1/3}$.
Hence $u^{\prime }(x)=2x$ and $f(u(x))=f(x^{2})=(1-x^{4})^{1/3}$.

Thus

$$h^{\prime }(x)=2(1-x^{4})^{1/3}x.$$

A generalization is to find the derivative of an integral where both limits are functions of $x$, as in this question.

Title: Derivative question


问:
Linearize ODE given by $x'=x^2+y^2-4$, $y'=x^2-y^2$

Linearize ODE given by $x'=x^2+y^2-4$, $y'=x^2-y^2$

Find all critical points, find the linearization at each critical point.

So I computed the jacobian matrix, $\begin{pmatrix}2&2\\2&-2\\\end{pmatrix}$

Then I found the eigenvalues to be $2\sqrt{2},-2\sqrt{2}$

For a critical point I need $x'(x,y)=0$ and $y'(x,y)=0$

$x^2-y^2=0$, then $x^2=y^2$ then $x^2+y^2-4=x^4-4=0$, so $x^2=2$

So I believe then the critical points are $(\sqrt{2},\sqrt{2}),(-\sqrt{2},-\sqrt{2})$

Do I know these are all critical points?

And then my linearization of the critical points is just $Jx$?

答:
The linearization of the system $$\left\{ \begin{align} \frac{dx}{dt}=f(x,y)\\ \frac{dy}{dt}=g(x,y)\end{align}\right.$$ at a point $(x_0,y_0)$ is

$$\left\{ \begin{align} \frac{dx}{dt}=f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)\\ \frac{dy}{dt}=g(x_0,y_0)+\frac{\partial g}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial g}{\partial y}(x_0,y_0)(y-y_0)\end{align}\right.$$ Thus, if $(x_0,y_0)=(\sqrt 2,\sqrt 2)$ it is

$$\left\{ \begin{align} \frac{dx}{dt}=2\sqrt 2(x-\sqrt 2)+2\sqrt 2(y-\sqrt 2)\\ \frac{dy}{dt}=2\sqrt 2(x-\sqrt 2)-2\sqrt 2(y-\sqrt 2)\end{align}\right.$$


答:
Hint:

The relation

$$\frac1{j+1}\binom nj=\frac1{n+1}\binom{n+1}{j+1}$$ will turn your second identity to the first.

Title: Tackling a slightly modified sum identity involving binomials

Body:
Summing binomials with alternating signs is known in certain cases, for instance:
$$
\sum_{j=0}^{n} (-1)^j\binom nj=0 \tag{1}
$$
I am trying to show that when the prefactor of the binomial is divided by a first order polynomial of type $j+1$, the sum becomes $\frac{1}{1+n}$. In other words, that

$$
\sum_{j=0}^{n} \frac{(-1)^j\binom nj}{j+1}=\frac{1}{1+n} \tag{2}
$$
I've tried first tackling the case of $j$ instead of $j+1$ in the denom., but it didn't lead to anything simpler than the current problem. Any ideas for how to make progress showing such identity would be much appreciated, or referring to similar problems that are solved.


问:
matrix operator 2 norm inequality involving product of three matrices
Let $A$ be a symmetric positive definite matrix, is it true that
$$
\| BAB^T \|_2 \geq C \| B\|^2_2\| A\|_2
$$
for some constant $C$? Assuming all matrices are real and the constant may depend on the size $n$.

Here $\Vert \cdot \Vert_2$ is the induced/operator 2-norm defined as

$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$

答:
Yes, you can find a constant that is dependent on $A$ alone.

Based on your comment you can write
$$
\begin{aligned}
\|BAB^T\|_2 &= \|(BA^{1/2}) (B A^{1/2})^T\|_2 \\
&= \|BA^{1/2}\|^2_2 \\
&\geq \|B\|_2^2 \sigma_{\min}(A^{1/2})^2 \\
& = \|B\|_2^2 \sigma_{\min}(A)\\
\end{aligned}
$$
where on the second line we used the fact that $\|XX^*\|_2 = \|X\|_2^2$ (transposition gives the adjoint operator if we view the vector space as a real inner product space and note that the $2$-norm of the vector can be defined via this inner product). On the third line we used the inequality found in this question. Noting that $\|A\|_2 = \sigma_{\max}(A)$ we can take $C = \frac{\sigma_{\min}(A)}{\sigma_{\max}(A)}$ which is non-trivial as $A$ is positive definite.


问:
$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_2\cdot\aleph_1^{\aleph_0}$
I've seen this statement in multiple posts (e.g. here and here), but I can't seem to understand it. I can see why

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0},$$

by noting that every $\alpha<\omega_2$, so $|\alpha|\leq\aleph_1$, meaning

$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\max_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0}.$$

So how does $\aleph_2$ enter the equation?

答:
We can prove something quite more general. Given a regular cardinal $\kappa$, noticing that each function $\lambda\rightarrow \kappa$ for $\lambda<\kappa$ ($\lambda$ cardinal) must have bounded range in $\kappa$, we can write $$\kappa^{\lambda}=\bigcup_{\alpha <\kappa} \alpha^{\lambda},$$ where $\alpha$ ranges over ordinals. Therefore we get easily $$\kappa^{\lambda}=\sum_{\alpha< \kappa}\vert \alpha\vert^{\lambda}.$$ Using this fact, since each successor cardinal is regular, we can prove (for all $\alpha,\beta$ ordinals) the Hausdorff's Formula:
$$\aleph_{\alpha +1}^{\aleph_{\beta}}=\aleph_{\alpha +1}\cdot\aleph_{\alpha}^{\aleph_{\beta}}.$$
(The question is a particular case for $\alpha =1$ and $\beta =0$).


答:
$$F(s)=\prod_p (1+\sum_{k\ge 1} {-1/2\choose k}(-1)^kp^{-sk})=\sum_{n\ge 1} a_nn^{-s}$$
is the unique Dirichlet series such that $F(s)^2=\zeta(s)$ and $F(\infty)=1$. It converges for $\Re(s) > 1$. It has an awful branch point at $s=1$ making the asymptotic of $\sum_{n\le x} a_n$ complicated.


问:
A man said I am only four times my son's age.

A man said "I am only four times my son's age. My father is 10 years less than twice my age. And together we form age of 120 years old”. Find out age of the man

Someone please answer this question, I'm not able to figure out the solution. Please solve this problem.

Thanks & Cheers,
Pradeep Kumar.M

答:
Let x = the man's age

x/4 = the man's son's age

2x - 10 = the man's father's age

x/4 + 2x - 10 + x = 120

x/4 + 3x = 130

13x = 520

x = 40

thus, the age of the man is 40


问:
Why is a cylinder Lebesgue-measurable?
Let $C\subseteq \mathbb{R}^{n-1}$ be Lebesgue-measurable and $h\geq 0$. Prove that $Z:=C\times [0,h]$ is Lebesgue-measurable.

Sketch:

Since $C$ is measurable, I thought that I could lift the set $C$ up to different levels $z=h$ and the set would still be measurable. A countable union of measurable sets is still measurable and thus $C\times [0,h]$ is measurable.

Is that a working method?

EDIT: I just realised that $[0,h]$ with $h\geq 0$ is an uncountable set.

答:
Let $\mathcal M^n$ the set of Lebesgue measurable set of $\mathbb R^n$. Notice that $$\mathcal M^{n-1}\otimes \mathcal M=\sigma \{A\times B\mid A\in \mathcal M^{n-1},B\in \mathcal M^1\}\subset \mathcal M^{n}.$$
Since $C\times [0,h]\in \mathcal M^{n-1}\otimes \mathcal M$, it belong to $\mathcal M^n$.

Title: Why is a cylinder Lebesgue-measurable?


问:
Calculating in quotient ring of $\mathbb{R}[X]$
Part of an old Oxford exam (1992 A1)

We want to find which elements of the quotient ring $\mathbb{R}[X]/(x^3-x^2+x-1)$ are equal to their own square.

Now, we note first that $x^3-x^2+x-1=(x-1)(x^2+1)$

Let $f(x)=(x-1)(x^2+1)$. Clearly we have $[1+(f(x))]^2=[1+(f(x))]$ but I cannot see how to find others. Any guidance would be appreciated.

Thanks.

答:
Your previous question concerns generalized CRT. So you know that

$$\frac{{\bf R}[x]}{(x-1)(x^2+1)}\cong\frac{{\bf R}[x]}{(x-1)}\times \frac{{\bf R}[x]}{(x^2+1)}\cong {\bf R}\times{\bf C}$$

Find the idempotents in $\bf R$ and $\bf C$ to find the idempotents in ${\bf R}\times{\bf C}$, then pull them back through the isomorphisms implicit above.

Indeed the idempotents of ${\bf R}\times{\bf C}$ are $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$. The first and last correspond to $0$ and $1$ in $S={\bf R}[x]/(x-1)(x^2+1)$. To find what the others correspond to, solve e.g.

$$f(x)\equiv\begin{cases}0 & \mod x-1, \\ 1 & \mod x^2+1. \end{cases} $$

without loss of generality with $\deg f(x)\le 2$. Thus we can write $f(x)=(ax+b)(x-1)$ and solve for $a,b$ by distributing out this product, reducing mod $x^2+1$ and setting equal to zero. The other case proceeds in exactly the same way.


问:
How do I solve the following example using Kmap?
I am trying to solve the SOP function: Em(0,1,2,6,8,9,10). Here's what I got in the table:

```<code>    C'D' C'D CD CD'
A'B' 1   1   0  1
A'B  0   0   0  1
AB   0   0   0  0
AB'  1   1   0  1
</code>```

My output function is y = B'C'+A'CD'+AB'D'. Is it accurate? My confusion is primarily about the position AB'CD'. Do I group it with AB'C'D' or all positions at the corners that include positions A'B'C'D' & A'B'CD'?

答:
To simplify as much as possible, you always want to create as large as groups as possible.

So yes, use the expression $B'D'$ to cover all corners instead of using $AB'D'$

So, the ultimate expression is $B'C'+A'CD'+B'D'$ ... which is indeed a little simpler than what you got.


问:
how to calculate the steady state mc linear equations
i have the following Markov chains equations and not sure how to solve those:
a=.2 a +.5b+0.6d
b=.1a+.1b+.2d
c=0.7a+.1c
d=.4b+.9c+.2d
pi=a+b+c+d
how do you approach that? i tried
-.8a+.5b+.6d=0
but that just gave me zeros for everything. the teacher did not show me how to resolve those and i m not sure how to approach ..

答: You need to remove the singularity by removing one of the linearly dependent equations and use instead the probability condition: $a +  b + c + d = 1$.


问:
Verify that $U + W'$ is also given by $U + W$
Suppose that $U, W$ and $W'$ are all subspaces of $\mathbb F^3$, then if $ U = \{(x, 0, 0) : x \in\mathbb F\}$ and $W = \{(0, y, 0) : y \in \mathbb F$} their sum is equal to
$$ U + W = \{(x, y, 0): x,y \in \mathbb F\}$$
But if $ W' = \{(y, y, 0): y \in \mathbb F\}$, show that $U + W'$ is also given by
$$ U + W' = \{(x, y, 0): x,y \in \mathbb F\}$$
if $U$ is the same as above.

How should I show this? Im a bit confused since for $U + W'$ I would have $(x+y, y, 0)$. Should I somehow show that these two sets are the same and contain the same elements as in $U+W = U+W'$?

答:
Let $V=\{(x,y,0):x,y\in F\}$. We show that $U+W'=V$.

"$U+W'\subset V$": For any $(x,0,0)+(y,y,0)=(x+y,y,0)\in U+W'$, clearly we have $(x+y,y,0)\in V$. Hence $U+W'\subset V$.

"$U+W'\supset V$": For any $(x,y,0)\in V$, let $z=x-y$ and we have $(x,y,0)=(z,0,0)+(y,y,0)\in U+W'$.

Thus $U+W'=V$.


问:
find a function that describes win percentage based on skill
I'm searching for a function $C:D\to[0,1]$ (where $D=[0,1]^2\setminus\{(0,0),(1,1)\}$)  that can describe the outcome of a match between two players, $p_1$ and $p_2$ with skills of $s_1$ and $s_2$. the function would determine the chance p1 would win, and will be written as $C(s_1,s_2)$. therefore, it must have the following properties:

$C(1,x)=1  \space \forall x\in[0,1)$ ,since the perfect payer always wins (such player doesn't exist, of course)

$C(x,y)+C(y,x)=1 \space\forall (x,y)\in D$, because if you have a chance of $p$, the other person has a chance of $1-p$.

while those are the two essential rules, I've derived some rules that would make the function behave nicer.

$C(x,C(x,y))=y\space\forall (x,y)\in (0,1)\times[0,1]$

$C(1-x,1-y)=C(y,x)\space\forall (x,y)\in D$

$C$ is continuous and differentiable.

simple rules can be derived from those rules but it lead me nowhere close to figuring out a numeric description of such function, so if you can help me find a function that satisfies all 5 rules, preferably simple, I would very much appriciate it.
also note, that is assuming a player's skill can be described at a range from 0 to 1, which is very much not always the case. however, please ignore the logic of the calculation, and focus on finding an answer. thank you.

答:
Here it is:

$$C(x,y) = \frac{x(1-y)}{x(1-y)+y(1-x)}=\frac{x-xy}{x+y-2xy}$$

It satisfies all requirements from $1$ to $5$.

The function comes thinking about probabilities proportional to the skill, which would lead to $\frac{x}{x+y}$ and then adding the perfect player requirement.


问:
How to find the limit $\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$
Find the limit
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx
$$

By intuition, I can guess the answer is $0$, but I have no idea how to start to prove it.

What I have tried is using the mean value theorem. There exists $c \in (0, 1)$:
$$
\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx = \lim_{n \to \infty}\frac{c^ne^c}{1 +e^c}
$$
and I don't see how I should continue from here.

答:
Since the integrand is a continuous function, the mean value theorem for integrals tells us that there exists $\;0<c<1\;$ s.t.

$$\int_0^1x^n\frac{e^x}{1+e^x}\,dx=c^n\frac{e^c}{1+e^c}\xrightarrow[n\to\infty]{}0$$

since $\;c^n\to 0\;$ and $\;0\le \cfrac{e^x}{1+e^x}\le 1\;$ is bounded.


问:
Is it possible to find out whether two pairs of concealed numbers have at least one number in common by applying a single test on each of the 2 pairs?
Two friends, each has one box, and each box conceals two balls carrying 2 unique natural numbers. The two friends need to know whether they have at least one number in common. While not allowed to divulge their numbers to one another, they are allowed to come up with ONE equation/formula and apply it on both boxes, then compare the results. That is, friend-A has 2 balls carrying the numbers {n, m} and applies the formula by plugging in his two numbers, then shares the result he gets with friend-B. Similarly, Friend-B plugs his numbers {x, y} into the very same formula, then they compare the results and judge whether they have one number in common based on that. Using the formula is like a ticket, used once for each person and can't be trial and error.

Is it possible for them to find out whether there exists k $\in$ {n, m} and k $\in$ {x, y} such that k is a natural number, while sticking to these rules?

答:
I think I figured it out.

Alice (with $\{m,n\}$) should compute:  ${\rm Pri}[n]^{{\rm Pri}[m]}$, where ${\rm Pri}[n]$ is the $n$th prime.  For example ${\rm Pri}[4] = 7$, ${\rm Pri}[20] = 71$.  This is implemented in Mathematica by Prime[n].  So if Alice has $\{ 4,6 \}$ her function will produce $7^{16} = 33232930569601$.

Bob (with $\{ x, y \}$) should do likewise.

By the Fundamental Theorem of Mathematics (unique prime factorization), one can recover both unknown or secret values.  For Alice, FactorInteger[33232930569601] gives {{7,16}}.  And likewise for Bob.

It is then a trivial matter to compute if they share a secret number (or both secret numbers), and what its value is.


问:
Laurent Series Expansion of $f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$ at $z=0$
For my function $$f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$$ I have determined that it contains $3$ poles at $z=0,z=-a,z=-\frac{1}{b}$

I believe these to be the only singularities.

I found them by factoring the denominator

$$f(z) = \frac{bz^2+2z+b}{z(bz^2+(1+b^2)z+b)}$$

Then clearly there is a pole at $z=0$ and the remaining singularities are found by solving the quadratic

$$bz^2+(1+b^2)z+b = 0$$

$z = \frac{-(1+b^2) + \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$ and $z =\frac{ -(1+b^2) - \sqrt{(1+b^2)^2-4(b)(b)}}{2b}$

This simplifies to

$z = \frac{-1-b^2 -b^2 +1}{2b}$ and $z=\frac{-1-b^2+b^2-1}{2b}$

$z = \frac{-2b^2}{2b}$ and $z= \frac{-2}{2b}$

So,

$z=-b$ and $z=-\frac{1}{b}$

I want to Laurent Expand these singularities and determine the radius of convergence for each singular point.

All help is appreciated. I want to focus on the $z=0$, once I have clarified the process I will be able to expand the remaining myself.

答:
Before factoring take notice of the similarity of the coefficients in the numerator and denominator. This suggests that there is a short cut that can be taken before attempting to partial fraction that would make our lives easier.

$$\frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz} = \frac{bz^2+(1+b^2)z+b+(1-b^2)z}{bz^3+(1+b^2)z^2+bz} = \frac{1}{z}+\frac{1-b^2}{bz^2+(1+b^2)z+b}$$

Now partial fractions is automatic

$$\frac{1}{z}+\frac{1-b^2}{(z+b)(bz+1)} = \frac{1}{z}+\frac{1}{z+b}-\frac{1}{bz+1}$$

$z=0$ won't actually provide illumination on what to do for the other singularities, so instead I will show the procedure for $z=-b$. Every term needs to be centered at $z=-b$ like so

$$\frac{1}{(z+b)-b}+ \frac{1}{z+b} + \frac{1}{b(z+b)+1-b^2} = -\frac{1}{b}\frac{1}{1-\frac{(z+b)}{b}}+ \frac{1}{z+b} + \frac{1}{1-b^2}\frac{1}{\frac{b(z+b)}{1-b^2}+1}$$

The scaling is done because it is important to have the $+1$ to apply geometric series. Now we assume the quantity $z+b$ is small so we can use the Taylor expansion

$$-\frac{1}{b}\sum_{n=0}^\infty \frac{(z+b)^n}{b^n} + \frac{1}{z+b} + \frac{1}{1-b^2}\sum_{n=0}^\infty (-1)^n\left(\frac{b}{1-b^2}\right)^n(z+b)^n$$

The case $b=1$ needs to be handled separately since in that case $b=\frac{1}{b}$


问:
The equation $A x=b$ has a solution $x \in \mathbb{R}^{k}$ whenever $y \in Null(A^T)~ \Rightarrow y \in~ Null(b^T)$
If $b^{T} y=0$ for every $y \in \mathbb{R}^{n}$ such that $A^{T} y=0$, show that the equation $A x=b$ has a solution $x \in \mathbb{R}^{k}$.

Here if $b \neq 0$, how can we say that  $rank(A:b)=rank(A)$?

答:
Hint: Consider the relationship between the rank of $[A \mid b]$ and the nullspace of
$$
[A \ \mid \ b]^T = \pmatrix{A^T\\ b^T}.
$$


问:
What is the maximum possible value of $E[X_1 X_2 X_3]$?

Assume $X_1,X_2,X_3$ are discrete random varibles defined on a common probability space $\Omega$ and taking values in $\{-1,1\}$. Further, assume that $E[X_1]=E[X_2]=E[X_3]=E[X_1 X_2]=E[X_2 X_3]=E[X_3 X_1]=0$. Given this, what is the maximum possible value of $E[X_1 X_2 X_3]$?

It's easy to see that $P(X_i=\pm 1)=P(X_i X_j = \pm 1)={1 \over 2}$ for each $i,j \in I_3 (i \neq j)$. But how do I progress further? Any help would be appreciated.

答:
Let $a=E[X_1 X_2 X_3]$

Of course we have $-1 \le a \le 1$

Following this parametrization we can write the joint probability as

$$P(x_1,x_2,x_3)=\frac18( a \, x_1 x_2 x_3  +1)$$
which gives additional restrictions $$0\le P(x_1,x_2,x_3)\le 1$$
or $0\le \frac18 (1-a) \le 1$ and $0\le \frac18 (1+a) \le 1$

But this is verified by the original candidate for the maximum ($a=1$)

Hence the maximum is $E[X_1 X_2 X_3]=1$ which is attained by

$$P(x_1,x_2,x_3) = \frac18( x_1 x_2 x_3 +1)=
\begin{cases} \frac14 & \text{if } x_1 x_2 x_3 = 1 \\
0 &\text{if } x_1 x_2 x_3 = -1
\end{cases}$$


答:
Let there be four states each with probability $1 \over 4$: $(X_1,X_2,X_3)\in \{(1,-1,-1),(1,1,1),(-1,-1,1),(-1,1,-1)\}$.

You can check that the conditions hold. However,

$$E(X_1X_2X_3)=1,$$

which is clearly the highest value this expression can take.

Title: What is the maximum possible value of $E[X_1 X_2 X_3]$?

Body:
Assume $X_1,X_2,X_3$ are discrete random varibles defined on a common probability space $\Omega$ and taking values in $\{-1,1\}$. Further, assume that $E[X_1]=E[X_2]=E[X_3]=E[X_1 X_2]=E[X_2 X_3]=E[X_3 X_1]=0$. Given this, what is the maximum possible value of $E[X_1 X_2 X_3]$?

It's easy to see that $P(X_i=\pm 1)=P(X_i X_j = \pm 1)={1 \over 2}$ for each $i,j \in I_3 (i \neq j)$. But how do I progress further? Any help would be appreciated.


问:
How to prove this inequality?It's hard for me.
My classmate told me this problem yesterday.
The problem is that:
let $a,b,c,d>0$,
prove that: $\sum_{cyc}{\frac{c}{a+2b}}+\sum_{cyc}\frac{c+2d}{a}\geq8(\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}-1)$

I tried to expand it and used Cauchy-Schwarz-inequality,then I got the following result:
$$
\begin{aligned}\sum_{cyc}{\frac{c}{a+2b}}+\sum_{cyc}\frac{c+2d}{a}&={\frac{c}{a+2b}}+{\frac{d}{b+2c}}+{\frac{a}{c+2d}}+{\frac{b}{d+2a}}+\frac{c+2d}{a}+\frac{d+2a}{b}+\frac{a+2b}{c}+\frac{b+2c}{d}\\&\geq\frac{[4(a+b+c+d)]^2}{4(ab+ac+ad+bc+bd+cd)}\\&=4(\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd})\end{aligned}
$$
BUT this doesn't look right.

答:
Note that your inequality can be rearranged to
\begin{eqnarray*}
\sum_{cyc}\frac{a+2b+ c}{a+2b}+\sum_{cyc} \frac{a+2b+ c}{c} &=& \sum_{cyc}{\frac{(a+2b+ c)^2}{(a+2b)c}}  \\ & \geq & 8\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}
\end{eqnarray*}
Note also that
\begin{eqnarray*}
\sum_{cyc}(a+2b)c  =2(ab+ac+ad+bc+bd+cd).
\end{eqnarray*}
Multiply through by this & now apply CS.


问:
Book suggestions on classic control
I am looking for suggestions about introductory books on classic control theory. Books that begin by trying to develop intuition to the reader using basic principles of the notion of control and investigate a logical flow of thoughts with the mistakes that an inexperienced student might make, until the establishment of the formal mathematical theory.
Would you please have any such suggestions to make?</p>

答:
Classical control is a stable theory and most popular textbooks are similar.

Feedback systems : an introduction for scientists and engineers - Karl J Åström; Richard M Murray is a good text, perhaps making a bigger effort to be intuitive to non-specialists.

Modern Control Engineering by Katsuhiko Ogata is also a good choice, with a more standard treatment. The older editions are no worse than the new ones, and arguably have some advantages.

Feedback Control of Dynamic Systems, by G. F. Franklin, J. D. Powell and A. Emami–Naeini gets into more detail about the frequency-domain design techniques. I think it's an excellent reference.

Other sources exist, and each one has its fans, but you cannot go wrong with any of the 3 above.

Let me add The Art of Control Engineering, by Ken  Dutton, Steve  Thompson, and Bill Barraclough. Perhaps less frequently adopted than the 3 above, but at points makes a bigger effort to emphasize the intuition. Altogether however the 4 texts are not fundamentally distinct.

Title: Book suggestions on classic control


问:
Checking if jointly CRVs are independent via the marginal density
I'm trying to determine whether the following continuous random variables are independent or not given their joint probability density.

$f_{X,Y}(x,y) = 2(x+y)$ for $ 0\le y\le x\le 1$

I calculated the marginal densities as follows:

$f_x(x) = \int_0^1{2(x+y)} dy$ = $ 2x+1$

Using the same method for $f_y(y)$ I got $f_y(y) = 2y+1$

Since $f_{X,Y}(x,y)\neq f_X(x)f_Y(y) $ they are not independent.

My concern is that my calculation of the marginal densities is incorrect, could someone verify whether my method is correct? Or if this is the best way to check if the two are independent.

答:
My concern is that my calculation of the marginal densities is incorrect, could someone verify whether my method is correct? Or if this is the best way to check if the two are independent.

It is not correct. $$f_{\small Y}(y)=\int_y^1 2(x+y)\,\mathbf 1_{0\leqslant y\leqslant 1}\,\mathrm d x\\f_{\small X}(x)=\int_0^x 2(x+y)\,\mathbf 1_{0\leqslant x\leqslant 1}\,\mathrm d y$$

However, there is a easier way.   The support for the joint p.d.f. is $0\leqslant y\leqslant x\leqslant 1$.   That is all you need to know.

 Without even integrating, you can assert that: $$\begin{align}\mathsf P(Y>1/2,X<1/2)&=0\\\mathsf P(Y>1/2)&>0\\\mathsf P(X<1/2)&>0\end{align}$$


问:
Proving Planarity of a graph
I know the condition to be planar graph is that $E \leq 3V − 6$, so if we have a graph where the number of edges is smaller than $3\cdot (vertices) − 6$, can we say its planar? If not, how would I go about proving that a graph is planar? (other than using Kuratowski's Theorem)

答: The Petersen graph is non-planar and has $10$ vertices and $15$ edges.  The inequality $E\leq 3V-6$ is not a "condition for a graph to be planar."  It is a property of planar graphs.  That is, any planar graph will satisfy this inequality, but as the example shows, there are non-planar that satisfy it also.

Title: Proving Planarity of a graph


问:
If $f$ is a bounded measurable function $\Longrightarrow$ there is a sequence of step functions such that $s_n \longrightarrow f \; a.e$?
If $f:[0,1]\longrightarrow\mathbb{R}$ is a bounded measurable function $\Longrightarrow$ there is a sequence of step functions $\displaystyle s_n=\sum_{j=1}^{p} c_j \cdot \chi _{I_j}$ such that $s_n \longrightarrow  f \; a.e$ ?

$\chi _{I_j}$: characteristic function

$I_j$: disjoint intervals with $\bigcup_{j=1}^{p} I_j=[0,1]$

$c_j$: real numbers

Any hints would be appreciated.

答: The function $f$ is integrable. The step functions are dense in $L^1(\mathbb{R})$ (corollary 18 here). A fortiori in $L^1([0,1])$. So there exists a sequence $(s_n)$ of step functions converging to $f$ in $L^1$. Finally, there exists a subsequence of the latter converging a.e. pointwise to $f$. As it is the case in general for convergent sequences in an $L^p$ space.</p>


问:
Does there exist a $4\times4$ matrix $A$ such that $A$ and $A^2$ are not normal but $A^3$ is normal?
I'm asked to provide a proof or a counterexample for this claim:

Does there exist a $4\times4$ matrix $A$ such that $A$, $A^2$ is not normal, but $A^3$ is normal.

Any ideas are welcome Thanks in advance.

答: Here is a counterexample where $A$ is nonsingular. Let $C=\pmatrix{0&-1\\ 1&-1}$ be the companion matrix of the characteristic polynomial $x^2+x+1=0$. One can verify that $C$ and $C^2$ are not normal but $C^3=I_2$. Now take $A=\pmatrix{C\\ &I_2}$ or $A=\pmatrix{C\\ &C}$.


问:
Using recurrence relation to solve the given equation for $a_{n+2}= 7a_{n+1} -12a_n$.
Find the general solution of the resurrence relation $a_{n+2}-7a_{n+1}+12a_n=0$.

so what I did is as follows:
I am using recurrence to solve this.

So the equation for the above will be $r^2-7r+12=0$. Solving the same we get roots as $r=4, 3$

so we can write $a_n = C_1 (4^n) + C_2(3^n)$

Is this okay or do I have to do some more steps after this?

答: You could also express $C_1,\,C_2$ in terms of arbitrary initial conditions $a_0,\,a_1$ by determining $C_1,\,C_2$ from simultaneous equations $a_0=C_1+C_2,\,a_1=4C_1+3C_2$, viz.$$a_n=(a_1-3a_0)4^n+(4a_0-a_1)3^n.$$


问:
$\mathcal{L} \int_{[A,B)} fd\lambda = \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx.$
Let $ -\infty < A < B \leq +\infty$, and let $f:[A,B)\to \mathbb{R}$ be almost continuous everywhere (sucht that $f_{[A,b]} \in \mathcal{R}(A,b)$ for all $A<b<B$). Then $\mathcal{R} \int_A^B f(x)dx := \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx.$

The following is a condition when the improper Riemann integral is the same as the Lebesgue integral. How do I prove this?

If $g \geq 0$ with $g \in \mathcal{L}([A,B))$ and $|f| \leq g$, then

$$\mathcal{L} \int_{[A,B)} fd\lambda = \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx.$$

答:
If $ f $ is almost-everywhere continuous then $ \mathcal{R} \int_A^b{f(x) dx} = \mathcal{L} \int_A^b{f(x) dx} $ for all $ A < b < B $.  So the only thing you have to prove is
$$
(*)\quad \lim_{b \rightarrow B} {\mathcal{L}\int_A^b{f(x) dx}} = \mathcal{L}\int_A^B{f(x) dx}
$$

For this, note $ \mathcal{L}\int_A^b{f(x) dx} = \mathcal{L} \int_A^B{f(x) \chi_{[A,b)} dx} $.  Now letting $ b_n $ be any sequence $ b_n \rightarrow B $, and $ f_n(x) = f(x)\chi_{[A,b_n)} $, we have $ |f_n(x)| < g(x) $ and $ f_n(x) \rightarrow f(x) $ (almost everywhere) on $ [A,B] $.  The dominated convergence theorem immediately gives us the desired limit $ (*) $.


问:
Game Theory Solution to Brexit Negotiations?
Edit: This post is meant for fun: I was initially interested in seeing a wider range of mathematical techniques applied to computing the likelihood of a Deal or No-Deal outcome of the negotiations. Specifically, I was interested in how mathematics can be applied to a current real-world example, wanting to see how different fields of mathematics might approach the problem (i.e. Bayesian probabilists vs. Game Theorists).

Since there have been no answers so far, I have decided to narrow the "question" specifically to Game Theory, as I would be most interested in seeing how this particular discipline could be applied here. If any game theorists are interested in responding, I would be very interested in seeing an example.

Original post: This post is meant for fun: given that we have two days of negotiations left, use any mathematical technique or reasoning to compute the probability of a No-Deal. Be creative: game theory, probabilistic theory, Monte Carlo: it's all allowed.

I will post my own answer in a few hours.

As per the request in the comments below, here are some details:

Negotiations must end by Sunday, whether a deal has been reached or not (this post is written on Fri evening, i.e. two days before the deadline)

EU wants guarantees on aligned regulation ("level playing field"), to make sure that UK can't change it's regulatory environment in key areas to "undercut the EU" and gain a competitive advantage

UK rejects the above, because it argues the point of Brexit is to be able to pass its own laws (i.e doesn't wanna be kept in the EU's "regulatory orbit")

Both sides have much to loose: EU exports many goods to the UK (trade between Germany and UK alone was worth $139bn in 2019), whilst the goods exported by the UK to the EU amounted to £294bn in 2019 (43% of all UK exports): no deal means trade tariffs governed by WTO (mathematically, treat no deal as economically damaging to both sides)

A deal would be economically beneficial to both sides, but EU demands regulatory alignment, which is rejected on political grounds by the UK (and there is an argument for economic benefits in the longer run if own laws can be passed in the future)

Mathematically, you can look at current odds (and their evolution over the past week and months), current market moves in the GBP currency, or use other techniques: use anything you like to come up with a mathematically-backed answer.

Try to stay apolitical.

答:
There have been no answers from Game Theorists (at least not so far): so I have decided to implore google. A bit of searching has led me to the following two Game-Theoretic papers on Brexit (both in a PDF format), I share them here in case others are interested in this topic (both papers are relatively short and relatively easy to follow):

1) Brexit and Europe's future: A game theoretical approach

2) Brexit and game theory: A single-case analysis


问:
Bounding sum of probabilities of n events given a bound on the probabilities of their pairwise intersections in a general probability space

Let $A_1,A_2,\cdots,A_n$ be events in some probability space. State and prove the inclusion-exclusion formula for the probability $\mathbb{P}(\cup_{i=1}^{n}A_i)$. Show also that
$$ \mathbb{P}\Big(\displaystyle\bigcup_{i =1}^{n} A_i \Big) \ge \sum_{i}\mathbb{P}(A_i) - \sum_{i<j}\mathbb{P}(A_i \cap A_j) $$
Suppose now that $n \geq2$ and that whenever $i \neq j$, we have $\mathbb{P}(A_i \cap A_j) \leq  \frac{1}{n}$. Show that there is a constant $c$ independent of n such that $\displaystyle\sum_{i}\mathbb{P}(A_i) \leq c\sqrt{n}$.

My question is about the last inequality we are asked to prove in the problem.

Apart from the direct application of the upper bounds on the probabilities of pairwise intersections on the events that yields a bound of order n, which is not good enough for the purposes of the question (we want a 'square root n bound'), I am not sure how to proceed. The problem seems to suggest to use the inequality derived therein (the first one mentioned, the Bonferroni inequality with k = 2 incidentally), but this seems too weak. It also seems that the inequalities mentioned above cannot both be 'sharp' for large n, for otherwise we would be violating the law of total probability. Also, the inclusion/exclusion principle that was derived in the first part of the question does not seem to be too relevant for this part.

Perhaps one will need to consider a finer partition of the union of the events, but with the bounds at hand, it is not clear how to obtain some improvement on the bounds of the intersections of the events.

Additionally, I tried to recast the problem in terms of indicator variables and their expectations (for instance replacing the probabilities of the events by the expectations of the indicator variables corresponding to these events) and tried using some probabilistic inequalities, such as Markov or Chebychev, but to no avail.

Any hints would be greatly appreciated.

Thanks in advance.

答:
(Fill in the gaps as needed. If you're stuck, explain what you've tried and show your work. There's a possibility that this is wrong, and if so, state any issues/concerns that you have.)

Let $ k = \lfloor \sqrt{n} \rfloor $.

Hint:  Show that $$\sum_{j=1}^k P ( A_{i_j} ) < 2. $$

 The question hints that you should apply the PIE Inequality that they provided.

Corollary: By splitting the events into $k+1$ groups each of size $k$ or smaller,

$$ \sum_{i=1}^n P( A_i)  < 2 (k+1 ) \leq 4 \sqrt{n}. $$


问:
Find $\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$
Evaluate $$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$$
My attempt:
Consider
$$I=\int_{0}^{1}x^ne^{x^2}$$
using Taylor's series expansion for $e^{x^2}$ we get
$$I=\int_{0}^{1}x^n\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)$$
So we get
$$I=\int_{0}^{1}x^n+\int_{0}^{1}\frac{x^{n+2}}{1!}+\int_{0}^{1}\frac{x^{n+4}}{2!}+\cdots$$
Integrating we get
$$I=\frac{1}{0!(n+1)}+\frac{1}{1!(n+3)}+\frac{1}{2!(n+5)}+\cdots$$
Now we have,
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}nI$$
Thus we obtain
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}\frac{n}{0!(n+1)}+\frac{n}{1!(n+3)}+\frac{n}{2!(n+5)}+\cdots=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+..$$
So we get
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=e$$
is this a valid approach?

答:
I think there is an easier way around this, just integrate by parts to get,
$$\int_{0}^{1} (x^n)'e^{x^2}x dx = e-\int_{0}^{1} (e^{x^2}x)'x^{n}dx$$
and by the dominated convergence theorem the right side tends to $e$ and we are done.


问:
Show that $|PQ|=|P'Q'|$
Let $P, P', Q, Q'$ be four different points in the euclidean plane, such that $PP'$ and $QQ'$ have different midpoints and the line $g$ passes through these midpoints and is perpendicular to $PP'$ and $QQ'$.

(a) Show that $|PQ|=|P'Q'|$.

(b) Let $P=(r,s)$ and let $g: \ ax+by+c=0$. Calculate $P'$.

$$$$

So that the line $g$ passes through the two midpoints and is perpendicular to $PP'$ and $QQ'$, the $PP'$ and $QQ'$ must be parallel, or not?

Does this mean then that $|PQ|=|P'Q'|$ ?

答:
So line $g$ is perpendicular to lines $PP'$ and $QQ'$ hence these lines are parallel. Moreover, $g$ is perpendicular bisector for both $PP'$ and $QQ'$ which means all points on the line $g$ are equidistant from $P$ and $P'$ and from $Q$ and $Q'$. From this we can conclude that $|PQ|=|P'Q'|$ (if $M$ is midpoint of $QQ'$, $\triangle PMQ \cong \triangle P'MQ'$).

Regarding b):

Let $P'(p,q)$ then we have that $\frac{q-s}{p-r}=\frac{b}{a}$  or $aq-bp=as-br$

We also know that midpoint belongs to line $g$ so $a\frac{r+p}{2}+b\frac{s+q}{2}+c=0$ or $ap+bq=-2c-ar-bs$.
Solving the system yields: $$q=\frac{a^2s-2abr-2bc-b^2s}{a^2+b^2}$$ $$p=\frac{b^2r-2abs-2ac-a^2r}{a^2+b^2}$$

Title: Show that $|PQ|=|P'Q'|$


问:
Birkhoff spectrum is not compact
Let $f:X\to X$ be a homemmorphim on a compact metric space $X$. Assume that $T:X \to \mathbb{R}$ is continuous. We consider
$$
\sigma_{f}(x):=\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}T(f^{k}(x).$$

$\textbf{Question}$: Why is
$$W(\alpha):=\{x \in X, \sigma_{f}(x)=\alpha\}$$
not compact?

$W(\alpha)$ is the set of point where the limit of $\sigma_{f}(x)$ exists.
I know the set might be an empty set(consider the identity functions), so consider the question in case the set is not empty.

答:
I assume you mean "not always compact" (since for the identity map $\sigma_{Id}(x)=T(x)$ and so $W(\alpha)=T^{-1}(\alpha)$ which is a (possibly empty) compact set).

I don't know what kind of answer in terms of "why" you may be looking for, but one type of example is that $T$ has a repelling fixed point $x_0$ and all the nearby $x$ proceed to a different Birkhoff average, while $x_0$ stays and gets its own average. Consider:

Let $X=[0,1]$ and $f(x)=x^2$, and $T(x)=x$. We then have $\sigma_f(x)=\lim \frac{1}{n}(x+x^2+x^4+\ldots+x^{2^{n-1}})$ which is $1$ if $x=1$ and is  zero for all other $x$. This means $W(0)=[0,1)$ is not compact.

Title: Birkhoff spectrum is not compact