问:
In a primitive symmetric association scheme, why does $E_j$ occur in some power of $E_i$ for each $i,j$?
I am having some trouble in the proof of the Absolute Bound Condition for primitive symmetric association Schemes in the book Algebraic Combinatorics I by Bannai and Ito (Chapter 2, Section 4, Theorem 9):

Let $\chi=(X,\lbrace{R_i\rbrace}_{0\leq i \leq d})$ be a primitive symmetric association scheme of class $d$. Let the $\lbrace E_i\rbrace_{0\leq i\leq d}$ be the primitive idempotents of the adjacency algebra of $\chi$. Denote by $E_i^h$ to be the Hadamard product $\circ$ of $E_i$ with itself $h$ times.
Since $\chi$ is primitive, we have for any $j$ and $i$ an $0\leq h \leq d$ such that $E_j$ occurs in the linear decomposition of $E_i^h$ with respect to the basis $E_0,\ldots,E_d.$

I do not understand how the emboldened statement is justified. It says in the book that this is because imprimitivity is a equivalent to there being a suitable rearrangement of indices $1,\ldots,d$ and an index $0<t<d$ such that the Hadamard product $E_i \circ E_j$ is a linear combination of $E_0,E_1,\ldots,E_t$ for all $0\leq i,j \leq t.$

To get an idea, I tried to show this by contradiction on the special case when $d=3$ and the assumption that $E_2$ does not occur in the decomposition of $E_1^h$ for $0\leq h \leq 3.$ With this assumption, we get $q_{11}^2=0$ so
\begin{align*}
E_1^2 = &q_{11}^0 E_0 + q_{11}^1 E_1 + q_{11}^3E_3,\\
E_1^3 =&q_{11}^0(q_{01}^0 E_0 +  q_{01}^1 E_1 + q_{01}^2 E_2 + q_{01}^3 E_3) + q_{11}^1(q_{11}^0 E_0 +  q_{11}^1 E_1 + q_{11}^2 E_2 + q_{11}^3 E_3)\\
      &+q_{11}^3(q_{13}^0 E_0 +  q_{13}^1 E_1 + q_{13}^2 E_2 + q_{13}^3 E_3).
\end{align*}
Since $E_2$ does not occur in $E_1^3$, the latter gives us $q_{11}^0q_{01}^2 + q_{11}^1 q_{11}^2 + q_{11}^3q_{13}^2 = 0.$ Since $q_{01}^2=0=q_{11}^2,$ we obtain $q_{11}^3q_{13}^2=0.$

If $q_{11}^3=0$, then since $q_{11}^2=0$, we have the linear span of $E_0,E_1$ as a subalgebra of the adjacency algebra, violating primitivity. However, I no longer know what to do when $q_{13}^2=0.$ If I could show $q_{33}^2=0$, then the linear span of $E_0,E_1,E_3$ would form a subalgebra of the adjacency algebra and would violate primitivity. However, I don't know how to show $q_{33}^2=0.$

答:
A Schur polynomial in the matrix $E$ is a linear combination of Schur powers of $E$.
If $p(t)$ is a polynomial, write $p\circ E$ to denote the corresponding Schur polynomial in $E$.

If $E_i$ is a spectral idempotent of an association scheme and $p\circ E_i$ is invertible, then $p\circ E_i$ is a linear combination of the Schur idempotents with no coefficient in
the linear combination equal to zero.

Now suppose $E$ is a spectral idempotent and there is no polynomial $p$ such that $p\circ E$ is invertible. We aim to show the scheme is primitive. Let $W$ be the subspace of the Bose-Mesner
algebra of the scheme spanned by the Schur powers of $E$. Note that $A_i(p\circ E)$ is a Schur polynomial in $E$ (because $p\circ E$ is a linear combination of spectral idempotents
and so $A(p\circ E)$ must be a linear combination of a subset of these). Therefore $W$ is invariant
under the full Bose-Mesner algebra. In particular $W$ is closed under matrix multiplication.

Since $W$ is Schur-closed, it has a basis of Schur idempotents. Let $B$ denote their sum.
Then $B$ is the adjacency matrix of a graph and, since the powers of $B$ all lie in $W$,
this graph is not connected. Therefore the scheme is imprimitive.

[Obviously the claim that Bannai and Ito are making is not trivial. It is dual to the fact that if $A_i$ is connected then there is a polynomial $p$ in $A_i$ such that $p(A_i)\circ A_j\ne0$
for all $j$.]


问:
Vector norm of integral inequality
On the Wikipedia page for the mean value theorem there is this Lemma (Lemma 2 under Mean value theorem for vector-valued functions) that says:
Let $v:[a,b]\rightarrow\mathbb{R}^m$ be a continuous function on $[a,b]$, then we have that
$$ \left\Vert\int_a^b v(t)\,dt\right\Vert\leq\int_a^b\Vert v(t)\Vert dt. $$
I found a paper that uses/states the following similar result
$$ \left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\int_0^1\Vert v(t)\Vert^2 dt. $$
I can follow the proof from Wikipedia, but I'm not sure how to proof the 'squared' version above. I guess the fact the integral is taken from 0 to 1 is important. I did find this post which asks a similar question, but it seems rely on it being a scalar function there.

答:
Squaring the first inequality gives
$$
\left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\left(\int_0^1\Vert v(t)\Vert \, dt \right)^2 \, .
$$
Now apply Inequality releating squared absolute value of an integral to the integral of the squared absolute values of the integrand to the scalar-valued function $f(t) = \Vert v(t)\Vert$ to conclude that the right-hand side is
$$
 \le \int_0^1\Vert v(t)\Vert^2 \, dt \,  .
$$

ContentLicense: CC BY-SA 4.0

Title: Vector norm of integral inequality


问:
Show that the fibre $\pi^{-1}(p)$ is a regular submanifold
Exercise 6.4 of the book Manifolds and Differential Geometry (Jeffrey M. Lee, 2009) states the following:

Show that if $(E,M,\pi, F)$ is a (smooth) fiber bundle, then
$\pi: E \rightarrow M$ is a submersion and each fiber $\pi^{-1}(p)$ is a regular submanifold
which is diffeomorphic to $F$. Show that if both $F$ and $M$ are connected,
then E is connected.

I think I can proof the first statement: Because $(E,M,\pi,F)$ is a fibre bundle, for each $p\in M$ there exists a neighborhood $U$ which contains $p$ and a diffeomorphism $\Psi:\pi^{-1}(U) \rightarrow U\times F$ such that $\pi|_U=\Psi\circ \mbox{proj}_U$ (composition is by the right). Hence $d\pi|_U = d\Psi\circ d\,\mbox{proj}_U$; since $\Psi$ is a diffeomorphism, $d\Psi$ is bijective. Clearly $d\,\mbox{proj}_U$ has constant rank equal to $\mbox{dim }(M)$ and thus it follows that $\pi$ is a submersion.

In order to see that $\pi^{-1}(p)$ for $p\in M$ is a regular submanifold of $E$ of dimension $k$, I have to find a chart $(A,\tau)$ of the smooth manifold $E$ such that

$$ \tau\big(A\cap\pi^{-1}(p)\big) = \tau(A)\cap)\big(\mathbb{R}^k \times\{c\}\big), $$

where $c$ is an element of $\mathbb{R}^{d-k}$ (typically $c=0$) and $d=\mbox{dim }(E)$.

So, my questions are:

1- Is right my proof?

2.- How I can see the fibre over $p$ is a regular submanifold? How proof that is diffeomorphic to $F$?

3.- If $M$ and $F$ are connected, then $U\times F$ is connected for all open set $U\subset M$ and thus the open set $\pi^{-1}(U)$ is connected in $E$ thanks to the diffeomorphism $\Psi$. So it is clear that $E$ is locally connected. However, the connectedness of $E$ has not proof yet. How is the proof?

答: Just to give an additional reference: In the book Manifolds, Groups, Bundles, and Spacetime by R. Dale Gray, there is a proof that $\pi$ is a submersion (Theorem $8.1.13$) and that the fibers are regular submanifolds diffeomorphic to $F$ (Theorem $8.1.14$).


Title: Summation of Infinite Series as given below

Body:
SERIES:

$1+(1+a)b + (1+a+a^2)b^2 + (1+a+a^2+a^2)b^3..........$

Given: $|a|<1;  |b|<1$

Approach:

$1.1+(1+a)b + (1+a+a^2)b^2 + (1+a+a^2+a^2)b^3..........$

$T_n = (1 * a^{n-1} + S_{n-1})b^n-1$

$S_{n-1} = \sum_{k=1}^n a^{n-2}$

$S_n = \sum_{n=1}^{\infty} {(1 * a^{n-1} + S_{n-1})b^n-1}$

Kindly help me in further simplification.

Body:
$$1+a+a^2+\ldots+a^n=\frac{1-a^{n+1}}{1-a}$$
The series can be written as
$$\sum_{n=0}^{\infty}\frac{1-a^{n+1}}{1-a}b^n=\frac{1}{1-a}\sum_{n=0}^{\infty}\left(1-a^{n+1}\right)b^n=$$
$$=\frac{1}{1-a}\left(\sum_{n=0}^{\infty}b^n-\sum_{n=0}^{\infty}a^{n+1}b^n\right)=$$
$$=\frac{1}{1-a}\left(\frac{1}{1-b}-a\cdot\frac{1}{1-ab}\right)=\frac{1}{(1-b) (1-a b)}$$


答:
You have
$$\left(||\int_0^1v(t)dt||\right)^2\leq\left(\int_0^1||v(t)||dt\right)^2$$
by the first result and the monotonicity of $x\mapsto x^2$ on $[0,\infty)$. If $t\mapsto||v(t)||$ is summable then by the convexity of $x\mapsto x^2$ and Jensen's inequality it follows
$$\left(\int_0^1||v(t)||dt\right)^2\leq\int_0^1||v(t)||^2dt$$
and thus the desired result.

Title: Vector norm of integral inequality

Body:
On the Wikipedia page for the mean value theorem there is this Lemma (Lemma 2 under Mean value theorem for vector-valued functions) that says:
Let $v:[a,b]\rightarrow\mathbb{R}^m$ be a continuous function on $[a,b]$, then we have that
$$ \left\Vert\int_a^b v(t)\,dt\right\Vert\leq\int_a^b\Vert v(t)\Vert dt. $$
I found a paper that uses/states the following similar result
$$ \left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\int_0^1\Vert v(t)\Vert^2 dt. $$
I can follow the proof from Wikipedia, but I'm not sure how to proof the 'squared' version above. I guess the fact the integral is taken from 0 to 1 is important. I did find this post which asks a similar question, but it seems rely on it being a scalar function there.


问:
Finding $\lim_{x\to 0}\frac{\sin x}{x} $
How to find </p>

<p>$$\lim_{x\to 0}\frac{\sin x}{x} $$</p>

答: The proof that this limit is $1$ can be found in most calculus textbooks. It’s also available online, e.g., here and here.</p>

ContentLicense: CC BY-SA 3.0


答:
If $\{e_1, \dots, e_m\}$ is the canonical basis of $\mathbb R^m$ and $v : t \mapsto v_1(t) e_1 + \dots + v_m(t) e_m$ then by definition of the vector integral

$$\int_0^1 v(t) \ dt = \sum_{i=1}^m \left(\int_0^1 v_i(t) \ dt \right) e_i$$ and therefore if $\Vert \cdot \Vert$ stands for the Euclidean norm

$$\left\Vert \int_0^1 v(t) \ dt \right\Vert^2 = \sum_{i=1}^m\left(\int_0^1 v_i(t) \ dt \right)^2 \le \int_0^1 \left(\sum_{i=1}^m v_i^2(t) \right) \ dt = \int_0^1 \Vert v(t) \Vert^2 \ dt$$

Title: Vector norm of integral inequality


问:
Does every number divide some $n^2+1$?
For every $m \in \mathbb{N}$, does there exist some $n \in \mathbb{N}$ such that $m$ divides $n^2+1$?

This question came up as a small side question as part of some research I'm doing. I don't know a lot about number theory, and this looks like one of those questions that could either be really simple or totally beyond the limits of our current knowledge.

答:
No.

$4$ cannot divide such an $n^2+1$, because modulo $4$, you have $n^2+1=1$ or $2$.

Title: Does every number divide some $n^2+1$?


问:
In a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ is larger than $s_2$
While solving a totally different problem , I made observation that in a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ say to $θ_1$ is larger than angle opposite to $s_2$ say $θ_2$.

I verified by drawing several triangles . Can someone tell if it's right and prove it ?

答:
In the diagram given above, $AB>BC$. Now, if this is true, then there must be a point $D$ between points $A$ and $B$  on segment $AB$ such that $AD=AC$. Draw $DC$.

Now, in $\triangle ADC$, since $AD=DC$, $\angle ADC=\angle ACD$[opposite angles of equal sides are equal].

$\angle ACD<\angle ACB$, since point $D$ is between $A$ and $B$.

Also, $\angle ADC=\angle DBC+\angle BCD>\angle DBC=\angle ABC$.

Thus, $\angle ACB>\angle ACD=\angle ADC>\angle ABC$.

$\angle ACB$ and $\angle ABC$ are the opposite angles of side $AB$ and $AC$ respectively. Hence, we have proved that $\angle ACB=$ opposite angle of longer side $AB>\angle ABC=$ opposite angle of shorter side $AC$.


问:
Compact Fourier multiplier?
Let us take some continuous $m\in L^\infty(\mathbb R^d)$ and suppose $m(\xi)\to 0$ as $|\xi|\to \infty$. Let $T_m:L^2(\mathbb R^d)\to L^2(\mathbb R^d)$ be the associated Fourier multiplier operator, i.e. $\widehat{T_m f} = m \widehat{f}$. As we know in general these Fourier multipliers are NOT compact operators. The reason for this is that $\mathsf R(m)$ might contain some open set, and for compact operators we require the spectrum to have at most one accumulation point, namely zero.

Set $\mathcal X := [-\frac{1}{2},\frac{1}{2}]^d$. Now I am trying to find a way to make compact the restriction of $T_m$ to the subspace $L^2(\mathcal X) := \{ f \in L^2(\mathbb R^d) : \mathrm{supp}(f) \subseteq \mathcal X\}$. I have a few reasons to believe it might be compact on this subspace, but I feel like I am missing some crucial background knowledge. My reasons are the following:

We can represent any $f\in L^2(\mathcal X)$, for $x\in\mathcal X$ by

$$ f(x) = \sum_{k\in\mathbb Z^d} \hat{f}(k) e^{2\pi i x \cdot k}, $$

with $\hat{f}(k) = \int_{\mathcal X} f(x) e^{-2\pi i x\cdot k} \ dx$. Now for Fourier series (on the torus), compactness of Fourier multipliers is much easier: $T_m$ is compact whenever $m\to 0$.  This looks a lot like our assumption above. However, it is not immediately clear to me how we can use the assumption above.

If we had the additional assumption that $m \in L^2(\mathbb R^d)$, then we could take the IFT of $m$ to find that in fact $\mathcal F^{-1} m \in L^2(\mathbb R^d)$, so the operator $T_m$ would act on $L^2(\mathcal X)$ like a convolution with an $L^2(\mathcal X\times\mathcal X)$ function. This operator is then a Hilbert-Schmidt operator, since we are only convolving with functions with support on $\mathcal X$. Note however that this assumption is not necessary, as this approach only would require $\mathcal F^{-1} m \in L^2_{\mathrm{loc}}(\mathbb R^d)$, not $L^2(\mathbb R^d)$.

If instead we had the additional assumption that $m \in L^1(\mathbb R^d)$, we could say $\mathcal F^{-1} m\in L^\infty(\mathbb R^d)$, and we could again view $T_m$ on $L^2(\mathcal X)$ as a convolution with an $L^2(\mathcal X \times \mathcal X)$ kernel. Again this assumption is too strong, as we would in this case only need $\mathcal F^{-1}m\in L^\infty_{\mathrm{loc}}(\mathbb R^d)$.

Now, looking at $[\mathcal F^{-1} m] 1_{\mathcal X}$ also does not give anything useful (as far as I can tell). If we take the FT of $[\mathcal F^{-1} m] 1_{\mathcal X}$, we end up with

$$ (m \star \widehat{1_{\mathcal X}})(\xi) = \int_{\mathbb R^d} m(x)\prod_{j=1}^d \mathrm{sinc}(\xi_j - x_j) \ dx. $$

And these sinc integrals are terrible to deal with, because $\mathrm{sinc}$ is not $L^1$, there is almost no way to estimate these integrals without losing important information. I have tried searching on the internet and in the literature for ways to solve these integrals and the approach people use is... a Fourier transform to get the indicator function...

I really am at a loss here and feel like I am missing something. Surely someone must have tried something similar before?

Any help is appreciated, thanks.

答:
The operator is indeed compact on $\mathcal X$. To prove this, we need to use the $L^2$ version of the theorem of Ascoli and Arzelà, and while answering this question I found a nice little paper of Robert Pego with it;

https://www.ams.org/journals/proc/1985-095-02/S0002-9939-1985-0801333-9/S0002-9939-1985-0801333-9.pdf

Let us see the details. As in the question, let $m\colon \mathbb R^d \to \mathbb C$ be such that $m(\xi)\to 0$ as $\lvert \xi \rvert \to \infty$, and define
$$
T_m f = \mathcal F^{-1}( m \hat f),$$
where $\mathcal F$ and $\widehat{}$ denote the Fourier transform. We claim that this operator of $L^2(\mathbb R^d)$ into itself is compact on
$$\mathcal X=\{f\in L^2\, :\, \text{supp}(f)\subset B_1\}.$$
(Here $B_1$ denotes the unit ball of $\mathbb R^d$, but any subspace of functions supported in some common bounded set will yield the same result).

To prove this, we consider a sequence $f_n\in \mathcal X$ such that $\lVert f_n\rVert_2\le 1$. As stated in the linked paper, to prove that $\{T_m f_n\}$ is a relatively compact subset of $L^2$ we need to check two conditions;

Equiboundedness. the norm $\lVert T_m f_n \rVert_2$ must be uniformly bounded;

Equicontinuity. $$\lim_{R\to \infty}\sup_{n\in \mathbb N}\int_{\lvert \xi\vert \ge R} \lvert \widehat{T_m f_n}(\xi)\rvert^2\,d\xi=0.$$

(There is also a third condition - equivanishing -, but it is automatically satisfied since all $f_n$ are supported in the same bounded set. EDIT This is true but it is not as trivial as I am putting it. See the other answer for details on this important point).

Now, the first condition is obviously satisfied, since $T_m$ is bounded. The second is satisfied because of the assumption on $m$, since
$$\int_{\lvert \xi\vert \ge R} \lvert \widehat{T_m f_n}(\xi)\rvert^2\,d\xi = \int_{\lvert \xi\vert \ge R} \lvert m(\xi)\rvert^2 \lvert \hat f_n(\xi)\rvert^2\,d\xi\le \sup_{\lvert \xi \rvert \ge R} \lvert m(\xi)\rvert^2, $$
and the right-hand side tends to $0$. $\Box$

Title: Compact Fourier multiplier?


答: If $m$ divides $n^2+1$, then $n$ is a unit mod $m$ of order $4$. No prime of the form $4k+3$ can divide $n^2+1$ because in this case the group of units mod $p$ has order $4k+2$, which is not a multiple of $4$.

Title: Does every number divide some $n^2+1$?

Body:
For every $m \in \mathbb{N}$, does there exist some $n \in \mathbb{N}$ such that $m$ divides $n^2+1$?

This question came up as a small side question as part of some research I'm doing. I don't know a lot about number theory, and this looks like one of those questions that could either be really simple or totally beyond the limits of our current knowledge.


答:
Let $p$ be an odd prime. Then $n^2\equiv -1 \bmod p$ is possible only if $-1$ is a quadratic residue modulo $p$.

This occurs precisely when $p\equiv 1 \bmod 4$ ie for $p=5, 13$ but not $p=3, 7, 11$

Now suppose $p$ is a factor of $m$ and that you want to find $n$ with $n^2+1=km$. Modulo $p$, this becomes $n^2\equiv -1$, so any odd prime factor of $m$ must be of the form $p=4r+1$.

There is more to do to specify precisely which $m$ are possible, but this is a start on the general problem.

Title: Does every number divide some $n^2+1$?


问:
How to use the squeeze lemma to prove a sequence converges and its limit?
I am solving the following problem.

Let $a_n = (n-\sin(n))/n$.

Now, using the squeeze lemma I am to show that $a_n$ converges and find it's limit.

I understand the squeeze lemma as a lemma that says the single limit of two sequences are equivalent implying the equivalence of another sequences limit (e.g., $\lim a_n = \lim c_n = \lim b_n$).

I would appreciate any help, thanks.

答:
Hint

$$-1 \le \sin n \le 1$$ for all $n \in \mathbb N$.


问:
Natural logarithm
Can someone please suggest how one proves:

$(1+2x)\ln(1+\frac{1}{x}) -2 >0$ where $x>0$.

I plotted the function in a program and the inequality should be correct.

答: Here is one approach.  Let $f(x)=(1+2x)\ln(1+\frac{1}{x})$.  You can show that $\lim\limits_{x\to+\infty}f(x)=2$, for example using l'Hôpital's rule, with a rewrite as $\dfrac{\ln\left(1+\frac{1}{x}\right)}{\left(\dfrac{1}{1+2x}\right)}$.  You can show that $f$ is always decreasing by showing that $f'$ is always negative.  In case working directly with $f'$ is too cumbersome, it turns out to be easy to show that $f''$ is always positive, and then it would suffice to show that $\lim\limits_{x\to+\infty}f'(x)=0$.</p>

Title: Natural logarithm


问:
Relation between different types of elliptic functions
Disclamer: all formulas in this post are intended to be schematic, coefficients signs etc. are mostly wrong.

I do various physics-motivated computations involving elliptic functions and the more sources I look at the bigger grows my confusion. It seems there are many different parametrizations of elliptic functions, for example

Elliptic integrals such as $K(x)=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-xt^2)}}$.

Jacobi theta functions like $\vartheta_3=\sum_{n\in\mathbb{Z}}q^{n^2}$

Double-periodic functions such as Weierstrass function $\wp(z)\sim\sum_{n,m}\frac{1}{(z-n-m\tau)^2}$

Modular forms like $E_2(q)=1-24\sum_n\frac{nq^n}{1-q^n}$ or $\eta=q^{1/24}\prod_n(1-q^n)$

For example, some relations between them are

$K(x)=\frac{\pi}{2}\vartheta_3^2(q)$ for $q=\exp\Big(-2\pi\frac{K(1-x)}{K(x)}\Big)$

$\vartheta_1'=\eta^3$

$\wp(z)\sim \frac1{z^2}+z^2E_2+z^4E_4+\dots$

Is it possible to give a concise summary of how different types of functions are related to each other? Make these relationships less surprising? For example I've known for a long time that $\vartheta_3^2(q)=K(x)$ but still learning that $\vartheta_2^2(q)=xK(x)$ surprised me.

Of course a deeper theory behind all these objects exist but learning it seems quite an investment. Are there some exposition intermediate between full-blown modern algebraic geometry and simple statement of definitions and formulas relating them? Especially I'm interested in some resource that would connect many of these thigs. For example Mumford's lectures on theta-functions are great but they mostly don't touch other types of functions mentioned above.

答:
There is a "simple" analytic definition of an elliptic function: it is a doubly-periodic meromorphic function of complex variable. Number of poles (a.k.a. order) of such a function is important - every such function may be expressed as a rational function of an elliptic function of 2nd order, and its derivate.

So, algebraically, it looks a bit like trigonometry, where $\sin(x)$ and $\cos(x)=\sin(x)'$ play a similar role, and inverse trigonometric functions are akin to elliptic integrals - inverses of elliptic functions.
But it's more complicated, as instead of just $\sin(x)$ you have infinitely many such "basic" functions.

Then you can start deriving relations between these functions and their inverses; this is an approach in e.g. (in Russian only, it seems) M. Lavrentiev and B. Shabat, Methods of the theory of functions of a complex variable.


问:
Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$.

Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.

I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5)
\\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$
I got stuck over here.
Am I right till here? Can someone tell what to do next?

答:
If you insist on doing it by induction, look at
$$p(k+1)-p(k)=5(k+2)(k+3)(k+4)(k+5).$$
Among the four consecutive numbers $k+2,k+3,k+4,k+5$, one number is a multiple of four, and another (different) number is a multiple of two. There is also at least one multiple of three. So $5(k+2)(k+3)(k+4)(k+5)$ is divisible by $5\times4\times2\times3=120$, which implies the result.

With very little adjustment, you can rephrase this argument without induction, which is probably the cleaner way to do it.


问:
Negative Binomial Random Variable Question
I am stuck with the question below.

If X is a negative binomial random variable, then $$ Y=r+x $$ is the total number of trails necessary to obtain r S's. Obtain the moment generating function of Y and then it's mean value and variance. Are the mean and variance intuitively constant with the expressions for E(X) and V(X)? Explain.

I know that the moment generating function of a negative binomial is:
$$ M_X(t)=\frac{p^r}{[1-e^t(1-p)]^r}. $$
How do I use relate this to the moment generating function I need when I add a constant? I can find the mean by deriving this moment generating function and plugging in 0 for t. I also know how to use the shortcut to find variance. Does the last part mean I just say that the mean and variance are as expected because I derive away the constant?

答:
I think you are looking for a connection to the geometric random variable? If $X\sim Geo(p)$ counts the failures and the first final success, we get
$$M_X(t)=\frac{pe^t}{1−e^t(1−p)}$$
But also
$$M_{X-1}(t)=\mathbf E[exp(Xt-t)]=\frac{pe^t}{1−e^t(1−p)}e^{-t}$$

For the negative binomial $Y\sim NB(r,p)$ you got
$$M_Y(t)=\left(\frac{p}{1−e^t(1−p)}\right)^r=\left(\frac{pe^t}{1−e^t(1−p)}\cdot e^{-t}\right)^r$$

So $$Y\sim \sum_{i=1}^r(X_i-1)=\left(\sum_{i=1}^rX_i\right)-r$$ and $X_i\sim Geo(p)$ independent. This $Y$ does not count the r successes, only the failures.


问:
If $f$ is uniformly continuous on the dense set $E \subset X$ then a continuous extension of $f$ exists on $X$.

Let $E \subset X$ which is a dense subset of $X$.The function $f$ has its range in $\mathbb{R}$.We need to show that there exists a continuous extension from $E$ to $X$.

My attempt :

let $x \in X$ and we choose a cauchy sequence {$e_n$} $\to x$.Now, since $\{e_n\}$ is a convergent sequence so it is a cauchy sequence. As $f$ is a uniformly continuous function, then $\{f(e_n)\}$ is also a cauchy sequence and so it is convergent to a point,we define that to be $f(x)$[As $\mathbb{R}$ is a complete metric space].

Now, we choose another sequence $\{(e_n)'\} \to x$.We see that
$d(e_n,x)<\frac{\delta}{2}$ for all $n \ge K_1$ and $d((e_n)',x)) < \frac{\delta}{2}$ for all $n \ge K_2$
Then, we see that $d_y(f((e_n)'),f(e_n)) \le \frac{\epsilon }{2}$ when $d(e_n,(e_n)')\le \delta$ for all $n \ge max\{K_1,K_2\}$.Using this equation we can show that $f((e_n)') \to f(x)$.

To prove:$f$ is uniformly continuous. We choose $d(x_m,x_n) < \frac{\delta}{3}$.

Let $\{(e_n^m)\} \to x_m$ and $\{(e_n^k)\} \to x_k$.Now,  $\{f(e_n^m)\} \to f(x_m)$ and $\{f(e_n^k)\} \to f(x_k)$.So,

$d_y(f(e_n^k),f(x_k)) < \frac{\epsilon}{3}$ for all $n \ge K_1$ and

$d_y(f(e_n^m),f(x_m)) < \frac{\epsilon}{3}$ for all $n \ge K_2$.

Now by some manipulations we can find a $N$ such that when $n_1,n_2 \ge N$ we have $d((e_{n_1}^m), (e_{n_2}^k))  < \delta $ and by choosing such a delta we can have $d_y(f(e_{n_1}^k),f(e_{n_2}^m) )< \frac{\epsilon}{3}$ for all $n_1,n_2\ge N$.Also we see that,

$d_y(f(x_m),f(x_k)) \le d_y(f(e_n^k),f(x_k)) + d_y(f(e_n^m),f(x_m)) + d_y(f(e_n^k),f(e_n^m))< \epsilon$ for all $n \ge max\{K_1,K_2,N\}$

This has been an attempt. Can someone go through my proof?I am not sure about the part where I am proving uniform continuity.

答:
I state the proposition and prove it.

Proposition: Let $(X,d)$ be a metric space. Let $E\subseteq X$ be
a dense subset. Suppose that $f:E\rightarrow\mathbb{R}$ is uniformly
continuous. Then $f$ has a unique uniformly continuous extension $\bar{f}:X\rightarrow\mathbb{R}.$
(Note that $(X,d)$ needs not be complete.)

Proof: Existence of extension: Let $x\in X.$ Since $E$ is dense,
we may choose a sequence $(x_{n})$ in $E$ such that $x_{n}\rightarrow x$.
We go to show that $\lim_{n}f(x_{n})$ exists. Let $\varepsilon>0$
be given. Since $f$ is uniformly continuous, there exists $\delta>0$
such that $|f(t)-f(t')|<\varepsilon$ whenever $t,t'\in E$ with $d(t,t')<\delta$.
Choose $N\in\mathbb{N}$ such that $d(x_{n},x)<\delta/2$ whenever
$n\geq N$. For any $m,n\geq N$, we have that $d(x_{m},x_{n})\leq d(x_{m},x)+d(x,x_{n})<\delta$.
Therefore, $|f(x_{m})-f(x_{n})|<\varepsilon$ whenever $m,n\geq N$.
This shows that $(f(x_{n}))_{n}$ is a Cauchy sequence in $\mathbb{R}$
and hence $\lim_{n}f(x_{n})$ exists by the completeness of $\mathbb{R}$.

Next, we go to show that the limit $\lim_{n}f(x_{n})$ is independent
of the choice of the sequence $(x_{n})$, i.e., if $(x_{n}')$ is
another sequence in $E$ such that $x_{n}'\rightarrow x$, then $\lim_{n}f(x_{n})=\lim_{n}f(x_{n}')$.
Let $\varepsilon>0$. Choose $\delta>0$ such that $|f(t)-f(t')|<\varepsilon$
whenever $t,t'\in E$ with $d(t,t')<\delta$. Choose $N$ such that
$d(x_{n},x)<\delta/2$ and $d(x_{n}',x)<\delta/2$ whenever $n\geq N$.
For $n\geq N$, we have $d(x_{n},x_{n}')\leq d(x_{n},x)+d(x,x_{n}')<\delta$.
Therefore $|f(x_{n})-f(x_{n}')|<\varepsilon$. Letting $n\rightarrow\infty$,
we obtain $|\lim_{n}f(x_{n})-\lim_{n}f(x_{n}')|\leq\varepsilon$.
Since $\varepsilon>0$ is arbitrary, we conclude that $\lim_{n}f(x_{n})=\lim_{n}f(x_{n}')$.

Define a function $\bar{f}:X\rightarrow\mathbb{R}$ by $\bar{f}(x)=\lim_{n}f(x_{n})$,
where $(x_{n})$ is an arbitrarily chosen sequence in $E$ such that
$x_{n}\rightarrow x$. From the above discussion, $\bar{f}$ is well-defined.
Observe that if $x\in E$, we may choose $(x_{n})$ such that $x_{n}=x$
for all $n$, then $\bar{f}(x)=\lim_{n}f(x_{n})=f(x)$. That is, $\bar{f}$
is an extension of $f$. Lastly, we go to show that $\bar{f}$ is uniformly
continuous. Let $\varepsilon>0$ be given. Choose $\delta>0$
such that $|f(t)-f(t')|<\varepsilon$ whenever $t,t'\in E$ with $d(t,t')<\delta$.
Let $x,y\in X$ such that $(x,y)<\delta/3$. By density of $E$, we choose
sequences $(x_{n})$ and $(y_{n})$ in $E$ such that $x_{n}\rightarrow x$
and $y_{n}\rightarrow y$. Choose $N$ such that $d(x_{n},x)<\delta/3$
and $d(y_{n},y)<\delta/3$ whenever $n\geq N$. For $n\geq N$, we
have that $d(x_{n},y_{n})\leq d(x_{n},x)+d(x,y)+d(y,y_{n})<\delta$.
Hence, $|f(x_{n})-f(y_{n})|<\varepsilon.$ Letting $n\rightarrow\infty$
yields $|\bar{f}(x)-\bar{f}(y)|\leq\varepsilon$. This shows that
$\bar{f}$ is uniformly continuous.

Uniqueness of extension: Suppose that $f_{1}$ and $f_{2}$ are continuous
extension of $f.$ Let $x\in X$. Choose a sequence $(x_{n})$ in
$E$ such that $x_{n}\rightarrow x$. We have that
\begin{eqnarray*}
f_{1}(x) & = & \lim_{n}f_{1}(x_{n})\\
 & = & \lim_{n}f(x_{n})\\
 & = & \lim_{n}f_{2}(x_{n})\\
 & = & f_{2}(x).
\end{eqnarray*}
Therefore $f_{1}=f_{2}$.


问:
How to use the Gram Schmid procedure in this particular case?
I have to find an orthonormal basis of the subspace $V= \mathbb R^4$ generated by $e_1, e_1 + e_2$ and $e_1 + 2e_4$

I am familiar with the Gram Schmidt procedure. In this case I am confused due to the generators. How can I interpret them? Or how can I use the Gram Schmidt procedure to find the orthonormal basis in this case? Any help or tip would be grateful. Thanks in advance.

答:
Note that$$\operatorname{span}\bigl(\{e_1,e_1+e_2,e_1+2e_4\}\bigr)=\operatorname{span}\bigl(\{e_1,e_2,e_4\}\bigr).$$So (assuming that the $e_k$'s are the vectors from the standard basis and that your inner product is the usual one), just take $\{e_1,e_2,e_4\}$.

But, yes, you can also apply Gram-Schmidt to those vectors.


问:
Grover Algorithm Orthogonal vectors
I'm study the Grover algorithm. About this picture my lecture say that the expression $|s'\rangle$ in the computational basis is $$|s'\rangle = \dfrac{1}{\sqrt{N-1}}\sum_{x\neq w}|x\rangle \tag{$*$}.$$ Why this expression?

I trying using the diagonal state in the computational basis. For example if $N=3$

$$|D\rangle = \dfrac{1}{\sqrt{3}}(|0\rangle+|1\rangle+|2\rangle),$$

Let be $|s'\rangle = |2\rangle/\sqrt{3}$ and $|w\rangle = |1\rangle/\sqrt{3}$ then $$\dfrac{1}{\sqrt{3}}|2\rangle=|D\rangle - \dfrac{1}{\sqrt{3}}(|0\rangle+|1\rangle).$$
But it is not the same expression as $(*)$.

答:
First, a notational issue. The vector you call $|w\rangle$ "double u" is actually called "omega" i.e. $|\omega\rangle$ (backslash-omega is the $\rm\TeX$ command).

The vector $|s'\rangle$ is determined as the only (up to an arbitrary overall normalization including the phase) linear combination of $|\omega\rangle$ and $|s\rangle$ – note that you use the symbol $|D\rangle$ for $|s\rangle$ (the equal superposition of all basis vectors) – that is orthogonal to $|\omega\rangle$. We're talking about a 2-dimensional plane here and it's convenient to have an orthonormal basis in that plane; the basis vectors are $|s\rangle$ we started with and $|s'\rangle$ which we add.

Consider a general
$$|s'\rangle = A|s\rangle + B |\omega\rangle $$
and demand $\langle \omega|s'\rangle = 0$. Because $\langle \omega|\omega\rangle = 1$ and $\langle \omega|s\rangle=1/\sqrt{N}$, we may see that $B=-A\sqrt{N}$. This gives us
$$|s'\rangle = A( |s\rangle -\sqrt{N} |\omega\rangle ) $$
The subtraction of $\sqrt{N}|\omega\rangle$ simply removes the term $|\omega\rangle$ from the sum over the basis that defines $|s\rangle$. So we have a sum over $x\neq \omega$ and the basis vectors are multiplied by $A/\sqrt{N}$.
$$ |s'\rangle =\frac{A}{\sqrt N} \sum_{x\neq \omega}|x\rangle$$

The factor $1/\sqrt{N-1}$ is simply added to make $|s'\rangle$ normalized to unity. Any phase could be added, too. We obtain this right factor in front of the $x\neq \omega$ basis vectors for $A=\sqrt{N/(N-1)}$. The coefficient $A$ of this magnitude erases the original $1/\sqrt{N}$ coefficient and replaces it with the new one, $1/\sqrt{N-1}$.

Title: Grover Algorithm Orthogonal vectors


问:
Matrix of a vector with respect to basis of different vector space
I'm following Linear Algebra Done Right by S. Axler. One of the theorems he provides is the following:

I don't understand how $M(v_k)$ ($v_k$ is a basis vector for $V$) can be expressed with respect to the basis of $W$. $v_k$ isn't even in $W$, it's in $V$ by definition, and even their dimensions are different. Can someone explain this theorem/proof more verbosely?

For reference, the definitions he refers to are here:

答:
There are typos in 3.64 and its proof: all $M(v_k)$ should be $M(Tv_k)$.

See the official errata of the book: https://linear.axler.net/LADRErrataThird.html


问:
$\Bbb{R}^3$ and a three dimensional subspace of $\Bbb{R^4}$
Lets say we have $\Bbb{R}^3$ and a three dimensional subspace of $\Bbb{R^4}$, call it $E$. Why is it that these spaces are not the same vector space? Does it have something to do with bases?

答:
A three dimensional subspace of $\mathbb{R}^4$ is the span of 3 linearly independent vectors, say $\{v_1, v_2, v_3\}$. We know at least a bare minimum about these vectors, namely that they are 4-tuples.

Meanwhile, $\mathbb{R}^3$ is the span of 3 linearly independent 3-tuples. They cannot be the same space because the vectors are different; a 3-tuple is not a 4-tuple. The intuition you're getting at is that these spaces are isomorphic; for all intents and purposes their structure is the same.


问:
The Soup Problem: how to asymptotically fairly split a geometric series and a constant one using a single pattern?
Literally every time I'm serving some soup I'm thinking of this little mathematical problem I devised.

Imagine you have a very large (= infinite, for the purposes of the actual problem) bowl of soup and want to divide it into two halves using a (finite) ladle. The trouble is that some of the good stuff floated to the surface and every scoop takes a part $p$ of it, $p < \frac12$ (so its amount goes down geometrically with a quotient of $1-p$). The other good stuff is evenly dissolved in the volume. The protocol is that you always take a full scoop in this manner, so the only freedom is in what order do you empty the ladle into $A$ or $B$'s plate. What order would you take so that, asymptotically, both parties get fair shares in both respects?

Obviously the answer can't be $ABABAB\ldots$, because $A$ would always get $1/(1-p)$-times more than $B$. $\overline{ABBA}$ does not work either because $1+(1-p)^{-3}$ is always greater than $(1-p)+(1-p)^2$. In fact (except for a countable number of special values of $p$ (*)) it can't be any periodic infinite word of $A$ and $B$ for a slight variant of the same reason. One more thing I found is that the greedy algorithm (go $A$ until it has more than $B$, then swap) won't work either because while it would always satisfy splitting of the geometrical part, it would break the linear one (number of scoops). That's all I know at the moment.

Rephrased: for a given $1-p =: q \in (\frac12, 1)$, find a subset $M ∈ \mathbb{N}$ such that both

$\displaystyle\lim_{N→∞} \frac{\#\left( M \cap \{1,\ldots,N\} \right)}{N} = \frac12$,

$\displaystyle\sum_{n=1}^∞ q^n \chi_M(n) = \frac12 \sum_{n=1}^∞ q^n$.

I'm pretty sure the answer will lie somewhere in the theory of non-integer base systems or combinatorics on words or both, but I know little of those.

(*) For example, for $q = φ - 1$ the equality $q + q^2 + q^3 = 1 + q^4 + q^5$ holds, so for $p = 2 - φ \approx 0.381966$ the sequence $\overline{ABBBAA}$ would work.

答:
There is an uncountable set of $q\in (1/2,1)$ for which there are only two sets $M$ (which are complements of each other) such that $\sum_{m\in M}q^m=\frac{1}2\cdot \frac{1}{1-q}$ - and uncountably many of these forced sets are not of asymptotic density $1/2$.

To prove this, let us define, for any subset $M$ of the natural numbers, the summing function
$$f_M(q)=\frac{\sum_{m\in M}q^m}{\sum_{n\in\mathbb N}q^n}=(1-q)\cdot \sum_{m\in M}q^m$$
Note that for any $M$, the function $f_M$ is continuous where $|q|<1$ - and, in fact, the family of such functions is equicontinuous on this range.

Let's also define the left shift operator
$$L(M)=\{m' \in \mathbb N : m'+1\in M\}$$
and note that
$$f_{M}(q) = (1-q)\cdot \chi_{M}(0) + qf_{L(M)}(q)$$
where $\chi_M$ is the indicator function of $M$.

Let's say a set $M$ has $n$-bounded runs if there are no values $k$ such that $k\in M$ but $k+1,k+2,\ldots,k+n$ are all not in $M$ or such that $k\not\in M$ and the following $n$ integers are all in $M$.

We can now state a lemma:

Lemma: For any $n$, there is some $\varepsilon > 0$ such that if $M$ has $n$-bounded runs, then for all $q\in[1/2,1/2+\varepsilon)$ and all $M'\subseteq \mathbb N$ we have $f_M(q)=f_{M'}(q)$ implies $M=M'$.

Proof. First, we need to choose $\varepsilon$ so that $\left|f_{M}(q) - \frac{1}2\right| > \varepsilon$ for all sets $M$ with $n$-bounded runs and all $q\in [1/2,1/2+\varepsilon)$ - this is possible since $|f_M(1/2)-\frac{1}2|$ can never be less than $\frac{1}{2^{n+1}}$ for such a set, considering binary representations, and the family of functions $f$ is equicontinuous, so the values $|f_M(q)-\frac{1}2|$ must remain bounded away from zero for all $q$ close enough to $\frac{1}2$ - which allows us to choose a $\varepsilon$ satisfying our condition. Note that this, more helpfully, ensures that $f_{M}(q)$ cannot ever be in $[1-q, q]$.

One can now fix any $M$ with $n$-bounded runs and any $M'$ with $f_M(q)=f_{M'}(q)$ for some $q\in [1/2,1/2+\varepsilon)$ and prove inductively that $M=M'$. Consider the equation:
$$f_M(q)=f_{M'}(q)$$
$$(1-q)\chi_{M}(0)+qf_{L(M)}(q) = (1-q)\chi_{M'}(0)+qf_{L(M')}(q)$$
Note, however, that this expression is either less than $1-q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=0$) or greater than $q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=1$) by choice of $\varepsilon$. Thus, $0$ is either in both or neither of the sets. However, this also immediately then simplifies as:
$$f_{L(M)}(q)=f_{L(M')}(q)$$
which then allows us to repeat this argument inductively to see that $M=M'$, proving the lemma.

Then we finish with another lemma:

Lemma 2: Let $M$ be a non-empty set with $\ell$-bounded runs such that $0\not\in M$. Define $M_n = \{0\} \cup (M + n + 1)$ to be a right shift of $M$ with an initial element appended. For all large enough $n$, there exists some $q\in [1/2,1]$ such that if $f_{M'}(q)=1/2$ and $0\in M'$, then $M'=M_n$.

Proof. Note that $L(M_n)$ is always a set with $\ell$-bounded runs since the initial large run of elements not in the set is not preceded by an element that is in our set, as our definition would require - and all other runs must be in $M$ itself. Moreover, note
$$f_{M_n}(q)=(1-q) + q^{n+1}f_{M}(q).$$
Note, in particular, that $f_{M_n}(1/2)\geq \frac{1}2$ and that if $q^{n+1} \leq q - \frac{1}2$, then $f_{M_n}(q)\leq \frac{1}2$ - and note that, for any $\varepsilon>0$ there is some large enough $n$ so that this is true of some $q$ in $(1/2,1/2+\varepsilon)$. In particular, suppose we choose some $\varepsilon$ so that Lemma 1 applies to sets with $\ell$-bounded runs with that $\varepsilon$ and choose $n$ big enough so that $q_0^{n+1}\leq q_0 - \frac{1}2$ for some $q_0\in (1/2,1/2+\varepsilon)$. Then, by the intermediate value theorem, there has to be some $q\in [1/2,q_0]$ such that $f_{M_n}(q) = \frac{1}2$.

However, if we let $M'$ be such that $f_{M'}(q)=\frac{1}2$ and $0\in M'$, we can then realize that $$f_{L(M')}(q) = f_{L(M_n)}(q)$$ and apply Lemma 1 to see that $L(M')=L(M_n)$, which then, since both $M'$ and $M_n$ contain zero, implies that $M'=M_n$, completing Lemma 2.

To then reach our end goal: fix any pair $(q,M)$ with the property from Lemma 2 that if for any $M'$ with $0\in M'$ we have $f_{M'}(q)=\frac{1}2$ implies $M=M'$. Note that if we had some arbitrary set $M''$ with $f_{M''}(q)=\frac{1}2$, we would also have that $f(\mathbb N\setminus M'')=\frac{1}2$ - and either $M''$ or its complement must contain $0$ - thus must equal $M$.

Together, this gives us a method to construct a $q$ for which fairly splitting soup is impossible from any set with bounded runs but not with asymptotic density $1/2$. This gives an uncountable set of $q$ for which fairly splitting soup is not possible. Note that if you choose the set $M$ to be have a periodic pattern, the produced counterexamples will be algebraic.


问:
one doubt on big oh notation
I see an example that $f(n)=O(n)$ and $\sum_{i=1}^nf_i(n)=O(n^2) $ in which $f_i$ is a function from natural to natural number.

I ran into a challenge, that above notation is differ from the
following or not? my misunderstanding point is why the above fact is
false?

答: $\Theta$ is the two sided version of big $O$.  If something is $O(n^2)$ it can be anything smaller, like $n$.  $\Theta$ says it is within a constant factor, so $n=O(n^2)$ but $n \neq \Theta(n^2)$  The statement you quote is true if the leading $f$ is replaced by $f_i$, meaning that each $f_i(n)=O(n)$

Title: one doubt on big oh notation


问:
Quick Vectors with Motion Question
So I was given the following prompt:

"The position of a car moving along a flat surface at time $t$ is modeled by $(x(t), y(t))$ with velocity vector $v(t)=<3+6\sin(3t), 1+e^{2t}>$ for $0 \leq t \leq 2$. Both $x(t)$ and $y(t)$ are measured in feet, and $t$ is measured in seconds. At time $t=0$, the car is at position $(0,0)$. Write, but do not evaluate, an integral expression that gives the total distance traveled by the car from time $t=0$ to time $t=2$."

I guess I'm a bit confused about what that integral might look like. I understand that the bounds of the integral would be from $0$ to $2$, but I'm a bit confused about what expression the question is looking for to be integrated. Any clarification would be appreciated!

答: Since $\mathbf{v}(t) = \langle \frac{dx}{dt}(t), \frac{dy}{dt}(t)\rangle$ and you were given $\mathbf{v}(t)$, you should have enough to plug into the arc length formula now.

Title: Quick Vectors with Motion Question


答:
=== new answer ====

Actually it is easy to prove or more general case of $(n+1)(n+2) ..... (n+k)$ is divisible by $k!$.

And this is a case of proving the induction step by .... induction.  A proof within a proof as it were....

Base case:  $n+1$ is divisible $1! = 1$.  Always.

Induction step:  Assume $(n+1)..... (n+(k-1))$ is divisible by $(k-1)!$ for any $n$ then we will prove $(n+1)....(n+k)$ is divisible by $k!$ for any $n$.  And we will prove it by induction!

Base case: $1*2*......*k = k!$.

Induction step:  Suppose $(n+1) ......(n+k)$ is divisible by $k!$.

then $(n+2)......(n+k)(n+k+1) =$

$(n+2)......(n+k)([n+1]+k)=$

$(n+2)......(n+k)[n+1] + k(n+2)......(n+k)=$

$\color{blue}{(n+1)(n+2)......(n+k)} + k\color{red}{(n+2)......(n+k)}$.

Now we began by assuming $\color{blue}{(n+1)(n+2)......(n+k)}$ is divisible by $k!$.

And $\color{red}{(n+2)......(n+k)}$ is the product of $k-1$ consecutive numbers.  And we are doing this entire proof within a prook under the assumption that the product of any $k-1$ consectualve numbers is divisible by $(k-1)!$.  So $\color{red}{(n+2)......(n+k)}$ is divisible by $(k-1)!$ and $k\color{red}{(n+2)......(n+k)}$ is divisible by $k*(k-1)! = k!$.

So $\color{blue}{(n+1)(n+2)......(n+k)} + k\color{red}{(n+2)......(n+k)}$ is the sum of two numbers divisible by $k!$ so is itself divisible by $k!$.

That proves the induction step of our proof within a proof, and so proves our proof within a proof:  If the product of any $k-1$ consecutive numbers is divisible by $(k-1)!$ then the product of any $k$ consecutive number is divisible by $k!$.

And that proof within a proof was itself the induction step for our claim $(n+1)(n+2).....(n+k)$ will be divisible by $k!$.

....

And so $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $5! = 120$.

===old answer===

If $(n+1)(n+2)(n+3)(n+4)(n+5) = 120k$ then

$(n+2)(n+3)(n+4)(n+5)(n+6) = [(n+2)(n+3)(n+4)(n+5)]([n+1] + 5)=$

$[(n+1)(n+2)(n+3)(n+4)(n+5)] + 5[(n+2)(n+3)(n+4)(n+5)]=$

$120 k + 5\color{blue}{(n+2)(n+3)(n+4)(n+5)}$

So it remains to show that $(n+2)(n+3)(n+4)(n+5)$ is divisible by $24$.

Claim: $(n+1)(n+2)(n+3)(n+4)$ is divisible by $24$.

Induction:  If $(n+1)(n+2)(n+3)(n+4) = 24m$ then

$(n+2)(n+3)(n+4)(n+5) = (n+2)(n+3)(n+4)[(n+1) + 4] = $

$(n+1)(n+2)(n+3)(n+4) + 4(n+2)(n+3)(n+4)=$

$24m + 4\color{orange}{(n+2)(n+3)(n+4)}$

So it remains to show $(n+2)(n+3)(n+4)$ is divisible by $6$.

Claim:  $(n+1)(n+2)(n+3)$ is divisble by $6$.

Induction:  If $(n+1)(n+2)(n+3) = 6j$ then

......

......

......

======

Okay....

Do it by induction

First prove (by induction)  that $(n+1)(n+2)$ is always divisible by $2$.

(Easy:  Base case: $1*2$ is divisible by $2$.  Induction: If $(n+1)(n+2) = 2m$ then $(n+2)(n+3) = (n+2)(n+1) + 2(n+2) = 2m + 2(n+2)$ is divisible by $2$.

Then prove by induction that $(n+1)(n+2)(n+3)$ is always divisible by $6$.

Base case: $1*2*3=6$.  Induction; if $(n+1)(n+2)(n+3) = 6k$ then $(n+2)(n+3)(n+4) = (n+1)(n+2)(n+3) + 3(n+2)(n+3) = 6k + 3(n+2)(n+3)$ and $(n+2)(n+3)$ is divisible by $2$ so $3(n+2)(n+3)$ is divisible by $6$.

Then prove by induction that any four consecutive .....

....

Look..... just prove any product of $k$ consecutive numbers are divisible by $k!$.

Base case:  Any product of $1$ consecutive number is divisible by $1! = 1$.

Induction.  If any the product of any $k$ consecutive integers is divisible by $k!$ then  we can prove any product of any $k+1$ consecutive integers is divisble by $(k+1)!$ by induction.

Base case:  $1*2*....*(k+1) =(k+1)!$

Induction.   If $(n + 1)( n+ 2) ...... (n+k+1)$ is divisible by $(k+1)!$ then

$(n+2)(n+2)(n+3) ...... (n+k+1)(n + k+ 2) = $

$(n+2)(n+2)(n+3) ...... (n+k+1)(n + 1+(k+1)) = $

$(n+1)(n+2)....(n+k)(n+k+1) + (k+1)(n+2)(n+3).....(n+k+1)$.

And $(n+1)(n+2)....(n+k)(n+k+1)$ is divisible by $(k+1)!$ and $(n+2)(n+3).....(n+k+1)$ is the product of $k$ consecutive ingtegers so it's divisible by $k!$.  So $(k+1)(n+2)(n+3).....(n+k+1)$ is divisible by $(k+1)k!=(k+1)!$.

Title: Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$.

Body:
Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.

I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5)
\\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$
I got stuck over here.
Am I right till here? Can someone tell what to do next?


答: Your screenshot proves a $\Theta(n)$ function's values from $1$ to $n$ have a $\Theta(n^2)$ sum. The sum of $n$ functions in $O(n)$ being $O(n^2)$ (if that's what you were asking about) requires a different treatment. Indeed, since each of the $f_i(n):=in\in O(n)$ but $\sum_{i=1}^nf_i(n)\sim\tfrac12n^3$, it's not true in general. If however some $f\in O(n)$ upper-bounds all $f_i$, you can use a strategy almost identical to the screenshotted one.

Title: one doubt on big oh notation

Body:
I see an example that $f(n)=O(n)$ and $\sum_{i=1}^nf_i(n)=O(n^2) $ in which $f_i$ is a function from natural to natural number.

I ran into a challenge, that above notation is differ from the
following or not? my misunderstanding point is why the above fact is
false?


问:
Estimate probability with the central limit theorem
I have a task to estimate probability using CLT. $P(52<X<98)$, $X=\sum_{i=1}^{50}X_i$ and  $X_i\in U([1,2])$ $X_i$ are independent.

So $X$ parameters are: $\sigma = \sqrt\frac{25}{6}, EX = 75$

This is what i did:
$$P\Biggl(\frac{X-75}{\sqrt\frac{25}{6}}<\frac{98-75}{\sqrt\frac{25}{6}}\Biggl) = \Phi(11,3)$$
$$1-P\Biggl(\frac{X-75}{\sqrt\frac{25}{6}}<\frac{52-75}{\sqrt\frac{25}{6}}\Biggl) = 1-\Phi(-11,3)$$
$$\Phi(11,3)-(1-\Phi(-11,3)) = 1-(1-0) = 0$$
But the answer in my book is $0,9672.$ I have no clue where i did a mistake.

答:
One way to check this by yourself would be by simulating the distribution of $\sum X_i$.
In R (with tidyverse) this can be done by

```<code>require(tidyverse)
set.seed(1234)
mins <- 1
maxs <- 2
n <- 50
x1  <- seq(from = mins*n, to = maxs*n, length.out = 500)
replications <- 100000
df <- as_tibble(replicate(replications,sum(runif(n, mins, maxs))))

dens <- data.frame(value = x1, y = dnorm(x1, mean = n*(maxs+mins)/2, sd = sqrt(n*(maxs-mins)^2/12)))

ggplot(df, aes(x = value)) +
  geom_histogram(aes(y = stat(count) / sum(count)), binwidth  = 1, col = "red", fill = "grey75")+
  geom_line(data=dens,aes(y=y), colour="blue")+
  labs(title = "Histogram of sum(X_i) with normal approx. density") + theme_bw(16, "serif") +
  theme(plot.title = element_text(size = rel(1.2), vjust = 1.5))
</code>```

where the plot looks like this:

Another quite easy way is to use the $\sigma$-rules of the normal distribution (e.g. on Wikipedia1 ; Wikipedia2), that is aproximatly 99.73% of the "mass" are within $\pm 3 \sigma$ around $\mu$.

Body:
One way to check this by yourself would be by simulating the distribution of $\sum X_i$.
In R (with tidyverse) this can be done by

require(tidyverse)
set.seed(1234)
mins <- 1
maxs <- 2
n <- 50
x1  <- seq(from = mins*n, to = maxs*n, length.out = 500)
replications <- 100000
df <- as_tibble(replicate(replications,sum(runif(n, mins, maxs))))

dens <- data.frame(value = x1, y = dnorm(x1, mean = n*(maxs+mins)/2, sd = sqrt(n*(maxs-mins)^2/12)))

ggplot(df, aes(x = value)) +
  geom_histogram(aes(y = stat(count) / sum(count)), binwidth  = 1, col = "red", fill = "grey75")+
  geom_line(data=dens,aes(y=y), colour="blue")+
  labs(title = "Histogram of sum(X_i) with normal approx. density") + theme_bw(16, "serif") +
  theme(plot.title = element_text(size = rel(1.2), vjust = 1.5))

where the plot looks like this:

Another quite easy way is to use the $\sigma$-rules of the normal distribution (e.g. on Wikipedia1 ; Wikipedia2), that is aproximatly 99.73% of the "mass" are within $\pm 3 \sigma$ around $\mu$.


问:
How to find the minimum value of cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(20x)
I want to find the minimum value of the series cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(2nx). x could be 2pit. Anyone can share a method of how to determine the minimum value of the series?

答:
You can use
$\displaystyle \sum_{k=1} \cos(kx)=R\left(\sum_{k=1} e^{2ikx}\right)$


答: Let $D$ the operator on $\mathbb{Z}[x]$ given by $P\to D(P)(x)=P(x+1)$. It is easy to see that ($I$=identity) $D(P)-P=(D-I)(P)$ is of degree strictly less than $P$. Hence if $P$ is of degree $s$, $(D-I)^{s+1}=0$ ; in your case, $P$ is of degree $5$, hence $(D-I)^6(P)=0$, and you can see that $P(n+6)$ is a combination with constant coefficients in $\mathbb{Z}$ of $P(n),...P(n+5)$. Now your induction is easy (if for $m, m+1,m+2,m+3,m+4,m+5$ you have the good property, show it works for $m+1,...,m+6$) (note that you can begin with $-5,-4,-3,-2,-1, 0$)


答: Welcome to MSE. As a hint, you might want to look into Lagrange's trignometric identities. (They can be derived with complex numbers like Luis is suggesting.</p>

Title: How to find the minimum value of cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(20x)

Body: I want to find the minimum value of the series cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(2nx). x could be 2pit. Anyone can share a method of how to determine the minimum value of the series?


问:
$E(|X+Y|^p)\ge E(|X|^p)$
If $X$ and $Y$ are independent, $E(|X|^p)<\infty$ for some $p\ge 1$, and $E(Y) = 0$, then $E(|X+Y|^p)\ge E(|X|^p)$.

Is it maybe true that for each fixed $x$ that the following inequality is true?
$$\int|x+y|^p \mu_Y(dy) \ge |x|^p$$

So that integrating both sides would give

$$\int\int|x+y|^p \mu_Y(dy)\mu_X(dx) \ge \int|x|^p\mu_X(dx)$$

答:
As $E(Y)=0$ you can rewrite your first display equation as $$E(|x+Y|^p)\geq |x+E(Y)|^p$$

For $p\geq 1$ the function $f(y) = |x+y|^p$ is convex, hence the result follows from Jensen's inequality.

Title: $E(|X+Y|^p)\ge E(|X|^p)$


问:
What is the dimension of $U\cap W$?
Let $U.W \subset V$ be two subspaces different from $V$, a vector space whose dimension is $n$, if the dimensions of both $U$ and $W$ are equal to $n-1$ then, what is the dimension of $U \cap W$?

Intuitively, I know that $\dim(U \cap W)\leq n-1$. I thought that perhaps Grassman's formula ($\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$) could help me solve this problem, but then I realized that I would need more information about $\dim(U+W)$ which I only intuitively know to be less or equal to $n$.

Edit:
I think I have an idea, using Grassman's formula: dim(U+W)=dim(U)+dim(W)−dim(U∩W)
. Because dim(W)=dim(U)=n−1 and dim(V)=n then n−1≤dim(U+W)≤n. Therefore n−1≤dim(U)+dim(W)−dim(U∩W)≤n ⇒ n−1≤(n−1)+(n−1)−dim(U∩W)≤n ⇒ −n+1≤−dim(U∩W)≤−n+2 ⇒ n−2≤dim(U∩W)≤n−1 Is there a way to decide if dim(U∩W)=n−2 or dim(U∩W)=n−1 or is the inequality the best possible answer?

答: Consider two lines $L_1$ and $L_2$ through the origin in $\mathbb R^2$. If they are different lines, then the intersection is 0-dimensional. But if they are the same line, then the intersection is 1-dimensional. So, without more information about $U$ and $W$, we can only get a range of values for $\dim(U \cap W)$.

Title: What is the dimension of $U\cap W$?


问:
Equality of triple integrals over unit sphere $ \iiint_{\text{unit ball}} x e^{ax + by + cz} dV$
I have to calculate
$$ \iiint_{\text{unit ball}} x e^{ax + by + cz} \,dV,$$

where by "unit ball" I mean the region $x^2 + y^2 + z^2 \leq 1$.

I know how to calculate this (rotation matrix that takes $(a,b,c)$ to $(0,0,\sqrt{a^2 + b^2 + c^2})$ and then spherical coordinates). The answer gives
$$\frac{4\pi a}{r^5}((3 + r^2)\sinh(r) - 3r \cosh(r)), \,\,\,\,\,r = \sqrt{a^2 + b^2 + c^2}$$

Now this would suggest that we have the following equality:

$$ \iiint_{\text{unit ball}} \frac{x}{a} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{y}{b} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{z}{c} e^{ax + by + cz} \,dV$$

Is it true? How would one quickly prove it if it is? It would be relevant because if one could quickly spot and prove this, then by naming the value of the above integrals $I$, we would have
$$(a^2 + b^2 + c^2)I = \iiint_{\text{unit ball}} (ax+by+cz) e^{ax + by + cz} \,dV$$

This integral would be much easier to calculate than the first since it wouldn't require to calculate the rotation explicitly (just considering an arbitrary rotation that orients $(a,b,c)$ along the $z$-axis suffices.)

Any ideas?

答:
@Quanto has provided an elegant answer to your question, but probably the following ideas could also be useful.
Let's consider $I(r,R_0)=\iiint_{\text{ball radius Ro}} x e^{ax + by + cz}dV=\frac{\partial}{\partial{a}}\iiint_{\text{ball radius Ro}}e^{ax + by + cz}dV$.

We choose $\vec{r}=(a,b,c)$, $r=|\vec{r}|=\sqrt{a^2+b^2+c^2}$, then $ax+by+cz=(\vec{r},\vec{R})$, where $\vec{R}=(x,y,z)$ and $R=|\vec{R}|$.

In order to integrate over $d^3{\vec{R}}$ ($dV$) we have to choose a coordinate system (which is not defined apriory). So, we can orient it in such a way that the polar axis $Z$ has the direction of the vector $\vec{r}$. At this convenient choice we get $ax+by+cz=(\vec{r},\vec{R})=rR\cos\theta$ (polar angle).

Integration is straightforward:

$I(r,R_0)=\frac{\partial}{\partial{a}}\int_0^{2\pi}d\phi\int_0^{R_0}{R}^2{dR}\int_0^\pi{e}^{rR\cos\theta}\sin\theta{d}\theta=2\pi\frac{\partial}{\partial{a}}(\int_0^{R_0}{R}^2{dR}\int_{-1}^1{e}^{rRt}{d}t)=$$=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\int_0^{Ro}(e^{rR}-e^{-rR})RdR)=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\frac{\partial}{\partial{r}}\int_0^{Ro}(e^{rR}+e^{-rR})dR)=2\pi\frac{\partial}{\partial{a}}(\frac{1}{r}\frac{\partial}{\partial{r}}\frac{e^{rR_0}-e^{-rR_0}}{r})$

Given the fact that $\frac{\partial}{\partial{a}}f(r)=(\frac{\partial}{\partial{r}}f(r))\frac{\partial}{\partial{a}}r=(\frac{\partial}{\partial{r}}f(r))\frac{\partial}{\partial{a}}\sqrt{a^2+b^2+c^2}=(\frac{\partial}{\partial{r}}f(r))\frac{a}{r}$ we get
$$I(r,R_0)=\frac{2\pi{a}}{r}\frac{\partial}{\partial{r}}\left(\frac{1}{r}\frac{\partial}{\partial{r}}(\frac{e^{rR_0}-e^{-rR_0}}{r})\right)$$

Finally, $$I(r,R_0)=\frac{2\pi{a}}{r^5}\Biggl((r^2R_0^2+3)\sinh(rR_0)-3rR_0\cosh(rR_0)\Biggr)$$

At $R_0=1$ we get your answer.


问:
Combinatorical problem with k-subsets of [n] with no consecutive elements
How many $k$-substes of $[n]$ do not contain consecutive elements.

I saw the resolution to this problem, but I didn't understand quite well.

Here is what they have:

Line up k $S$ and  $(n-k)$  $R$ so that no $S$ are consecutive.

The problem is ,now, equivalent to choose $k$ spaces between $n-k+1$  ($k$ spaces between $R$, in the front and in the end of the sequence)

The final answer is ${n-k+1 \choose k}$

Can someone give me a hint to what the author might have thought when solved this problem, or if you have a similarly good answer to this problem, expose it here.

Thanks

答:
Let the elements in the subset in increasing order be labeled $x_1,x_2,\dots,x_k$.  Let $[n]=\{1,2,3,\dots,n\}$

Let us consider the variables $y_1,y_2,\dots,y_{k+1}$ defined as $y_1=x_1-1$, $y_2=x_2-x_1$, $y_3=x_3-x_2$, ..., $y_k=x_k-x_{k-1}$ and finally $y_{k+1}=n-x_k$.  That is to say, the variables $y_i$ count the number of empty spaces between each of the $x_i$, taking the extremes into account as well.

The condition that none of the elements in your set are consecutive is equivalent to $y_i\geq 2$ for all $i\in\{2,3,\dots,k\}$.  $y_1$ may equal zero as can $y_{k+1}$.

Notice that $y_1+y_2+\dots+y_k+y_{k+1} = (x_1-1)+(x_2-x_1)+(x_3-x_2)+\dots+(n-x_k) = n-1$

Now, one final change of variable, since we don't like working with lowerbounds of $2$, we prefer to work entirely with lowerbounds of zero or lower bounds of $1$, let us let $z_i = y_i-2$ for each $i\in\{2,3,\dots,k\}$ and let $z_1=y_1$ and $z_{k+1}=y_{k+1}$.

We have then $z_1+z_2+\dots+z_{k+1}=n-1-(k-1)\cdot 2$ and each $z_i\geq 0$.

This is in a known form for a stars-and-bars problem of counting number of non-negative integer diophantine solutions to a system with answer

$$\binom{(n-1-2\cdot(k-1)+(k+1)-1}{(k+1)-1}$$

Simplifying, $n-1-2k+2+k+1-1 = n-k+1$ giving:

$$\binom{n-k+1}{k}$$


问:
How to get sign right for parametric area for circle
For finding the area of a circle of radius $r$ using $x=r \cos(t)$ and $y=r \sin(t)$ with $t$ from $0$ to $2 \pi$, it seems that the formula usually given for parametric area, which is $\int_a^b y(t)x'(t)dt$, gives a negative result and all the examples I could find use the formula of $\int_a^b x(t)y'(t)dt$

From what I've read, for the area formula to work, you'd need a closed curve but the sign depends on the direction? From what I read on brilliant (the above paragraph, as well as the circle example), my negative sign is due to the using the first formula and not the second... however, if I use my formula with $\int_{2\pi}^{0} r \sin(x) \cdot -1 r \sin(x) \ dx$ then it works out well and I get $\pi r^2$ with the positive sign

Then reading this answer to a user with a similar issue, the suggestion is due to the sign of the product of the terms inside more than anything.

Which is the right formula to use, and is there a good way to predict which one would give the correct sign?

答:
For a length $l$ between points $A$ and $B$ of a curve we have $AB^2=\Delta x^2+\Delta y^2$. As $\Delta x, \Delta y \to 0$, $AB \to l$ and we have $(dl)^2=(dx)^2+(dy)^2.$ So then $(dl)^2=\left(1+(\frac{dy}{dx})^2\right)dx^2\Leftrightarrow dl=\sqrt{1+(\frac{dy}{dx})^2}dx$. The equation of the circle is of course $x^2+y^2=r^2.$ So we have $2xdx+2ydy=0\Leftrightarrow \displaystyle\frac{dy}{dx}=-\frac{x}{y}.$ Let $l$ be the circle's circumference. You have $l = \displaystyle\int_{0}^{R}{\sqrt{1+\left(\frac{x}{y}\right)^2}dx}=\int_{0}^{R}{\sqrt{\frac{R^2}{R^2-x^2}}}dx=\int_{0}^{R}{\frac{1}{\sqrt{1-\displaystyle\frac{x^2}{R^2}}}}dx=R\int_{0}^{R}{\frac{1}{\sqrt{1-\displaystyle\frac{x^2}{R^2}}}}d\left(\frac{x}{R}\right)= \\
=R\cdot arcsin\left(\displaystyle\frac{x}{R}\right)\Biggr|_{0}^{R}=2\pi R.$ So now integrating it is easily seen that the area of a circle is $\pi R^2.$ I know it isn't quite what you wanted but I thought it was interesting to consider.

But now of course Green's theorem comes in handy when you need to integrate parametric curves. The answer above provides a very nice example and shows how careful you need to be choosing the limits. Notice here that $t \in [0,2\pi]$ and the $x(t), y(t)$ describe parametric curves in $\mathbb{R}^2$. So obviously you can have $\displaystyle\int{y\hspace{0.11cm}dx}=\int_{0}^{2\pi}{y(t)x'(t)dt}=r^2 \int_{0}^{2\pi}{cos^2(t)dt}=4r^2\int_{0}^{\frac{\pi}{2}}{cos^2(t)dt}=I \hspace{0.5cm} (1)$, which is always an incredibly useful trick in dealing with curves... Let's solve for $I$! Now let $\theta=\frac{\pi}{2}-t\Leftrightarrow du = -dt$. So $I = 4r^2\displaystyle\int_{0}^{\frac{\pi}{2}}{sin^2(t)dt}\hspace{0.5cm} (2)$. Now $(1)+(2)=2I=4r^2\displaystyle\int_{0}^{\frac{\pi}{2}}{1d\theta}\Leftrightarrow 2I = 2\pi r^2\Leftrightarrow I = \pi r^2.$ As desired in the first place.


问:
A semisimple Lie algebra has finitely many representation of a given finite dimension.
Let $\mathfrak{g}$ be a finite dimensional, semisimple Lie algebra, with Cartan subalgebra $\mathfrak{h}$. I want to show that $\mathfrak{g}$ has finitely many representations of dimension $n$, up to isomorphism, where $n \in \mathbb{N}$.

I know that finite dimensional representations of $\mathfrak{g}$ are the direct sum of their weight spaces. But since $\mathfrak{h}$ is finite dimensional, also $\mathfrak{h}^*$ is, thus there are only finitely many ways we can decompose a representation $V$, of dimension $n$, as the sum of its weight spaces.

I think that two representations with same root space decomposition are isomorphic,
thus concluding the argument, but I'm not quite sure how to prove it.

答: For finite-dimensional repns, we know there is a unique highest weight (and weight space). By irreducibility, this weight space generates the whole repn. In particular, there is a chain of lowering operators (actions by elements in the negative weight spaces) taking the highest weight vector to the lowest weight vector (which we know has the negative of the weight of the highest). Thus, the dimension of the whole is certainly at least the number of steps it takes to get from highest to lowest weight, which is something like $\sum_i |\lambda_i|$ where (in some normalization) the $\lambda_i$ are the coordinates of the highest weight $\lambda$. Since highest weights are "integral" and "dominant", there are only finitely many with $\sum_i |\lambda_i|$ below a given bound.


问:
Prove That $n(n+1)(n+2)(n+3)$ is never a square of a positive integer.
$n \in \mathbb{N}$, bigger than zero.

My proof:(A Proof by Contradiction)

Alert: There is a lot of variables.

Suppose that:

$n(n+1)(n+2)(n+3)=x^{2}$, with $x \in \mathbb{N}$
$$n(n+3)(n+2)(n+1)=x^{2}$$
$$(n^{2}+3n)(n^{2}+3n+2)=x^{2}$$
let $n^{2}+3n = a$.
$$a(a+2)=x^{2} \iff a^{2} +2a=x^{2} \iff a^{2} +2a-x^{2}=0$$

And this is a quadratic equation in $a$, So:

$$a=\frac{-2 \pm \sqrt{4+4x^{2}}}{2} \implies \sqrt{4+4x^{2}} \in \mathbb{N} \iff 4+4x^{2}=b^{2}$$
$$ 4(1+x^{2})=b^{2} \iff 4 \mid b^{2} \iff 2 \mid b \iff b=2c.$$

Now let's plug this result to this equation:

$$4(1+x^{2})=(2c)^{2} \iff 4(1+x^{2})=4c^{2} \iff 1+x^{2}=c^{2} \iff (c-x)(c+x)=1$$

And $1=1 \cdot 1$, so :$(c-x)=1$ and $(c+x)=1$
$$c+x=c-x \iff x=-x \iff 2x=0 \iff x=0$$

By plugging this result into this equation : $a(a+2)=x^{2}.$

$a(a+2)=0 \iff a=0$ or $a=-2$ but $-2 \notin \mathbb{N}$ So : $x=0$

And finally:
$$n(n+1)(n+2)(n+3)=0 \iff n=0$$
And the other values $\notin \mathbb{N}$

This proof is too huge for me so perhaps I've made some mistakes, And if you have an easier proof please post it.

答:
Hint: Show that $ n(n+1)(n+2)(n+3) + 1$ is a perfect square.

Hence, the original expression cannot be the square of a positive integer (but could be the perfect square 0).


问:
First countable compact spaces are sequentially compact, how weaker assumption ease the proof?
A first countable, countably compact space is sequentially compact

José Carlos Santos's answer

Let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $X$. There are two possibilities:

There is some $x\in X$ such that $x_n=x$ for infinitely many $n$'s. Then it is obvious that $x$ is a limit of a subsequence of the sequence $(x_n)_{n\in\mathbb N}$.

For every $x\in X$, the equality $x_n=x$ holds for finitely many $n$'s only. In that case, the set $\{x_n\,|\,n\in\mathbb N\}$ is infinite. Since $X$ is limit point compact, there is some $x\in X$ which is a limit of a sequence of elements of $\{x_n\,|\,n\in\mathbb N\}$. Therefore, $x$ is a limit of some subsequence of the sequence $\{x_n\,|\,n\in\mathbb N\}$.

I cannot really see if we have one more assumption such as Compactness not Countably Compactness, how this assumption ease the proof?

答:
The proof for a compact space is exactly the same, we need only that for a countably infinite subset $A$ there is a strong limit point i. e. a $p \in X$ such that for every neighbourhood $O$ of $p$ we have that $A\cap O$ is infinite. It is well-known that countable compactness is equivalent to this fact.

In either case we then proceed to exploit the first countability at $p$ to construct the required convergent subsequence to $p$.

Compactness does not help the proof at all. But as compactness implies countable compactness the theorem is stated with the weakest assumption only.


问:
Prove by induction that $(n+1)(n+2)(n+3)(n+4)(n+5)$ is divisible by $120$.

Prove by induction that $$(n+1)(n+2)(n+3)(n+4)(n+5)$$ is divisible by $120$.

I tried to solve it, but could do so only till the inductive step.
I assumed: $$p(k)=(k+1)(k+2)(k+3)(k+4)(k+5), and $$ $$p(k+1)=(k+2)(k+3)...(k+6)$$
Then, I distributed it as: $$p(k+1)=(k+2)(k+3)...(k+5)(k+1+5)
\\=(k+1)(k+2)...(k+5)+5(k+2)(k+3)...(k+5)\\=p(k)+5(k+2)(k+3)...(k+5).$$
I got stuck over here.
Am I right till here? Can someone tell what to do next?

Body:
$120=2\cdot 3\cdot 4\cdot 5$.  $\Rightarrow   120|1\cdot 2\cdot 3\cdot 4\cdot 5$ and $120|2\cdot 3\cdot 4\cdot 5\cdot 6$.

Now, if $120|k(k+1)(k+2)(k+3)(k+4)$ and $120|(k+1)(k+2)(k+3)(k+4)(k+5)$, then,

$120|(k+1)(k+2)(k+3)(k+4)\{(k+5)-k\}$

$\Rightarrow 24|(k+1)(k+2)(k+3)(k+4)$

Now, use normal induction and proceed.


问:
Asymptotics of $\int_1^\infty a^x x^{-x} dx$
Let
$$I(a) = \int_1^\infty \frac{a^x}{x^x}\,dx.$$
What is the asymptotic growth of $I(a)$? I am primarily interested in seeing if there is an upper bound of the form $I(a) = O\big(\exp(a^c)\big)$, with $c<1$.

答: Hint (basically Laplace method): substitute $x=(a/e)+t\sqrt{a}$ (note that $x=a/e$ is the maximum of the integrand) and take $a\to\infty$. With all the necessary checks for DCT to apply, we get $$\lim_{a\to\infty}e^{-a/e}I(a)/\sqrt{a}=\int_{-\infty}^\infty e^{-et^2/2}\,dt=\sqrt{2\pi/e}.$$


问:
Exchanging limit with integral for functions of two variables
Assume $F(0, \infty) \times (0,\infty) \rightarrow [0,\infty)$ to be a measurable function such that for every $x>0$
$$
\lim_{z \rightarrow \infty } F(x,z) =0.
$$
I also know that
$$
\int_0^z F(x,z) dx \leq C \quad \forall z >0.
$$
Can I conclude that
$$
\lim_{z \rightarrow \infty } \int_0^z F(x,z) dx  =0 ?
$$

答:
The statement is not true. Take the counterexample

$$F(x,z) = \frac{z}{z^2+x^2}$$

The limit goes to $0$ in $z$ and the integral in $x$ is bounded but

$$\int_0^z \frac{z}{z^2+x^2}dx = \arctan 1 = \frac{\pi}{4} \not\to 0$$


问:
continuous and residual spectrum of the right shift
Let $1\le p<q\le\infty$, $L:\ell^p\to\ell^p,x\mapsto(x_2,x_3,\ldots)$ and $R:\ell^q\to\ell^q,y\mapsto(0,x_1,x_2,\ldots)$.

It's easy to show $L'=R$, $\sigma(L)=\sigma(R)=\overline B_1(0)$, $\sigma_p(L)=B_1(0)$, $\sigma_r(L)=\emptyset$, $\sigma_c(L)=\partial B_1(0)$, $\sigma_p(R)=\emptyset$ and $B_1(0)\subseteq\sigma_r(R)$.

But I'm failing to show that $\sigma_c(R)=\emptyset$. How can we do that?

Clearly, if $\lambda\in\sigma_c(R)$, then $\mathcal R(\lambda-R)$ is dense and hence $\lambda\not\in\sigma_p(L)$, i.e. $\lambda\in\partial B_1(0)$ ... But I'm not able to derive a contradiction from that.

答:
Since $q>1$, you have $R=L'$, considering $L$ acting on $L^p$ for $p$ conjugate to $q$.  From
$$
\|(R-\lambda I)x\|_p=\|Rx-\lambda x\|_p\geq\|Rx\|_p-|\lambda|\,\|x\|_p=(1-|\lambda|)\,\|x\|_p,
$$
we get that $R-\lambda I$ is bounded below for all $\lambda\in B_1(0)$. This precludes $\lambda$ from being an approximate eigenvalue. Combined with the fact that the boundary of the spectrum consists of approximate eigenvalues shows that $\sigma_{\rm ap}(R)=\partial B_1(0)$.

From $\sigma(R)=\sigma_r(R)\cup\sigma_{\rm ap}(R)$ we conclude that $B_1(0)\subset\sigma_r(R)$.

If $\lambda\in\partial B_1(0)$, then $ \lambda\not\in\sigma_p(L)$ (here we use $p<\infty$). This means that $\ker(L-\lambda I)=\{0\}$. It is well-known (proof at the end) that if an operator is injective, then its adjoint has dense range. So $\overline{\operatorname{ran}(R-\lambda I)}=\ell^q$,
showing that $\lambda\in\sigma_c(R)$. That is, $\partial B_1(0)\subset \sigma_c(R)$. We have
$$
\overline{B_1(0)}=\sigma(R)=\sigma_r(R)\cup\sigma_c(R)
$$
(since $\sigma_p(R)=\emptyset$). As $\sigma_r(R)$ and $\sigma_c(R)$ are disjoint, with $B_1(0)\subset\sigma_r(R)$ and $\partial B_1(0)\subset\sigma_c(R)$, it follows that
$$
\sigma_r(R)=B_1(0),\qquad \sigma_c(R)=\partial B_1(0).
$$

Proof that $T:X\to Y$ is a bounded operator and $T'$ is injective, then $T$ has dense range.

If $\operatorname{ran}T$ were not dense, there exists $y\in Y\setminus\overline{\operatorname{ran}T}$. By Hahn-Banach there exists $g\in Y'$ with $g(y)=1$ and $g(Tx)=0$ for all $x\in X$. This means that
$$
(T'g)(x)=g(Tx)=0
$$
for all $x$, so $T'g=0$. As $T'$ is injective, we get that $g=0$ a contradiction.


问:
Uniqueness of the solution of a linear system of equations
QUESTION

Consider the positive integers, $N,T,J$, where $N>T$, $N>J$. $1_{a}$ denotes the $a\times 1$ vector of ones.

Consider the system of equations
$$
\begin{pmatrix}
D'Y\\
F'Y
\end{pmatrix}= \underbrace{\begin{pmatrix}
D'D & D'F\\
F'D & F'F
\end{pmatrix}}_{\equiv W}
\beta
$$
where $Y$ is an $NT\times 1$ vector; $D$ is an $NT\times N$ matrix; $F$ is an $NT\times J$ matrix; $\beta$ is an $(N+J)\times 1$ vector.

Could you help me to find some necessary (and, if possible, also sufficient) conditions such that the system has a unique solution wrto $\beta$. That is, such that the matrix $W$ is invertible?

I report below some properties of the system that I analyse. I'm asking your help, because perhaps you have already seen linear systems with the same structure and hence you may be able to advise.

SOME PROPERTIES OF THE SYSTEM

$Y$ does not have elements equal to zero. The matrices $F$ and $D$ have specific structures.

With regards to the matrix $F$:

Let me decompose $F$ in $N$ submatrices, but cutting it every $T$ rows. Then, with some abuse of notation, let me write $F$ as the collection of such $N$ submatrices, $F\equiv \{F_1,F_2,...,F_N\}$, where each submatrix $F_i$ has size $T\times J$.

Property 1: For each $i\in \{1,...,N\}$, the submatrix $F_i$ is such that in each row, one and only one element takes value in $(0,1)$ and all the other elements takes value zero.

Property 2: If the submatrix $F_i$ has each of its nonzero elements positioned in the same columns as the submatrix $F_k$, then it should be that $F_i=F_k$.

Construct the bipartite graph such the nodes in one side (hereafter, side 1) represent the column indices $j\in \{1,...,J\}$ and the the nodes in the other side (hereafter, side 2) represent the submatrix indices $i\in \{1,...,N\}$. Draw an edge between nodes $j\in \{1,...,J\}$ and $i\in \{1,...,N\}$ if at least one element of the $j$-th column of $F_i$ has a nonzero element.

Property 3: The aforementioned bipartite graph should be such that each node in side 1 is indirectly connected to at least another node in side 1. Further, the bipartite graph should be such that each node in side 2 is indirectly connected to at least another node in side 2. This is called a connected bipartite graph.

For example, for $N=4$, $T=2$, $J=3$, we could have
$$
F\equiv \Big\{\underbrace{\begin{pmatrix}
0.3 & 0 & 0\\
0 & 0.1 & 0\\
\end{pmatrix}}_{F_1},
\underbrace{\begin{pmatrix}
0.7 & 0 & 0\\
0.2 & 0 & 0\\
\end{pmatrix}}_{F_2},
\underbrace{\begin{pmatrix}
0 & 0 & 0.5\\
0 & 0.3 & 0\\
\end{pmatrix}}_{F_3},
\underbrace{\begin{pmatrix}
0 & 0 & 0.6\\
0 & 0 & 0.4\\
\end{pmatrix}}_{F_4}\Big\}
$$
with bipartite graph

With regards to the matrix $D$:

As done for $F$, let me write $D\equiv \{D_1,D_2,...,D_N\}$, where each submatrix $D_i$ has size $T\times N$.

Property 4:  For each $i\in \{1,...,N\}$, the submatrix $D_i$ has its $i$-th column equal to $1_{T}$ and all the other elements equal to zero.

For example, for $N=4$ and $T=2$, we have
$$
D\equiv \Big\{\underbrace{\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 0 & 0& 0\\
\end{pmatrix}}_{D_1},
\underbrace{\begin{pmatrix}
0 & 1 & 0& 0\\
0 & 1 & 0& 0\\
\end{pmatrix}}_{D_2},
\underbrace{\begin{pmatrix}
0 & 0 & 1& 0\\
0 & 0 & 1& 0\\
\end{pmatrix}}_{D_3},
\underbrace{\begin{pmatrix}
0 & 0 & 0&1\\
0 & 0 & 0&1\\
\end{pmatrix}}_{D_4}\Big\}
$$

Given, the properties above, $D'D$ and $F'F$ are diagonal matrices with strictly positive diagonal elements, where

$$
\underbrace{D'D}_{N\times N}=   \begin{pmatrix}
    T & & \\
    & \ddots & \\
    & & T
\end{pmatrix}, \underbrace{F'F}_{J\times J}=\begin{pmatrix}
    \sum_{k=1}^{NT}F(k,1)^2 & & \\
    & \ddots & \\
    & &     \sum_{k=1}^{NT}F(k,J)^2
\end{pmatrix}
$$

Further,
$$
F'D=\begin{pmatrix}
\sum_{t=1}^T F(1,t) & \sum_{t=T+1}^{2T} F(1,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(1,t)\\
\vdots & \vdots & \vdots & \vdots\\
\sum_{t=1}^T F(J,t) & \sum_{t=T+1}^{2T} F(J,t) & ... & \sum_{t=(N-1)T+1}^{NT} F(J,t)\\
\end{pmatrix}
$$

答:
As I note in my comment, we can write
$$
W = \pmatrix{D & F}' \pmatrix{D & F}.
$$
As such, $W$ is invertible if and only if the columns of $\pmatrix{D & F}$ are linearly independent.

Now, rearrange the rows of the matrix $\pmatrix{D & F}$ such that we end up with a matrix of the form
$$
M_1 = \pmatrix{I_N & G_1\\
I_N & G_2\\
\vdots & \vdots \\
I_N & G_T},
$$
where $I_N$ denotes the size $N$ identity matrix.
For the example that you gave, we could end up with
$$
M_1 = \left[\begin{array}{cccc|ccc}1&0&0&0& 0.3 & 0 & 0\\
0 & 1 &0 & 0 & 0.7 & 0 & 0\\
0 & 0 & 1& 0 & 0 & 0 & 0.5\\
0 & 0 & 0 & 1 & 0 & 0 & 0.6\\
\hline
1 & 0 & 0 & 0 & 0 & 0.1 & 0\\
0 & 1 & 0 & 0 & 0.2 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0.3 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0.4
\end{array}\right].
$$
Now, apply row operations on the matrix $\pmatrix{D&F}$. In particular, subtract the first block-row from each of the others to get the matrix
$$
M_2 =
\pmatrix{I_N & G_1\\
0 & G_2 - G_1\\
\vdots & \vdots \\
0 & G_T - G_1}.
$$
We see that $M_2$ has linearly independent columns if and only if the smaller matrix
$$
M_3 = \pmatrix{G_2 - G_1\\ \vdots\\ G_T - G_1}
$$
has linearly independent columns.

In terms of the original problem, we can reach the following conclusion.
For each of the blocks of $F$, obtain the matrix $H_i$ by removing the frist row and subtracting the first row of $F_i$ from the other rows. For your example, we would have
$$
H_1 = \pmatrix{-0.3 & 0.1 & 0}, \dots, H_4 = \pmatrix{0 & 0 & -0.2}.
$$
$W$ will be invertible if and only if the block-matrix $H = \{H_1,\dots,H_4\}$ has linearly independent columns. We can see this by noting that $H$ can be attained by rearranging the rows of the matrix $M_3$ obtained via the the process described above.


问:
What are the requirements on a matrix $A$ for $A^TA$ to have an inverse?
I know there's a theorem stating that if the columns of $A$ are independent, then the columns of $A^TA$ are also independent ($A$ is invertible $\Rightarrow A^TA$ is positive symmetric definite is the most common way to prove this. That $A^TA$ is invertible follows immediately from the fact that $0$ is not an eigenvalue)

But is that the minimum requirement for $A^TA$ to be invertible, or can $A^TA$ be invertible even in cases where $A$ is not? I have not been able to prove in general that "$A$ is not invertible $\Rightarrow A^TA$ is not invertible".

答: If $A$ is not invertible, then$$\det(A^TA)=\det(A)^2=0,$$and therefore $A^TA$ is not invertible too.


问:
Symmetric positive definitine matrix $I-X^TX$ => $X^TX<1$
If $I-X^{\top}X$ is symmetric positive definite, why is $\lVert X \rVert < 1$?
It seems logically to me but why does it follow?

It's
$v^{ \top} ( I - X^{ \top } X ) v
=
\lVert v \rVert^2 - \lVert X v \rVert^2      > 0$....

(An attempt was $- \lVert X v \rVert^2   \geq - \lVert X \rVert^2  \cdot  \lVert v \rVert^2 $ but it doesn't help..)

答:
Warm-up (Non-strict inequality). $\quad$ If $I-X^T X \succeq 0$, then $X^T X \preceq I$.

Therefore, we conclude that for each $\xi$ such that $\|\xi\| = 1$,
$$
\|X\xi\|^2 = \xi^T X^T X \xi \leq \xi^T I \xi = \|\xi\|^2 = 1.
$$
Consequently, $\|X\| = \sup_{\|\xi\| = 1} \|X \xi \| \leq 1$.

Strict inequality: $\quad$Suppose now that $I - X^T X \succ 0$. Then $X^T X \prec I$.
By compactness, there exists a $\xi_\star$ of unit norm such that
$$
\|X\| = \|X \xi_\star\| = \sqrt{\xi_\star^T X^T X \xi_\star}
< \sqrt{\xi_\star^T I \xi_\star} = 1.
$$
This gives the strict inequality as required.


答:
Let's assume that $U\ne W$, otherwise the answer is obvious.

Neither can be contained in the other, because they have the same dimension. Thus there is a vector in $W$ not in $U$ and so $U+W=V$.

Now Grassmann's formula tells us that
$$
\dim(U\cap W)=\dim U+\dim W-\dim(U+W)=n-1+n-1-n=n-2
$$

Title: What is the dimension of $U\cap W$?

Body:
Let $U.W \subset V$ be two subspaces different from $V$, a vector space whose dimension is $n$, if the dimensions of both $U$ and $W$ are equal to $n-1$ then, what is the dimension of $U \cap W$?

Intuitively, I know that $\dim(U \cap W)\leq n-1$. I thought that perhaps Grassman's formula ($\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$) could help me solve this problem, but then I realized that I would need more information about $\dim(U+W)$ which I only intuitively know to be less or equal to $n$.

Edit:
I think I have an idea, using Grassman's formula: dim(U+W)=dim(U)+dim(W)−dim(U∩W)
. Because dim(W)=dim(U)=n−1 and dim(V)=n then n−1≤dim(U+W)≤n. Therefore n−1≤dim(U)+dim(W)−dim(U∩W)≤n ⇒ n−1≤(n−1)+(n−1)−dim(U∩W)≤n ⇒ −n+1≤−dim(U∩W)≤−n+2 ⇒ n−2≤dim(U∩W)≤n−1 Is there a way to decide if dim(U∩W)=n−2 or dim(U∩W)=n−1 or is the inequality the best possible answer?


答:
Well,
$$
I - X^\top X \succ 0 \implies v^\top(I-X^\top X)v > 0 \implies \lVert v \rVert^2 > \lVert Xv\rVert^2 \implies \frac{\lVert Xv\rVert^2}{\lVert v \rVert^2} < 1, \forall v \neq 0.
$$

But by the definition of the norm
$$
\lVert X \rVert = \sup_{v\neq 0} \frac{\lVert X v\rVert}{\lVert v\rVert} = \max_{\lVert v\rVert=1} \{\lVert Xv\rVert\}< 1.
$$

For the equivalence
$$
\sup_{v\neq 0} \frac{\lVert X v\rVert}{\lVert v\rVert} = \sup_{\lVert v\rVert = 1} \{\lVert Xv\rVert\}
$$ see Equivalent Definitions of the Operator Norm,
The equivalence
$$
\sup_{\lVert v\rVert} \{\lVert Xv\rVert\} = \max_{\lVert v\rVert} \{\lVert Xv\rVert\},
$$
follows from the fact that $X:V\to W$ is continuous, and $\{v\in V~\vert~ \lVert v \rVert = 1\}$ is a compact set (assuming that $V$ finite dimensional), as pointed out by Drew Brady.

Title: Symmetric positive definitine matrix $I-X^TX$ => $X^TX<1$

Body:
If $I-X^{\top}X$ is symmetric positive definite, why is $\lVert X \rVert < 1$?
It seems logically to me but why does it follow?

It's
$v^{ \top} ( I - X^{ \top } X ) v
=
\lVert v \rVert^2 - \lVert X v \rVert^2      > 0$....

(An attempt was $- \lVert X v \rVert^2   \geq - \lVert X \rVert^2  \cdot  \lVert v \rVert^2 $ but it doesn't help..)


问:
Quick Vertical Tangent Line Question
So I was given the following prompt:

"The position of a car moving along a flat surface at time $t$ is modeled by $(x(t), y(t))$ with velocity vector $v(t)=<3+6\sin(3t), 1+e^{2t}>$ for $0 \leq t \leq 2$. Both $x(t)$ and $y(t)$ are measured in feet, and $t$ is measured in seconds. At time $t=0$, the car is at position $(0,0)$. At what time $t$, for $0≤t≤2$, is the line tangent to the path of the car parallel to the $y$-axis?"

Albeit this question is a bit similar to the last one I asked, I was given the same prompt with a different ending. This ending is also a bit confusing to me, I understand that the vertical tangent line in an equation can be found by setting the denominator of a fraction to $0$, but I'm confused about how I might go about doing that here. Any help would be appreciated!

答:
Note that when the tangent is vertical
$$v_x(t)=3+6\sin(3t)=0$$
which yields $t = \frac{7}{18}\pi,\>\frac{11}{18}\pi$.

Title: Quick Vertical Tangent Line Question


问:
Why is Cauchy's integral formula always written with the function as the subject?
Scouring textbooks, lecture notes, Wikipedia, etc., I notice that the standard presentation of Cauchy's integral formula is $$f(w)=\frac1{2\pi i}\int_L\frac{f(z)}{z-w}\,\mathrm dz\tag1$$ rather than $$\int_L\frac{f(z)}{z-w}\,\mathrm dz=2\pi i f(w).\tag2$$

I'm new to complex analysis, and have so far directly utilised only form $(2)$ (that is, that equation as an integral formula҂), which, to boot, is a little more compact and reads more intuitively than form $(1).$

Or is the canonical form $(1)$ actually more natural? Please shed light on its intuitive interpretation that I must be missing.

҂A formula is typically a rule for expressing a subject in terms of some other variable(s), so the circle-area formula being canonically presented as $$r=\sqrt{\frac A\pi}$$ would similarly cause a double take: is this a quirky misnomer, is $r$ more fundamental than $A$ when discussing circles, does this square-root form lend itself better to typical applications of the equation, etc.?

答: I guess the focus of the formula is not that you can evaluate certain integrals by plugging in function values, but that you can recover a function's values by an integral along a circle.</p>


问:
Find $\{\bar{u}_{k}\}$ in $\mathbb{R}^{2}$ s.t. $\{\|\bar{u}_{k}\|\}$ is bounded increasing but $\{\bar{u}_{k}\}$ not convergent.
I'm trying to figure out how a sequence's distance can be increasing and bounded? Would the sequence $\{(-1)^n)\}$ satisfy this?

Find a sequence $\{\bar{u}_{k}\}$ in $\mathbb{R}^{2}$ with the property that $\{\|\bar{u}_{k}\|\}$ is a bounded increasing sequence but with $\{\bar{u}_{k}\}$ not convergent.

答: Imagine a spiral from inside a unite disk which gets closed and closer to the unit circle from inside but never crosses it. Pick $u_{k}$ on this spiral.


问:
"$f(x)$ decays AT LEAST as $1/x$: Is this Big-Oh notation or Little-oh notation rather?
If I want to show that, as $x\to x_0$, the function $f(x)$ decreases to some value at least as fast as $1/x$, is this saying that one wants to show that $f(x)\in O(1/x)$ as $x\to x_0$?

Or does the "at least" rather mean the little-oh idea, i.e. $f(x)\in o(1/x)$ as $x\to x_0$?

答: “At least” means non-strict inequality ($\le$) so it means Big Oh ($\mathcal{O}$).


答:
This can be shown without invoking determinants as follows: Note that $A^TAx=0$ iff either $Ax=0$ or $A^Tz=0$ for some $z\in\operatorname{Im}(A)$. Consequently:

$$ \ker(A^TA) = \ker(A) \oplus A^{-1}\Big(\ker(A^T)\cap \operatorname{Im}(A))\Big)$$

Here, $A^{-1}$ denotes the pre-image. However, $\ker(A^T) = \operatorname{Im}(A)^\perp$ (this is a fundamental Lemma in linear algebra, remember it/ try to prove it on your own!), hence $\ker(A^T)\cap \operatorname{Im}(A)=\{0\}$. So

$$\ker(A^TA) = \ker(A)\oplus A^{-1}\big(\{0\}\big) =\ker(A) \oplus \ker(A) = \ker(A)$$

In particular, if $A$ is square, then $A$ is invertible if and only if $A^T A$ is invertible.

Title: What are the requirements on a matrix $A$ for $A^TA$ to have an inverse?

Body:
I know there's a theorem stating that if the columns of $A$ are independent, then the columns of $A^TA$ are also independent ($A$ is invertible $\Rightarrow A^TA$ is positive symmetric definite is the most common way to prove this. That $A^TA$ is invertible follows immediately from the fact that $0$ is not an eigenvalue)

But is that the minimum requirement for $A^TA$ to be invertible, or can $A^TA$ be invertible even in cases where $A$ is not? I have not been able to prove in general that "$A$ is not invertible $\Rightarrow A^TA$ is not invertible".


问:
Proving that $|\operatorname{Aut}(C_n)|=2n$.

Question: Prove that $|\operatorname{Aut}(C_n)|=2n$.

This the question $2$ chapter $2$, in the book Algebraic Graph Theory, Godsil and Royle.

I've managed to prove that $\left|\operatorname{Aut}(C_n)\right|\geq2n$ considering that the subgroup generated by the permutation $(12\ldots n)$ and the reflexion.

I haven't been able to prove equality.

答:
Let $v_1, v_2, \dots, v_n$ be the vertices of $C_n$, with edges $v_i v_{i+1}$ for $i=1, \dots, n-1$ as well as $v_1 v_n$.

To pick an automorphism of $C_n$, we need to:

Pick $k \in \{1,2,\dots,n\}$ and decide that $v_1 \mapsto v_k$. Of course, one of these possibilities has to be true, and there's $n$ of them.

Because $v_1 v_2$ has to map to another edge, we know that $v_2 \mapsto v_{k-1}$ or $v_2 \mapsto v_{k+1}$. Choose one of these options, and already you have $2n$ possibilities.

So now it remains to show that each of these choices can be completed to only one automorphism of $C_n$. (You've already shown that each choice can be completed to at least one automorphism.) To do this, check that there's only one possibility for $v_3$, and then that leaves only one possibility for $v_4$, and so on by induction until you get to $v_n$.

(Note that in general this strategy also requires us to check that at the end, what we've defined really is an automorphism. We can skip that here, because you've already proved that at least $2n$ automorphisms exist, and here we see that there are at most $2n$.)


问:
Folland 1.17: The intersection of any set with a sequence of measurable set implies additivity.
The original problem is pretty easy to understand:

Let $A$ be any set in $\mathbb R^n$ and $\{E_i\}_{i = 1} ^{\infty}$ be a sequence of disjoint and measurable sets. Let $m_{*}$ denote the Lebesgue outer measure. We would like to prove the following:

$$
m_{*}(A\cap(\cup_{i=1} ^ {\infty} E_{i})) = \sum_{i = 1} ^ {\infty} m_{*}(A \cap E_i).
$$

I referred to the solution here (page 8, Proposition 1.9), and in this solution, there's one line that confuses me a lot. My equivalent translation is as follows:

Let $B_n := \displaystyle \cup_{i = 1} ^ {n} E_i $, then $$m_{*}(A \cap B_n) = m_{*}(A \cap B_n \cap E_n) + m_{*}(A \cap B_n \cap E_n^C).$$

I would like to know how we get this equation. My confusion is that the outer measure does not support the additivity of disjoint sets whose distances are not positive. Here, I can only have $A \cap B_n \cap E_n$ and $A \cap B_n \cap E_n^C$ are disjoint, how can we directly apply that additivity?

For additivity does not apply to JUST disjoint sets: https://math.stackexchange.com/a/832402/686556

答:
As noted in the comments, the equation is the Caratheodory condition for measurability: namely, $E_n$ is considered to be $m_\ast$-measurable if and only if every subset $\hat A$ of $\mathbb R^n$,

$$m_\ast(\hat A) = m_\ast(\hat A \cap E_n) + m_\ast(\hat A \cap {E_n}^C).$$

Replace $\hat A$ with $A \cap B_n$ and you're done.


问:
Number of roots of the equation $f(xf(x)) = \sqrt{9 - x^2f^2(x)}$

Given the following graph of a $5$th degree polynomial $y = f(x)$:

Find the number of solutions of the equation: $$f(xf(x)) = \sqrt{9 - x^2f^2(x)}$$

My attempt was to figure out what $f(x)$ is, by noticing the roots of its derivative based on the graph to know $f'(x)$, then find $\int f'(x)$, but the function I found did not match the original graph. (Even if I founded, I think it would be very messy to plug it in the equation)

One other idea is that let $t = xf(x)$, then the equation become:

$$f(t) = \sqrt{9-t^2}$$

But I can't process from here.

Is my idea correct? If yes, then how do I continue? Or is there a better way to solve this?

Body:
The number of roots seems to be very dependent on the $f(x) = a_0+a_1x+a_2x^2+a_3x³+a_4x^4+a_5x^5$ coefficients. Follows a MATHEMATICA script showing the results for an adjusted polynomial following the given figure.

f[x_] := a0 + a1 x + a2 x^2 + a3 x^3 + a4 x^4 + a5 x^5
equs = {f[0] == 2, f[3] == 0, f[-1] == 1, f[1] == 3, f[6] == 4, (D[f[x], x] /. {x -> 3}) == 0}
sol = Solve[equs, {a0, a1, a2, a3, a4, a5}][[1]]
f0 = f[x] /. sol
Plot[f0, {x, -2, 8}, AspectRatio -> 1, PlotRange -> {{-2, 7}, {-2, 8}}, PlotStyle -> {Thick, Blue}]

According to the required question the polynomial found is

$$
f(x) = -\frac{1219 x^5}{30240}+\frac{477 x^4}{1120}-\frac{32747 x^3}{30240}-\frac{477 x^2}{1120}+\frac{1189 x}{560}+2
$$

Now following

F[x_] := 2 + (1189 x)/560 - (477 x^2)/1120 - (32747 x^3)/30240 + (477 x^4)/1120 - (1219 x^5)/30240
error = F[x F[x]] - Sqrt[9 - x^2 F[x]^2]
Plot[error, {x, -2, 4}, PlotRange -> {{-1.5, 4}, {-3, 1}}, PlotPoints -> 50, PlotStyle -> {Thick, Blue}]

so we can count $10$ real roots located at

$$
\{-1.09006, 0.242585, 0.472128,\cdots,3.43159, 3.62582 \}
$$


问:
Question about an ODE Theorem from Marsden and Hoffman
So theorem $5.7.2$ states:

Under the above assumptions (the assumptions are a contraction mapping I believe), there is a $\delta>0$ such that the equation:

$$\frac{dx}{dt}=f(t,x),\space x(t_0)=x_0$$

Has a unique $C^1$ solution with $x(t)=\phi(t)$, with $\phi(t_0)=x_0$ for $t_0-\delta<t_0<t_0+\delta$, i.e $\phi'(t)=f(t,\phi(t))$.

This theorem on it's surface make sense to me, but then I'm given the following problem:

Consider $\frac{dx}{dt}=x^2$ with $x(0)=1.$  Solve this explicitly to see that $\delta$ is finite.

This differential equation is easy to solve either by separating the differential equation or proceeding via Picard Iteration, but what is $\delta$? Is it not $0$? since $\phi(0)=1$ for only $t=0$? And now I'm not really sure what this theorem is saying, should $\delta$ always be zero when you solve the differential equation?

答: Restrict the consideration to some neighborhood of the initial point, $x\in[0,2]$. Then $M=\max_{x\in[0,2]}|x^2|=4$. Solution can not grow faster than with slope $4$ in that set, so restrict the time to the interval $|t|\leδ=\frac14$. The Lipschitz constant of $x^2$ on $[0,2]$ is likewise $L=\max_{x\in[0,2]}|2x|=4$. As $Lδ=1$, you get a contraction constant $q=Lδ'<1$ for any $δ'<δ$, so that a solution can be extended to at least the interval $[-\frac14,\frac14]$.


答:
If the matrix $A$ is square, then invertibility of $AA^T$ is the same as $A$ being invertible. Indeed, if $B$ is an inverse of $AA^T$, then $A(A^TB)$ is the identity, so this provides a right inverse for $A$. A single-sided inverse to a square matrix is sufficient to prove it is invertible.

The converse is obvious, because if $A$ is invertible then $(A^T)^{-1}=(A^{-1})^T$.

However, the argument for general matrices is much more interesting. The fact to use is that $A$ and $AA^T$ have the same rank. Once you know this, you see that $AA^T$ is invertible if and only if $A$ has full row rank (rank equals the number of rows).

Why do the matrices $A$ and $AA^T$ have the same rank? It's easier to prove this for $B$ and $B^TB$ and then use $B=A^T$.

We can prove that $B$ and $B^TB$ have the same null space. If $v\in N(B)$ (that is, $Bv=0$), then obviously $v\in N(B^TB)$. Conversely, from $v\in N(B^TB)$ we have $B^TBv=0$, so also
$$
0=v^TB^TBv=(Bv)^T(Bv)
$$
implying $Bv=0$.

Now the rank-nullity theorem allows us to finish:
\begin{align}
n&=\operatorname{rk}B+\dim N(B) \\
&=\operatorname{rk}B^TB+\dim N(B^TB)
\end{align}
so we reach $\operatorname{rk}B=\operatorname{rk}B^TB$ as desired.


问:
Acceptance sampling schemes for binomial distribution
Two acceptance sampling schemes, A and B, are proposed for deciding whether or not
to accept a large batch of items from a production process in which 5% of the items
produced are defective.
Scheme A: take a random sample of size 20 and accept the batch if no more than 2
defectives are found, otherwise reject it.
Scheme B: take a random sample of size 10; accept the batch if no defectives are found and
reject the batch if 2 or more are present. If 1 defective is found, take a further sample of
size 10, and accept the batch if the number of defectives in this second sample is no more
than 1; otherwise reject the batch.

(a) Find the probability of acceptance for each of the two schemes.

(b) Find the expected number of items sampled for each scheme.

It cost Rs. 10.00 to test each item sampled and for every batch rejected a further cost of
Rs.2000.00 is incurred. Do not consider the cost of testing any replacement batches.

(c) Suggest, giving your reason, which scheme should be adopted.

*here I got some answers for***

(a) For A scheme the probability of acceptance = 0.9245

```<code>For B scheme the probability of acceptance = 0.8867
</code>```

(b) For A the expected number of items = 18.49

```<code>For B the expected number of items = 11.747 ( I am not sure about this answer,if I am wrong please correct me)
</code>```

(c) this part is some sort of confusing I require some help from you guys here..

答:
You basically need to find the expected cost of each scheme.

The cost of that scheme now will depend on the cost incurred for each item sampled, and the cost for each batch rejected (sum of both expected costs).

The cost of each item sample in turn depends on the expected number of items sampled under the scheme.

The cost of each rejected batch depends on the acceptance of the scheme.


问:
Prove: every strongly connected graph has a single stationary distribution
Let $G= \langle V,E ~\rangle$ be a finite graph. We define $A_g$ to be the matrix with $[A_g]_{i,j}=1$ if there is an edge $e=(i,j)$ and $0$ otherwise.

Let $P$ be the transition matrix for a random walk on the graph $G$. So $P_{i,j}=P(x_t=j|x_{t-1}=i)$

$P$ is a normalize version of $A_g$ such that every row sums to $1$.

We define $\pi$ to be a stationary vector if it holds  that $\pi_i \geq 0$ and $\lVert \pi \rVert_1=1$ and satisfy $\pi^T=\pi^TP$

For ergodic graphs, I know there is a single stationary vector $\pi$ and for every vector $x$ we have $\lim_{t\to\infty}x^TP^t=\pi^T$

What I want to prove, is that for not ergodic graphs, but still strongly connected (for every vertex $i$ and vertex $j$ there is a way to get from $i$ to $j$, or in other word there is some $k$ that $[P^k]_{i,j} > 0$), there is only a single stationary vector $\pi$.

I know I won't be promised that for every $x$ we get $\lim_{t\to\infty}x^TP^t=\pi^T$, but still there is only 1 $\pi$

I can find a stationary vector with $\pi_i=\frac{deg(i)}{2\cdot |E|}$. One can multiply it by $P$ and see it works. But I am struggling to show only this $\pi$ can work with strongly connected graph.

Thanks

答:
For a finite graph there is only one stationnary vector $\pi$

You are looking for eigenvector for the matrix $P^T  $ of eigenvalue $1$ (that is solving $P^T  \pi= \pi$. But $P^T$ is a non negative matrix and primitive so the Perron-Frobeinus theorem state that there is only one eigenvector for the spectral radius eigenvalue (here $1$).
https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem#Multiplicity_one for more details.


问:
$f: (a, b] \to \mathbb{R}$ continuous. Prove: $f$ continuously resumable in $a \iff f$ is uniformly continuous
Let $f: (a, b] \to \mathbb{R}$. Prove that: $f$ is continuously resumable in $a \iff f$ is uniformly continuous.

My approach:

"$\implies$:"

Since $a$ is continuously resumable, then $a$ has to be an accumulation point and $\lim_{x \to a} f(x)$ exists. I define another function $g$:

$$g: [a, b] \to \mathbb{R}, \left\{\begin{array}{lr}
        x \to f(x), & \text{for } x \in (a, b]\\
        \lim_{z \to x} f(z), & \text{for } x = a\\
        \end{array}\right\}$$

$g$ is continuous on $[a, b]$ so $g$ is also uniformly continuous on $[a, b]$ and therefore also uniformly continuous on $(a, b]$. This implies $f$ is also uniformly continuous on $(a, b]$.

"$\impliedby$:"

I think this direction is quite trivial to prove. Edit: This direction is not so trivial, as Danny Pak-Keung Chan's proof shows.

My question is: Is my proof correct? Did I made a too big jump from "$g$ is uniformly continuous in $[a, b]"$ to "$g$ is uniformly continuous in $(a, b]$"?

答:
$\Leftarrow:$ Suppose that $f$ is uniformly continuous on $(a,b]$.
We go to prove that we can define $f(a)$ such that $f:[a,b]\rightarrow\mathbb{R}$
is continuous. Let $(x_{n})$ be an arbitrary sequence in $(a,b]$
such that $x_{n}\rightarrow a$. We go to show that $(f(x_{n}))_{n}$
is a Cauchy sequence. Let $\varepsilon>0$ be given. Choose $\delta>0$
such that $|f(x)-f(y)|<\varepsilon$ whenever $x,y\in(a,b]$ with
$|x-y|<\delta$. Choose $N\in\mathbb{N}$ such that $|x_{n}-a|<\delta/2$
whenever $n\geq N$. For any $m,n\geq N$, we have that $|x_{m}-x_{n}|\leq|x_{m}-a|+|a-x_{n}|<\delta$,
so $|f(x_{m})-f(x_{n})|<\varepsilon$. Therefore $(f(x_n))_n$ is a Cauchy sequence. By completeness of $\mathbb{R}$,
$\lim_{n}f(x_{n})$ exists.

We go to show that $\lim_{n}f(x_{n})$ is independent of the choice of $(x_{n})$.
For, let $(x_{n}')$ be another sequence in $(a,b]$ such that $x_{n}'\rightarrow a$.
Choose $N_{1}$ such that $|x_{n}-a|<\delta/2$ and $|x_{n}'-a|<\delta/2$
whenever $n\geq N_{1}$. Note that for $n\geq N_{1}$, we have $|x_{n}-x_{n}'|\leq|x_{n}-a|+|a-x_{n}'|<\delta$.
Therefore $|f(x_{n})-f(x_{n}')|<\varepsilon$. Letting $n\rightarrow\infty$
yields $|\lim_{n}f(x_{n})-\lim_{n}f(x_{n}')|\leq\varepsilon$. Hence,
$\lim_{n}f(x_{n})=\lim_{n}f(x_{n}')$ because $\varepsilon>0$ is
arbitrary.

Define $f(a)=\lim_{n}f(x_{n})$, where $(x_{n})$ is any sequence
in $(a,b]$ such that $x_{n}\rightarrow a$. From the above discussion,
$f(a)$ is well-defined. For the function $f:[a,b]\rightarrow\mathbb{R}$,
we already know that it is continuous on $(a,b]$, so it remains to
show that it is continuous at $a$. By Heine Theorem, it suffices
that for any sequence $(x_{n})$ in $(a,b]$ such that $x_n\rightarrow a$, $f(x_{n})\rightarrow f(a)$,
but this follows from the very definition of $f(a)$. Therefore, $f$
is continuous at $a$.

$\Rightarrow:$ This direction is trivial. By assumption, we can define
$f(a)$ such that $f:[a,b]\rightarrow\mathbb{R}$ is continuous. Since
$[a,b]$ is compact, $f$ is uniformly continuous on $[a,b]$ (a result
due to Cantor), and hence trivially uniformly continuous on $(a,b]$.