问:
Perfect square involving the exponential law
If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?

I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?

答:
We can factor out the $2^{10}$ first:

$$2^{10}(1+2^3+2^{n−10})$$

We can simplify $2^3 = 2\cdot{2}\cdot{2} = 8$. Also, because $2^{10}$ is a perfect square, we can remove it from the equation.

$$1+8+2^{n-10}=9+2^{n-10}$$

Now we can add a new variable, $m$, where $m$ can be any positive integer. We can rewrite our equation as:

$$9+2^{n-10}=m^2$$

We can now test values for $n-10$. If we do this, we'll find that $9 + 2^4 = 9+16=25=5^2$.

So $n-10=4$, and $n=14.$


问:
Moving an integral to the inside of a sum
I am deriving the fourier coefficient formula, and was wondering under what conditions I can move an integral from the outside of a sum to inside the sum? (as I have done below)
$$ \frac{1}{2\pi} \int_{-\pi}^{\pi}\sum_{-\infty}^{\infty}a_n e^{i(n-m)x} \; dx = \frac{1}{2\pi} \sum_{-\infty}^{\infty} \int_{-\pi}^{\pi}a_n e^{i(n-m)x} \; dx$$

答: If the series $\sum_{-\infty}^{\infty}a_n e^{i(n-m)x}$ is uniformly convergent on $[- \pi , \pi]$, then  you can move the integral from the outside of a sum to inside the sum.

ContentLicense: CC BY-SA 4.0


问:
The interval [0,1] with additional modulo 1
The interval $[0, 1]$ is an abelian group with addition modulo $1$.
Let $H$ be a proper subgroup of $[0, 1]$, which is closed as subset of $[0,1]$.
Show $H$ is finite.

I assumed $H$ is infinite: since $H$ is closed, all limit points of $H$ are in $H$.

Then let $x$ be in $[0, 1]\setminus H=S$ ($S$ is open since $H$ is closed) so there exists an open set $U_x$ which contains $x$ in it and $U_x \subset S$.

Let $\{x_n \}_n$ be a sequence converging to $x$. Thus for all $\epsilon>0$ there exists a $N$ natural number so that for all $n>N$ $|x_n-x|<\epsilon$.

For some $\epsilon_1 >0$ $B(x,\epsilon_1)$ is contained in $U_x$, then if I prove that $x$ is an limit point of $H$ then I can say $H$ is dense (because I express a generic element of $[0, 1]$ as belonging to $H$ or as limit point of $H$) then closure of $H$ will equal to $[0, 1]$ and this would contradict $H$ being proper subgroup.

But I couldn't find a way to show that.

答:
Hint : Consider the function $\pi : \mathbb{R} \rightarrow [0,1)$ defined for all $x \in \mathbb{R}$ by $\pi(x)=x-\lfloor x \rfloor$. Then :

Show that $\pi^{-1}(H)$ is a subgroup of $\mathbb{R}$.

Deduce that either $\pi^{-1}(H)=a\mathbb{Z}$ for a real number $a$, either $\pi^{-1}(H)$ is dense in $\mathbb{R}$.

Show that the fact that $H$ is closed in $[0,1)$ implies that $\pi^{-1}(H)$ cannot be dense, so it has to be of the form $a \mathbb{Z}$.

Deduce that $H$ is finite.


问:
Is this $\epsilon$-$\delta$ limit correct?
I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$  and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?

答:
Normally, before or while I present an answer, I am supposed to respond to the OP's question(s) and point out any errors or omissions.  I simply can't do that here.  I have to agree with Ted Shifrin's comment.

Let $f(x) = x^2 + x + 1.$

To Prove:

$\forall \epsilon > 0 ~\exists ~\delta > 0~$ such that

$|f(x) - 7| < \epsilon~$ whenever $~0 < |x - 2| < \delta.$

I will present this as a full solution, showing how you craft the relationship between $\delta$ and $\epsilon$.

Suppose that $(2-\delta) < x < (2 + \delta)$.  Then

$4 - 4\delta + \delta^2 < x^2 < 4 + 4\delta + \delta^2.$

$2 - \delta < x < 2 + \delta.$

Therefore, $7 - 5\delta + \delta^2 < f(x) < 7 + 5\delta + \delta^2.$

The first thing to do is establish that one of the constraints that will be imposed on $\delta$ is that $\delta < 1.$

This will guarantee that $0 < \delta^2 < \delta.$

With this constraint imposed,

$$7 - 5\delta < f(x) < 7 + 6\delta.$$

This means that for any $0 < \delta < 1,$ if $\epsilon$ happens to be greater than $6\delta$, then
$$|f(x) - 7| < \epsilon.$$

So one easy solution is (for example) set $\delta = \min\left[0.9, (\epsilon/10)\right].$


问:
How is $\frac{-2(-1)^n}{n}$ equal to $\frac{2(-1)^{n+1}}{n}$
Just as the title asks, how is

$\frac{-2(-1)^n}{n}$

equal to

$\frac{2(-1)^{n+1}}{n}$

Body: $\frac{-2(-1)^n}{n}=\frac{(-1)2(-1)^n}{n}=\frac{2(-1)^{n+1}}{n}.$


问:
1D Fourier transform of 2D function with shift
Say you have a 2D function $f\left(x,t\right)$ whose 1D Fourier transform in time is
\[F\left(x,\omega\right)=\int_{-\infty}^{\infty}f\left(x,t\right)e^{-i \omega t}dt.\]
And you now introduce a time-dependent translation $x \rightarrow x + vt$ where $v$ is a constant. How is the 1D Fourier transform of the translated function,
\[\int_{-\infty}^{\infty}f\left(x+vt,t\right)e^{-i \omega t}dt,\]
related to the original 1D Fourier transform, $F\left(x,\omega\right)$?

I know that for 2D Fourier transforms the effect of the translation in the time domain is a translation in the frequency domain:
\[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f\left(x+vt,t\right)e^{-i \omega t}e^{-i k x} \,dt \, dx = F\left(k,\omega + vk\right).\]
But I'm not sure how to do the 1D case.

答:
Look at this simple trick: denoting $G(x,v,\omega) \equiv \int_{-\infty}^{\infty}f(x+vt,t)e^{-i \omega t} \,dt$, your formula for two dinensional Fourier transformation may be rewritten as
$$\int_{-\infty}^{\infty} G(x,v,\omega) e^{-i k x} \, dx = F\left(k,\omega + vk\right).$$

Considering $v$ and $\omega$ as parameters, $F_{v,\omega}(k)$ is just a 1D Fourier transformation of $G_{v,\omega}(x)$. Thus you may recover $G_{v,\omega}$ as inverse Fourier transformation of $F_{v,\omega}$ for any parameters $v$ and $\omega$.


问:
How does PatrickJMT do variation of parameters without the Wronskian?
In this PatrickJMT video, he doesn't seem to use the Wronskian. He produces a set of equations and solves the example with them, but I'm not exactly sure where the equations come from.

Everywhere else I look seems to solve these kinds of problems using a Wronskian (and I think I have to, too, as I don't have the fundamental set of solutions given to me in the question like Patrick does).

Could anyone explain to me where the PatrickJMT method comes from?

答:
$y_1,y_2$ are solutions of the differerntial equation:
$$y''+py'+qy=0$$

Of course he used the Wronskian:
$$c'_1y_1+c'_2y_2=0$$
$$c'_1y'_1+c'_2y'_2=f$$
With matrix notation we have:
$$\pmatrix { y_1 & y_2 \\ y'_1 &y'_2}\pmatrix {c'_1 \\c'_2}=\pmatrix {0 \\ f}$$
$$\pmatrix {c'_1 \\c'_2}=\pmatrix { y_1 & y_2 \\ y'_1 &y'_2}^{-1}\pmatrix {0 \\ f}$$
$$\pmatrix {c'_1 \\c'_2}=\dfrac 1 W\pmatrix { y'_2 & -y_2 \\ -y'_1 &y_1}\pmatrix {0 \\ f}$$

Where $W$ is the Wronskian.


问:
Linearly independent subset?
If $u,v,w$ and $z$ are distinct elements in $R$, then

 {$(1,u,u^2,u^3), (1,v,v^2,v^3), (1,w,w^2,w^3), (1,z,z^2,z^3)$} is a linearly independent subset of $R^4$.

Is that true or false?
I even can't start the first step..

答:
If you haven't seen the Vandermonde matrix, then this exercise is a bit tricky. A usual strategy is to consider the 'same' problem, but in lower dimension. That is, instead of consider four vectors and ponder about independence in $\mathbb R^4$, consider constructing just three vectors, $(1,u,u^2),(1,v,v^2), (1,w,w^2)$, and ponder about independence in $\mathbb R^3$. Better yet, consider just two vectors, $(1,u)$ and $(1,v)$ first. Are these independent in $\mathbb R^2$? Yes, assuming that $u\ne v$. This is clear enough, but let's look at the determinant of the matrix with these two vectors as rows. It is easily seen to be equal to $v-u$. This shows again that the vectors are independent (since the determinant is non-zero). </p>

<p>Now move on to consider the three vectors as above. Form the corresponding matrix and compute its determinant. While not immediate it's also not too difficult and you'll get a nice form for the determinant that will show easily it is non-zero. This will give you a hint as to what will happen with four vectors. The computation gets a bit messier but still doable by brute force. </p>

ContentLicense: CC BY-SA 3.0

Title: Linearly independent subset?

Body:
If you haven't seen the Vandermonde matrix, then this exercise is a bit tricky. A usual strategy is to consider the 'same' problem, but in lower dimension. That is, instead of consider four vectors and ponder about independence in $\mathbb R^4$, consider constructing just three vectors, $(1,u,u^2),(1,v,v^2), (1,w,w^2)$, and ponder about independence in $\mathbb R^3$. Better yet, consider just two vectors, $(1,u)$ and $(1,v)$ first. Are these independent in $\mathbb R^2$? Yes, assuming that $u\ne v$. This is clear enough, but let's look at the determinant of the matrix with these two vectors as rows. It is easily seen to be equal to $v-u$. This shows again that the vectors are independent (since the determinant is non-zero).

Now move on to consider the three vectors as above. Form the corresponding matrix and compute its determinant. While not immediate it's also not too difficult and you'll get a nice form for the determinant that will show easily it is non-zero. This will give you a hint as to what will happen with four vectors. The computation gets a bit messier but still doable by brute force.


问:
A lemma (possibly of Atiyah) in a proof of the fundamental theorems of invariant theory
A classical problem of invariant theory is to describe the space $((V^*)^{\otimes r} \otimes V^{\otimes s})^G$ in terms of generators and relations, or equivalently to determine generators and relations of $\text{End}_GV^{\otimes r}$, for $V$ some $K$-dimensional vector space (often $K = \mathbb{C}$).
The classical groups $GL(V)$, $O(V)$, and $Sp(V)$ have "fundamental theorems" that describe the above mentioned spaces, proven via symmetric group representations or via Brauer diagrams and functorial correspondences.
An example formulation of the first fundamental theorem I am working with is

The unique covariant functor $F$ from the Brauer category (with respect to a certain fixed element of $\mathbb{C}$) to the subcategory of $G$-modules $V^{\otimes r}$ is full, i.e. induced maps from Brauer algebras $B_r^s \to \text{Hom}_G(V^{\otimes r}, V^{\otimes s})$ are surjective for all $r,s$.

There are many proofs for this statement (and its equivalent formulations, e.g. via matrices giving a generating set of $((V^{\otimes 2r})^*)^G$), one of which uses the following idea/lemma:

Let $E^+$ be the subset of $\text{End}(V)$ which consist of self-adjoint operators (for $f \in \text{End}(V)$, its adjoint is the unique $f^\alpha$ for which $(fv,w) = (v,f^\alpha w)$ for all $v,w \in V$ - obviously this definition depends on if the form on $V$ is symmetric or skew-symmetric).
Let $\omega: \text{End}(V) \to E^+$ be the map given by sending $f$ to $f^\alpha f$.
Let $\phi : \text{End}(V) \to \mathbb{C}$ be such that $\phi(gf) = \phi(f)$ for all $g \in G$.
Then there exists some $\psi \in \mathbb{C}[E^+]$ for which $\phi(f) = \psi(\omega(f))$ for all $f \in E$.

My notes say this lemma is due to an idea of Atiyah's, possibly appearing in a paper of Atiyah and two others (initials ABP - possibly Bott and Patodi?), but I cannot find exactly where, nor a proof of it.

答:
This is in fact suggested in "On the heat equation and the Index Theorem" by Atiyah, Bott, and Patodi (1973) in Inventiones Mathematicae, Vol. 19 pp 279-230.

The result in particular is the lemma in Appendix I.
Although they only prove it for the orthogonal group, very similar arguments can be used to prove it for the symplectic group.


问:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction?
I've been trying to solve the following problem:

Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.

I had the following idea, we write:

$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$

Let's pretend the identity we want to prove is true, then:

$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$

We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that

$$x= \frac{2n (n+1) (2 n+1)}{3} $$

We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?

Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.

答: Hint: notice that $$\sum(2k)^2=\sum4k^2=4\sum k^2.$$


问:
a calculus of variations question
I want to maximize $\int_0^1(g^2(x)+3g^2(x)g'(x)+2[g'(x)]^2)dx$ subject to g(0)=1, g(1)=0. I can find Euler equation but I can not find g(x) that is maximizes this integral. I have to use calculus of variations. Can you help me?

答:
The Euler Eq. becomes
$$2g+6gg'-6gg'-4g''=0 \implies g''(x)-\frac{1}{2} g(x)=0$$
Sp $$g(x)=A \cosh (x/\sqrt{2})+ B\sinh(x/\sqrt{2})$$
$g(0)=1$ gives $A=1$. $g(1)=0$ gives
$$A\cosh(1/\sqrt{2})+B \sinh(1/\sqrt{2})=0 \implies B=-\coth(1/\sqrt{2})$$
FGinally we get $$g(x)=\cosh(x/\sqrt{2})-\coth(1/\sqrt{2}) \sinh(x/\sqrt{2})$$

Title: a calculus of variations question


问:
Definition of a "Surface" -- an open ball or an open subset of $R^2$?
I have learnt that surface is a topological space such that every point has a neighborhood which is homeomorphic to an open subset of the Euclidean plane $R^2$. --- (a)

But then, it seems that people use another definition that "a surface is ... that has a neighborhood homeomorphic to an open ball in $R^2$". --- (b)

Clearly, (b) implies (a), but does (a) also imply (b)?

答: Yes, it does. Let $p$ be the point on the surface and $\varphi$ the homeomorphism. The open subset of $\mathbb R^2$ contains an open ball around each point by definition. Take an open ball around $\varphi(p)$. This ball is homeomorphic to an open neighborhood of $p$ via $\varphi^{-1}$ restricted to the ball.


问:
Why are addition and multiplication commutative, but not exponentiation?
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation
where

$H_0(a,b) = b+1$   (successor op)

$H_1(a,b) = a+b$   (addition op)

$H_2(a,b) = ab $   (multiplication op)

$H_3(a,b) = a^b$   (exponentiation op)

$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )

etc.

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?

Thanks.

答:
A question "why..." in number theory is difficult to answer if it is not just an algebraic but a qualitative one, but let's give a try to improve at least intuition.

For me it is more intuitive to see the (hyper-)operators as (possibly repetatively/iteratively applied) operation at a startvalue, say $x_0$ , having a "base"-parameter $b$ and an iteration-counter $h$.

Then we can also reflect the left- or right-associativity in the completed table:

$\displaystyle \qquad \begin{eqnarray}
 G_{1,L}(x_0,b,h) &=& ((((x_0) \underset{\text{h times}}{ \underbrace{+ b) + b) \ldots + b)} }
&\quad= \quad&
G_{1,R}(x_0,b,h) &=&  \underset{\text{h times}}{ \underbrace{  (b + \ldots  (b + (b +} } (x_0)))) \\
 G_{2,L}(x_0,b,h) &=& ((((x_0) \underset{\text{h times}}{ \underbrace{\cdot b) \cdot b)  \ldots \cdot b)} }
&\quad =^{1)} \quad&
G_{2,R}(x_0,b,h) &=&  \underset{\text{h times}}{ \underbrace{  (b \cdot  \ldots (b \cdot (b \cdot} } (x_0)))) \\
 G_{3,L}(x_0,b,h) &=&(((( x_0) \underset{\text{h times}}{ \underbrace{ \uparrow b) \uparrow b ) \ldots \uparrow b)} } &\quad \quad & G_{3,R}(x_0,b,h) &=&  \underset{\text{h times}}{ \underbrace{  (b \uparrow  \ldots  (b \uparrow (b \uparrow} } (x_0 ))))
\end{eqnarray}$

$\qquad \qquad ^{1)}$ only over certain (for instance:scalar) fields

Here we have symmetry in the left- or right-associativity $G_{1,L} = G_{1,R} (=G_1)$ and over a scalar field $G_{2,L} = G_{2,R} (=G_2) $

In $G_1$ we have even the interchangeability of the iterator and the base $G_1(x_0,b,h) = G_1(x_0,h,b)$

We can now look at elements with special values in $x_0$ and $b$ to find that the operation is "neutral" ($b=0$ for the addition and $b=1$ for the multiplication and the "power" ($G_{3,L}$)) and becomes "absorbing" for some elements ($b=0$,$x_0=0$ for multiplication and $x_0=1$ for "power" ($G_{3,L}$) and $b=1$ for "exponentiation" ($G_{3,R}$)) and also at the  absorbing effects of $\infty$ and of fixpoints in $G_{3,R}$ etc. ...

If I look at all this and even a bit further then the miracle of the "broken symmetry" after the two operations addition and multiplication becomes less mysterious and only one small between other changes of properties (although it is surely remarkable...) .

An additional remark: although I expand here on a hierarchical scheme based on the iteration of operations, I find it worth to remember what Quiaochu Yuan says, that the different operations should be seen in some other view focused at some  qualitative(?)/topological(?) properties, for instance the exponential map in the complex plane.


Title: Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction?

Body:
I've been trying to solve the following problem:

Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.

I had the following idea, we write:

$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$

Let's pretend the identity we want to prove is true, then:

$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$

We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that

$$x= \frac{2n (n+1) (2 n+1)}{3} $$

We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?

Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.

Body:
Let $$S=1^1+3^2+\cdots+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2=\sum_{k=1}^{n}[4 k^2-4k+1]$$
$$S=4\sum_{k=1}^n k^2- 4\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$
$$S=4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2}+n=\frac{n(4n^2-1)}{3}$$


问:
Can this integral be written in terms of special functions?
I'm trying to find some nice expression for integrals of the form
$$
I(a,b,c)=\int_{-\infty}^{\infty} dx \frac{1}{({a x^4+ b x^2+c})^{3/2}}.
$$

So far I failed to find a rewriting of this in terms of special function (elliptic, hypergeometric, Bessel..) but something tells me that this should be possible. Does anyone know if this is possible?

答:
If the denominator has a real zero, the integral diverges.  So assume
the denominator has no real zeros.  Factor it.  For example,
we get this if $u,v>0$, $u \ne v$:
$$
\int_{-\infty }^{\infty }\! \frac{{\rm d}x}{\left(  \left( {x}^{2}+u \right)  \left( {
x}^{2}+v \right)  \right) ^{3/2}}\,
={\frac {2}{\sqrt {v}\;
 \left( u-v \right) ^{2}\;u} \left[  \left( u+v \right) {\rm E}
 \left( \sqrt {1-{\frac {u}{v}}} \;\right) -2u\;{\rm K} \left(
\sqrt {1-{\frac {u}{v}}} \;\right)  \right] }
$$
and in case $u=v$ we get $(3/8)\pi u^{-5/2}$.
The complete elliptic integrals E and K may be found here.


答:
Consider the identity $$(2k+1)^2=4k^2+4k+1$$ which relates the sum of the odd squares to the sum of the squares.

The extra terms $4k+1$ are easily dealt with with a telescoping,

$$(k+1)^2-k^2=2k+1.$$

Altogether,

$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$


问:
Differentiable function $f'\left( x \right) - 2f\left( x \right) > 0$
If $f: ℝ → ℝ $ is a differentiable function such that $f'\left( x \right) - 2f\left( x \right) > 0$ for all $x ∈ ℝ$, and $f(0) = 1$, then

[A] $f(x)$ is increasing in $(0, ∞)$

[B] $f(x)$ is decreasing in $(0, ∞)$

[C] $f(x)>e^{2x}$ in $(0, ∞)$

[D] $f'(x)<e^{2x}$ in $(0, ∞)$

The official answer is [A] and [C]

My approach is as follow

$f'\left( x \right) - 2f\left( x \right) > 0$

$\frac{d}{{dx}}\left( {f\left( x \right){e^{ - 2x}}} \right) > 0$

$t(x) = f\left( x \right){e^{ - 2x}}$

For $x\in(0,\infty)$

$t(x) = f\left( x \right){e^{ - 2x}}>t(0)$

$t(0)=1$ as $f(0)=1$ & $e^{-2*0}=1$

Hence for $x\in(0,\infty)$ , $f\left( x \right)>{e^{  2x}}$, hence $[C]$

As $e^{-2x}>0$

And $t(x)$ is increasing function therefore $f(x)$ is increasing for $x\in(0,\infty)$ , hence [A]

How do we satisfy option [D] is incorrect

答: You know that $f'(x) > 2f(x) > 2e^{2x}$ for every $x \in (0,\infty)$ which is incompatible with $f'(x) < e^{2x}$ for any $x \in (0,\infty)$.


问:
How to prove this inequality by using induction?
If $x,y$ are distinct real numbers such that $x+y>0$ and $n\ge 1$, then $2^{n-1}(x^n+y^n)\ge (x+y)^n$.

It is obvious for $n=1$. How to do the rest by using induction?

答:
Hint: convexity of $f\colon t\in\mathbb{R}_+\mapsto t^n$.

Observe that the result is equivalent to
$$
\frac{x^n+y^n}{2} \geq \left(\frac{x+y}{2}\right)^n
$$
that is
$$
\frac{f(x)+f(y)}{2} \geq f\!\left(\frac{x+y}{2}\right)
$$

However, this method does not rely on induction.

For one with induction: assume the result is known for $n\geq 1$:
$$
\frac{x^n+y^n}{2} \geq \left(\frac{x+y}{2}\right)^n
$$
Then (as $x+y>0$)
$$
\left(\frac{x+y}{2}\right)^{n+1} \leq \frac{x+y}{2}\cdot\frac{x^n+y^n}{2} =
\frac{x^{n+1}+y^{n+1}+xy^n+yx^n}{4}
$$
Remains to prove that
$$
\frac{x^{n+1}+y^{n+1}+xy^n+yx^n}{4} \leq\frac{2x^{n+1}+2y^{n+1}}{4}
$$
i.e. that
$$
x^{n+1}+y^{n+1} - (xy^n+yx^n) \geq 0
$$
But the LHS can be rewritten
$$
x^{n+1}+y^{n+1} - (xy^n+yx^n) = (x^{n}-y^{n}) (x-y)
$$
which is indeed always positive as $x+y>0$ (whether $x>y$ or $y>x$, at least one of them is positive; a further analysis shows that, independently of the parity of $n$, both factors will have the same sign).


答: $ 1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ multiplying this by $ 2^2 $ gives the sum $ 2^2+4^2+\dots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3} $ and subtracting this from the given identity results int the required equality.


问:
Perfect squares using 20 1's, 20 2's and 20 3's.
How many perfect squares can be formed using 20  1's, 20  2's and 20  3's. This is a recent exam question, which I had no clue how to solve?

There is some kind of trick here, since time allotted to solve it was just 4 minutes.

答: Hint: The resulting number is divisible by $3$ but not by $9$, since the digit sum is $120$.


问:
Cylindrical Shell Method of finding volume around non-standard axes
I am given a question to find the volume of the solid formed by a region in the second quadrant bounded above by the curve $y=-x^3$, below by the x-axis, and on the left by the line $x=-1$, revolved around the axis $x=-2$. I intend to solve this by the cylindrical shell method.

By drawing out the curve, I have come up with a few conclusions to help me set up my integral:

The radius of each layer of the "shell" is $x+1$, because, crudely, the interval between $x=-1$ and $x=-2$ is not bounded.

Height of each "shell" is given by $y=-x^3$, as given by the curve itself.

The region extends from $x=-1$ to $x=0$.

Hence, my integral should be $2\pi \int_{-1}^0 (x+1)(-x^3)dx$. However, it is now known to me that this is incorrect. May I know what is wrong in my thought process?

In general, is there any formula/procedure when calculating for volumes of solids formed by revolutions around non-standard axes (x and y-axes)?

Any advice is appreciated!

答:
The cylindrical shell radius you are looking for is $(2 + x)$ and not ($1 + x)$.

As the rotation is of area between $x = -1$ and $x = 0$, around $x = -2$,

At $x = -1$, radius $ = 1$. At $x = 0$, radius $ = 2$.

So the correct integral should be -

$\displaystyle 2\pi \int_{-1}^{0} (2 + x) (-x^3) dx$

EDIT: added a diagram for the region (marked in yellow) that you are rotating around $x = -2$.


问:
How to prove that a given disjunctive or conjuctive normal form is minimal?
what is the argument that a given canonical normalform cannot be reduced any further?

For example have a look at this $dnf(f) = (\neg a \wedge b \wedge c) \vee (a \wedge \neg  b \wedge c) \vee(a \wedge b \wedge \neg c)$?

Every term in this $dnf(f)$ is complete (there are only variables $a,b,c$).

Link to wolfram alpha

Can somebody help?

答:
The statement being in canonical disjunctive normal form means that every one of its disjuncts makes reference to each of the variables exactly once. And, since that disjunct is in fact a conjunction of those variables or negations thereof, that disjunct is true for exactly one truth-assignment of those variables. Finally, since the disjuncts are all different from each other, they will all 'pick out' a different truth-assignment. That is, any two different disjuncts $\phi$ and $\psi$ from the CDNF cannot be true at the same time. Therefore, removing any one of its disjuncts will mean that the resulting formula is no longer true for the truth-assignment 'picked out' by that very disjunct, meaning that the resulting formula is not equivalent to the original. So that is why it can not be reduced any further while still in CDNF.

Of course, it is perfectly possible for statements in CDNF to be reduced to 'smaller' statements, but the result will no longer in CDNF.


Body: You have a typo. Call the known sum $P_n$. The desired sum is$$\begin{align}P_{2n+1}-4P_n&=\frac{(2n+1)(2n+2)(4n+3)-4n(n+1)(2n+1)}{6}\\&=\frac{(n+1)(2n+1)}{3}(4n+3-2n)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}.\end{align}$$


答:
Suppose the claim is true for $n$. Then, multiply $2^{n-1}(x^{n}+y^{n})\ge (x+y)^n$ by $(x+y)$ on both sides to obtain (note that here we are using $x+y>0$)
$$ 2^{n-1}(x^{n}+y^{n})(x+y) \ge (x+y)^{n+1}$$
To prove the inductive hypothesis it is enough to show that
$$ 2^{n}(x^{n+1}+y^{n+1}) \ge 2^{n-1}(x^{n}+y^{n})(x+y)$$
which boils down to proving (I will let you proceed the intermediate steps on your own):
$$ x^{n+1}+y^{n+1}\ge x^n y + xy^{n}$$
Without loss of generality, assume $x\ge y$. Then:
$$ x^{n+1}+y^{n+1}-x^{n}y-xy^{n}=(x^{n}-y^{n})(x-y)\ge 0$$
as desired.

Edit: as pointed out by Ross Millikan, we do not need the assumption that $x$ and $y$ are distinct.

Title: How to prove this inequality by using induction?

Body:
If $x,y$ are distinct real numbers such that $x+y>0$ and $n\ge 1$, then $2^{n-1}(x^n+y^n)\ge (x+y)^n$.

It is obvious for $n=1$. How to do the rest by using induction?


问:
Why $\cos(t-s)$ is a "positive function"?
Problem: Let $c: \mathbb R\longrightarrow \mathbb R$ be a finite support function, i.e. the set $\{t: c(t)\neq 0\}$ is finite. Prove that
\begin{align}\label{eq1}
\sum_{t, s\in \mathbb R} c(t)c(s) \cos(t-s)\geq 0.
\end{align}

Reason: The reason I ask this question is because $\cos(s-t)$ can be considered as a covariance function (see the answer here). And we know that every covariance function is positive (in the sense that above inequality holds for all finite support functions). However, I don't know how to prove above inequality independently. Could anyone give me a hint, thank you!

答:
\begin{align}
\sum_{s,t} c(s)c(t)\cos(s-t) &=
\sum_{s,t}\big(c(s)c(t)\cos(s)\cos(t)+c(s)c(t)\sin(s)\sin(t)\big)
\\ &=
\left(\sum_s c(s)\cos(s)\right)^2 + \left(\sum_s c(s)\sin(s)\right)^2 \ge 0.
\end{align}


问:
Is this $\epsilon$-$\delta$ limit correct?
I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$  and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?

答:
It really is as simple as you were making it, although there are mistakes and it is badly written. You correctly realized that it is important to bound one of $|x-2|, |x+3|$, namely $|x-2|$.

It is standard to start with $\delta := 1$ as this gives us more information to work with for small $\epsilon$, in particular when we have $0 < \epsilon < 1$.

Remember that we are considering the limit as $x$ approaches $2$, so we need to first look at how to restrict $x$, sometimes you must do this and in this case in particular, because we need to add $5$ for $|x+3|$ while still leaving it bounded in some way (otherwise $x$ could just be negative number where $|x|$ is very large).

That being said, we can restrict $x$ around $2$ such that $1 < x < 3$ so that $|x-2| < 1:\delta_{1} $ and $4 < x+3 <6 \implies |x+3| < 6$.

Hence, if we choose $\delta_{1} := 1$, we see what our other term is bounded by.

Now we can choose $\delta_{2}$ so that $|x-2| < \delta_{2} = \frac{\epsilon}{6}.$

Therefore, for some arbitrary $\epsilon > 0$ if we take $\delta:=\min\{\delta_{1},\delta_{2}\}$, then $|x^{2}+x-6| =|x-2||x+3| < \frac{\epsilon}{6}\cdot 6 = \epsilon$.


问:
Why does $\displaystyle \frac{1}{100} \sum_{k=1}^{n} \frac{1}{k}$ represent the worm on the rubber band ratio?
Here's the description from Wikipedia:

Suppose that a worm crawls
along an infinitely-elastic one-meter rubber band at the same time as
the rubber band is uniformly stretched. If the worm travels 1
centimeter per minute and the band stretches 1 meter per minute, will
the worm ever reach the end of the rubber band? The answer,
counterintuitively, is "yes", for after n minutes, the ratio of the
distance travelled by the worm to the total length of the rubber band
is $\displaystyle \frac{1}{100} \sum_{k=1}^{n} \frac{1}{k}$.

Now my problem is that for say $n=2$ we get $\frac{1}{100}(1+\frac{1}{2})=\frac{3}{200}$. But since only two minutes passed the worm could only have traveled 2 centimeters while the band was stretched to 2 meters giving a ratio of $\frac{2}{200}$. Therefore there must be an error with my reasoning (or on Wikipedia), where is it?

答: If you crawl 1 cm while the rubber has length $k$ meters (i.e., for simplicity we assume that worm crawls at the beginning of ecvery full second and the band stretches in-between), you improve your ratio by $\frac1{100k}$. Adding the first $n$ "improvements" gives the result.


问:
Binome-like sum with even integers
I am trying to compute the following sum
$$ \sum_{k=0}^n C^{2k}_{2n} \frac{(2k)!}{k! 2^k} x^{2n - 2k} y^{2k}. $$
I don't have many ideas, I tried to play a little with the factorials, but it did not lead me anywhere. I am supposed to recognized something here ?

I should mention that my notation of binomial coefficient is
$ C^i_j = \frac{j!}{i! (j-i)!}$ hence $C^i_j = {j \choose i}$.

(P-s: It appears that I can not make a research on MSE this morning, for some reason only the asking question feature works. Hence, apologies if the question has been asked somewhere else.)

Title: Binome-like sum with even integers

Body:
$$S_n=\sum_{k=0}^n\binom{2 n}{2 k}\,\frac{ (2 k)! }{2^{k}\,k!}\,  x^{2 n-2 k}\,y^{2 k}$$
Let $x^2=2y^2\, z$ to make
$$S_n=2^n y^{2n}\sum_{k=0}^n\binom{2 n}{2 k}\,\frac{(2 k)! }{2^{2 k}\,k!} z^{n-k}$$
$$\sum_{k=0}^n\binom{2 n}{2 k}\,\frac{(2 k)! }{2^{2 k}\,k!} z^{n-k}=(-1)^n\, U\left(-n,\frac{1}{2},-z\right)$$ where appears Tricomi's  confluent hypergeometric function


问:
Distribution Theory. How to solve $x^{m}{u}=\delta$?
I just got to study this topic and trying to get familiar with new sources.
I want to find all $u$ in $D'(\mathbb R)$ such :

$$ {x}^{m}{u}=\delta.$$
where $m$ is a positive integer.
My attempt was to try $u=\operatorname{p.v.} \frac{1}{x^m}$.

Any hint?

答: Paul Garrett's comment gives a particular solution, then you'll need all the solutions to $x^m w=0$ that you'll find from the same method as in your previous question (take $\psi$ constant $=1$ on $[-1/2,1/2]$ to simplify the computation).


问:
Conditional probability, Are my answers right?
Hi there I just want to check my answer is wright, could you please help me ?

```<code>Customers are asked to evaluate hotels at an international hotel booking website. In the past, 95% of
5-star hotels received good reviews,
60% of 4-star hotels received good reviews, and 10% of 3-star hotels received
good reviews. In addition, 40% of hotels at the website are 5-star hotels, 35%
of hotels at the website are 4-star hotels, and 25% are 3-star hotels.
1. What is the probability that a hotel on the website has a good review?
2. If a hotel attains a good review, what is the probability that it will be a
5-star hotel?
3. If a hotel does not attain a good review, what is the probability that it will
be a 3-star hotel?
</code>```

My answers are;

```<code>Probability of a hotel on the website has a good review.
All the good reviews; 380 + 210 + 25 = 615
Pr (G) = 615/1000 = %61.5

2) Pr (Good | 5-Star) = [Pr ( Good ∩ 5-Star) ] / [ Pr (5-Star)]
[38/100] / [95/100] = 38/95  %40

3) Number of hotels with Not-Good review.
615 out of 1000 has good  385 Not-Good
Pr ( Not-Good | 3-Star) =
[Pr ( Not-Good ∩ 3-Star)] / [ Pr( 3-Star)]
[225/1000] / [25/100] = 225/250 = %90
</code>```

答:
If a hotel attains a good review, what is the probability that it will be a
5-star hotel?

If $A$ is the event of a hotel being $5$ star and $B$ is the event of a hotel attaining good review

$P(A \cap B) = 38\% \,, \,$ as you calculated in your first answer. That is $38$ hotels out of $100$ are good as well as a $5$ star.

$P(B) = 61.5\% \, , \,$ that is $61.5$ hotels are expected to be good out of $100$.

So $P(A|B) = \frac{38}{61.5} = \frac{76}{123}$

If a hotel does not attain a good review, what is the probability that it will
be a 3-star hotel?

Similar to previous one -

$100 - 61.5 = 38.5 \% \, $ hotels in total do not attain a good review. This is your $P(B)$.

$25\% - 2.5\% = 22.5 \% \, , \,$ $3$ star hotels do not attain good review. This is your $P(A \cap B)$.


问:
Proof of inequality in probability theroy using Cauchy-Schwarz inequality
The domain of random variable $X$ is the set of nonnegative integers. (We assume that all sums are well-defined.)

I do not understand why the following inequality holds (or not) $$\bigg( \sum_{i=1}^\infty i \cdot Pr[X=i] \bigg)^2 \leq \bigg( \sum_{i=1}^\infty i^2 \cdot Pr[X=i] \bigg) \bigg( \sum_{i=1} Pr[X=i] \bigg). \quad\quad (*)$$

I apply Cauchy-Schwarz inequality on the left-hand side of $(*)$ to obtain $$\bigg( \sum_{i=1}^\infty i \cdot Pr[X=i] \bigg)^2 \leq \bigg( \sum_{i=1}^\infty i^2 \bigg) \bigg( \sum_{i=1}^\infty Pr[X=i]^2 \bigg).$$

Now, the problem is reduced to show $$\bigg( \sum_{i=1}^\infty i^2 \bigg) \bigg( \sum_{i=1}^\infty Pr[X=i]^2 \bigg) \leq \bigg( \sum_{i=1}^\infty i^2 \cdot Pr[X=i] \bigg) \bigg( \sum_{i=1} Pr[X=i] \bigg).$$

I try to figure it out why this is true but with no success. Thank you for any suggestions.

答: You just need to apply Cauchy-Schwartz a bit differently the first time round. You have decomposed $i \cdot P[X=i]$ as $i$ times $P[X = i]$. Try writing it as $i \cdot P[X=i]^{\frac12}$ times $P[X=i]^{\frac12}$ and apply Cauchy-Schwartz.


问:
Punctured torus with antipodal boundary points identified
Given a torus with an open disk removed and its antipodal boundary points identified (note that the boundary refers that of the removed disk, which is a $S^1$), what space is it?

At first I viewed the torus as a square with opposite edges identified, cut along a diagonal, and tried to glue the two parts together in a nice way, but it ended up more confusing. Then I tried to view the torus as a square glued with two strips (consider torus as two cylinders glued together over a square patch, [shown by the following picture from Basic Topology by Armstrong], I simply cut off the square and get three pieces), and the procedure amounts to replacing the two strips with two Mobius strips and identifying several points. With some struggle, I got the result: a real projective plane with two tangent disks replaced by two Mobius strips, which is not satisfactory.

I've been stuck since then. Please help. I'm not sure what kind of description can be considered good and clear. (Maybe the cell structure will do, but I couldn't determine it either.)

答:
Claim: Let $X$ denote the space that you described. It's homeomorphic to $\Bbb{R}P^2\# T^2$, the connected sum of a real projective plane and a torus. (If I understand your question correctly)

I'll prove this by two ways:

Cut-And-Paste Argument:

Or simply using algebraic notations about labelling scheme (of equivalence relation),

$$afab^{-1}fb^{-1}\mapsto b^{-1}afab^{-1}f \text{  by permutation}$$
$$\mapsto b^{-1}afx^{-1},\text{ }xab^{-1}f \text{ by cutting}$$
$$\mapsto x^{-1}b^{-1}af,\text{ }f^{-1}ba^{-1}x^{-1} \text{ by permuting the first part and reversing orientation of the second}$$
$$\mapsto x^{-1}b^{-1}aba^{-1}x^{-1}\text{ by pasting}$$
$$\mapsto b^{-1}aba^{-1}x^{-1}x^{-1}\text{ by permuting}$$

And this is equivalent to $\Bbb{R}P^2\# T^2$.

Euler Characteristics and Orientability:

Theorem: Two compact closed 2-manifolds $X$ and $Y$ are homeomorphic iff their orientability agree and $\chi(X)=\chi(Y)$ (equal euler characteristics).

Let $U$ be a small open nhbd around the central $S^1/{(x\sim -x)}$, and $V$ be an open nhbd of the complement of $U$ s.t. their intersection deformation retract onto $S^1$.

Then, Apply Mayer-Vietoris sequence, which will give you
$$
H_p(X)\cong
\begin{cases}
\Bbb{Z} & p=0\\
\Bbb{Z}\oplus\Bbb{Z}\oplus\Bbb{Z}/2 & p=1\\
0 & \text{otherwise}
\end{cases}
$$

So, by the definition of Euler characteristic in terms of betti number, we have
$$\chi(X)=\sum_{p}(-1)^p\operatorname{rank}H_p(X)/T_p(X)=1-2=-1$$ where $T_p(X)$ denotes the torsion of $H_p(X)$.
By the classification theorem of surfaces, we know that $X$ must be the connected sum of projective planes (must) and tori (may). Hence, we have the relationship
$$\chi(X)=\chi(M_g)+\chi(N_k)-\chi(S^2)=2-2g+2-k-2=-1$$
for $k,g\in\Bbb{Z}_+\cup\{0\}$. Here, note that $M_g$ is the connected sum of $g$ tori, and $N_k$ is the connected sum of $k$ projective spaces.

There are two sets of solutions to this equation

$g=1$, $k=1$. This is $\Bbb{R}P^2\# T^2$

$g=0$, $k=3$. This is $\Bbb{R}P^2\# \Bbb{R}P^2\# \Bbb{R}P^2$

An it turns out that these two surfaces are homeomorphic, so we get the answer by the theorem stated before.


问:
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.

I solved it in the following way:

I state that $a$ is the number in front of $2010$, in other words if the number we're looking for is $A$, then $A=10^4\cdot a+2010$.

Hence, $a\cdot10^4\equiv1\bmod2011$.

$1956a\equiv1\bmod2011$

$55a\equiv2010\bmod2011$

So $55a=2010+2011k$ for some positive integer $k$.

So $30+31k\equiv 0 \bmod55$

After doing some trial and error for $k\in[0,54]$

$k_{min}=15$

So $55a=32175$, $a=585$.

Hence the smallest number possible is $5850000$

I solved this question using classical number theory and trial-and-error. However, I realize that this question can be solved using the euclidean algorithm. When I looked it up on the internet the only thing I could find was this: https://en.wikipedia.org/wiki/Euclidean_algorithm which is how to find the gcd using euclid's algorithm. Could you please explain to me how to solve the question using the euclidean algorithm, and also please explain to me what the euclidean algorithm is, how it can be applied and some resources to study from?

Body:
We have $$30+31k \equiv 0 \pmod{55}$$

$$31k \equiv -30 \equiv 25 \pmod{55}$$

Let us find $31^{-1}\pmod{55}$

\begin{align}
55 &= 31 + 24\\
31 &= 24 + 7 \\
24 &= 3(7)+3 \\
7 &= 2(3)+1
\end{align}

Hence we can write
\begin{align}
1 &= 7 - 2(3)\\
&= 7 - 2(24-3(7))\\
&=7(7)-2(24) \\
&= 7(31-24) - 2(24)\\
&= 7(31)-9(24)\\
&= 7(31) - 9(55-31)\\
&= 16(31) - 9(55)
\end{align}

Hence we can conclude that $31^{-1} \equiv 16 \pmod{55}$.

From $31k \equiv 25 \pmod{55}$, we have

$$k \equiv 31^{-1}(25) \equiv 16(25) \equiv 400 \equiv 15\pmod{55} $$

which agrees with your solution. They key is to learn to find the modulo inverse.


答:
The Wikipedia article add, immediately after the text you cite, "In fact the actual ratio is a little less than this sum as the band expands continuously".
They are approximating the process as a staggered, discrete march:

worm moves

band stretch

goto 1)

First minute, 1cm over 1 m, and the term is $\frac{1}{100}$, second minute, 1 cm over 2 m,  $\frac{1}{200}$, these are the increments to be added.
Your $\frac{2}{200}$ refers to total forward movement over actual length after 2 mind, which is a different topic.

Title: Why does $\displaystyle \frac{1}{100} \sum_{k=1}^{n} \frac{1}{k}$ represent the worm on the rubber band ratio?

Body:
Here's the description from Wikipedia:

Suppose that a worm crawls
along an infinitely-elastic one-meter rubber band at the same time as
the rubber band is uniformly stretched. If the worm travels 1
centimeter per minute and the band stretches 1 meter per minute, will
the worm ever reach the end of the rubber band? The answer,
counterintuitively, is "yes", for after n minutes, the ratio of the
distance travelled by the worm to the total length of the rubber band
is $\displaystyle \frac{1}{100} \sum_{k=1}^{n} \frac{1}{k}$.

Now my problem is that for say $n=2$ we get $\frac{1}{100}(1+\frac{1}{2})=\frac{3}{200}$. But since only two minutes passed the worm could only have traveled 2 centimeters while the band was stretched to 2 meters giving a ratio of $\frac{2}{200}$. Therefore there must be an error with my reasoning (or on Wikipedia), where is it?


问:
Pre Calculus Math Equation With Logarithms
Please Help me with this I think i figured out question 1... but I get no solution...
please help me start number 2 or if you can show full solution that be sick thanks.

$\log_{3x}(81)=2$

$\log_3(\log_x(\log_4 16))=-1$

my attempt for number 1 ! :)

$$\frac{1}{81}\log_{3x}(81)=\frac{2}{81}$$
$$\log_{3x}=2/81$$
$$10^{\large{\log_{3x}}}=10(2/81)$$
$$3x=10 (2/81)$$
$$x=10 (2/81) /3 $$

Body:
$$\log_{3x}(81)=2$$
$$(3x)^{\log_{3x}(81)}=(3x)^{2}$$
$$81=9x^2$$
$$9=x^2$$
$$3=x$$


问:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction?
I've been trying to solve the following problem:

Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.

I had the following idea, we write:

$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$

Let's pretend the identity we want to prove is true, then:

$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$

We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that

$$x= \frac{2n (n+1) (2 n+1)}{3} $$

We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?

Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.

Body:
Let $S=1^2+3^2+5^2+\cdots+(2n+1)^2$$.

Next, consider the sums
$$S_1=2^2+4^2+6^2+\cdots+(2n)^2$$
$$S_2=1^2+2^2+3^2+\cdots+(2n+1)^2$$
Thus, we can calculate $S$ as $S=S_2-S_1$.

Let's evaluate $S_1$ and $S_2$ using the identity we are given.

Clearly,
$$S_2=\frac{1}{6}(2n+1)(2n+2)(4n+3)$$
and
$$S_1=2^2(1^2+2^2+\cdots+n^2)$$
$$S_1=2^2\cdot\frac{1}{6}n(n+1)(2n+1)$$

Therefore,
$$S=S_2-S_1$$
$$S=\frac{1}{6}\left[2(n+1)(2n+1)(4n+3)-4n(n+1)(2n+1)\right]$$
$$S=\frac{(n+1)(2n+1)}{6}\left[8n+6-4n\right]$$
$$S=\frac{(n+1)(2n+1)(2n+3)}{3}$$
as desired.

Hope this helps :)


问:
How many vertices does a planar graph need to have a complete $k$-colouring?
A proper colouring $V(G) \to \{1,...,k\}$ of a graph $G$ is complete if every distinct pair of colours is connected by an edge. What is the least $n_k$ such that there exists a planar graph on $n_k$ vertices with a complete $k$-colouring?

Such a graph must have at least $\binom k2$ edges, and but because it is planar it can have no more than $3n-6$ edges, so immediately we get a naive bound $n_k\ge\bigl\lceil\tfrac{k(k-1)}6+2\bigr\rceil$ (for $k\ge3$). But this bound isn't always attainable:

Suppose there exists a complete 6-colouring on a graph with $\bigl\lceil\tfrac{6\cdot5}6+2\bigr\rceil=7$ vertices. By the pidgeonhole principle, some pair of vertices must have the same colour, and upon their removal you'd have a complete 5-colouring of a 5-vertex graph—but this can only be the complete graph $K_5$, which is nonplanar. So $n_6>7$.

The first few values of $n_k$ are 1, 2, 3, 4, 6, 8, 10, 12. Can $n_k$ be determined exactly for all $k$? Failing that, are there better bounds than the quick and dirty ones below? $$\bigl\lceil\tfrac16k(k-1)+2\bigr\rceil\le n_k\le\bigl\lfloor\tfrac14k^2\bigr\rfloor \quad \text{(for $k\ge4$)}$$

答:
Your question your may be rewritten as:

For each $k$, what is the smallest number $n$, such that there exists a planar graph $G$ having $n$ vertices and $\psi(G)=k$, where $\psi$ indicates the graph's pseudochromatic number.

As far as I know there does not exist an explicit formula for the sequence you asked about. However, some stronger bounds may be found here. For example, using your notation, in the article we may find that:
$$k \leq \Big\lfloor \sqrt{2n_k+\frac{1}{4}} +\frac{1}{2}\Big\rfloor$$


问:
Three player combinatorial game - minimizing communication between Alice and Bob
Consider the following game between three parties: Alice, Bob and a Referee. The game starts with $n$ closed boxes. For some fixed $k<n$ known to all parties, the Referee freely chooses $k$ distinct boxes and places a prize inside each of them. Alice sees the Referee do this and thus, she knows which boxes have a prize. Bob now enters the room. He can talk to Alice, then choose one box and if this box has a prize, he can keep it.

What is the smallest number of bits of information that Alice needs to convey to Bob such that Bob is guaranteed to win a prize?

The naive strategy is for Alice to pick a prize-containing box at random and give the ID of this box to Bob. This costs $\log n$ bits of information since she picks one box out of $n$ choices. But given that there are $k$ choices and any of them would do the job, can Alice and Bob win the game with less than $\log n$ bits of communication? If yes, what should their strategy be?

EDIT: I am happy to assume the Referee is not adversarial if that admits a better strategy for Alice and Bob.

答:
What @antkam suggested is a good starting point, but not entirely correct (according to me). The question asks how long the longest message should be that Alice gives to Bob so that they are guaranteed to win. If we admit messages with at most $b$ bits, we actually have $2^{b+1}-1$ messages since they could also be shorter.

Now lets go deeper on @antkam's point. Assume his third assumption holds. Now for each of the $2^{b+1}-1$ messages Alice can give, Bob needs to have a response. If this response were non-deterministic, then just select an arbitrary (but fixed) response from Bob's considerations, since each possible decision of Bob must lead to a prize. Now if $2^{b+1} -1+ k \leq n$ we see that we have a problem if the referee chooses the $k$ boxes that this strategy cannot reach. For this it is sufficient that the referee plays each of the possible $\binom{n}{k}$ combinations with non-zero probability (but need not be 'adverserial'). Notice that this condition is also necessary: if there is just one combination that the referee never plays, Alice and Bob can do better! We also see that Alice playing a random strategy or not has no effect on the counting argument.

Conclusion: to be certain of a win, Alice and Bob need at least one message with $\lceil\log_2(n-k+2)\rceil - 1$ bits of information (and this bound can trivially be achieved).

This is however only minimizing the maximal message size. If you want the message size to be the lowest-possible on average against for example a referee playing uniformly, you need the techniques that @leonbloy suggested. If we consider a deterministic strategy $f$ for Alice, meaning on input $X=(X_1,\dots,X_k)$ she redirects Bob to the box with ID $f(X)$, then it might well be that not every ID has equal probability of being chosen. By agreeing on a renumbering of the boxes, Alice and Bob can make advantage of this by giving boxes with a high probability of being $f(X)$ a short message length.

The average number of bits needed with this strategy $f$ is best measured with the classical entropy-function $\sum{-p_i \log_2(p_i)}$ (if I'm not mistaken). Now we should minimize the entropy of the image of $f$. Some candidates for $f$ could be: $f(X) = \min(X)$, $f(X) = \text{median}(X)$,...
My experiments for small $n$ and $k$ show that the minimum-function is best on average among several $f$, but I don't know if this could be proven analytically.


问:
Inconsistent System of Linear Equations
Let $A ∈ M_{n\times n}(F)$. Suppose that the system of linear equations $AX = B$
has more than one solution. Prove that there is a column $C ∈ F^n$ such that
the system of linear equations $AX = C$ is inconsistent.

I thought it'd be clever to use this (underlined by my pencil):

答:
Lemma: Let $AX=C$ be a system of linear equations. Then the system is consistent if and only if $\text{dim Im}~(A)=\text{dim Im}~(A \mid C)$.

Let $Y$ and $Y'$ be solutions to the system $AX=B$. Then

$$ A(Y) = A(Y')=B,$$

or

$$ A(Y)-A(Y')=A(Y-Y')=0,$$

so $\dim \text{Ker}~(A)\geq 1$, which means $\dim \text{Im}~(A)<n$, or simply that

$$\text{span}~(\{A^{(1)},\dots,A^{(n)}\})\neq F^n.$$

In other words, there exists at least one $A^{(i)}$ such that

$$A^{(i)}=k_1A^{(1)}+\cdots +k_{i-1}A^{(i-1)}+k_{i+1}A^{(i+1)}+\cdots + k_nA^{(n)}\neq ce_i,$$

for some $e_i\in \{e_1,\dots,e_n\}$ the standard basis for $F^n$ with $k_i,c\in F$. Thus, by the lemma, we clearly see that such an $e_i$ is the desired $C$ for which $AX=C$ has no solutions, namely

$$\text{dim Im}~(A)\neq \text{dim Im}(A \mid e_i)=\text{dim Im}(A \mid C).$$

This is so because if $\text{dim Im}(A)=m$, then $\text{dim Im}~(A \mid e_i)=m+1$ for such an $e_i$ described above.

Title: Inconsistent System of Linear Equations


问:
relation, function, "Abbildung", injective, surjective
I am teaching theoretical foundations of computer science, which uses many concepts of set theory. But I learned, that many of my students are not really firm with set theory, and so I decided to write a little booklet, so that they can read it and learn the most important concepts of set theory by reading this booklet. (I do not teach set theory in my course.)

Because I did not study mathematics, and because I want to make everything as correct as possible, I spend a lot of time researching the terms and concepts of set theory, but I found, that many terms and concepts are explained not just only in different ways, but also contradictory.

In German very often the term "Abbildung" is used, which I believe to be "mapping" in English, but I am not sure. I also are not sure, if this is a synonym for relation or for function or for something else.

I am pretty sure, that every subset of the Cartesian product is a relation.

I found Definitions of "Abbildung" that say that every subset of the Cartesian product is an "Abbildung" (which would make it a synonym for relation). But I also found Definitions of the same term that say that only a left-total and right-unique relation is an "Abbildung", which I thought was the definition of "function".

But then I also found resources where they say, that a left-total and right-unique relation is only called a function, if all elements of both sets are numbers. So, a left-total and right-unique relation between two non-numeric sets would not be a function (but still an "Abbildung")

So, here are my questions:

How exactly are the terms relation and function defined (related to the Cartesian product of two sets)?

If you also know the German term "Abbildung": What is its English name? How is it defined?

Let A and B be sets. A is the Latin alphabet (26 letters = elements) and B the Greek alphabet (24 elements). Is a left-total and right-unique subset of the Cartesian product $A \times B$ a function?

Look at the following picture which shows 16 different relations:

Which of them are functions? Which of them are injective, which of them are surjective and which of them are bijective?

Columns are about set A:

column 1-5-9-13: In A are elements that send 0, 1 or 2 arrows to B

column 2-6-10-14: In A are elements that send 1 or 2 arrows to B (no element of A sends 0 arrows)

column 3-7-11-15: In A are elements that send 0 or 1 arrow to B (no element of A sends 2 arrows)

column 4-8-12-16: All elements in A send exactly 1 arrow to B

Rows are about set B:

row 1-2-3-4: In B are elements hit by 0, 1 or 2 arrows

row 5-6-7-8: In B are elements hit by 1 or 2 arrows (no element of B is hit by 0 arrows)

row 9-10-11-12: In B are elements hit by 0 or 1 arrow (no element of B is hit by 2 arrows)

row 13-14-15-16: All elements in B are hit by exactly 1 arrow

Would it make a difference, if the elements in my picture were not Glagolitic and Armenian letters but rational numbers?

答:
"Would it make a difference if the elements in my picture were not Glagolitic and Armenian letters but rational numbers?"  No, not at all.

"How exactly are these terms defined?"

A relation with domain $X$ and codomain $Y$ is very simply any subset of the cartesian product $X\times Y$.

A function is specifically a relation where every element in the domain appears exactly once as the first entry in an ordered pair in the relation.  That each occurs at least once might be worded as the relation being "everywhere defined" and that each occurs at most once might be worded as the relation being "well defined."  A function must be both everywhere defined and well defined.

A surjective function is specifically a function where every element in the codomain appears at least once as the second entry in an ordered pair in the relation.

An injective function is specifically a function where every element in the codomain appears at most once as the second entry in an ordered pair in the relation.

A bijective function is specifically a function who is both injective and surjective.

As for which relations are functions and which are not... all of the first three columns are not functions as they have at least one element in the domain with a number of arrows leaving it different than $1$.  (0 in the case of first and third column making it not everywhere defined, 2 in the case of first and second column making it not well defined).  All of the relations in the fourth column are functions.

As for which functions are injective / surjective / bijective, the first and second row are not injective since there is at least one element in the codomain who has a number of arrows pointing to it strictly greater than $1$.

The first and third row are not surjective since there is at least one element in the codomain who have strictly fewer than $1$ arrows pointing to it.

That leaves $4$ as a function which is neither injective nor surjective, $8$ as a function which is injective but not surjective, $12$ as a function which is surjective but not injective, and $16$ as a function which is both surjective and injective and hence bijective.


问:
How to know that a digraph has odd number of hamiltonian paths
Is there any way to know that a digraph has odd nubmer of hampaths if hamiltonian paths exist?

Just curious.

答:
In some cases, deducing the number of Hamilton paths modulo $2$ is easy, such as when the graph has an order $2$ automorphism.

Observation: If a digraph has an automorphism $\alpha$ of order $2$, then the number of Hamilton paths is even.

Proof: We act on the set of Hamilton paths using $\langle \alpha \rangle$, thereby partitioning the set into orbits of size either $2$ or $1$.  By the Orbit-Stabiliser Theorem, orbits of size $1$ contain Hamilton paths that are stabilised by $\alpha$, but this cannot happen (as it would imply the directed path has an automorphism, which it doesn't). $\square$

(By the same argument, we can deduce that $|\mathrm{Aut}(G)|$ divides the number of Hamilton paths of $G$.)

In general, though, I suspect it would require enumeration.  Working modulo $2$, it would be possible to reduce the search space using some tricks.  E.g. if during the enumeration we encounter the two partial Hamilton paths (a,b,c,d) and (c,a,b,d), then any completion of (a,b,c,d) also results in a completion of (c,a,b,d), and vice versa, together contributing $0 \pmod 2$ to the overall total, and thus we needn't continue extending these paths.


问:
Inequality about $(\sum\limits_{i=1}^\infty a_i)^2 $ and $\sum\limits_{i=1}^\infty a_i^2$
When I do some thing about infinite matrix, I want a inequality which is the infinite version of
$$
\frac{(a_1+...+a_n)^2}{n}\le a_1^2+...+a_n^2.
$$
Namely, if $a_i\ge 0$ and  $\sum\limits_{i=1}^\infty a_i=A $ , whether there is inequlity like
$$
(\sum\limits_{i=1}^\infty a_i)^2  \le f(A) \sum\limits_{i=1}^\infty a_i^2  ~~~~~?
$$
But seemly, the Holder inequation implies the inequality like the above does not exist, so I ask here, thanks for any help.

答:
Indeed, there is not. To see this, simply note that there exists a sequence of sequences $a_{i,n}$ such that $\sum_{i} a_{i,n}=1$ for every $n$ but $\lim_{n\to\infty} \sum_{i} a_{i,n}^2=0$.

Let $a_{i,n}=\frac{1}{n}$ for $i\leq n$ and $0$ else. Then, $\sum_{i} a_{i,n}=1$ by definition, but
$$
\sum_{i} a_{i,n}^2=\frac{1}{n}
$$
Thus, there is no general $\ell^2$-inequality of the form, you desire, even for series which are both $\ell^2$- and $\ell^1$-convergent.

Note, of course, that you would also get a counter example if $a_i$ is any sequence which is in $\ell^2$ but not in $\ell^1$. I just assumed it would be more interesting to see that it fails even if we condition on both sides of the inequality being finite.


问:
Having dificulty to figure out event and other variables in binomial distribution
A company is selling 105 tickets to a plane that has 98 seats because only 85% of the buyers actually take the flight. What is the probability that even taking this approach 3 seats will be unused?

For me the sucess would be 15, the number of the sample 105 and the event would be 1, 2 and 3. But it seems its wrong.

答:
The probability that exactly 3 seats will be unused is:

$$\binom{105}{95}0.85^{95}0.15^{10}$$


问:
Is there any function such that $f(x)\to0$ as $x\to \infty$ but $x\log(x) f'(x)\to +\infty$?
My question es exactly the one in the title. I am trying to find any smooth function $f\in C^\infty((1,\infty))$ such that $$
f(x)\xrightarrow{x\to+\infty}0 \qquad \hbox{but} \qquad x\log(x)f'(x)\xrightarrow{x\to+\infty}+\infty.
$$
So far I've tried with $1/\log(x)$ and $1/\log(\log(x))$, etc, with no success. Now I am wondering if that kind of function can actually exists. A priory looks like a quite reasonable requirement, but surprisingly I cannot find any such function.

答:
If $x\log(x)f'(x)\to+\infty$ as $x\to+\infty$, then there exists $M$ such that
$f'(x)\ge\frac{1}{x\log(x)}$ whenever $x>M$. Integrating gives
$$f(x)\ge f(M)+\int_M^x\frac{dt}{t\log(t)}$$
whenever $x>M$. And this integral can be evaluated as
$$\log\log (x)-\log\log(M)$$
which tends to $+\infty$ as $x\to+\infty$, contradicting $\lim_{x\to+\infty}f(x)=0$.


问:
Rank of the incidence matrix of a directed graph.
Can someone please help me prove that the rank of the incidence matrix of a 'simple' directed graph with $n$ nodes and $m$ edges is $n-1$? The nodes are labelled $\{1,2,...,n\}$ and the edges are labelled $\{1,2,...,m\}$.

The element $a_{ij}$ of the incidence matrix is determined in the following way,

$a_{ij}=0$ if the node $i$ is not on edge $j$.

$a_{ij}=-1$ if the edge $j$ ends on node $i$.

$a_{ij}=1$ if the edge $j$ starts on node $i$.

答:
Let $(G,E)$ refer to our simple, directed graph. Let $(G',E')$ denote the corresponding undirected graph.

One proof is as follows: begin with the case where $G'$ is a tree, which can be handled in the manner described in this post.

Now, consider the case where $G'$ is any connected graph. We use the following facts:

Every undirected connected graph has a spanning tree,

Every tree on $n$ nodes  has $n-1$ edges,

Relabeling the nodes/edges (or equivalently, permuting the rows/columns of the incidence matrix) does not change the rank of the incidence matrix.

Relabel the edges of the graph so that the edges $1,\dots,n-1$ are the edges of our spanning tree. The first $n-1$ columns of the matrix form the incidence matrix of a tree, so these are linearly independent. It follows that the span of these $n-1$ columns is given by the subspace $S \subset \Bbb R^n$, defined by
$$
S = \{(x_1,\dots,x_n) : x_1 + \cdots + x_n = 0\}.
$$
Indeed, it suffices to observe that the span of these colums is a subspace of $S$ and that the dimension of the span is equal to that of $S$. Now, we see that the remaining columns of the incidence matrix are each elements of $S$. Thus, the column span of the entire incidence matrix of $G$ is $S$, which means that this incidence matrix has rank $n-1$.

Finally, consider the case of an arbitrary graph $G$. Let $G_1',\dots,G_k'$ denote the connected components of $G'$. Relabel the vertices so that the vertices of $G_1'$ come first, followed by the vertices of $G_2'$, and so forth. Similarly, relabel the edges so that the edges corresponding to $G_1'$ come first, followed by the edges of $G_2'$, and so forth.

I claim that the incidence matrix of $(G,E)$ (under this relabeling) has the block-diagonal form
$$
B = \pmatrix{B_1 \\ & B_2 \\ && \ddots \\ &&& B_k},
$$
where $B_j$ is the incidence matrix of $G_j$. Conclude that the rank of $B$ is the sum of the ranks of $B_1,\dots,B_k$, and is therefore given by $n - k$, where $k$ is the total number of connected components.


问:
Binomial type sum
Does the following reduce to something simpler

$\sum_j {k \choose 2j} x^{2j}$

I have perused the combinatorial identities by could but I did not find anything that fits my case.

答:
HINT

we say $\sum_j{k\choose 2j} a_j$ to be the $even $ $aerated $ $binomial $ $transformation$ of the sequence ${a_j}$. Now if $b_j$ is another sequence such that $∆b_j$$=a_j$ for $j>0$

we can easily show that

THEOREM $1$

if ${g_k}$ is the binomial transform of ${a_j}$ $∆g_k-a_0[n=-1]$ is the binomial transform of $a_{j+1}$, binomial transform is defined by $g_k=\sum_j{k\choose j} a_j $

Let $G_k$ and $H_k$ be the even aerated binomoial transforms of $a_j$ and $b_j$ respectively, then for all $k$ consider the sequence {$b_0,0,b_1,0,b_2,....$}, theorem $1$ and get the new sequence {$b_1,0,b_2,....$}. the binomial transform of  {$b_1,0,b_2,....$} is given by {$H_{k+2}-2H_{k+1}+H_n+b_0[k=-1]-b_0[k=-2]$}.Since $∆b_j$$=a_j$ for $j>0$ so for all $k$ $G_k=H_{k+2}-2H_{k+1}+H_n+b_0[k=-1]-b_0[k=-2]-H_n$=$H_{k+2}-2H_{k+1}+b_0[k=-1]-b_0[k=-2]$

I wish this helps you.

if $P$ is a statement $[P]$ is $1$ for $P$ is true and $0$ otherwise.

$∆b_j=b_{j+1}-b_j$

Title: Binomial type sum


问:
If $|f''| \leq M|f |$, how to prove that $f(x) \equiv 0$
This time I come across a problem.
A function  $f(x)$ is defined on an  interval $[a,b]$ , and given that $f(a) = f'(a) = 0$ , there exists a also const $M$ which satisfies $$|f''(x)| \le M|f(x)|$$
How to obtain $f(x) \equiv 0$ ?

Edit: the content of the related section is about the differential form of Grönwall's inequality here:

If $f(x)$ is differential on $[a,b]$, $f(a)=0$ and there exists a const $M$ which satisfies $$|f'(x)| \le M|f(x)|$$ we can easily obtain  $$f(x) \equiv 0$$.

I tried to use a similar method in the book to prove the second order derivative situation, but without success.

答:
Here is an approach using the fundamental theorem. If $x_0 \in (a,b)$ and $x \in [a,x_0]$ the hypothesis $f(a) = f'(a)$ gives you
$$f(x) = \int_a^x f'(t) \, dt = \int_a^x \int_a^t f''(y) \, dy dt $$ so that $$|f(x)| \le \int_a^x \int_a^t |f''(y)| \, dy dt \le \max_{[a,x_0]} |f''| \int_a^x \int_a^t \, dydt \le M \left( \max_{[a,x_0]}|f| \right) \frac{(x_0-a)^2}{2}.$$
Now take the maximum of all $x \in [a,x_0]$ on the left-hand side to get
$$\max_{[a,x_0]}|f| \le M \left( \max_{[a,x_0]}|f| \right) \frac{(x_0-a)^2}{2}.$$
Choose $x_0$ sufficiently close to $a$ so that
$$\frac{(x_0 - a)^2}{2} < \frac 1M$$
to find that
$$\max_{[a,x_0]} |f| = 0.$$
Thus $f$ is identically zero on a nondegenerate interval $[a,x_0]$ whose length depends only on $a$ and $M$.

Can you proceed from here to show the function is identically zero on $[a,b]$?


问:
Is this space normed or not?
This particular question was part of my analysis assignment I was unable to prove / disprove it.

Let $V$ denote the vector space of all sequences $a=(a_1 , a_2,...,a_n,...)$ of real numbers such that $\sum_{n\geq 1} 2^n |a_n|$ converges. Define ||.||: $V \to \mathbb{R}$ by $||a|| =\sum_{n\geq 1}2^n |a_n|$. Then is $V$ complete normed space or not.

There were four questions to be answered regarding $V$ . I have done 3 but unable to do this. I know about completely normed space but I was unable to do this . The  idea is to show that every sequence of sequence converges to a limit and that limit is to be constructed or guessed but I am unable to do it

Kindly guide on how this should be done

答:
Let $l_1=\{(\alpha_1,\alpha_2,...): \sum_{n=1}^{\infty}|\alpha_n|<\infty\}$ and let $T:l_1\to V$ with
$$T(\alpha_1,\alpha_2,...,\alpha_n,...)=\bigl(\frac{1}{2}\alpha_1,\frac{1}{2^2}\alpha_2,...,\frac{1}{2^n}\alpha_n,...\bigr)$$

By the identity
$$||T(\alpha)||_V=\sum_{n=1}^{\infty}2^n|\frac{\alpha_n}{2^n}|=\sum_{n=1}^{\infty}|\alpha_n|=||\alpha||_{l_1}$$

$T$ is a well defined linear and bounded operator from $l_1$ to $V$. Now observe that $T^{-1}(x_1,x_2,...,x_n,...)=(2x_1,2^2x_2,...,2^nx_n,...)$, which means that $T$ is one to one and onto.

Now, for $x=(x_n)_n \in V$ we have
$$||T^{-1}(x)||_{l_1}=\sum_{n=1}^{\infty}2^n|x_n|=||x||_V$$
which means that $T^{-1}$ is also bounded. Hence, $V$ and $l_1$ are isometrically isomorphic. Which means that $V$ is complete.

Title: Is this space normed or not?


问:
Proving a Cyclic group $G$ of order n is isomorphic to $\mathbb{Z_n}$
My impression was that $\mathbb{Z_n} = \{0,1,...,n-1\}$ and so to prove this result, you would only have to define an isomorphism from the set $\{0,1,...,n-1\}$ to $G$. However in the proof I was given, the author defines isomorphism $\theta: \mathbb{Z_n} \rightarrow G$ s.t $\theta([x]) = g^x$ where $g$ is generator for cyclic group $G$. He then goes on to show initially that the  map is well defined by showing that for any $x' \in [x]$, (Where the equivalence relation must be $x \sim y \iff x \equiv y$ mod $n$) the function is well defined. This confused me because my impression was that $\mathbb{Z_n}$ was only the set $\{0,1,...,n-1\}$ so I can't see why you have to worry about equivalence classes here. Any integer that is not contained in set  $\{0,1,...,n-1\}$ is surely irrelevant right?

答:
If $G$ is cyclic of order $n$, then the underlying set is $G=\{g,\dots,g^n(=e)\}$. Consider the map $\varphi\colon G\to \Bbb Z_n$, defined by $g^i\mapsto [i]_n$.

Injectivity: $[i]_n=[j]_n\Rightarrow j-i\equiv 0\pmod n\Rightarrow j=kn+i\Rightarrow g^j=g^{kn+i}=(g^n)^kg^i=e^kg^i=g^i$.

Surjectivity: by construction.

Homomorphism: $\varphi(g^ig^j)=\varphi(g^{i+j\pmod n})=[i+j]_n=[i]_n+[j]_n=\varphi(g^i)+\varphi(g^j)$.

You have to firstly take care about good-definition in case you consider the other way around, namely $\psi\colon \Bbb Z_n\to G$ defined by $[i]_n\mapsto g^i$. In fact, here we are using a class representative, $i$ (yes, the elements of $\Bbb Z_n$ are equivalence classes) to define a map whose argument is an equivalence class as such; so, we want $j\in [i]_n\Rightarrow \psi([j]_n)=\psi([i]_n)$; but this is the case, because $j\in [i]_n\Rightarrow [j]_n=[i]_n$.


问:
Lower bound for tail probability
I need help to prove the following lower bound for tail probability. I have tried using well-known inequalities like Chebyshev and Paley-Zygmund, but cannot get the required bound.

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and $X$ be a random variable with $\mathbb{P}(X\in [-M,M])=1$ for some positive constant $M$. Show that $\mathbb{P}\left(|X-\mu|\geq \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$, where $\mu= \mathbb{E}(X)$ and $\sigma^2=\text{Var}(X)$.

Thank you.

Additional question: How can I also prove that $\mathbb{P}\left(X\geq \mu+ \frac{\sigma^2}{4M} \right)\geq \frac{\sigma^2}{8M^2}$ ?

答:
The fourth moment bound says:
$$
E[|X|] \ge \frac{E[X^2]^{3/2}}{E[X^4]^{1/2}}.\\
$$
Let $s = \max\{|\mu+M|, |\mu-M|\} \le 2M$, that is the furthest the mean can be from $X$.
Then we have $E[(X-\mu)^4] \le s^2 E[(X-\mu)^2]$.

Thus, by Paley–Zygmund's inequality:
$$
\Pr[|X-\mu| \ge \alpha\sigma^2]
\ge (1-\alpha)^2\frac{E[|X-\mu|]^2}{E[|X-\mu|^2]}
\ge (1-\alpha)^2\frac{E[|X-\mu|^2]^2}{E[|X-\mu|^4]}
\ge (1-\alpha)^2\frac{E[|X-\mu|^2]}{s^2}.
$$

Setting $\alpha=1/(4M)$ we have
$$
\Pr[|X-\mu| \ge \frac{\sigma^2}{4M}]
\ge (1-1/(4M))^2\frac{\sigma^2}{s^2}
\ge \frac{\sigma^2}{8M^2}
,
$$
where the last bound requires $M\ge 1$. However, that requirement could probably be removed by using a slightly stronger version of Paley–Zygmund.

Title: Lower bound for tail probability


答:
Since the function is everywhere differentiable, it can have local extrema only at the points where $f'=0$.  The formula for the derivative is
$$\begin{split}
f'(x) &= \sum _{i=1}^n (z_i-x)_+ + c_i \{(z_i - x)_+- (y_i - x)_+  \}  \\
&= \sum_{i=1}^n (1+c_i)(z_i-x)_+  - \sum_{i=1}^n c_i (y_i - x)_+
\end{split} \tag 1 $$
if I parsed your formula correctly (it appears to have a few unnecessary parentheses).
This function is piecewise linear; its graph has corners at the points $z_i$ and $y_i$.

I don't think there is a very clever way to deal with such  piecewise linear functions. I would list the numbers $z_i,y_i$ in increasing order, plug each into $f'$, and noted the signs. The maximum will be found within an interval where $f'$ is positive at the left endpoint and is negative at the right endpoint. The precise location is found by solving linear equation (since $f'$ is linear within the interval of interest).

For example, on the interval $(1.68227033, 1.91068738 )$ we have
$$\sum_{i=1}^n (1+c_i)(z_i-x)_+ = (1+0.5670724) (1.91068738-x)$$
and
$$\sum_{i=1}^n c_i (y_i - x)_+ =  0.1032198 (2.4177490-x)+0.000045619 (4.7770777-x) + 0.5670724(2.3840472-x)$$
So, you can simplify $f'(x)=\sum_{i=1}^n (1+c_i)(z_i-x)_+ - \sum_{i=1}^n c_i (y_i - x)_+$ by expanding the parentheses, and check if $f'$ turns to zero or not.

Title: Find local maxima of this quadratic function

Body:
How can I find local maxima of this quadratic function?

$$f(x) = \sum _{i=1}^n -\frac{(z_i - x)_+^2}{2} - \left\{((\frac{(z_i - x)_+^2}{2})-(\frac{(y_i - x)_+^2}{2}) ) * c_i\right\} $$

which + means If $z_i$ is bigger than u its equal $z_i - x$ and else it's equal zero.
Also $z_i$, $y_i$ and $c_i$ is a  vector of values with same lengths.$z_i$ and $y_i$ is an interval.

$$ z = (0.04124633, 0.05404163, 0.30939022, 0.57690859, 0.70211949, 0.73597262, 0.92130814,1.05226194, 1.68227033, 1.91068738)$$
$$y = (0.2112338, 0.4172289, 0.3991616, 0.8755037, 1.2448935, 1.3493179, 1.6522002, 2.4177490, 4.7770777, 2.3840472)$$
$$ c = (5.337958, 2.275704, 10.64689, 2.873863, 1.293402, 1.03867, 0.6372883, 0.1032198, 0.000045619,  0.5670724)$$
I plot this quadratic function in terms of x which can be change between max finite value of $z_i$ and $y_i$ and see that theres is one local maxima but I do not know how can I find it matematically?


Title: For a positive integer $ n$, $ P(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{2^n-1}$, then

Body:
Four given options are:
$$A) P(100)<100$$ $$B) P(100)>100$$
$$C) P(200)<100$$ $$D) P(100)=150$$

Despite much try I am unable to find which option holds true. I am just unable to understand the question.
$$1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+....+1/15)+....+1/{(2^n)-1}$$
= $$1+(1/2+1/2²-1)+(1/4+1/5+1/6+1/2³-1)+.....+1/{(2^n) -1}$$
What to do next ?

Body:
Let me show you a pattern,
P(1) = 1

P(2) = 1 + 1/2 + 1/3 = P(1) + 1/2 + 1/3

P(3)                 = P(2) + 1/4 + 1/5 + 1/6 + 1/7
.
.
.

now, we need to find the limit of P(n), from observation.

P(1) =1  <=1

P(2) = P(1) + (<= 1/2 + 1/2)   <= 2

P(3) = P(2) + (<= 1/4 + 1/4 + 1/4 + 1/4)   <= 3

P(n) < n [we can remove the equality sign actually]
=> P(100) < 100 and also P(200) < 200

Similarly, we can also conclude P(n) >= 1 + 0.5*n + 1/($2^n$ - 1)


问:
Limit of convex increasing function
If $f$ is strictly increasing and strictly convex (or $f'>0$ & $f''>0$), then $$\lim_{x\rightarrow∞}{f(x)}=∞$$

Is this statement true?

If this statement is true, how can I prove?

答: Let $m=f'(0)$ and set $g(x)=mx+f(0)$.  We have $f(x)\ge g(x)$ for all $x>0$ and $\lim_{x\rightarrow \infty}g(x)=\infty$.</p>

Title: Limit of convex increasing function


问:
Matrix inconsistency
I've been looking into the following deriviation of a parameterization of a green assets Markowitz problem. One author completes the following step in a calculation that I can not follow.

\begin{align}
a&=(1-\lambda)\frac{d}{a^2}\left[ \frac{1}{\eta^2} \left(\underbrace{ I_N-\frac{1}{\eta^2/\sigma^2+N} \mathbf{1}_{N\times 1}\mathbf{1}_{N\times 1}' }_{Y}\right)\right]g\\
&=(1-\lambda)\frac{d}{a^2\eta^2}g
\end{align}
where $I_N$ is the identity matrix in N dimensions; $\mathbf{1}$ is a matrix of all ones; $\lambda, \eta, d, a$ are scalars; g is an $N\times1$ matrix.

The authors mention $\eta^2\approx(0.7/0.3)\sigma^2$, however this feel irrelevant in this context.

The above implies that the entire paranthesised part is $Y=1$? And I do not really understand why? I would be super grateful if anyone could point me in the right direction.

EDIT 1: Additional information

In the list of assumptions these two were shortly mentioned:
$$
(\mathbf{1}_{N\times 1})'g=0, g'g =1
$$

答:
The key point is that $Yg = g$. To see this, note that
$$
Yg = [I_N - K \cdot \mathbf 1 \mathbf 1^T]g =
I_N g - K \cdot \mathbf 1 (\mathbf 1^T g) = g - 0 = g.
$$

Title: Matrix inconsistency


问:
Real symmetric matrix of rank $n-1$ has a submatrix of rank $n-1$
Let $A$ be a symmetric real matrix of dimension $n \times n$ and rank $n-1$. Prove that there is a $k \in \{1,2,...n\}$ such that on deletion of the $k$th row and column the resulting matrix has rank $n-1$.

I think we would have to use adjugate of the matrix here since that is the space of all $(n-1) \times (n-1)$ submatrices, but I am not very sure of how to proceed

答:
This is usually proved by using matrix congruence, but yes, you can prove the statement by using the adjugate matrix.

As $A$ has rank $n-1$, it adjugate matrix has rank one. Since $A$ is also symmetric, so must be its adjugate matrix. Therefore $\operatorname{adj}(A)={\pm  vv}^T$ for some nonzero vector $v$. Thus $\operatorname{adj}(A)$ has some nonzero diagonal entries. As the diagonal entries of $\operatorname{adj}(A)$ are the principal $(n-1)$-rowed minors of $A$, the result follows.


问:
For which functions is $\sqrt{\int_1^2 f(x)^2*xdx} = 0$ true
I have to find all functions for which the following statement is true
$$\sqrt{\int_1^2 f(x)^2*xdx} = 0$$
$f(x)$ can be anything: polynomial, trigonometric, exponential, ...
I know I can leave the square root and I have to find an antiderivative for which $F(2)-F(1) = 0$ is true, but I can't seem to get any further than that... any tips or answers would help a lot!

答: Observe that $x>0, f(x)^2\ge0$. Therefore for the integral to be $0$ you need $f(x)=0$ almost everywhere on $[1,2]$.


问:
Why is my solution wrong to $\,99x \equiv 18 \mod 30$
ive had to solve $\,99x \equiv 18 \mod 30$

Firstly $\gcd(99,30)=3$ and $3  | 18$ so the equation has solutions.

I can devide by $3$. So we get:

$$33x \equiv 6 \mod 10$$

Since $\,\gcd(33,10)=1$ we can find an inverese to $33$ with the extended euclidean algorithm.

I calculated: $(-3)33+10*10=1$, so the inverse element should be $-3$

Now we write:

$$(-3)33x \equiv (-3)6 \mod 10$$

which should give:

$$x \equiv -18 \mod 10$$

So $x=10*p + 8$ with $p\in \mathbb{Z}$

But when I try out the found $x$ the solution is simply wrong..

Actually I found by trying out, $x = 12$ does work.. I am really confused where I messed it up :(

Can someone help? Thanks

答:
Just reduce $\pmod {30}$ to $9x \equiv 18 \pmod {30}$ so that $x=2$ is a solution.

Even further, dividing by $3$ we get $3x \equiv 6 \pmod {10}$ to see that $x\equiv 2 \pmod {10}$ is the set of solutions.

Your solution isn't working because you have $-99x \equiv -18 \pmod {10}$ and to make this clear to you, add $100x$ to both sides so that $x \equiv -18 \equiv 2 \pmod {10}$. You could get to this instantly by noticing $99 \equiv -1 \pmod {10}$ as well.


答: $f(x)$ has to be identically zero namely $f(x)=0$., because the integrand of the area-integral $A=\int_{1}^{2} x f^2(x) dx$ is positive definite in $(1,2)$. Hence $A$ cannot be zero and if it is zero then $f(x)=0$.

Title: For which functions is $\sqrt{\int_1^2 f(x)^2*xdx} = 0$ true

Body:
I have to find all functions for which the following statement is true
$$\sqrt{\int_1^2 f(x)^2*xdx} = 0$$
$f(x)$ can be anything: polynomial, trigonometric, exponential, ...
I know I can leave the square root and I have to find an antiderivative for which $F(2)-F(1) = 0$ is true, but I can't seem to get any further than that... any tips or answers would help a lot!


问:
A conjecture about $\mu(n)$
Möbius function $\mu(n)$ is defined as $\mu(n) = (-1)^k$ when $n$ is square free and $n$ contains $k$ distinct prime factors, and 0 otherwise. Also $\mu(1)=1$.

There are nice properties of this function like $$\sum_{d|n} \mu(d) = 0$$ unless $n=1$.

Let us consider $$S(x) = \sum_{n\leq x} \frac{\mu(n)}{n}$$
We see that $S(1)=1$ and $S(2) = \frac{1}{2}$ and so on.
What I observed by examples is $|S(x)| \leq 1$ for all $x$ and equality is satisfied only when $x=1$ but I am not able to prove or disprove this statement for any arbitrary $x$. If this is true then there is a possibility that $\sum_{n=1}^{\infty} \frac{\mu(n)}{n}$ converges.

答:
As mentioned in one of the comments $$\sum_{n \leq x} \mu(n) \left\lfloor \frac{x}{n} \right\rfloor = 1$$ is helpful in showing
your inequality is true.

Observe that

$$
1 = \sum_{n\leq x} \mu(n) \left\lfloor \frac{x}{n} \right\rfloor =
\sum_{n\leq x} \mu(n) \left( \frac{x}{n} - \left\{ {\frac{x}{n}} \right\}\right) =
xS(x) - \sum_{n\leq x} \mu(n) \left\{ {\frac{x}{n}} \right\}
$$

where $\left\{ y \right\}$ is fractional part of $y$ which is always between 0 and 1.

So $$x|S(x)| \leq 1 + \sum_{n \leq x} \left\{ {\frac{x}{n}} \right\} \leq 1 + \left\{ x \right\}+  \sum_{2 \leq n \leq x} \left\{ {\frac{x}{n}} \right\} \le 1 + \left\{ x \right\}+ \lfloor x\rfloor -1 = x$$

Dividing by $x$ we get your strict inequality.

Title: A conjecture about $\mu(n)$


问:
When is the system of first order differential equations asymptotically stable?

Given the system $$\mathbf{x}'=A\mathbf{x} $$
where $A=\begin{bmatrix}a &0 & 4 \\ -1 & -1 & 0 \\ -2-a & 0 & -3 \end{bmatrix}$

In what interval of $a$ is the system asymptotically stable, and for what value of $a$ is the system stable/unstable if such a case even exists?

Attempt

For a system to be asymptotically stable the real part of all eigenvalues must be negative. $$\Re(\lambda)<0, \: \: \: \text{for all} \: \: \lambda $$

Since we are dealing with a $3\times3$ matrix, I used maple to find the eigenvalues from the system matrix. I also tried to use Maple to solve the inequality case for the eigenvalues, but I don't think I'm getting correct results.

The results I get from Maple state that the system is asymptotically stable when $$-8 < a \leq5-4\sqrt{3} $$
I spoke with my peers, and they said that this interval of $a$ is wrong. I posted my Maple document below. Can anyone see what is going wrong?

答:
I agree with your peers. Simply because the range indicated means $a$ must be smaller than 0, while 0 is a feasible solution (roots at $-\frac{3}{2} \pm \frac{1}{2}\sqrt{23}i$). Since only the real part is indicating stability, it is important to figure out when the square root becomes imaginary:
$$a^2-10a-23=0$$
$$(a-5)^2=48$$
$$a = 5\pm 4\sqrt{3}$$
Knowing these roots, you can deduct that this is less than zero if:
$$5-4\sqrt{3}<a<5+4\sqrt{3}$$
Within this range, the stability is entirely relying on $ \frac{1}{2}a-\frac{3}{2}$. from here it can be seen that the upperbound of $a$ should be: $a<3$.
The lower bound is a bit tricky as it exceeds the range on which the root is imaginary, but luckily that has been already given by your calculator: $a>-8$, which is the value on which the second eigenvalue will yield 0. So the actual range of $a$ is:
$$-8<a<3$$

Why maple failed to give this answer might have something to do with the imaginary part. the given range is where the answer does not have an imaginary part and is strictly less than 0. I am not really familiar with Maple, but maybe make it try to solve real(your equation < 0){a}.


问:
Calculating banking rates - Equation
[Note: I have no mathematical background at all and as this is my first time posting here, let me know if I forget anything in my post, especially regarding the tags as I can't just pick up "equation" and I don't understand the popular tags' meaning.]

I try to calculate a rate for banking purpose.
This formula gives me the monthly payment I'll be entitled to reimburse each month :

$M = \frac{C\cdot\frac{t}{12}}{1-\left(1+\frac{t}{12}\right)^{-12n}}$

Given :

C = 200000 (if I loan 200k€)

t = 0,02 (at a rate of 2%)

n = 20 (over 20 years)

M = ? (I'll have to pay M €/months)

$M = \frac{200000*\frac{0,02}{12}}{1-(1+\frac{0,02}{12})^-12*20} = 1011.77 \approx 1012€/\text{month}$

Now my problem is : if I change the $C$ value to $199000$ how can I get the new $t$ from the following equation ?

$1012 = \frac{199000\cdot\frac{t}{12}}{1-(1+\frac{t}{12})^{-12\cdot20}}$

I know the answer is $t = 2,05%$ but I have no idea how to calculate it so I can't integrate it into my code. Thanks for your help.

答:
HINT:

We can develop an approx formula by logarithmic differentiation

$$ \text{Let}  \; t/12= x,m= (-12 n)\;$$
$$ \dfrac{Cx}{1-(1+x)^m} =M$$
$$ \log C + \log x - \log (1-(1+x)^m) = \log M$$

$$\dfrac{ \Delta C}{C} + \dfrac{ \Delta x}{x}+ \dfrac{  m (1+x)^{m-1} \cdot \Delta x}{(1-(1+x)^m)}=0$$

$$\dfrac{ \Delta C}{C} + \Delta x \cdot \left( \dfrac{1 }{x}+ \dfrac{  m (1+x)^{m-1} }{1-(1+x)^m}\right) =0$$

It does not depend on $M$

$$ \Delta C= 0.01* 10^5 , \Delta t = 12 \Delta x $$

which relation allows to you to find $\Delta x$ and so the new $ \Delta t.$

Title: Calculating banking rates - Equation


问:
Diagonal dominance versus positive semi-definiteness
I know that for a symmetric matrix $A$, diagonal dominance, i.e. $$A_{ii} \ge \sum\limits_{j \ne i} |A_{ij}|$$ implies positive semi-definiteness.

How about the other way? Does positive semi-definiteness imply diagonal dominance? Could you point to a proof or a counter example?

答:
You can see link http://www.win.tue.nl/~aeb/srgbk/node16.html
In that link, the author said that if ma trix symmetric and stricly diagonal dominant then A positve define.</p>


Title: Why is my solution wrong to $\,99x \equiv 18 \mod 30$

Body:
ive had to solve $\,99x \equiv 18 \mod 30$

Firstly $\gcd(99,30)=3$ and $3  | 18$ so the equation has solutions.

I can devide by $3$. So we get:

$$33x \equiv 6 \mod 10$$

Since $\,\gcd(33,10)=1$ we can find an inverese to $33$ with the extended euclidean algorithm.

I calculated: $(-3)33+10*10=1$, so the inverse element should be $-3$

Now we write:

$$(-3)33x \equiv (-3)6 \mod 10$$

which should give:

$$x \equiv -18 \mod 10$$

So $x=10*p + 8$ with $p\in \mathbb{Z}$

But when I try out the found $x$ the solution is simply wrong..

Actually I found by trying out, $x = 12$ does work.. I am really confused where I messed it up :(

Can someone help? Thanks

Body: $-18\equiv 2\bmod 10$, not $8$.


问:
Understanding the definition of the $ACC$ condition.
My professor gave us the following definition:

Let $R$ be a commutative ring. $R$ has the ascending chain condition $(ACC)$ if, for every ascending chain of ideals $$I_{1} \subset I_{2} \subset I_{3} \subset ...$$ in $R,$ there is an $N > 0$ so that $I_{n} = I_{n+1}$ for $n \geq N.$

My question is:

Does that condition means they are equal up to associates? i.e. $I_{n} = u I_{n+1}$ for some unit $u \in R^{\times}$

If that is correct, can anyone show me a rigor proof to that please?

答:
What the equation $I_n = I_{n+1}$ means, i.e. what it is defined to mean, is equality of sets; in other words, it means that $x \in I_n$ if and only if $x \in I_{n+1}$.

On the other hand, if $u$ is a unit, then you should be able to check that $I_{n+1} = u I_{n+1}$, which means that $x \in I_{n+1}$ if and only if $x \in u I_{n+1}$. (Hint on the proof: $x = u (u^{-1} x)$).

Therefore, $I_n = I_{n+1}$ is equivalent to $I_n = uI_{n+1}$ for some unit $u \in R^{\times}$.


问:
Three 'equivalent' definitions of a holomorphic vector bundle
I'm aware that similar questions have been asked here and here but neither of these seems to have settled my issue. I have the following three definitions of a holomorphic vector bundle (in all cases $M$ is a complex manifold and a complex vector bundle means a smooth vector bundle whose fibres are complex vector spaces):

A holomorphic vector bundle $\pi:E\rightarrow M$ is a complex vector bundle together with the structure of a complex manifold on $E$ for which there exist trivialisations
      \begin{equation}
 \phi_U:\pi^{-1}(U)\rightarrow U\times\mathbb{C}^k
 \end{equation}
      which are biholomorphic maps of complex manifolds.

A holomorphic vector bundle $\pi:E\rightarrow M$ is a complex vector bundle for which the transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta \rightarrow GL_k(\mathbb{C})$ are holomorphic maps of complex manifolds.

A holomorphic vector bundle $\pi:E\rightarrow M$ is a complex vector bundle together with the structure of a complex manifold on $E$ for which the projection $\pi:E\rightarrow M$ is a holomorphic map of complex manifolds.

The wikipedia article here seems to merge all three definitions into one! Whereas Griffiths' and Harris' book goes with the first definition. I am unsure how to see whether these definitions are equivalent, or whether this is actually the case.

Moreover, I am actually trying to prove that the holomorphic tangent bundle $T^{1,0}M$ is indeed a holomorphic vector bundle - although I have no idea where to start with this, with any of the three definitions... any help on either of these problems would be much appreciated!

答:
Consider $E$'s transition functions as a smooth manifold, induced by its transition functions as a complex vector bundle:

$$\tau_{UV}(p, v) = \phi_V \circ \phi_U^{-1}(p, v) = (p, g_{UV}(p) \cdot v)$$

where $p \in U \cap V$ and $v \in \mathbb C^k$. By definition,

$\pi : E \to M$ is a holomorphic vector bundle in the first sense if the $\phi_U$'s are biholomorphisms.

$\pi : E \to M$ is a holomorphic vector bundle in the second sense if the $g_{UV}$'s are holomorphic maps.

$\pi : E \to M$ is a holomorphic vector bundle in the third sense if the $\tau_{UV}$'s are biholomorphisms.

Let $\mathscr U$ be the open cover of $M$ on which the local trivializations are defined. (WLOG we may assume that each distinguished open $U \in \mathscr U$ is biholomorphic to an open subset of $\mathbb C^n$.) Since $\mathscr F = \{ \phi_U \}_{U \in \mathscr U}$ is a smooth atlas on $M$, the following statements are equivalent:

$\mathscr F$ is a holomorphic atlas on $M$.

$\mathscr F$'s charts $\phi_U$ are biholomorphisms.

$\mathscr F$'s transition functions $\tau_{UV}$ are biholomorphisms.

Hence the first and third definitions are equivalent.

Since $\tau_{UV}$ is already a diffeomorphism, the following statements are equivalent

$\tau_{UV}$ is a biholomorphism.

$\tau_{UV}$ is a holomorphic map.

$g_{UV}$ is a holomorphic map.

Hence the second and third definitions are equivalent.


问:
Why is this step being taken when calculating a fourier series for $e^t$
I am trying to compute the fourier series for $e^t$ in range $-\pi<t<\pi$ and I am stuck at the marked part of the picture.

In the solution it is written that you get $\frac{1+in}{1+n^2}$, why do you have to do it?

答:
That step is just rewriting $\frac{1}{1-in}$ by multiplying top and bottom by $1+in$. This is a routine trick for writing $\frac{a+bi}{c+di}$ as $x+yi$ where $a,b,c,d,x,y$ are all real numbers.

You don't technically need to do this, but it helps when it comes time to add the $n$ term and the $-n$ term to get a representation in terms of sines and cosines instead of complex exponentials.


问:
Factor simple cubic polynomial $ax^3 + cx + d = 0$
I'm working on a self-made problem where I have a cubic equation I need to factor. I have some flexibility in choosing my own coefficients for the polynomial, so I am strategically trying to choose my coefficients within the admissible set so that the cubic I end up with is simple to factor. What are the forms of some simple cubic equations to factor? For example, if the second coefficient is zero, so it is

$$ax^3 + cx + d = 0$$
is that simpler to factor than the general case? Any other examples? I'm not trying to solve for the general cubic, which is hard, I want to modify my cubic so that it is easy to factor. Thank you!

答:
For the cubic polynomial $ax^3+c x+d$ to factorize, its coefficients $a$, $c$ and $d$, assuming all rational, have to satisfy
$$d=-r(ar^2+c)$$
where $r$ is rational. Then, the polynomial factorizes as
$$ax^3+c x +d=a(x-r)(x^2+ r x +r^2 +\frac ca)$$


问:
Probability of determinants being coprime
I have a question that is not of particular significance, but I would love to understand the underlying principles.

Suppose we have two square $3 \times 3$ matrices, $M_1$ and $M_2$ with

$$M_1 = \begin{pmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix} \qquad\text{and}\qquad
M_2 = \begin{pmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{pmatrix}$$

with the coefficients $a_n,b_n \in \mathbb{Z}$ and $1 \leq a_n,b_n \leq 9$

What is the probability that the matrices' determinants are coprime, when uniformly random coefficients satisfying the conditions are chosen.

I am familiar with the Riemann's $\zeta$ function way to find out the probability of two random integers being coprime, but I have no clue how to apply that here with additional conditions on the numbers given.

I did test it mechanically, using Mathematica and the result is around 30%, but I would like to see a proper way to do it.

I would love to at least get a few pointers as what to research to tackle this problem.

Thank you very much!

答:
This paper actually shows that the probability two large matrices with integer entries are co-prime is $\frac{1}{3}$, which actually is a bigger version of your question.

http://www.mccurley.org/papers/relprime.pdf


问:
Linearly independent subset?
If $u,v,w$ and $z$ are distinct elements in $R$, then

 {$(1,u,u^2,u^3), (1,v,v^2,v^3), (1,w,w^2,w^3), (1,z,z^2,z^3)$} is a linearly independent subset of $R^4$.

Is that true or false?
I even can't start the first step..

答:
If they are linearly dependent, then the equation
$$
\underbrace{\pmatrix{1&u&u^2&u^3\\ 1&v&v^2&v^3\\ 1&w&w^2&w^3\\ 1&z&z^2&z^3}}_{A}
\pmatrix{a\\ b\\ c\\ d}=\pmatrix{0\\ 0\\ 0\\ 0}
$$
has a nontrivial solution $(a,b,c,d)$, i.e. there is a nonzero real polynomial $a+bx+cx^2+dx^3$ of degree $\le3$ that has at least four distinct roots $u,v,w,z$. This is impossible.

As pointed out by the others, The matrix $A$ in the above displayed equation is called a Vandermonde matrix. As we have demostrated, showing that it is invertible is easy (and the above proof works for any field, not just $\mathbb{R}$). The tricky part, though, is to find its determinant or inverse.

Title: Linearly independent subset?


问:
Application of Mean Value/Rolle's Theorem?
Question

Let $f$ be a differentiable function such that $f'$ is continuous on $[0, 1]$ and $M$ the maximum value of $|f'(x)|$ on $[0, 1]\ $. Prove that, if $f(0) = f(1) = 0$, then $$\int^{1}_{0} |f(x)| \mathrm {d}x \leq \frac M 4\ .$$

I have stared long and hard at this question for quite some time, but I am not sure where to even begin. Since I see that the endpoints are given, I think of MVT, but since they are also equal, I think of Rolle's Theorem too. These are just my intuitions - they could be wrong. Moreover, the moduli of $f'(x)$ and $f(x)$ are throwing me off. Any hints/suggestions on how to approach this question would be greatly appreciated :)

P.S. I am only taking an introductory calculus module in college, so the tools at my disposal are, roughly, IVT, MVT and Rolle's Theorem.

答:
I use Kelenner's hint to solve this problem.

We have
$$|f(x)| = |f(x) - f(0)| = |f(1)-f(x)|\text{.}$$
and by the MVT, for any $0 < x \leq 1$, there exists a $c_1 \in (0, 1)$ satisfying
$$f^{\prime}(c_1) = \dfrac{f(x)-f(0)}{x - 0} = \dfrac{f(x)-f(0)}{x} \implies |f(x)-f(0)| = |f^{\prime}(c_1) \cdot x| = |f^{\prime}(c_1)||x|\leq M|x|\text{.}$$
Thus we have
$$|f(x)| = |f(x) - f(0)| \leq M|x|\text{.}$$
Additionally, by the MVT, for any $0 \leq x < 1$, there exists a $c_2 \in (0, 1)$ satisfying
$$f^{\prime}(c_2) = \dfrac{f(1)-f(x)}{1-x} = \dfrac{-f(x)}{1-x} = \dfrac{f(x)}{x-1} \implies |f(x)| = |f^{\prime}(c_2) \cdot (x-1)| = |f^{\prime}(c_2)||x-1| \leq M|x-1|\text{.}$$
Hence,
$$|f(x)| \leq M|x-1|\text{.}$$
Thus
$$\int_{0}^{1}|f(x)|\text{ d}x \leq \int_{0}^{1/2}M|x|\text{ d}x+\int_{1/2}^{1}M|x-1|\text{ d}x = \dfrac{M}{8} + \dfrac{M}{8} = \dfrac{M}{4}\text{.}$$


问:
What is the limit of sum of split Wiener process?
I wish to find the limit in $L^2$ for $n\rightarrow\infty$ of

$A_n = \sum_{i=1}^nW^a(it/n)-W^a((i-1)t/n))(W^b(it/n)-W^b((i-1)t/n))$

How can I go about that?

答:
I assume that $W^a,W^b$ are independent $1$ dimensional BMs.

Let $Y_n = \sum_{i=1}^n \left(W^a(\frac{it}n) - W^a\left(\frac{(i-1)t}{n}\right)\right)\left(W^b(\frac{it}n) - W^b\left(\frac{(i-1)t}{n}\right)\right)$. Note that for different values of $i$, the random variables inside the bracket are independent by the independence of increments of a BM, along with the independent blocks theorem.

Therefore, $E[Y_n^2] = \sum_{i=1}^n E\left[\left(W^a(\frac{it}n) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\left(W^b(\frac{it}n) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]$

again by independence, the inner expectation breaks :
$$
E\left[\left(W^a\left(\frac{it}n\right) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\left(W^b\left(\frac{it}n\right) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]\\ = E\left[\left(W^a\left(\frac{it}n\right) - W^a\left(\frac{(i-1)t}{n}\right)\right)^2\right]E\left[\left(W^b\left(\frac{it}n\right) - W^b\left(\frac{(i-1)t}{n}\right)\right)^2\right]  \\ = \frac 1{n} \cdot \frac 1n = \frac 1{n^2}
$$

and when we take the sum, we just get $E[Y_n^2] = \frac 1n$. Hence it is clear that $Y_n \to 0$ in $L^2$.

Note that if $W^a$ and $W^b$ are correlated random variables and jointly Gaussian (note : if uncorrelated and jointly Gaussian, then they are independent by a well known property of Gaussian random variables), it is still possible to perform this kind of calculation, by creating BMs using appropriate functionals of $W^a$ and $W^b$.


问:
Graph of a Sine Function Increased on One Side
If we want something looks like this sine wave, what is a function that will satisfy this:

Further, how can we make this go both ways, meaning either 1) a sine wave staying the same at the bottom but going upwards linearly as it progresses (as shown) or 2) a sine wave staying the same at the bottom but going downwards linearly as it progresses?

I have tried, for example, $x/2(\sin x)$, which gives us a linear increase, but I want it to increase on the top while staying the same on the bottom.

Thanks!

答:
Plot[ x (1 - Sin[x]), {x, -1, 45}, GridLines -> Automatic]

You can adjust the slope $m$ of envelope, time period etc as:

$$ y= m x \;(1- \sin \dfrac{2 \pi x }{\lambda})$$

To reflect about x-axis making it go down, equation can be changed as:

```<code>Plot[ x (-1 + Sin[x]), {x, -1, 45}, GridLines -> Automatic]
</code>```

Body:
Plot[ x (1 - Sin[x]), {x, -1, 45}, GridLines -> Automatic]

You can adjust the slope $m$ of envelope, time period etc as:

$$ y= m x \;(1- \sin \dfrac{2 \pi x }{\lambda})$$

To reflect about x-axis making it go down, equation can be changed as:

Plot[ x (-1 + Sin[x]), {x, -1, 45}, GridLines -> Automatic]


问:
Composition of a vector measure and a linear functional
Assume that $X$ is topological vector space over the field $\mathbb{R}$. Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a nonempty set $T$. We say that
$$m: \mathcal{M}\to X$$
is a vector measure if
$$m(E_1\cup E_2)=m(E_1)+m(E_2)$$
whenever $E_1$ and $E_2$ are disjoint members from $\mathcal{M}$.

Question. Why is it that for each $x'\in X'$, the composition map
$$x'\circ m: \mathcal{M}\to \mathbb{R}$$
is a measure on $\mathcal{M}$ whenever the vector measure $m:\mathcal{M}\to X$ is countably additive?

In the above, $X'$ denotes the space of all continuous linear functionals on $X$. So far, I can show that $(x'\circ m)(\varnothing)=0$. What bothers me is to show that $x'\circ m$ is countably additive. Any tips? Thanks in advance...

答:
Let $\{E_i:i\in \mathbb{N}\}\subseteq \mathcal{M}$ where the $E_i's$ are pairwise disjoint. Since $m:\mathcal{M}\to X$ is countably additive,
$$m\left(\bigcup_{i=1}^{+\infty}E_i\right)=\sum_{i=1}^{+\infty}m(E_i).$$
Because $x'\in X'$, as $n\to +\infty$ we get
$$\sum_{i=1}^{n}(x'\circ m)(E_i)=x'\left(\sum_{i=1}^{n}m(E_i)\right)\to x'\left(\sum_{i=1}^{+\infty}m(E_i)\right)=(x'\circ m)\left(\bigcup_{i=1}^{+\infty}E_i\right).$$ Hence
$$\sum_{i=1}^{+\infty}(x'\circ m)(E_i)=(x'\circ m)\left(\bigcup_{i=1}^{+\infty}E_i\right).$$
This proves that $x'\circ m$ is countably additive.</p>