问:
Linear Algebra: Matrices Contained in Subspaces
I am not sure I understand this question:

Describe the smallest subspace of the matrix space M that contains:

$$ \left[ \begin{array}{ccc}   1&0\\   0&0 \end{array} \right] $$

and $$ \left[ \begin{array}{ccc}   0&1\\   0&0 \end{array} \right] $$

The answer is:

All matrices

$$ \left[ \begin{array}{ccc}   a&b\\   0&0 \end{array} \right] $$

Why is it not
$$ \left[
\begin{array}{ccc}
  a&a\\
  0&0
\end{array}
\right] $$, since the two column values of both matrices contain 1's?

答: Because neither $\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ nor $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$ is of the form $\left[\begin{smallmatrix}a&a\\0&0\end{smallmatrix}\right]$ for some number $a$. On the other hand, if $a,b\in\Bbb R$, then$$a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&b\\0&0\end{bmatrix}.$$So, every subspace of $\Bbb R^{2\times2}$ to which both $\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$ belong must contain every matrix of the form $\left[\begin{smallmatrix}a&b\\0&0\end{smallmatrix}\right]$. On the other hand, the matrices of this type form a subspace of $\Bbb R^{2\times2}$, and therefore the smallest subspace of $\Bbb R^{2\times2}$ containing $\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$ is the space of the matrices of the form $\left[\begin{smallmatrix}a&b\\0&0\end{smallmatrix}\right]$.


问:
If vector $a = b$ and $a = c$, then $b = c$?
I have three vectors: $\vec{u}$, $\vec{v}$ and  $\vec{w}$ and this equality:
$\vec{u} ·  \vec{v} = \vec{u} ·  \vec{w}$. Then the question is if $\vec{v} =  \vec{w}$.
At first i tried to write the equality representing components:
$$\vec{u} · \vec{v} = u_1·v_1+\ldots+u_n·v_n$$
$$\vec{u} · \vec{w} = u_1·w_1+\ldots+u_n·w_n$$
Then:
$$u_1·v_1+\ldots+u_n·v_n = u_1·w_1+\ldots+u_n·w_n.$$
But I got stuck right there; I thought if $c_i, i \in [1,n]$ could be canceled from both terms, but don't know if that is possible somehow.

答:
You can rearrange and use associativity of the dot product:

$\vec u \cdot \vec v = \vec u \cdot \vec w \implies \vec u \cdot (\vec v - \vec w) = 0$

Which implies that:

The vector (of nonzero norm) formed by the difference $\vec v - \vec w$ is orthogonal to $\vec u$ (also of nonzero norm)

OR

$\vec u$ is a null vector

AND/OR

$\vec v - \vec w$ is a null vector.

The final condition implies $\vec v = \vec w$, but it is clearly not the only possibility.

ContentLicense: CC BY-SA 4.0


问:
Show that the single matrix that commutes with all invertible matrices is the scalar matrix
Show that the scalar matrix $\lambda E$, with $E$ being the identity matrix and $\lambda \in K$ a constant, is the only matrix $A\in K^{n\times n}$ that commutes with all invertible matrices.

I am not sure how to tackle this question. I would appreciate if you could help me. Thanks!

答:
For any nonzero vector $u$, if $u$ and $Au$ are linearly independent, there exists some invertible matrix $B$ such that $Bu=u$ and $B(Au)=Au-u$. But then $0=(AB-BA)u=u$, which is a contradiction. Hence $u$ and $Au$ must be linearly dependent, i.e. $Au=c_uu$ for some scalar $c_u$.

This scalar $c_u$ actually does not depend on $u$. For, given any two nonzero vectors $u$ and $v$, if we pick an invertible matrix $B$ such that $Bu=v$, we obtain $0=(AB-BA)u=(c_v-c_u)v$, i.e. $c_u=c_v$. If we call this common value of all $c_u$s as $c$, we get $Au=cu$ for all nonzero vectors $u$. Hence $A=cI$.


问:
Proving that for any 3 infinite sequences $\{a_n\},\{b_n\},\{c_n\}$ in $\Bbb N,\exists p,q\in\Bbb N$ s. t. $a_p\ge a_q,b_p\ge b_q,c_p\ge c_q.$

Prove that for any three infinite sequences $\{a_n\},\{b_n\},\{c_n\}$ in $\Bbb N$, there are distinct $p,q\in\Bbb N$ s. t. $$a_p\ge a_q\space\land\space b_p\ge b_q\space\land\space c_p\ge c_q.$$

The task appeared on the $1961$ All Soviet Union Olympiad.

My attempt:

I tried to use the following lemma:

Every sequence in $\{x_n\}$ in $\Bbb R$ has a monotonic subsequence.

and its corollary:

For finitely many sequences $a^{(1)},a^{(2)},\ldots,a^{(k)}$ there is a strictly increasing subsequence $p:\Bbb N\to\Bbb N$ s. t. all of the subsequences $a^{(1)}\circ p,a^{(2)}\circ p,\ldots,a^{(k)}\circ p$ are monotonic.

and the proof of the corollary is:

From the lemma, we know there a strictly increasing sequence $q^{(1)}:\Bbb N\to\Bbb N$ s. t. $a^{(1)}\circ q^{(1)}$ is a monotonic subsequence of $a^{(1)}$. We now look at the subsequences $a^{(1)}\circ q^{(1)},a^{(2)}\circ q^{(1)},\ldots,a^{(k)}\circ q^{(1)}$ where the first one is monotonic, while others don't have to be. Again, from the lemma, there is a strictly increasing sequence $q^{(2)}:\Bbb N\to\Bbb N$ s. t. $a^{(2)}\circ q^{(1)}\circ q^{(2)}$ is a monotonic subsequence of $a^{(2)}\circ q^{(1)}$. Since a subsequence of a monotonic sequence is also monotonic, among the subsequences $a^{(1)}\circ q^{(1)}\circ q^{(2)},a^{(2)}\circ q^{(1)}\circ q^{(2)},\ldots,a^{(k)}\circ q^{(1)}\circ q^{(2)}$, the first two are monotonic, while others don't have to be. We repeat the process $(k-2)$ more times and obtain the subsequences $a^{(1)}\circ p,a^{(2)}\circ p,\ldots,a^{(k)}\circ p$, where $p=q^{(1)}\circ q^{(2)}\circ\cdots\circ q^{(k)}$.

To avoid confusion, let $d:\Bbb N\to\Bbb N$ take the role of the strictly increasing sequence $p:\Bbb N\to\Bbb N$ from the corollary of the lemma.

Now, every sequence in $\Bbb N$ has a minimum element, so, if some of the sequences $a\circ d,b\circ d,c\circ d$ happen to be decreasing, at some point they should become constant and hence monotonically increasing so, e.g., if $a'=a\circ d$ is decreasing $$\space m_a=\min\{m\in\Bbb N\mid\space a'_l\ge a'_m,l> m\},$$ and if we similarly define $m_b,m_c$ (if necessary), $p=\max\{m_a,m_b,m_c\}$, and any $q\ge p$ should work as we end up with three monotonically increasing sequences $a\circ d,b\circ d,c\circ d$ starting from the index $p$.

May I ask if this argument is valid and if there is anything wrong?

Thank you in advance!

答:
I'm posting this as a community wiki so that this question doesn't appear in the unanswered queue.

As verified by @HansEngler in the comments, the last paragraph of my proof can be summarized by noting that:

All monotonic sequences in $\Bbb N$ are either non-decreasing and unbounded or eventually constant.


问:
If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$?
If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$

(a) $1/2e^{-2x+2}$

(b) $e^{-x-1}$

(c) $e^{1-x}$

(d) $e^{-x}$

(e) $-e^{x}$

If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.

答:
This is a separable differential equation.  We have $\frac{dy}{dx} = -y$, so

$$\int \frac{dy}{y} = \int -dx$$

and we get get the general solution $y = e^{-x + k}$.  The initial condition $y(1) = 1$ implies $k = 1$ in this case.

ContentLicense: CC BY-SA 3.0


问:
A Straightforward optimization problem, but without calculus
I am helping a middle schooler out with the following Art of Problem Solving competition problem:

An ant travels from the point $A (0,-63)$ to the point $B (0,74)$ as follows. It first crawls straight to $(x,0)$ with $x \ge 0$, moving at a constant speed of $\sqrt{2}$ units per second. It is then instantly teleported to the point $(x,x)$. Finally, it heads directly to $B$ at 2 units per second. What value of $x$ should the ant choose to minimize the time it takes to travel from $A$ to $B$?

A straightforward but somewhat tedious solution involves using calculus to optimize the time taken as a function of x (with this method we get $\approx 23.3$). However, the student I am helping has not been introduced to calculus, and I was beating my head against a wall trying to find a clever way to solve this with simpler methods like algebra with quadratics or geometry. For instance, The function of x which represents the time of the ants travel is:

$$T = \frac{\sqrt{x^2 + 63^2}}{\sqrt{2}} + \frac{\sqrt{x^2 + (74-x)^2}}{2}$$

I thought that, instead of minimizing $T$ w.r.t $x$, I could minimize $T^2$
, which I would then be able to algebraically massage into the form of a quadratic equation. Simple knowledge of the properties of these equations would allow for a location of the minimum. This approach stalled out due to the cross term that results from the RHS. Any help would be appreciated

答:
We rephrase the problem: the bug does not teleport, but instead heads from $(0,-63)$ to $(x,0)$ at $\sqrt{2}$ units per second and then heads from $(x,0)$ to $(0,74-x)$ at $2$ units per second, since the time taken is the same for the bug.

Now, the key is that the points $(0,74-x)$, $(x,0)$, and $(37,37)$ always form an isosceles right triangle! So, we say that the bug heads from $(0,-63)$ to $(x,0)$ at $\sqrt{2}$ units per second, and then travels from $(x,0)$ to $(37,37)$ at $\sqrt{2}$ units per second as well (since the time taken for the bug is again the same).

Now, to minimize the time, we want $(0,-63)$, $(x,0)$ and $(37,37)$ to be on a line (since the shortest distance between 2 points is a line.) The equation for the line through $(0,-63)$ and $(37,37)$ is $y = \frac{100}{37}x - 63$, so the bug heads through $(\frac{2331}{100},0)$, giving our answer of $\frac{2331}{100}$.


问:
Representation of the action of $ S_3 $ on $ (1,2,-3) \in M$ On the plane
Consider the action of the permutation group  $S_3$ on the vector space $M =\lbrace (\lambda_1, \lambda_2, \lambda_3)\in \mathbb{R}^3,  \lambda_1+\lambda_2+\lambda_3=0\rbrace$.

The problem I have is the following: represent the action of $S_3$ on $\lambda = (1,2,-3) \in M $ on the plane $\mathbb{R}^2$ (wich we  identifiy with M).

The hints I've been given is first to choose an orthonormal basis of M and then to draw the two axis of symmetry wich represent the two generators of $S_3$ namely the permutations $(1,2)$ and $(2,3)$.

My questions are:

why do we have to choose an orthonormal basis of M, is it not enought to work with an arbitrary basis ?

what are the equations of the two axis of symmetry in $\mathbb{R}^2$ that we use to illustrate the image of  $\lambda $ under the action of $S_3$ ?

答:
why do we have to choose an orthonormal basis of $M$, is it not enough to work with an arbitrary basis ?

I guess in general you don't have to choose an orthonormal basis. You could still choose any basis and identify $M$ with $\Bbb{R}^2$ in such a way. Maybe though, since all permutation matrices have determinant $\pm 1$, and are thus orthogonal, it is nice to choose an orthonormal basis so that they remain orthogonal transformations in the new basis.

what are the equations of the two axis of symmetry in $\Bbb{R}^2$ that we use to illustrate the image of $\lambda$ under the action of $S^3$?

This will depend on the basis you choose for $M$. In $\Bbb{R}^3$, the action of $(1,2)$ is the reflection over the plane $y=x$. This plane intersects $M$ in the line spanned by the vector $(1,1,-2)$. So the action of $(1,2)$ on $M$ is the reflection over this line. After choosing a basis, you can try to express this line in those coordinates.

Similarly, the permutation $(2,3)$ acts as reflection over the plane $y=z$, which intersects $M$ in the line spanned by $(-2,1,1)$. Again, you'll want to find an equation for this line in the coordinates corresponding to the basis you choose for $M$.


答:
This question can be solved 'in a flash' when looked at in the right way.

If you look at the second term in your expression for time, you can simplify it to $$\frac{\sqrt{(x-37)^2+37^2}}{\sqrt2}.$$

We have now, in effect, made it a route with two legs travelled at the same speed, $\sqrt2$.

Furthermore, you will be able to draw this as a route travelling from $(0,0)$ to $(37,100)$  via a point at $(x,63)$.

For a straight line route make $x/37=63/100$ and the problem is solved.

Title: A Straightforward optimization problem, but without calculus

Body:
I am helping a middle schooler out with the following Art of Problem Solving competition problem:

An ant travels from the point $A (0,-63)$ to the point $B (0,74)$ as follows. It first crawls straight to $(x,0)$ with $x \ge 0$, moving at a constant speed of $\sqrt{2}$ units per second. It is then instantly teleported to the point $(x,x)$. Finally, it heads directly to $B$ at 2 units per second. What value of $x$ should the ant choose to minimize the time it takes to travel from $A$ to $B$?

A straightforward but somewhat tedious solution involves using calculus to optimize the time taken as a function of x (with this method we get $\approx 23.3$). However, the student I am helping has not been introduced to calculus, and I was beating my head against a wall trying to find a clever way to solve this with simpler methods like algebra with quadratics or geometry. For instance, The function of x which represents the time of the ants travel is:

$$T = \frac{\sqrt{x^2 + 63^2}}{\sqrt{2}} + \frac{\sqrt{x^2 + (74-x)^2}}{2}$$

I thought that, instead of minimizing $T$ w.r.t $x$, I could minimize $T^2$
, which I would then be able to algebraically massage into the form of a quadratic equation. Simple knowledge of the properties of these equations would allow for a location of the minimum. This approach stalled out due to the cross term that results from the RHS. Any help would be appreciated


问:
On the partitions of a number
I need to find the generating function for number of partitions of $n$ into parts that are divisible by $2$ and every one of them appears at most twice.
I think that solution is $f(x)=(1+x+x^2)(1+x^2+x^4)(1+x^4+x^8)\cdots$
Is this correct?

答:
That is the right idea, but your first factor corresponds to part size $1$, which is not divisible by $2$ and hence should appear.  The correct generating function is
$$\prod_{k \ge 1} (1+x^{2k}+x^{4k})$$

Title: On the partitions of a number


问:
Prove that $3^{2n} +7$ is divisible by 8

Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$

So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?

If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$

Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$

If $n=k+1$
$$3^{2(k+1)} +7$$
$$3^{2k+2} +7$$
If $3^{2n} +7$ is divisble by 8 then $3^{2n} +7=8A$ where $A \in \Bbb Z$, and thus $3^{2k+2} +7=8B$ where $B \in \Bbb Z$
$$3^2 \times 3^{2k}+7$$
$$3^2 \times (8A)=72A$$
$$72A =8(9A)=9B$$

So by induction $3^{2n} +7$ is divisibe by 8 $\forall n \in \Bbb N$

Body:
We can prove it like this :
$$ 9^n + 7 = 9^n + \sqrt[n]{7}^n $$
And by using factorization :
$$ 9^n + \sqrt[n]{7}^n = (9+7)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ 9^n + \sqrt[n]{7}^n = (16)(9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1})$$
$$ \frac{9^n + 7}{16} = 9^{n-1} -9^{n-2}\cdot7 + ... + 7^{n-1}$$

We have $16$ divides $9^{n}+7$ and $8$ divides $16$, so $8$ divides $9^{n}+7$ which is the same as' $8$ divides $3^{2n}+7$'


问:
Numerical differentiation - find bound for the error
I have be given a set of points $(-3.0, -2.8, -2.6, -2.4, -2.2, -2.0)$ and a function $f(x)=e^{x/3} + x^2$ and asked to find the bound for the error in each case. I already find $f'(x)$ for each point but I don't know how we should find the bound for the error

(a) Determine as accurately as possible approximations for each entry in the following table. (hint: employ the formulas for numerical differentiation)
$$\begin{array}{ccc}\hline
 x   & f(x)   & f'(x)\\\hline
-3.0 & 9.3678 & \\
-2.8 & 8.2332 & \\
-2.6 & 7.1803 & \\
-2.4 & 6.2093 & \\
-2.2 & 5.3203 & \\
-2.0 & 4.5134 & \\\hline
\end{array}$$
(b) Given that $f(x)=e^{x/3}+x^2$ find the bound for the error in each case.

答:
If you want to keep the error $O(h^2)$, you have presumably used centered finite differences for the inner points and something like
$$
f'_i \approx \dfrac{\mp (3 f_i - 4 f_{i\pm 1} + f_{i\pm2})}{2h}
$$
for the first and last points. In this case, since you are given the exact expression for $f$, you can compute the exact value of the error.

The result, using the process I described is
$$
\left(
\begin{array}{cccc}
 -3 & 9.3678 & -5.87725 & 1.2352\times 10^{-4} \\
 -2.8 & 8.2332 & -5.46875 & 1.6976\times 10^{-4} \\
 -2.6 & 7.1803 & -5.05975 & 1.33205\times 10^{-4} \\
 -2.4 & 6.2093 & -4.65 & 2.23679\times 10^{-4} \\
 -2.2 & 5.3203 & -4.23975 & 1.48233\times 10^{-4} \\
 -2 & 4.5134 & -3.82925 & 3.8904\times 10^{-4} \\
\end{array}
\right)
$$

If you do not want to depend on the expression of $f$ and want to use informations like $\displaystyle \max_{x\in[-3,-2]}|f'''(x)| < 0.02 $, you get get error bounds, instead of the actual error. For instance, for the interior points
$$
\left| f'(x_i)-\dfrac{f(x_{i+1})-f(x_{i-1})}{2h}\right|=\frac{h^2}{6}|f'''(\xi)|< \frac{0.2^2}{6}\times 0.02 = 1.33333\times 10^{-4}
$$

The apparent contradiction of having the actual error sometimes larger then the error bound is explained by the fact that the values in the table are not exact.


问:
Revenue Equivalence in Auction Theory: how does an English auction generate the same revenue as First Price Sealed Bid?
The theroem:

The revenue equivalence theorem states that, if all bidders are
risk-neutral bidder and have independent private value for the
auctioned items, then all four of the standard single unit auctions
have the same expected sales price (or seller's revenue).The four
standard single unit auctions are the English auction, the Dutch
auction, first-price sealed-bid auction, and the second-price sealed-bid
auction.

Given the preconditions, how is it possible for an English auction to genereate the same revenue as a first-price sealed auction?

Per my understanding, an English auction, when played optimally, generates the same revenue as a second price sealed auction (Vickrey auction).

On the other hand, a First Price Sealed Bid, when played optimally, results in a revenue equivalent to the highest bid.

So how can the two be revenue equivalent?

答:
At first sight, if one person bids 10\$, another 20\$, and a third one 30\$, then it seems that first-price auction would result in 30\$ and second-price in 20\$. However, under different rules -  that are of course known in advance - the participants would make different bids. E.g., under a second-price auction, you tend to bid more than what you are willing to pay because that's not what you are going to pay. Then again, you won't bid that much more and risk the second-price be above your original evaluation.  This may explain how the "obvious" difference is not that obvious after all. But can we expect in this scenario that the  second-highest price is much higher than what the second bidder estimates? Shouldn't they - by the same argument - bid more than 20, but only slightly more, perhaps 22\$ (so still way below 30\$)?

The key part in the full theorem is also that it is about "privately-known signal[s] independently drawn from a common strictly-increasing, atomless distribution" (emphasis mine). So it is not the case that one participant inherently values the object more than the others. I didn't look up the proof though, but under these circumstances, it doesn't look that much infeasible any more.


问:
Does this liminf characterization hold true?
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that
$$0<a:=\liminf_{|t|\to\infty} tf(t) <\infty.$$
I am interested in writing the above relation by using the definition. On by lecture notes, I have written that it corresponds to say that, fixed $\varepsilon>0$, it is
$$ tf(t)\geq(a-\varepsilon) - c_1 \quad\mbox{ for all } t, $$
where $c_1$ denotes a positive constant. Now, I would like to obtain a similar relation but using the "reverse" inequality, I mean, it is true that
$$ tf(t)\leq (a+\varepsilon) +c_2\quad\mbox{ for all } t $$
holds too?

Thank you in advance!

答:
Let $g(t)=\inf \{x f(x): |x|\ge t\}$. This is an increasing function with value in ${\mathbb R}\cup \{-\infty\}$ and we have $\lim_{t\to\infty} g(t) = a$.
The condition that $0<a<+\infty$ is equivalent to the fact that $0<g(t_0)$ for some $t_0$ and $g$ is bounded from above by some constant $A$. This entails the two conditions which characterize the fact that $a \in (0, \infty)$

$\exists m > 0,\exists t_0 \in {\mathbb R}, \forall t, |t|\ge t_0 \Rightarrow t f(t)\ge m$

$\exists A\in{\mathbb R}, \exists t_n,  |t_n|\to \infty, t_n f(t_n)\le A$


问:
Question about finite/infinite intersection

$A= \bigcap_{i \in \mathbb{N}} (-1+\frac{1}{i}, \frac{1}{i})$

$B= \bigcap_{i=1}^{n} (-1+\frac{1}{i}, \frac{1}{i})$ with $n<\infty$

In case of $A,$ I think the intersection should be empty.

but I am unsure about $B$

the first few intervals would look like this

$i=1: (0,1)$

$i=2: (-1/2,1/2)$

$i=3: (-2/3, 1/3)$

$\vdots$

$i=n: (\frac{1-n}{n}, \frac{1}{n})$ $n<\infty$

so the value on the left would get smaller and 0 would be included in all intervals, but how about the right side?, the last value $\frac{1}{n}$ would be the smallest of all, could I just say the finite intersection would be $B=(0,\frac{1}{n})$?

答:
As $0 \not \in (-1 +\frac 11, \frac 11)= (0,1)$ we can't have $0$ in any of either the finite or infinite intersections.

If $r > 0$ then there is an $m > \frac 1r$ and $\frac 1m < r$ so $r \not \in (-1 +\frac 1m, \frac 1m)$ and so $r$ will not be in any intersection that includes an index $m > \frac 1r$.  As the infinite intersection includes that index, no $r>0$ will be in the infinite intersect.

As for whether $r$ is in the finite intersection, we'll get to that in a moment.

If $r < 0$.... then $r < 0 = -1 + \frac 11$ so $r \not \in (-1 +\frac 11, \frac 11)=(0,1)$ so $r$ is not in any of the intersections finite or infinite.

So there is no values that are in the infinite intersection.  So $\bigcap_{m\in \mathbb N}(-1+\frac 1m, \frac 1m) =\emptyset$.

And if $\bigcap_{m=1}^n (-1+\frac 1m, \frac 1m) \subset (0,\infty)$.

Okay.... back that that observation that if $m > \frac 1r$ then $r\not\in (-1+\frac 1m, \frac 1m)$.

Well..... let's step back.   If $0< r < \frac 1n$ then $r < \frac 1n < \frac 1{n-1} <....$ and $r < \frac 1m$ for all $\frac 1m$ (and, of course, $r > 0 \ge -1+\frac 1m$) so $r\in (-1+\frac 1m, \frac 1m)$ for all $m \le n$ so $r$ is in the finite intersection.

But if $r\ge \frac 1n$ then $r\not \in (-1+\frac 1n, \frac 1n)$ and $r$ is not in the finite intersection.

So $\bigcap_{m=1}^n (-1+\frac 1m, \frac 1m) = (0, \frac 1n)$.

=======

Here's a slicker way to view it.

Consider $r \in \cap_{\alpha \in A} (a_\alpha, b_\alpha)$.  Then $r $ is in every interval $(a_\alpha, b_\alpha)$ so $r > \max a_\alpha$ if a maximum element exists and $r \ge \sup \max a_\alpha$ if a maximum element doesn't exist.

And the same reasoning to conclude $r < \min b_\alpha$ if a minimum element exist or $r \le \inf b_\alpha$ if it doesn't.

ANd thus the intersection can be determined by considering whatever interval can be made from $\max,\sup a_\alpha, \min,\inf b_\alpha$.

(If $\max,\min$ exist that side of the interval will be open and not include the $\max, \min$; otherwise it will be closed and will include the $\sup, \inf$; If $\max,\sup  a_\alpha > \min,\inf b_\alpha$ then the intersection will be empty; and if $\sup a_\alpha = \inf b_\alpha$ it will be the single point.

So $\max_{m\in \mathbb N} (-1+\frac 1m)=\max_{m\le n}(-1+\frac 1m) = (-1+\frac 11)=0$.

And $\min_{m< n}\frac 1m = \frac 1n$. So $\cap_{m=1}^n (-1+\frac 1m,\frac 1m) =(0,\frac 1n)$.

And $\min_{m\in \mathbb N}\frac 1m$ does not exist and $\inf_{m\in \mathbb N}\frac 1m = 0$.

So if $r\in \bigcap_{m\in \mathbb N}(-1+\frac 1m,\frac 1m)$ then $r > \max -1 + \frac 1m = 0$ and $r \le \inf \frac 1m = 0$.  So $0 < r \le 0$ which is impossible.  So  $ \bigcap_{m\in \mathbb N}(-1+\frac 1m,\frac 1m)= \emptyset$.


问:
Finding integers $s$ and $t$ such that $s+t=\frac{\alpha}{b}+b$ and $st=\frac{\beta}{b}+b$ for integers $\alpha$, $\beta$, $b$
In my research I need to find integers $s$ and $t$ with the following properties (respect to $\alpha, \beta, b$) :
\begin{align}
s+t=\frac{\alpha}{b}+b\\
st=\frac{\beta}{b}+b,
\end{align}
where $\alpha, \beta, b\in\mathbb{Z}$ and $b\neq0$. Do there exist any integers $s$ and $t$ with the above properties?

Anyone can help me? Thanks in advance.

答:
$s,t$ are the roots of
$$ X^2-(\tfrac\alpha b+b)X+\tfrac\beta b+b$$
and the formula to find these is well-known:
$$s,t=\frac{\tfrac\alpha b+b\pm\sqrt{(\tfrac\alpha b+b)^2-4(\tfrac\beta b+b)}}{2}
=\frac{\alpha+b^2\pm\sqrt{(\alpha +b^2)^2-4(\beta b+b^3)}}{2b}
$$
and for these to be integers, the radicand at least needs to be a perfect square. (and even then, dividing by $2b$ might spoil it).


问:
Deciding Convergence of the following improper integral of $e^{3x}$
Let's consider the following improper integral:

$$\int_4^\infty e^{3x}dx$$

I was thinking that the integral would converge when compared to $e^x$. Why does this integral diverge? Do you compare it to $e^x$?

答:
To show it diverges formally, what if you calculate the integral $\int_4^b e^{3x} \, dx$? What happens when $b$ grows?

Intuitively, though, you can just draw the graph of $y = e^{3x}$. Is the area under the curve for the region to the right of $x = 4$ finite?


问:
Calculate the maximum of $\sum_{1\le i<j<k\le 100}a_ia_ja_k$ where $\sum a_i = 2007$.

Calculate the maximum of $$\sum_{1\le i<j<k\le 100}a_ia_ja_k$$ where $\sum a_i = 2007$.

This is a question that I've seen long before and not been able to solve it.

My initial thought is that the more even $a_i$ is, the larger the sum will be. Hence, when there are $93$ $20$s and $7$ $21$s, the sum will supposedly become the biggest.

But I got stuck on proving this point. I'd appreciate some help.

PS. Can we extend the problem, e.g. $100$ to $n\in\mathbb{N}_+$ and $2007$ to $k>n\land k\in\mathbb{N}_+$?

Body:
The value could reach infinity as we can take $a_{1}=-n ,a_{2}=-n , a_{3}=2007,a_{4}=2n$ and $a_{j}=0$ for all $j\geq 5$ then $$S=a_{1}a_{2}a_{3}+a_{1}a_{2}a_{4}+a_{1}a_{3}a_{4}+a_{2}a_{3}a_{4}=2007n^{2}+2n^{3}-8028n^{2}$$ Choosing $n\rightarrow \infty$
We have $$S\rightarrow \infty$$


问:
How to calculate this sum using orthonormal closed system?
Let $\left\{\phi_{n}\right\}_{n=1}^{\infty}$ be a closed orthonormal system of real valued
functions in the inner product space V of continuous piecewise continuously
differentiable functions on [0, 1] with the inner product $$
\langle f, g\rangle=f(0) \overline{g(0)}+\int_{0}^{1} f^{\prime}(t) \overline{g^{\prime}(t)} d t
$$
Calculate : $$
\sum_{n=1}^{\infty}\left(\alpha \phi_{n}(0)+\int_{0}^{a} \phi_{n}^{\prime}(t) d t\right)\left(\beta \phi_{n}(0)+\int_{0}^{b} \phi_{n}^{\prime}(t) d t\right)
$$
while $\alpha , \beta$ are complex numbers and $0\leq a,b \leq1$

My idea was to open the integrals values in the sum and get $$\int_{0}^{a} \phi_{n}^{\prime}(t) d t = \phi_{n}(a) - \phi_{n}(0)$$
$$\int_{0}^{b} \phi_{n}^{\prime}(t) d t = \phi_{n}(b) - \phi_{n}(0)$$
But it doesn't go well and I am losing the similarity for the inner product.
I thought on another idea define a piece wise smooth function which get the value 1 in some interval and $0$ otherwise but I am not sure how to perform this idea (if it is true)

答:
Hint: Consider the coordinates of the following functions:
$$f(x)=\left\{\matrix{ \alpha+x& \text{if }x<a\\
\alpha+a& \text{if }x\ge a}\right.\\
g(x)=\left\{\matrix{ \beta+x& \text{if }x<b\\
\beta+b& \text{if }x\ge b}\right.
$$


问:
Hunt for trapezoid area & arithmetic progression sum formula similarities.
When I was looking into Arithmetic progression sum formula, I found out that it is similar to Trapezoid are formula.

Arithmetic progression sum formula

$S_n = \frac{(a_1+a_n)n}{2}$

Trapezoid area formula

$A = \frac{(a+b)h}{2}$

I was wondering: is there any connection between them? Maybe a geometric representation which I'm not able to image or is it just a nonrelevant imagination?

答: Sure:  Look at this figure and see if you can reason it out:


问:
How and why does $\int\frac{m^kx^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx$ become $(\frac{m}{m-s})^k\int\frac{(m-s)^k x^{k-1}e^{-(m-s)x}}{(k-1)!}dx$?
How does
$$\int\frac{m^kx^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx$$
become
$$\left( \frac{m}{m-s}\right)^k \int \frac{(m-s)^k x^{k-1}  e^{-(m-s)x}}{(k-1)!} dx $$

I understand how they collected the like terms with the exponent and that they took the $m^k$ outside the integral at the start, but have absolutely no idea where and why the $$ (m-s)^k$$ came from, both in the denominator outside the integral and in the numerator inside the integral.

Can explain what they did there??

Note, the integral is from $0$ to infinity which equals 1, not sure if that's important.

答:
Remember that for any integral with integrand $f(x)$ and where $a$ is a constant,
$$\int af(x)~ {dx}=a\int f(x)~dx$$
In your case, as
$$\begin{align}\int\frac{m^kx^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx&=\int m^k\times\frac{x^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\\
&=\int m^k\times\frac{(m-s)^k}{(m-s)^k}\times\frac{x^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\\
&=\int \frac{m^k}{(m-s)^k}\times(m-s)^k\times\frac{x^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\\
&=\int \left(\frac{m}{m-s}\right)^k\times(m-s)^k\times\frac{x^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\\
&=\left(\frac{m}{m-s}\right)^k\int (m-s)^k\times\frac{x^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\\
&=\left(\frac{m}{m-s}\right)^k\int \frac{(m-s)^kx^{k-1}e^{-mx}e^{sx}}{(k-1)!} dx\end{align}$$

Does that help? Please tell me if you don't understand something.


答:
Hint:

For functions $f,g \in V$ we have
$$\sum_{n=1}^\infty \langle f, \phi_n\rangle \langle \phi_n,g\rangle = \langle f,g\rangle.$$

Title: How to calculate this sum using orthonormal closed system?

Body:
Let $\left\{\phi_{n}\right\}_{n=1}^{\infty}$ be a closed orthonormal system of real valued
functions in the inner product space V of continuous piecewise continuously
differentiable functions on [0, 1] with the inner product $$
\langle f, g\rangle=f(0) \overline{g(0)}+\int_{0}^{1} f^{\prime}(t) \overline{g^{\prime}(t)} d t
$$
Calculate : $$
\sum_{n=1}^{\infty}\left(\alpha \phi_{n}(0)+\int_{0}^{a} \phi_{n}^{\prime}(t) d t\right)\left(\beta \phi_{n}(0)+\int_{0}^{b} \phi_{n}^{\prime}(t) d t\right)
$$
while $\alpha , \beta$ are complex numbers and $0\leq a,b \leq1$

My idea was to open the integrals values in the sum and get $$\int_{0}^{a} \phi_{n}^{\prime}(t) d t = \phi_{n}(a) - \phi_{n}(0)$$
$$\int_{0}^{b} \phi_{n}^{\prime}(t) d t = \phi_{n}(b) - \phi_{n}(0)$$
But it doesn't go well and I am losing the similarity for the inner product.
I thought on another idea define a piece wise smooth function which get the value 1 in some interval and $0$ otherwise but I am not sure how to perform this idea (if it is true)


问:
Does $f(x)=x^{2}\sin(\frac{1}{x^{2}})$ satisfy the relation $f(x)+f(y)-2f(\frac{x+y}{2})=O(|x-y|^{2})$?
Does $f(x)=x^{2}\sin(\frac{1}{x^{2}})$ satisfy the relation $f(x)+f(y)-2f(\frac{x+y}{2})=O(|x-y|^{2})$?

I can't check it. Who will hint it? Please.

答:
Hint: $\displaystyle\sin(x)\approx x-\frac{x^3}{6}$ so $$f(x)\approx x^2\left(\frac{1}{x^2}-\frac{1}{6x^6}\right)=1-\frac{1}{6x^4}$$

(this is assuming $x$ is meant to be large)


问:
Proof involving consecutive coefficients

If $a_1,a_2,a_3,a_4$ are four consecutive coefficients in the expansion of $(1+x)^n$, prove that $$\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{2a_{2}}{a_{2}+a_{3}}$$

My solution:

Let $a_1 x^{r-1},a_2 x^{r},a_3 x^{r+1},a_4 x^{r+2} = {n\choose{r-1}} x^{r-1},{n\choose{r}} x^{r},{n\choose{r+1}} x^{r+1},{n\choose{r+2}} x^{r+2}$.

I sub this into the required proof and I have to do some laboured, long algebra process.

From my experience, someone on MSE will have a much more concise proof than this, so I am asking to enlighten myself.

答:
This is just another variation where we stick at binomial coefficients avoiding factorials. We recall two binomial identities:
\begin{align*}
\binom{n+1}{r} &= \binom{n}{r-1}+\binom{n}{r}\tag{1}\\
\binom{n+1}{r}&=\binom{n}{r-1}\frac{n+1}{r}\tag{2}
\end{align*}

Left-hand side:
\begin{align*}
\color{blue}{\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}}
&=\frac{\binom{n}{r-1}}{\binom{n}{r-1} + \binom{n}{r}}+\frac{\binom{n}{r+1}}{\binom{n}{r+1} + \binom{n}{r+2}}\\
&=\frac{\binom{n}{r-1}}{\binom{n+1}{r}}+\frac{\binom{n}{r+1}}{\binom{n+1}{r+2}}\tag{$\to (1)$}\\
&=\frac{\binom{n}{r-1}}{\binom{n}{r-1}\frac{n+1}{r}}+\frac{\binom{n}{r+1}}{\binom{n}{r+1}\frac{n+1}{r+2}}\tag{$\to (2)$}\\
&=\frac{r}{n+1}+\frac{r+2}{n+1}\\
&\,\,\color{blue}{=\frac{2(r+1)}{n+1}}
\end{align*}
Right-hand side:
\begin{align*}
\color{blue}{\frac{2a_{2}}{a_{2}+a_{3}}}&=\frac{2\binom{n}{r}}{\binom{n}{r}+\binom{n}{r+1}}
=\frac{2\binom{n}{r}}{\binom{n+1}{r+1}}\tag{$\to (1)$}\\
&=\frac{2\binom{n}{r}}{\binom{n}{r}\frac{n+1}{r+1}}\tag{$\to (2)$}\\
&\,\,\color{blue}{=\frac{2(r+1)}{n+1}}
\end{align*}
and the claim follows.


问:
Where is a mistake? Problem related to the Riemann zeta function.
Let
$$
\xi\left(z\right) = z\left(z - 1\right)
\pi^{-z/2}\,\,\Gamma\left(\frac{z}{2}\right)\zeta\left(z\right),
$$ where $\zeta(z)$ is the Riemann zeta function.

After some manipulations we may write:
$$\xi(z)=\int_{-\infty}^{+\infty}\Phi(u)\exp(izu)du$$
Where $\Phi(u)=2e^{2u}\{3\psi'(e^{2u})+2e^{2u}\psi''(e^{2u})\}e^{\frac{u}{2}}$

and $\psi(x)=\sum_{n=1}^{\infty}\exp(-\pi n^2x)$

And this is well known result.

Here is what i have done. Let $F(u)=\psi(e^{2u})$, then after few calculations we get: $\Phi(u)=(F'(u)+F''(u))e^{\frac{u}{2}}$.

Lets consider a Fourier transform of $F$. This is:
$$W(z):=\int_{-\infty}^{+\infty}F(u)\exp(izu)du$$

Again after some calculation and using properties of Fourier transform, we get:
$$\xi(z-\frac{1}{2i}) = \int_{-\infty}^{+\infty}(F'(u)+F''(u))e^{\frac{u}{2}}\exp(i(z-\frac{1}{2i})u)du=$$
$$=\int_{-\infty}^{+\infty}(F'(u)+F''(u))\exp(izu)=izW(z)+(iz)^2W(z)$$

This implies that $z= +\frac{1}{2i}$ and $z=i+\frac{1}{2i}$ are zeros of function $\xi(z)$

The question:

Where is a mistake?

Thank you in advance

Regards.

答:
Let $z=w-i/2$ then by your calculation,
$$\xi(w)=i(w-i/2)W(w-i/2)-(w-i/2)^2W(w-i/2),$$
and as you say it follows that
$$\xi(-i/2)=i(-i)W(-i)-(-i)^2W(-i)=W(-i)-W(-i)=0$$
and
$$\xi(i-i/2)=i(i-i)W(i-i)-(i-i)^2W(i-i)=0,$$
so algebraically $-i/2$ and $i-i/2$ do appear to be roots of $\xi$.

Maybe the problem lies in the convergence of the integral $W(z)$. For example,
$$W(-i)=\int_{-\infty}^\infty\sum_{n=1}^\infty\exp(-\pi n^2 e^{2u})\exp(u)du=\int_{-\infty}^\infty\frac{1}{2} e^u \left(\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)-1\right)du,$$ where $\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)$ is an Elliptic Theta function. Plotting this integrand (below) over $\mathbb{R}$ indicates that the integral may diverge. I guess you'd need to prove that $\frac{1}{2} e^u \left(\vartheta _3\left(0,e^{-e^{2 u} \pi }\right)-1\right)\to1/2$ as $u\to-\infty$.


问:
Proof that the canonical mapping of $V$ into $V^{**}$ is not bijective
Let $V$ be an infinite vector space over a skew field $K$ and $V^{**}$ its bidual. Write $c_V$ for the canonical injection of $V$ into $V^{**}$. I want to show that $c_V$ is not surjective.

We can restrict to the case where $V:= K_s^{(L)}$ for some infinite set $L$. In that case there is an isomorphism of $\varphi$ of $V^*$ onto $K^L$. Let $(a_\lambda)_{\lambda\in L}$ be the canonical basis of $V$ and $(a^*_\lambda)_{\lambda\in L}$ be the corresponding family of coordinate forms in $V^*$. Let $F'$ be the span of the family $(a^*_\lambda)_{\lambda\in L}$ in $V^*$. Then $V^*\ne F'$, otherwise $K^{(L)}_d=\varphi(F')=\varphi(V^*)=K^L$. This implies that there exists a hyperplane $H'$ of $V^*$ such that $F'\subset H'$. Clearly $H'\ne V^*$; it follows that $(V^*/H')^*\ne0$ and $(V^*/H')^*\cong H''$ where $H''$ is the orthogonal of $H'$ in $V^{**}$. Write $F$ for the orthogonal of $F'$ in $V$. At this point the author says that
$$(H''\cap c_V(V))\subset c_V(F)=0.$$
I don't understand why $c_V(F)=0$? By defintion
$$F:=\{x\in V\ |\ (\forall x^*)(x^*\in F'\implies\langle x,x^*\rangle=0)\}.$$
Take $x\in F$ and $x^*\in V^{*}$. Why would $c_V(x)(x^*)=\langle x,x^*\rangle=0$?

答: This is just because $F$ itself is $0$.  Indeed, if $x\in F$, then in particular $\langle x,a_\lambda^*\rangle=0$ for each $\lambda$, which means every coordinate of $x$ is $0$ so $x$ is $0$.


问:
Dirac delta in polar coordinates
Given

$$x=r\,\cos\theta\\y=r\,\sin\theta$$ and
$$x'=r'\,\cos\theta'\\y'=r'\,\sin\theta'$$

how can I express

$$\delta(x'-x)\delta(y'-y)$$ in terms of the polar coordinates?

And the more general case:

$$\delta(x'-x-a)\delta(y'-y-b)$$

答:
By defintion the dirac delta function should satisfy the following condition.

$\int\limits_{-\infty} ^\infty \delta(\bar x - \bar x_{0}) \bar dx = 1$ . Now in polar coordinates $\bar dx = rdr d\theta$ which makes our integral $\int\limits_{0} ^\infty \int\limits_{0} ^{2\pi} \delta(\bar x- \bar x_{0}) rdrd\theta = 1$ For this integral to satisfy the defintion: $\delta(\bar x-\bar x_{0}) = \frac{1}{r} \delta(r-r_{0})\delta(\theta -\theta_{0})$. fixed it now

Title: Dirac delta in polar coordinates


问:
Solve $x - \ln(x) > 1$ without using derivatives
I would like to solve $x - \ln(x) > 1$ without studying the function $f(x) = x - \ln(x) - 1$ and computing its derivative.

I wonder if solving for $x$ in $x - \ln(x) = 1$ is possible.

It could be possible to consider the Lambert $W$-function, since we can rewrite the expression such that we have

\begin{align*}
  \ln(x) - x   &= 1  \\
  \ln(xe^{-x}) &= 1  \\
  xe^{-x}      &= e  \\
  -xe^{-x}     &= -e
\end{align*}

Yet it seems impossible and non-elementary to have a "nice" analytical solution to this.

答:
I think you made a mistake in your solution
\begin{align*}
  x-\ln(x)  &>1  \\
  \ln\left(\frac{e^x}{x}\right) &> 1  \\
  \frac{e^x}{x}    &= e  \\
  e^x    &= xe
\end{align*}
So, it has only one solution which is $x=1.$ I hope that helps


问:
$\arccos(1/2)$ products
I'm finding points of intersection between two functions, $y = \tan(x)$ and $y = 2\sin(x)$, which are bound between $-\pi/3$ and $\pi/3$.

Solving for the intersections points, I end up performing $x = \arccos\left(\frac12\right)$

I know this produces: $x = \pi/3$, but should it rather produce $x = \pm\pi/3$ ?

This seems to make sense as visually the two functions intersect at both $-\pi/3$ and $\pi/3$, but I want to verify if this is correct.

I'm usually on Stack Overflow. Let me know if any notation here is wrong. Additional apologies if this question is very low level, I haven't worked with trig in a while.

答:
Note that\begin{align}\tan(x)=2\sin(x)&\iff\frac{\sin(x)}{\cos(x)}=2\sin(x)\\&\iff\sin(x)=0\vee\cos(x)=\frac12.\end{align}And$$\cos(x)=\frac12\iff x=\frac\pi3+2n\pi\vee x=-\frac\pi3+2n\pi,$$with $n\in\Bbb Z$.

In particular, although $\arccos\left(\frac12\right)\left(=\frac\pi3\right)$ is indeed a solution of the equation $\cos(x)=\frac12$, it is not the only one.

Title: $\arccos(1/2)$ products


Title: $\arccos(1/2)$ products

Body:
I'm finding points of intersection between two functions, $y = \tan(x)$ and $y = 2\sin(x)$, which are bound between $-\pi/3$ and $\pi/3$.

Solving for the intersections points, I end up performing $x = \arccos\left(\frac12\right)$

I know this produces: $x = \pi/3$, but should it rather produce $x = \pm\pi/3$ ?

This seems to make sense as visually the two functions intersect at both $-\pi/3$ and $\pi/3$, but I want to verify if this is correct.

I'm usually on Stack Overflow. Let me know if any notation here is wrong. Additional apologies if this question is very low level, I haven't worked with trig in a while.

Body:
Let $ x\in[-\frac{\pi}{3},\frac{\pi}{3}] $.

$$\tan(x)=2\sin(x) \iff$$

$$\sin(x)(1-2\cos(x))=0\iff$$

$$\cos(x)=\frac 12 =\cos(\frac{\pi}{3})\iff $$

$$x=\pm \frac{\pi}{3}+2k\pi$$

but, if $ k\ne 0$, then $ x $ is out of the domain $ [-\frac{\pi}{3},\frac{\pi}{3}]$.
So, $ k=0 $ and$$x=\pm \frac{\pi}{3}$$


答:
The Cosine function is positive is Quadrants I and IV. Therefore, given $\cos(x)=\dfrac{1}{2}$, $x=\dfrac{\pi}{3}$ is the solution from Quadrant I and $x=-\dfrac{\pi}{3}$ (or $x=\dfrac{5\pi}{3}$) is the solution from Quadrant IV.

In high school, I remembered this using the acronym "All Students Take Calculus".

Title: $\arccos(1/2)$ products


问:
How to solve a second order partial differential equation
I am trying to solve this partial differential equation: $$\frac{\partial^{2}}{\partial x^{2}}y(x,t)-\frac{\partial^{2}}{\partial t^{2}}y(x,t)=\frac{A}{t^{2}}y(x,t)$$
With $A \in \mathbb{R}$. I would like to find the more general analytical solution to this with unfixed integration constants. I am not a mathematician so I am not even sure if this is possible. If it were not: is it possible to prove that such a solution exists? Thank you for the help!

(As I said above: I'm not a mathematician, so be kind!)

答:
$$\frac{\partial^{2}}{\partial x^{2}}y(x,t)-\frac{\partial^{2}}{\partial t^{2}}y(x,t)=\frac{A}{t^{2}}y(x,t)$$
An infinity of particular solutions can be found thanks to the separation of variables method. We look for particular solutions on the form $y(x,t)=X(x)T(t)$ where $X(x)$ and $T(t)$ are unknown functions.
$$X''T-XT''=\frac{A}{t^{2}}XT$$
$$\frac{X''}{X}-\frac{T''}{T}=\frac{A}{t^{2}}$$
$$\frac{X''(x)}{X(x)}=\frac{T''(t)}{T(t)}+\frac{A}{t^{2}}=z^2$$
$z$ is an arbitrary complex (or real) constant.

A function of $x$ and a function of $t$ can be equal any $x,t$ only if both are equal to a common constant.
$$X''(x)-z^2X(x)=0\quad\implies\quad X(x)=e^{\pm z\;x}$$
This includes a lot of functions of mixed exponential and sinusoidal functions.
$$T''(t)+\left(\frac{A}{t^2}-z^2\right)T(t)=0$$
This is an ODE of Bessel kind which solution is :
$$T(t)=c_1\sqrt{t}\:I_\nu(z\;t)+c_2\sqrt{t}\:K_\nu(z\;t)\quad ; \quad \nu=\sqrt{1-4A}$$
$I$ and $K$ denote the modified Bessel functions.

Now a lot of particular solutions are known with arbitrary complex or real constants $z,c_1,c_2$ :

$$y=e^{\pm z\;x}\left(c_1\sqrt{t}\:I_\nu(z\;t)+c_2\sqrt{t}\:K_\nu(z\;t \right)$$

A more wide solution is made of any linear combination of those particular solutions :
$$y(x,t)=\sqrt{t}\int f(z)e^{\pm z\;x}\:I_\nu(z\;t)dz+\sqrt{t}\int g(z)e^{\pm z\;x}\:K_\nu(z\;t)dz$$
$f(z)$ and $g(z)$ are arbitrary functions.

If some initial condition is specified one can expect to determine the functions $f,g$. In general this is a very difficult task far to be always possible. However, if you are lucky with the initial condition, it is possible but rarely that the solution be much simpler than the above equation.


问:
The forgetful functor $U: \text{Con}(\mathcal C) \to \mathcal C$ has a left adjoint.
Let $\mathcal C$ a category. We define the category of cones $\text{Con}(\mathcal C)$ in the following way:

Objects: quadruples $(\mathcal Z, \underline{M}, M, m)$ consisting of a small category $\mathcal Z$, a functor $\underline{M}: \mathcal Z \to \mathcal C$ and a cone $(M, m)$ of $\underline{M}$ (i.e. $M$ is an object in $\mathcal C$ and $m = (m_Z)_{Z \in \mathcal Z}$ is a collection of morphism $m_Z: M \to \underline{M}(Z)$ satisfying a compatibility condition) .

Morphisms: a morphism $\underline{\phi}: (\mathcal Z, \underline{M}, M, m) \to (\mathcal Y, \underline{N}, N, n)$ is a triple $\underline{\phi}:( \phi^0, \phi^1, \phi^2)$ where $\phi^0: \mathcal Y \to \mathcal Z$ is a functor, $\phi^1: M \to N$ is a morphism in $\mathcal C$ and $\phi^2: \underline{M}\circ \phi^0 \to \underline{N}$ is a natural transformation such that $\phi^2_Y \circ m_{\phi^0(Y)} = n_Y \circ \phi^1$ for all $Y \in \mathcal Y$.

I am trying to show that the forgetful functor $U: \text{Con}(\mathcal C) \to \mathcal C$ has a left adjoint, i.e. there is a functor $S : \mathcal C \to \text{Con}(\mathcal C)$ such that
$$\text{Hom}_{\text{Con}(\mathcal C)}(S(X), Y) \cong \text{Hom}_\mathcal C (X, U(Y)).$$
The more natural way to associate an element $C \in \mathcal C$ to a quadruple in $\text{Con}(\mathcal C)$ is to send $C$ to $(\varnothing, F_\varnothing, C, m)$ where $F_\varnothing$ is the unique functor that goes from $\varnothing$ to $\mathcal C$ and $(C, m)$ is a cone for $F_\varnothing$ (a cone for this functor is just an element, there is no need to consider such an $m$).

This seems to be a natural way to define the functor $S:\mathcal C \to \text{Con}(\mathcal C)$ as I did above but I do not really see how it goes for the morphism. If we consider a morphism $\underline{f} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), Y)$ (for $S(C)$ construct above and $Y = (\mathcal Y, \underline{N}, N, n)$ a quadruple in $\text{Con}(\mathcal C)$), $\underline{f}$ should be a triple $\underline{f} = (f^0, f^1, f^2)$ with $f^0: \mathcal Y \to \varnothing$ and $f^3: F_\varnothing \circ f^0 \to \underline{N}$. The problem is that $F_\varnothing \circ f^0 : \mathcal Y \to \mathcal C$ factors trough the empty set and therefore it has no image, $F_\varnothing \circ f^0$ is not even a functor.

My questions are the following: Is it really possible to consider a functor as $f^0$ such that $f^0: \mathcal Y \to \varnothing$ ? Is it the right way to construct the left adjoint $S$ ? I also tried to construct $S$ by sending $C$ to an other quadruble where the associate $\mathcal Z$ is not the empty set but it seems that in those cases there are many possiblities which are all very arbitrary (there is a unique functor $\varnothing \to \mathcal C$ but there are a lots a functor from $\mathcal Z = \underline{1} \to \mathcal C$, where $\underline{1}$ is the category with a unique element, for example).

答:
Your idea to take the cone over an empty category to represent objects is basically good, however, as was noted in the comments, a morphism from a cone over shape $\mathcal Z$ to a cone over $\mathcal Y$ involves a contravariant part, namely the functor $\phi^0:\mathcal Y\to\mathcal Z$.

This is because the hom functor (of the category of categories) is contravariant in the first argument.

So, instead of the initial category we need to consider the terminal category: the discrete category $\bf 1$ with one object $\ast$.

And, then for an object $X\in\mathcal C$, define $S(X):=({\bf 1},\ \ast\mapsto X,\ X,\ 1_X)$, and verify the claim.


问:
What is the point in showing that $u\cdot v=u \cdot v_u$ in linear algebra when talking about the dot product?
I am currently taking a course in linear algebra and in my linear algebra book at the chapter about scalar product / dot product it says

"Assume that $u=AB$ and $v=AC$. We now have that $|AC|cos(\theta)=|AD|$ and therefore $u\cdot v=|AB||AC|cos(\theta)=|AB||AD|$".

The figure that illustrates this is a line $AB$ that is visibly longer than a line $AC$. The line $AC$ going above the line $AB$ at an angle of what looks like about 35 degrees. Then there is a dotted line from the end of $AC$ straight down, creating a triangle at point $D$ which is between the points $A$ and $B$.

What is the point of showing that $u\cdot v = u \cdot v_u$? This is clearly $v=v_u$.

答:
In general, when you have
$$
x\cdot y=x\cdot z
$$
where $x,y,z$ are three vectors (say in the Euclidean space),
it does not follow that $y=z$.

This is because $x\cdot w=0$ does not imply "$x=0$ or $w=0$".


问:
Computing probabilities in a game of bridge
In the game of bridge, a player is given 13 cards from a deck of 52 cards, what is the probability that he/she gets exactly one king and one queen? Furthermore what is the expected value of the number of aces he/she gets?</p>

答:
Hypergeometric distribution

Probability of one king and one queen:
$$\frac{\binom{4}{1}\binom{4}{1}\binom{44}{11}}{\binom{52}{13}}=\frac{36556}{189175}$$

Expected number of aces:
$$\sum_{k=0}^4 k \frac{\binom{4}{k}\binom{48}{13-k}}{\binom{52}{13}} = 4 \cdot \frac{13}{52}=1$$


问:
Finitely number natural $n$ such that $2^{n}=xy-1$ with $x,y$ primes number and $x\not=y$

If $x\not=y$ with $x,y$ primes number. Are there finitely many $n\in \mathbb{N}$ such that
$2^n=xy-1$?

答:
If $n=2k+1$ , then we can write:

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

So x or y can always be equal to  3. Since n is odd the number of terms of polynomial is odd and can not have factor 3 but itself can be prime.For example for n=5 we get:

$xy=32+1=33=3\times 11$

Or for n=7 we get:

$xy=2^7+1=129=3\times 43$

And so on.


问:
Solid of revolution axis $y=5$
The problem goes like this

Rotate the indicated area around the given axis to calculate the volume of the solid of revolution $y=x^2+1$, $x=0$, $x=2$, $y=0$ , around the axis $ y = 5$.

My question is if the solution is given by the following integral (should I take the limits of integration from $0$ to $2$?)

$$2\pi\int_0^2 ((5-y)\sqrt{y-1})\,dy$$

答:
Be careful that you are calculating "volume", not "area".

The volume is given by
$$
\int_0^2 A(x)\,dx
$$

where
$$
A(x)=\pi[5^2-(5-f(x))^2]^2
$$
with $f(x)=x^2+1$.

Title: Solid of revolution axis $y=5$


问:
If $f \in \mathbf{C}^n(\mathbf{R})$ and $f(x_{0})=...=f(x_{n}) = 0$ for $x_{0} < ... < x_{n}$ then $f^{n}(\xi)=0$ for some $\xi \in (x_{0},x_{n})$
I was wondering if someone could look at my proof and tell me if this is sufficient enough.

Because $f$ is continuous on $\mathbf{R}$ and $f(x_{0})=f(x_{1})=0$ for $x_{0} < x_{1}$, we can apply rolle's theorem to show that $f'(a)=0$ for some $a \in (x_{0},x_{1})$. Similarly, we can apply rolle's theorem again since $f$ is continuous on $\mathbf{R}$ and $f(x_{1}) = f(x_{2})$ for $x_{1} < x_{2}$ to show that there exists a $f'(b)$ for some $b \in (x_{1},x_{2})$. Since $f'(a)=0$ for some $a \in (x_{0},x_{1})$ and $f'(b) = 0$ for some $b \in (x_{1},x_{2})$, we have $f''(c)=0$ for some $c \in (x_{0},x_{2})$ by rolle's theorem again. By applying rolle's theorem continuously n times, we have $f^{n}(\xi) = 0$ for some $\xi \in (x_{0},x_{n})$.

Thank you!

答:
You are using the correct tool. What remains is to write the answer in a manner that cannot be faulted by your enemy. Ideally, you try to make it clear to the reader what you are about to do and then you do it. Afterwards you should draw conclusions. There is no shame in writing in a mechanical style. The use of fixed terms and phrases allows the reader to concentrate on the mathematics.

In general, if $g : \mathbb{R} \rightarrow \mathbb{R}$ is differentiable and has at least $n+1$ distinct zeros, then $g'$ has at least $n$ distinct zeros that are interlaced with the zeros of $g$ much like the teeth of zipper.

Now let $a = \min x_j$ and let $b = \max x_j$. Let $S_n = \{1,2,3,\dotsc,n\} \subset \mathbb{N}$ and let $V \subseteq S_n$ be given by
$$ V = \{ k \in S_n \: | \: \text{$f^{(k)\:}$ has at least $n+1-k$ distinct zeros in $[a,b]$} \}.$$
We will now show that $V = S_n$ using the well-ordering principle. By assumption, $f$ has $n+1$ distinct zeros in $[a,b]$. By Rolle's theorem it follows that $f'$ has at least $n$ distinct zeros in $[a,b]$. This shows that $1 \in V$. Now assume that $V \not = S_n$. It follows that $S_n \setminus V \subseteq \mathbb{N}$ is not the empty set. Therefore, the well-ordering principle implies that $S_n \setminus V$ has a smallest element $k$. Since $1 \in V$, we must have $2 \leq k$. Now consider $k-1$. We must have $k-1 \in V$ or $k$ is not the smallest element of $S_n \setminus V$. This shows that $f^{(k-1)}$ has at least $n+2-k$ distinct zeros in $[a,b]$. By Rolle's theorem, $f^{(k)}$ has at least $n+1-k$ distinct zeros in $[a,b]$. This shows that $k \in V$. This is a contradiction because $k \in S_n \setminus V$. We conclude that $V = S_n$. In particular, $f^{(n)}$ has at least one zero in $[a,b]$ because $n \in V$. This completes the proof.


答:
Translate the parabola and the $x$-axis $5$ units down

The area to rotate is limited by

$y=x^2-4,\;y=-5;\;y=0$

Apply the formula here
$$V=\pi  \int_0^2 \left((-5)^2-\left(x^2-4\right)^2\right) \, dx=\frac{494 \pi }{15}$$

Title: Solid of revolution axis $y=5$

Body:
The problem goes like this

Rotate the indicated area around the given axis to calculate the volume of the solid of revolution $y=x^2+1$, $x=0$, $x=2$, $y=0$ , around the axis $ y = 5$.

My question is if the solution is given by the following integral (should I take the limits of integration from $0$ to $2$?)

$$2\pi\int_0^2 ((5-y)\sqrt{y-1})\,dy$$


问:
inverse trigonometry ranges
$$\arctan(x)+\arctan(y)=a+b\left(\arctan\frac{x+y}{1-xy}\right)$$

Find sum of all distinct possible values of $a$ and $b$.
Also when we plot this graph and keep on changing the values of $a$ and $b$ we get different curves for different values.Is there any deeper insight for that?

答:
If you do mean
$$\arctan(x)+\arctan(y)=a+b\left(\arctan\frac{x+y}{1-xy}\right)$$
then this is easily solved. For $\lvert xy\rvert<1$, we have
$$\arctan(x)+\arctan(y)=\arctan\frac{x+y}{1-xy}$$
However, if $\lvert xy\rvert\not<1 $ then either
$$\arctan(x)+\arctan(y)=\pi+\arctan\frac{x+y}{1-xy}$$
or
$$\arctan(x)+\arctan(y)=-\pi+\arctan\frac{x+y}{1-xy}$$
Can you solve it now?

See here for more details.

Title: inverse trigonometry ranges


问:
What is wrong with this procedure-writing Angular momentum operator in spherical coordinates
I am trying to write the first component of the quantum angular momentum operator  $L_1$ in spherical coordinates.
$$L_1=x_2p_3-x_3p_2=-i\hbar (x_2\dfrac{\partial}{\partial x_3}-x_3\dfrac{\partial}{\partial x_2})=$$

The defining equations of the spherical coordinates are:
$x_1=r \sin\theta \cos\varphi,x_2=r\sin\theta\sin\varphi, x_3=r\cos\theta\tag{1}$
Then using the chain rule:

\begin{align}
L_1
&=-i\hbar (x_2\dfrac{\partial}{\partial x_3}-x_3\dfrac{\partial}{\partial x_2})\\
&=-i\hbar \Big[r\sin\theta\sin\varphi\left(
\dfrac{\partial r}{\partial x_3}\dfrac{\partial}{\partial r}
+\dfrac{\partial \theta}{\partial x_3}\dfrac{\partial}{\partial \theta}
+\dfrac{\partial \varphi}{\partial x_3}\dfrac{\partial}{\partial \varphi}\right)\\
&\qquad\qquad\; -r\cos\theta\ \left(
\dfrac{\partial r}{\partial x_2}\dfrac{\partial}{\partial r}
+\dfrac{\partial \theta}{\partial x_2}\dfrac{\partial}{\partial \theta}
+\dfrac{\partial \varphi}{\partial x_2}\dfrac{\partial}{\partial \varphi}\right)
\Big] \tag{2}
\end{align}

Using (1):

$\dfrac {\partial r}{\partial x_3}=\dfrac {1}{\frac{\partial x_3}{\partial r}}=\dfrac {1}{\cos \theta}$

$\dfrac {\partial \theta}{\partial x_3}=\dfrac {1}{\frac{\partial x_3}{\partial \theta}}=\dfrac {-1}{r \sin \theta}$

$\dfrac {\partial \varphi}{\partial x_3}=0$

$\dfrac {\partial r}{\partial x_2}=\dfrac {1}{\frac{\partial x_2}{\partial r}}=\dfrac {1}{\sin \theta \sin \varphi}$

$\dfrac {\partial \theta}{\partial x_2}=\dfrac {1}{\frac{\partial x_2}{\partial \theta}}=\dfrac {1}{r \cos \theta \sin \varphi}$

$\dfrac {\partial \varphi}{\partial x_2}=\dfrac {1}{\frac{\partial x_2}{\partial \varphi}}=\dfrac {1}{r \sin \theta \cos \varphi}$

Plugging these results in (2):
$$L_1=-i\hbar [r\tan\theta\sin\varphi
\dfrac{\partial}{\partial r}
-\sin \varphi\dfrac{\partial}{\partial \theta}
-r \dfrac{\cot\theta}{\sin \varphi}\dfrac{\partial}{\partial r}
-\dfrac{1}{\sin \varphi}\dfrac{\partial}{\partial \theta}
-\dfrac{\cot \theta}{ \cos \varphi}\dfrac{\partial}{\partial \varphi}
] \tag{3}$$

But the result should be
$$L_1=+i\hbar [\sin \varphi\dfrac{\partial}{\partial \theta}
+\cos \varphi \cot \theta \dfrac{\partial}{\partial \varphi}
] \tag{4}$$

What am I doing wrong?

答:
When enough variables are involved, partial derivatives need a disambiguation as to what is held constant in their definition, e.g. $\left(\tfrac{\partial u}{\partial v}\right)_w$ holds $w$ constant. In particular, $\left(\tfrac{\partial u}{\partial v}\right)\left(\tfrac{\partial v}{\partial u}\right)\ne1$ when we use different conventions in these derivatives.

A simpler version of your mistake is found in planar polar coordinates with $4$ variables $x,\,y,\,r,\,\theta$. For example, compare$$\left(\tfrac{\partial x}{\partial r}\right)_\theta=\left(\tfrac{\partial r\cos\theta}{\partial r}\right)_\theta=\cos\theta=\tfrac{x}{r}$$with$$\left(\tfrac{\partial r}{\partial x}\right)_y=\left(\tfrac{\partial\sqrt{x^2+y^2}}{\partial x}\right)_y=\tfrac{x}{\sqrt{x^2+y^2}}=\tfrac{x}{r}.$$These aren't reciprocals; in fact, they're equal. (Physics has another famous example of what you'd think would be reciprocals in fact being equal; it's the basis of a lot of relativity "paradoxes".)

The triple product rule is another important example of such "gotchas". If $f(x,\,y,\,z)=0$ then$$\prod_\text{cyc}\left(\tfrac{\partial x}{\partial y}\right)_z=-1$$(note the minus sign).


问:
two-member relationship R maths
I have $A=\{a,b,c,d\}$ and we have the  two-member relationship :

$R=\{(a,a),(a,c),(c,c),(c,b),(b,b),(a,d),(d,d),(b,d)\}$

is it $R$ partial order?

is any way  that I  can solve it and give me the right answer?

答:
A partial order has to satisfy (1) $aRa$ for all $a$; (2) if $aRb$ and $bRc$, then $aRc$; (3) for $a\ne b$, at most one of $aRb$ and $bRa$ holds.

Your R has $aRc$ and $cRb$ but not $aRb$, so it is not a partial order (in the jargon, it is not transitive).

Title: two-member relationship R maths


问:
Spivak Calculus 4th Ed, Prologue, Problem 19a: what is the motivation behind the assumptions to this problem?
This question in Spivak seems to be about using algebraic manipulations to show the Schwarz Inequality.

Now, this is the Prologue so I am guessing the objective here is just to practice a bit with manipulating some expressions and maybe gaining some insight on some very basic derivations. However, in the case of this particular problem, I don't understand what the insight is.

Here is a picture of the relevant problem:

And here is the solution from the solutions manual.

I am particularly interested in the last part of a), where it is assumed that there is no $\lambda$ such that $x_1=\lambda y_1$ or $x_2=\lambda y_2$. The problem gives you a quadratic in $\lambda$ to work with that is stated as larger than zero, so all you have to do is apply the formula $b^2-4ac<0$ to make determine the the relationship between the values of $x_1, y_1, x_2, y_2$ that make the quadratic inequality true. The relationship that you end up with is the given Schwarz Inequality.

Question is: what is the deal with the assumptions on $\lambda$ here? I understand the work to get to the answer, but what is the motivation for this problem after all, with regards to starting with this setup of $\lambda$?

By the way, here is what Problem 18 cited in Problem 19a looks like:

答:
Spivak's statement of the inequality is the two-dimensional version of
$$\left|v\cdot w\right|^2\le(v\cdot v)(w\cdot w)$$
where $v$ and $w$ are vectors in a vector space and $\cdot$ is a dot product on that vector space. (For more on vector spaces, see e.g. this page.) So, in two dimensions, if $v=(x_1,x_2)$ and $w=(y_1,y_2)$, their dot product is $x_1y_1+x_2y_2$, which we see in Spivak's problem. The condition with $\lambda$ has to do with whether $v$ and $w$ are parallel, so if that condition holds we can say $v=\lambda w$ giving the equality case, but otherwise we get a strict inequality. Viewing the givens $x_1,x_2,y_1,y_2$ as vectors may help highlight how rare this equality case actually is: Spivak asks you to show that in every other case, so long as $v$ and $w$ are not parallel and neither is the zero vector, the inequality is strict.


问:
KKT points and nonlinear optimality (theory)
I have some questions about Karush Kuhn Tucker conditions. I am not clear about the theory. There are plenty of theorems and I am confused with it. Let us say we have the following problem: We have a general problem, not a convex problem. $$\min f(x)\\g_{j}(x)\le 0,\forall j\in J\\h_{k}(x)=0,\forall k \in K. $$
I will solve it.....Lagrange function, verifying primal feasibility, dual feasibility and complementary slackness and I will get some KKT points. Now what is the conclusion?

If I get no KKT point, can I say that the problem has no solution or what is the conclusion?

If I get only one KKT point, is it a local minimum or even a global minimum?

If I get more KKT points, what is the conclusion?

In which case I will verify SOSC condition? If I get more KKT points or even for one point is it necessary? And the conclusion is what? The KKT points which satisfies the SOSC condition are strick local minimum and the poins which does not satisfy it are not the solution of the problem?

I am so sorry for so many questions, but any help would be very appreciated. Thank you very much in advance.

答:
In answer to your first question: You can't conclude anything without making additional assumptions about the regularity of the problem- these assumptions are called "constraint qualifications."

For example, the problem

$\min x_{2}$

subject to

$(x_{1}-1)^{2}+x_{2}^{2}=1$

$(x_{1}+1)^{2}+x_{2}^{2}=1$

has only one feasible solution at $x_{1}=0$, $x_{2}=0$, but this point doesn't satisfy the KKT conditions. Thus there are no KKT points but $x_{1}=0$, $x_{2}=0$ is the unique optimal solution to the problem.


问:
Let $V$ be an inner product space and $W$ be a dense subspace of $V$ then $W^{\perp }={0}$.
Let $V$ be an inner product space and $W$ be a dense subspace of $V$ .

Claim: $W^{\perp }={0}$.

Let non-zero $\ x\in W^{\perp} \implies (\forall w \in W,\ \langle x ,w\rangle=0)\ \implies W \subseteq x^{\perp}$ and $x^{\perp}$ is closed subspace.

So $W \subseteq x^{\perp} \implies  \overline{W} \subseteq \overline{x^{\perp}} \implies V \subseteq x^{\perp}$. This is a contradiction because this is possible only when $x=0$.

So $W^{\perp}=0.$

Is it okay and
also, is there any example of an inner product space $V$ and subspace  $W \leq V$, such that $W$ is not dense in $V$, but $W^{\perp}=0$?

答:
Your proof seems fine to me. Quick alternative proof: ${W}$ dense in ${V}$ means that every vector in ${V}$ can be written as the limit of ${w_n}$ for some sequence in ${W}$. Take some ${x \in W^{\perp}}$, and take some ${w \in V}$. Then take the sequence ${w_n \in W\ |\ w_n\to w}$. We have
$$
\langle w_n,x\rangle = 0\ \forall\ n \in \mathbb{N}
$$
by the continuity of the inner product,
$$
\Rightarrow \langle w,x\rangle = 0
$$
but the choice of $w$ was arbitrary. In particular, if we had ${w=x}$ we have that
$$
\langle x,x\rangle = 0
$$
which of course implies ${x=0}$. As required.....................


问:
A simple graph must have at least one pair of vertices whose degrees are equal.
I came across this proof while studying graph theory and although it instinctively makes sense, I have a problem with a part of it.

Proposition: A non-trivial simple graph $G$ must have at least one pair of vertices whose degrees are equal.

Proof:
Let $G$ be a graph with $n$ vertices. Because $G$ is a simple graph, then by definition it doesn't contain multi-edges or loops. Therefore, there appears to be $n$ possible degree values for any vertex, namely $0, ..., n-1$. However, there cannot be both a vertex of degree 0 and a vertex of degree $n - 1$, since the presence of a vertex of degree 0 implies that each of the remaining $n - 1$ vertices is adjacent to at most $n-2$ other vertices. Hence, the $n$ vertices of $G$ can realize at most $n − 1$ possible values for their degrees [This is the part where I have troubles]. Thus, the pigeonhole principle implies that at least two of the $n$ vertices have equal degree.

Isn't this proof based on assumption that there exists a vertex with degree $0$. What happens if that is not case? Does the proof still hold? And why can the $n$ vertices of the graph realize at most $n-1$ possible values? I don't understand that step.

答:
Consider two cases:

There is a vertex of degree $0$. Then the other degrees are in $0,1,\ldots, n-2$. So the degrees are $n$ elements of ${0,1,\ldots, n-2}$ which has size $n-1$. Hence two of the degrees coincide.

There is no vertex of degree $0$. Then the other degrees are in $1,\ldots, n-2, n-1$. So the degrees are $n$ elements of ${1,\ldots, n-1}$ which has size $n-1$. Hence two of the degrees coincide.


问:
$n-$section formulas in goniometry without the calculus
We know that the formulas of bisection of an angle $\alpha$ it is:

$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$

In particular

$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}, \qquad \cos\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}$$

My question is:

For the formulae of $n-$section of an angle $\alpha$, i.e.

$$\sin\left(\frac{\alpha}n\right), \qquad \cos\left(\frac{\alpha}n\right)$$

Do I must necessarily use the development in series of Taylor or the general formulas can be obtained with methods that do not use the derivatives, or the numerical methods?

My doubts arise for the students of an high school that not use the derivate.

答:
Well, with very large values of $n$, you will get
$$\sin(x)=\text{polynomial in $\sin\left(\frac{x}{n}\right)$ or $\cos\left(\frac{x}{n}\right)$  or both with a very large degree}$$
where the degree of a polynomial is the highest power in it, for example the degree of $x^2+3x+5$ is $2$.

However, we don't general formulae for any general polynomial equation with degree greater than or equal to $5$ then I guess for $n\ge5$ what you ask is indeed impossible.

I have no rigorous response to Bumblebee's point, but the following formulae makes it seem unlikely for your question to have 'yes, it's possible at least sometimes (apart from the trivial cases)' for an answer.
$$\begin{align}
\sin(n\theta) &= \sum_{k\text{ odd}} (-1)^\frac{k-1}{2} {n \choose k}\cos^{n-k} \theta \sin^k \theta, \\
\cos(n\theta) &= \sum_{k\text{ even}} (-1)^\frac{k}{2} {n \choose k}\cos^{n-k} \theta \sin^k \theta \,,
\end{align}$$


问:
Differentiable function exercise
Regarding:

Let $f:[0,\infty) \to \mathbb{R}$ be a differentiable function, $f'$ is increasing, and $f(0)=0$.

Prove that for every $x>0$ we have $\frac{f(x)}{x} \le f'(x)$.

--

I have tried many things, like showing that $f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0} \frac{f(x)}{x}$, I have tried using monotone of limits, toast, showing that $0 \le f''$, looking at thorems like Darboux, Ferma, Rolle....

Unfortunately I do not know what to do with "$f'$ is increasing".

Please, if someone can only give me a hint I will very appreciate that!

Thanks!

答:
hint

Let $ x>0$.

By MVT, there exists $ c\in (0,x) $ such that

$$f(x)-f(0)=(x-0)f'(c)\le xf'(x)$$
because $ f'$ is increasing.

Title: Differentiable function exercise


答: HINT: Consider $$\int_0^1 f'(tx)\mathop{}\!\mathrm{d}t.$$

Title: Differentiable function exercise

Body:
Regarding:

Let $f:[0,\infty) \to \mathbb{R}$ be a differentiable function, $f'$ is increasing, and $f(0)=0$.

Prove that for every $x>0$ we have $\frac{f(x)}{x} \le f'(x)$.

--

I have tried many things, like showing that $f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0} \frac{f(x)}{x}$, I have tried using monotone of limits, toast, showing that $0 \le f''$, looking at thorems like Darboux, Ferma, Rolle....

Unfortunately I do not know what to do with "$f'$ is increasing".

Please, if someone can only give me a hint I will very appreciate that!

Thanks!


问:
Let $\Sigma=\{a,b,c\}$. Construct a minimal DFA recognizing the language constructed by the regular expression
Let $\Sigma=\{a,b,c\}$. Construct a minimal DFA recognizing the language constructed by the regular expression :

$$b ∘ a ∘ (a, b)\text{*} ∘\ ( (a ∘ c) ∩ (b, c)\text{*} )\text{*} ∘\ c $$

where $∘$ means concatenation

What is the best approach to constructing a minimal DFA in this case?

答: HINT: $(a\circ c)\cap(b,c)^*$ is empty, so the given regular expression is equivalent to the much simpler one $b\circ a\circ(a,b)^*\circ c$, for which you should not have much trouble designing a DFA. Once you have one, you can use standard techniques to convert it to a minimum DFA for the language, though you may well come up with a minimal DFA right off the bat.


问:
Linear combination of of $n+2$ vectors
Given $n+2$ vectors in an $n$-dimensional space, prove that there exists a non-trivial linear combination $$a_1 v_1 + a_2v_2 + \cdots + a_{n+2}v_{n+2}=0$$ with not all $a_i = 0$, yet with $\sum\limits_{i=1}^{n+2} a_i = 0$.

I tried to take $n+1$ vectors and build a barycentric combination of them, such that they would form the $(n+2)$nd vector and $\sum\limits_{i=1}^{n+1} a_i = 1$, and if I add $(n+2)$nd vector to this combination with $a_{n+2}=-1$, it will give the needed linear combination.

But I stuck proving this barycentric combination exists. So now I think I'm moving in the wrong direction, because it seems, that there is can be no barycentric combinations formed from these vectors.

答:
The set $\{v_1, \ldots, v_{n+2}\}$ is linearly dependent so there is a vector which can be expressed as a linear combination of the remaining $n+1$ vectors. WLOG assume
$$v_{n+2}=\sum_{i=1}^{n+1} \alpha_i v_i.$$

On the other hand, the set $\{v_1, \ldots, v_{n+1}\}$ is also linearly dependent so there are scalars $\beta_1, \ldots, \beta_{n+1}$ not all zero such that
$$\sum_{i=1}^{n+1} \beta_iv_i = 0.$$

Notice that if $\sum_{i=1}^{n+1} \beta_i = 0$ we are done as
$$\sum_{i=1}^{n+1} \beta_iv_i +0\cdot v_{n+2}= 0$$
would be our desired linear combination. Therefore we can assume $\sum_{i=1}^{n+1} \beta_i \ne 0$ and we have for all $\gamma \in \Bbb{R}$ that

$$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i )v_i - v_{n+2}=\underbrace{\sum_{i=1}^{n+1} \alpha_i v_i - v_{n+2}}_{=0}  + \gamma\underbrace{\sum_{i=1}^{n+1} \beta_iv_i}_{=0} = 0$$
where the sum of these scalars is
$$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i ) - 1 = \sum_{i=1}^{n+1}\alpha_i + \gamma \sum_{i=1}^{n+1} \beta_i - 1$$
so picking $\gamma = \frac{1-\sum_{i=1}^{n+1}\alpha_i}{\sum_{i=1}^{n+1} \beta_i}$ gives the desired linear combination.

Title: Linear combination of of $n+2$ vectors


问:
What is the Greek version of $\;\cal{quod~erat~demonstrandum}\;$?
What is the Greek version of "quod erat demonstrandum"?

Edit:

$\quad$The symbol $\square$ will denote the end of a proof; it may be read as "Q.E.D." abbreviating the Latin quod erat demonstrandum, which translates the Greek of Euclid of Alexandria "ὅπερ ἔδει δεῖξαι".

I found this in Bridge to Abstract Mathematics, but I was hoping to find something I could actually copy and paste into a tex file.

答:
"ὅπερ ἔδει δεῖξαι" was used by Euclid and Archimedes.

Wikipedia


答:
As a rule, I don't like writing proofs that go: "suppose $x \neq 0$ then ... and $x = 0$" if I don't have to. So I would just write: "let $x \in W^\perp$" then all that stuff you said so $x = 0$.

Is there any example of inner product space $V$ and subspace $W ≤ V$ , such that $W$ is not dense in $V$, but $W^⊥ = 0$?

Assuming the Hahn-Banach theorem (which is slightly weaker than the Axiom of Choice but still independent of ZF) then no.

Suppose $W$ is a closed, proper subspace of $V$ and $x \notin W$. Define a linear function $f : W \oplus \mathbb{R}x \to \mathbb{R}$ by $f(x) = 1$ and $f(w) = 0$ for all $w \in W$. Then $f$ is continuous and by the Hahn-Banach theorem, admits a continuous extension to all of $V$. By construction, this extension is in $W^\perp \setminus \{0\}$.

Title: Let $V$ be an inner product space and $W$ be a dense subspace of $V$ then $W^{\perp }={0}$.

Body:
Let $V$ be an inner product space and $W$ be a dense subspace of $V$ .

Claim: $W^{\perp }={0}$.

Let non-zero $\ x\in W^{\perp} \implies (\forall w \in W,\ \langle x ,w\rangle=0)\ \implies W \subseteq x^{\perp}$ and $x^{\perp}$ is closed subspace.

So $W \subseteq x^{\perp} \implies  \overline{W} \subseteq \overline{x^{\perp}} \implies V \subseteq x^{\perp}$. This is a contradiction because this is possible only when $x=0$.

So $W^{\perp}=0.$

Is it okay and
also, is there any example of an inner product space $V$ and subspace  $W \leq V$, such that $W$ is not dense in $V$, but $W^{\perp}=0$?


答:
There are (at least) two ways you can go about doing this, and in either case I would make sure you start with drawing a picture.

If we use the washer method we can think of think of splitting the region to be rotated into vertical slabs between $x = 0$ and $x = 2$, where the inner radius is $5 - x^2 - 1$ and the outer radius is $5$.  This gives us $$V = \pi\int_{0}^{2}\left(5^2 - (5 - x^2 - 1)^2\right)\,dx =\frac{494\pi}{15}.$$  If we use cylindrical shells, as it seems you are trying to do, we can think of splitting the region rotated into horizontal slabs between $y = 0$ and $y = 5$.  If we pick a slab at $y$ it will be a radius of $5 - y$ from the axis of rotation (as you said).  To get the height of each slab we need to think about splitting the  region into two pieces.  Between $y = 0$ and $y=1$ the height of the slab is a constant $2$.  Between $y = 1$ and $y = 5$ the height of the slab is the distance between the parabola and the line $x = 2$, which is $2 - \sqrt{y-1}$.  So, sing this method, we get $$V = 2\pi\int_{0}^{1}(5-y)(2)\,dy + 2\pi\int_{1}^{5}(5-y)(2-\sqrt{y-1})\,dy = \frac{494\pi}{15}.$$

Title: Solid of revolution axis $y=5$


答:
Another way to find the expected number $X$ of aces:

Let $X_i$ be the indicator that card $i$ is an ace, for $1\leq i\leq 13$. Then $\mathbb E[X_i]=\mathbb P(X_i=1)=1/13$ (if we imagine taking the top $13$ cards in a randomly shuffled deck, then clearly each card has a $1/13$ chance of being an ace). So by linearity of expectation
$$\mathbb E[X]=\mathbb E[X_1+\dots+X_{13}]=\mathbb E[X_1]+\dots+\mathbb E[X_{13}]=13\cdot\frac{1}{13}=1.$$

Title: Computing probabilities in a game of bridge

Body: In the game of bridge, a player is given 13 cards from a deck of 52 cards, what is the probability that he/she gets exactly one king and one queen? Furthermore what is the expected value of the number of aces he/she gets?</p>


问:
If two topologies have the same continuous automorphisms, must they be equal?
Let

$X$ be a set, and

$T_1,T_2$ be a pair of topologies on $X$.

Assume that, for all $f:X → X$, $f$ is continuous wrt $T_1$ iff $f$ is continuous wrt $T_2$. Must $T_1=T_2$?

If so, I'd like a proof and if not, a counterexample.

答:
If one is talking about functions $f:X → X$, I do not think it is true: consider a two-pointed space with

the trivial topology,

the power set topology.


答:
It originates from the Greek:  hóper édei deîxai (ὅπερ ἔδει δεῖξαι), meaning "which had to be demonstrated". So the Latin Q.E.D. is a variation on the translation of the the expression used by the Greeks, namely, we know it was used by Euclid and Archimedes, and likely other Greek scholars who followed "suit."

Further clarification in the Wikipedia entry is truer to Euclid's use of the expression:

Since the verb "δείκνυμι" also means to show or to prove, a better translation from the Greek would read, "what was required to be proved."

Added: See also Earliers Known Uses of Mathematical Terms.

Title: What is the Greek version of $\;\cal{quod~erat~demonstrandum}\;$?

Body:
What is the Greek version of "quod erat demonstrandum"?

Edit:

$\quad$The symbol $\square$ will denote the end of a proof; it may be read as "Q.E.D." abbreviating the Latin quod erat demonstrandum, which translates the Greek of Euclid of Alexandria "ὅπερ ἔδει δεῖξαι".

I found this in Bridge to Abstract Mathematics, but I was hoping to find something I could actually copy and paste into a tex file.


问:
Wrong/naive inductive hypothesis and tower of hanoi as an example
I am reading about how a wrong formulation of the tower of Hanoi and the inductive hypothesis can lead to a dead-end.

The example I am reading states the following:

The task is to move N discs from a specific pole to another
specific pole. Assume there are poles $A$, $B$ and $C$.   The base case is that when the number of discs is 0 then no steps are needed
to complete the tasks.      For the inductive step assume that we
can move $n$ discs from pole $A$ to pole $B$ and we are required to
show how to move $n + 1$ discs from $A$ to $B$

Then it highlights that this definition is a dead-end since the only 2 ways to use the induction hypothesis as set don't lead anywhere.

Specifically the options are:

Move the top $n$ discs from pole $A$ to $B$. After this point all possibilities of using the induction hypothesis have been exhausted
since $n$ discs are on pole $B$ and we do not have hypothesis about
moving discs from that pole.

Move the smallest disc from pole $A$ to $C$. Then move the remaining $n$ discs from $A$ to $B$. Once again we have exhausted all
possibilities of using the induction hypothesis, because $n$ discs are
now on pole $B$, and we have no hypothesis about moving discs from
this pole.

The reasoning for $1$ is clear to me. If we move the top $n$ discs i.e. the $n$ smaller discs to pole $B$ then at some point we would have to move them again since we would need to place the $nth + 1$ remaining largest disc bellow them. And there is no inductive hypothesis for that so I think I get it.

I need help with the second part. How I understand it is we move the smallest disc to pole $C$. Then we use the inductive hypothesis to move the remaining $n$ larger discs from pole $A$ to $B$. (At this part I am not sure how this could happen if pole $C$ is occupied by the smallest disc but I guess it is part of the logic of using induction in proof.)

Then at that state it seems to me that the only thing pending would be to move the smallest disc from $C$ to $B$ and finish the task.

Why does it state that we would need to move the $n$ discs from pole $B$ and that is not possible since we don't have an induction hypothesis?

Am I misunderstanding something about statements on induction here?

答: The reason that the second leads to a dead end is because having moved the smallest disk to $C$ you can't move the bottom $n$ disks to $B$ because you can't use $C$ to transfer them.  If you could, this would work fine.  The bottom $n$ disks are on $B$ and you just move the top disk onto them and are done.


问:
Solving $\cos2x-2\sin x-\cos^2x=-3$ with identities
I need to solve
$$\cos2x-2\sin x-\cos^2x=-3$$
I'm trying to find possible solutions in the interval $[0,2\pi]$, but I need to simplify the equation using trig identities first. I just can't seem to find the right ones to use.

答:
Generally with these types of questions, a good method is to transform your equation into an equation in terms of only one variable, such as only in terms of $\sin x$ or $\cos x$.

To do this, the following formulae sould be useful:
$$\sin^2x+\cos^2x\equiv1$$
$$\cos2x\equiv\cos^2x-\sin^2x\equiv2\cos^2x-1\equiv1-2\sin^2x$$
I leave it to you to choose which ones to use.

If you need any more help, please don't hesitate to ask :)


问:
existence of $W \subset X^*$ which separates the points in separable normed space $X$.
Let $X$ be a separable normed space. I have to see that there exists a numerable subset $W \subset X^*$ such that $W$ separates the points in $X$. I could start the proof but I have a problem: what does "separates the points in $X$" mean? Thanks

答:
Let $A, B$ be sets, and $\mathscr{F}$ a family of functions from $A$ to $B$. Then $\mathscr{F}$ separates points in $A$, if for all $a_1, a_2 \in A$ with $a_1 \neq a_2$ there is an $f \in \mathscr{F}$ such that
$$f(a_1) \neq f(a_2)\,.$$
Thus a point of $A$ is determined by the indexed family of its images under elements of $\mathscr{F}$, in other words the map $\Phi \colon A \to B^{\mathscr{F}}$ given by
$$\Phi(a) = \langle f(a)\rangle_{f \in \mathscr{F}}$$
is injective.

If $A$ and $B$ are vector spaces over the field $k$, and $\mathscr{F} \subset \mathscr{L}(A,B)$ is a family of $k$-linear maps, then $\mathscr{F}$ separates points if and only if for every $a \in A \setminus \{0\}$ there is an $f \in \mathscr{F}$ such that $f(a) \neq 0$.

Vector spaces and linear maps are more than we need for this characterisation by the way, it holds (with a slight modification of notation, replace "$0$" with the identity elements "$e$") for arbitrary groups and a family of group homomorphisms.

The proof that this characterises families separating points is easy. On the one hand, $a \in A \setminus \{e\}$ (I'm using the group setting, $e = 0$ for vector spaces) means $a \neq e$, so if $\mathscr{F}$ separates points there is an $f \in \mathscr{F}$ such that $f(a) \neq f(e) = e$. On the other hand, if $a_1 neq a_2$, then $a_1^{-1}a_2 \neq e$, so by assumption there is $f \in \mathscr{F}$ with
$$e \neq f(a_1^{-1}a_2) = f(a_1)^{-1}f(a_2)\,,$$
which is equivalent to $f(a_1) \neq f(a_2)$, and thus $\mathscr{F}$ separates points.

For your specific problem, pick a countable dense subset $D = \{x_n : n \in \mathbb{N}\}$ of $X$ (it may be convenient to assume $0 \notin D$) and for every $n$ choose an $f_n \in X^{\ast}$ (it can be convenient to choose them such that $\lVert f_n\rVert = 1$ for all $n$) based on $x_n$ alone (hint: Hahn-Banach). If you made the right choices, $W = \{ f_n : n \in \mathbb{N}\}$ will separate points in $X$.


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
In the end, the doctoral thesis has to contain enough to be passed by the readers
(and to satisfy any auxiliary rules of the institution to which it is submitted).  While a very slim thesis is unusual, if the results are good and the writing is comprehensible, I don't see that slimness is necessarily bad.

On the other hand,
there is a tradition of including a certain amount of expository materials in a thesis, and I think this is useful, both for the readers (who may not be as familiar with the area as the student and advisor, but who have to form a judgement on the thesis) and for the student writing it; it is good to practice writing clear expositional and motivational material, since this is similar to the kind of material that is required in introductions to papers, in grant proposals, in colloquium talks, and all the other things one has to write in one's career that will be read or listened to by people who are not as expert in one's area as oneself.

It is also quite likely that you will be sending a copy of the thesis, or at least a provisional version, to your letter writers, and so any expository or motivational material that will enhance their appreciation of the thesis is certainly worth including.

Overall, bear in mind that while a brevity can be a good thing, if it degenerates into opacity, that is bad.


问:
How many subsets of length 10 of the Latin alphabet don't contain the set $\{a, b, c, d\}$?
Let $A$ denote the set of the Latin alphabet. Let $B$ be a subset of $A$ such as $|B|=10$. How to claculate how many such subsets $B$ don't have for a subset the set $\{a, b, c, d\}$. In other words
$$
|S| = ?
\quad\text{for}\quad
S = \{B\mid B\subset A\land |B| = 10\land \{a, b, c, d\}\not\subset B\}
$$

My approach so far is to calculate how many such $B$'s exist, which I finded to be $C(26, 10) = 5.311.735$ and substract from them how many $B$'s contain the set $\{a, b, c, d\}$. The thing is I can't find a way to calculate that last quantity.

Any help is appreciated. Thanks in advance!

答:
The number of subsets where we don't care is indeed $\binom{26}{10}$ as you correctly found.  Further you are correct that to get the count that you are interested in, we can subtract the number of subsets which do contain all of $a,b,c,d$.

To find the number which do contain all of $a,b,c,d$... we know that each of them have those four letters and the only information missing is which other six letters are in the set out of the $22$ remaining letters.  That too is a standard problem solved by binomial coefficients, $\binom{22}{6}$.

$$\binom{26}{10}-\binom{22}{6}$$


问:
Maximize function consisting of Sum, Vectors and Signum function
Edit: Thanks for edit :-)

Can I ask please, how to maximize

$$\sum ({v_3 \text{sign}({xv_1+yv_2)}})\quad\text{where}$$
(edit: sum of i in 1..1000 ... 1 to length(v1) .. 3 vectors of same length
$$\sum_{i=1}^n ({v_3[i] \text{sign}({xv_1[i]+yv_2[i])}})\quad\text{where}$$

$v1,v2,v3,\cdots\space $-vector(s) are real numbers, generated around $0\space$
[x,y=...value(s)] to find

I even don't how to partially derivate it... that signum function

I need find vales $(x,y)$. (in future I will try to use $n$ number of vectors, but solution for $2$ is best start. For one vector it's easy but for $2$ vectors - high mathematician skill required)

thanks a lot if somebody know how to solve it

答:
I don't there exists as a closed form solution of this. So there might not be a formula that expresses the optimal solution.

However you express your problem as Quadratically Constrained Linear Program.

First we introduce a vector of dummy variables $s$ and rewrite our objective as:
$$\max \sum_i v_3[i]s[i]$$
subject to some additional constraints. We observer that $s$ has to have the same values as $\text{sign}$ term: $\forall i: \ s[i] \in \{-1,0,1\}$. This can be expressed as the constraint $s^2 = s^4$ which can be expressed as two quadratic constraints by yet another vector of dummy variables $j$. So we now got the constraints:
$$j[i] = s[i]^2$$
$$j[i]^2 = s[i]^2$$

Please observe that the value of $xv_1[i]+yv_2[i]$ is irrelvant only the sign matters, so we loose no information by constraining it's value to be $\in \{-1,0,1\}$. Using this we can formulate this constraint:

$$s[i] = x*v_1[i] + y*v_2[i]$$

The following QCLP gives you your solution:
$$\max_{s[1],...,s[n],j[1],...,j[n],x,y} \sum_{i=1}^n v_3[i]s[i]$$
subject to

$$\forall i \in \{1, ..., n\}: \ j[i] = s[i]^2$$
$$\forall i \in \{1, ..., n\}: \ j[i]^2 = s[i]^2$$
$$\forall i \in \{1, ..., n\}: \ s[i] = x*v_1[i] + y*v_2[i]$$

There is a large array of programs that can solve QCLP some of them might also be callable from R.


问:
Regular value of the determinant map
Let $\text{det}: M_n(\mathbb{R})\rightarrow \mathbb{R}$ be the determinant map from $n\times n$ matrices to $\mathbb{R}$, then how can I show that 1 is a regular value of this map? I am specifically having trouble to calculate the derivative of this map on matrices.

答:
Let $M$ such that $det(M)=1$, write $M=(m_1,...,m_n)$ where $m_i\in\mathbb{R}^n$.
consider $f(t)$ the matrix whose first column is $m_1+tm_1$ and whose colum $i$ is $m_i,i\neq 1$.
We have $det(f(t))=1+t$, this implies that $det'_M(f'(t))=1$, we deduce that $det$  is not degerated at $M$.

Title: Regular value of the determinant map


问:
Solid of revolution axis $y=5$
The problem goes like this

Rotate the indicated area around the given axis to calculate the volume of the solid of revolution $y=x^2+1$, $x=0$, $x=2$, $y=0$ , around the axis $ y = 5$.

My question is if the solution is given by the following integral (should I take the limits of integration from $0$ to $2$?)

$$2\pi\int_0^2 ((5-y)\sqrt{y-1})\,dy$$

答:
As you are integrating with respect to $y$, it should be $1 \leq y \leq 5$. But you are finding volume of region $2$ rotated around $y = 5$. You can do that but then you need to subtract it from the volume of the cylinder of radius $5$ to find volume of region $1$ that the question asks.

An easier way to do this is to directly find the volume of region $1$ as below -

$4-x^2 \leq r \leq 5$ (distance from $y = 5 \ $ - $ \ $ i) to $y = 0$ which is $5$ and ii) to parabola $y = x^2 + 1$ which is $(5 - (x^2+1) = 4 - x^2$).

Also, $0 \leq x \leq 2$

So the integral is $ \displaystyle \int_0^{2\pi}  \int_0^2  \int_{4 - x^2}^5   r \ dr \ dx \ d\theta = \frac{494 \pi}{15}$

Now coming to the integral that you have come up with is based on

$r = 5 - y, 0 \leq x \leq \sqrt{y-1}$

$V = \displaystyle \iiint r \ dx \ d\theta \ dr = \displaystyle \int_1^5 \int_0^{2\pi} \int_0^{\sqrt{y-1}} (5-y) \ dx \ d\theta \ dy$

$V = \displaystyle 2\pi \int_1^5 (5-y) \sqrt{y-1} \ dy = \frac{256 \pi}{15}$

Subtract it from the cylinder volume which is $\displaystyle \int_0^{2\pi}  \int_0^5  \int_0^2  y \ dx \ dy \ d\theta = 50 \pi$

So volume of region $1 \displaystyle = 50 \pi - \frac{256 \pi}{15} = \frac{494 \pi}{15}$

Title: Solid of revolution axis $y=5$


问:
If $f'(x) = -f(x)$ and $f(1)=1$, then what is $f(x)$?
If $f'(x) = -f(x)$ and $f(1)=1$, then $f(x)=$

(a) $1/2e^{-2x+2}$

(b) $e^{-x-1}$

(c) $e^{1-x}$

(d) $e^{-x}$

(e) $-e^{x}$

If you plug in $1$ for $x$ in option c it gives you $e^{1-1}$ or $e^0$ which is $1$ hence what is needed.

答:
If $f'(x) = -f(x)$,
then

$\frac{f'(x)}{f(x)}=-1 \implies \int \frac{f'(x)}{f(x)}dx=-\int dx \implies f(x)= e^{-(x-k)}\ $ (Here k is the integration constant)

so $e^x f(x)$ is constant.

Here, $e^x f(x) = e^k$
and set $x = 1$.
Then, since $f(1) = 1$,
$e f(1) =e^k $
so $f(x) = e e^{-x} = e^{1-x}$.

so, $f(x)= e^{1-x} \implies\ \ $ Option (b) is correct.


问:
Solve $x - \ln(x) > 1$ without using derivatives
I would like to solve $x - \ln(x) > 1$ without studying the function $f(x) = x - \ln(x) - 1$ and computing its derivative.

I wonder if solving for $x$ in $x - \ln(x) = 1$ is possible.

It could be possible to consider the Lambert $W$-function, since we can rewrite the expression such that we have

\begin{align*}
  \ln(x) - x   &= 1  \\
  \ln(xe^{-x}) &= 1  \\
  xe^{-x}      &= e  \\
  -xe^{-x}     &= -e
\end{align*}

Yet it seems impossible and non-elementary to have a "nice" analytical solution to this.

答:
Going on what Martin R said in the comments to the question, using the fact that $\ln(x)$ is defined only for $x>0$ and that $e^x$ is a strictly increasing function,

$x-\ln(x)>1,\ x>0 \implies e^{x-\ln(x)} > e^1,\ x>0 \implies \frac{1}{x}e^x > e,\ x>0 \implies e^x > ex,\ x>0.$

Substituting $x = y+1$ gives:

$e^{y+1} > e(y+1),\ y > -1 \implies e^y> y+1,\ y > -1.$

Various proofs of the final equation for all $y \in \mathbb{R}$ can be found here. In fact, you only need a proof that works for all $y > -1, $ which other people are welcome to do if they so desire (I'm not going to...).


问:
Can you spot the mistake here?
assume that :
$$x^{3} + y^{3} = z^{3}$$
so :
$$z^{3} - y^{3} = x^{3}$$
$$z^{3} ≡ y^{3}\bmod x$$
$$z ≡ y\bmod x $$
$$z-y = x $$
and now let's plug this result to the original equation:
$$(z-y)^{3} + y^{3} = z^{3}$$
$$z^{3}-3z^{2}y+3zy^{2}-y^{3}+y^{3}=z^3$$
by simplfying, we get this:
$$3zy^{2} = 3z^{2}y$$
dividing by $3zy$:
$$y=z$$
so this equation has solutions if and only if $y = z$, and now let's plug our results to the original equation:
$$x^{3}+y^{3} = y^{3}$$
$$x^{3} = 0$$
$$x = 0$$
Therefore :There is no solution when: $$x,y,z\in\mathbb{N}$$
and by mathematical induction we can prove it (instead of the power 3) for all $ n\in\mathbb{N}$

答: $z \equiv y \pmod x$ does not imply $z-y=x$.  We could have $z=y$ or $z-y=2x$ or many other possibilities.

Title: Can you spot the mistake here?


答:
Since $M \mapsto \det M$ is a function of more than one variable, finding a formula for its derivative is best done via directional derivatives. Let's denote and calculate the directional derivative of the determinant function at the matrix $M$ in the direction of the matrix $N$ by
\begin{align*}
\partial_N\det(M) &= \left.\frac{d}{dt}\right|_{t=0}\det(M+tN)\\
&= \left.\frac{d}{dt}\right|_{t=0}(\det M)\det(I+tM^{-1}N)\\
&= (\det M)\left.\frac{d}{dt}\right|_{t=0}\det(I+tM^{-1}N).
\end{align*}
On the other hand,
\begin{align*}
\det(I + tA) &= t^{n}\det(t^{-1}I + A)\\
&= t^{n}(t^{-n} + (\operatorname{tr} A)t^{-n+1} + \cdots)\\
&= 1 + t(\operatorname{tr}A) + t^2(\cdots)
\end{align*}
and therefore,
$$
\left.\frac{d}{dt}\right|_{t=0}\det(I + tA) = \operatorname{tr} A.
$$
Substituting this into the equation above, we get
$$
\partial_N\det(M) = (\det M)\operatorname{tr}(M^{-1}N).
$$
From here, it is straightforward to verify that $1$ is a regular value of the determinant function.

Title: Regular value of the determinant map

Body: Let $\text{det}: M_n(\mathbb{R})\rightarrow \mathbb{R}$ be the determinant map from $n\times n$ matrices to $\mathbb{R}$, then how can I show that 1 is a regular value of this map? I am specifically having trouble to calculate the derivative of this map on matrices.


问:
Question on forgetful functor from group to set is not full
I read that the forgetful functor $U: \textbf{Grp} \rightarrow \textbf{Set}$ is not full, i.e. given two objects $X$ and $Y$ in $\textbf{Grp}$, there are some arrows in the $\textbf{Set}$ category between $U(X)$ and $U(Y)$ that are not mapped to corresponding arrows in $\textbf{Grp}$ between $X$ and $Y$...

But does this mean that the category $\textbf{Set}$ includes all possible maps among its objects "by default" ? I tried looking for confirmation on this point in the text am reading but could not find any ...

答:
I am not sure if "by default" is a right term. It may be more precise to say that in the category of sets, or groups, or pointed sets, and many others, the maps are precisely all possible maps that "preserve the structure" of the objects in the category. (Group homomorphism being a concrete example of this idea).

So, in coming to your specific example, what structure a map in the category of sets has to satisfy? I think it would make the situation clear when you ask the same question in the category of groups. This is gives you a proof of statement that the forgetful functor $U: \textbf{Grp} \rightarrow \textbf{Set}$ is not full.


问:
Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$

What's the name of this identity? Is it the identity of the Pascal's triangle modified.

How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?

Thanks for your help.

EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.

答:
You remember that:
$$
(1+x)^m = \sum_k \binom{m}{k} x^k
$$
So the sum
$$
\sum_{m=0}^M \binom{m+k}{k}
$$
is the coefficient of $ x^k $ in:
$$
\sum_{m=0}^M (1+x)^{m+k}
$$ Yes?
So now use the geometric series formula given:
$$
\sum_{m=0}^M (1+x)^{m+k} = -\frac{(1+x)^k}{x} \left( 1 - (1+x)^{M+1} \right)
$$
And now you want to know what is coefficient of $x^k $ in there. You got it from here.</p>