问:
Prove that to fully connect $n$ nodes we need at least $n-1$ pairwise links
Suppose we have $n$ points/nodes, and we are given $m$ pairwise links/edges that we can use to connect arbitrary pairs of nodes.

Our goal is to arrange the edges in order to fully connect the nodes, that is, to have a configuration such that one can go from any node to any other node passing through the edges.

For example, for $n=4$ and $m=3$, denoting with $(ij)$ an edge between $i$-th and $j$-th node, such fully connected networks could be $(12)(23)(34)$, or $(12)(13)(14)$.

By trying this in a few cases, it seems clear that the smallest $m$ for which this is possible is $m=n-1$. What's a good way to prove that this is indeed the case?

答:
Define the connected components of a set of nodes and links (a graph) to be the maximal subsets of the nodes which are connected each other by paths. That is, two nodes are in the same component if they are joined by a path, and in different components otherwise.

If there are no links, then no node can reach any other, so there are $n$ connected components.

If a graph currently has $k$ connected components, and you add a single link to the graph, then the new graph has at least $k-1$ connected components. This is because, at best, the new link can join to components together, replacing those two with one and decreasing the over all number by $1$.

These two facts allow you to prove the following by induction: a graph with $n$ nodes and $m$ edges will have at least $n-m$ connected components. Therefore, if there are only $n-2$ links, there will be at least two connected components, so the graph is not connected.


问:
Integration by Parts proof for Lebesgue-Stieltjes Measure
I'm trying to complete exercise 35 b) on from chapter  3.5 from Folland's Real Analysis that goes like if $F,G$ are NBV and $- \infty < a< b< \infty $ and there are not points in $[a,b]$ where both $F$ and $G$ are discontinuous then
$$\int_{[a,b]} FdG + \int_{[a,b]} GdF = F(b)G(b) - F(a-)G(a-).$$
In part a of the problem we proved that

$$ \frac12\int _{[a,b]} (F(x) +F(x-)) dG(x) + \frac12 \int_{[a,b]} (G(x) +G(x-)) dF(x) = F(b)G(b) - F(a-)G(a-) $$

I proved this part and I am thinking to use to solve part b, perhaps by showing that $dF(x)$ and $dG(x)$ are $0$ on the points of discontinuity of the other function but I am not sure how to prove that because here we are working with the Lebesgue-Stieltjes measure and it is a signed measure. Perhaps there is something I am missing or an easier way to go about it?

答:
Indeed the discontinuities of $F$ forms a $\mu_G$-null set.  To prove this, recall that $G$ does not have discontinuities in common with $F$.  So if $F$ is discontinuous at $x$, then $G$ is continuous at $x$ and hence $\mu_G(\{x\})=G(x+)-G(x-)=0$.  Finally, since $F$ is NBV, there are only countably many points of discontinuity and $\sigma$-additivity of $\mu_G$ takes care of it.

Thus $F=(x\mapsto F(x-))$ $\mu_G$-a.e., and similarly with $F,G$ interchanged.  This gives the desired result.

ContentLicense: CC BY-SA 4.0


答:
This is a linear difference equation. You can get the solution as follows:

Compute the general solution of the homogenous equation $T_n = 4 T_{n-2}$, which is $H_n = c_1 \cdot 2^n +c_2 \cdot (-2)^n$.

Obtain a particular solution of the complete equation by trying something similar to the RHS, i.e. $P_n = k \cdot 4^n$, and compute $k$ substituting in the equation. One possibility is $P_n = \frac{4}{3} \cdot 4^n$.

The general solution to our equation is given by
$$
T_n = H_n + P_n = c_1 \cdot 2^n +c_2 \cdot (-2)^n + \frac{4}{3} \cdot 4^n
$$

From the conditions on $T_1, T_2$, compute $c_1, c_2$ to get the final answer:
$$
T_n = -\frac 72 2^{n}-\frac 43 (-2)^{n} + \frac 43 \cdot 4^n
$$

Title: Solving recurrence with substitution

Body: I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6)+2^{2(n-4)})+2^{2n-2}+2^{2n}=4^3T(n-6)+2^{2n-4}+2^{2n-2}+2^{2n}=...=4^k(T(n-2k)+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$.Substituting $T(n-2k)=2, i.e. T(2)=2$, $n-2k=2$ then $k=\frac{n-2}{2}$, so our equation looks like $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$ with $k=\frac{n-2}{2}$, $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}=2^{n-2}+2^{n+4}+...+2^{2n}=2^{n-2}+\frac{2^{n+4}(1-2^{\frac{n-4}{2}})}{1-2}=2^{n-2}+2^{\frac{3n+4}{2}}-2^{n+4}$. However I think this answer doesn't satisfy $T(2)=2$ and I really don't know where went wrong. Is there another way to do this problem?


问:
Set of bounded increasing functions is not totally bounded
Show that the set of bounded and increasing functions $\mathbb R \rightarrow [0,1]$ is not totally bounded in the supremum norm. First I am trying to show this for the simple case of any function $\mathbb R \rightarrow [0,1]$ rather than for increasing functions.

My approach is to find some $\delta$ such that there is no finite set that can cover $\mathcal F$ in sup norm but I can't see why such a universal $\delta$ would exist.

答:
Take any increasing function $f:\mathbb R \to (0,\frac{1}{2})$ you like the most, for example $f(x) = \frac{1}{2\pi}(\arctan(x) + \frac{\pi}{2}) $

For every $n$ define $f_n:\mathbb R \to [0,1]$ as follows:

$$ f_n(x) = \begin{cases} f(x) & x \in (-\infty,n] \\ f(n) + \frac{1}{2}(x-n) & x \in (n,n+1] \\ \frac{1}{2} + f(x) & x \in (n+1,\infty)\end{cases} $$

You can check yourself that it is indeed a continuous function for any $n$ (it looks like arctan, but at interval $[n,n+1]$ it has linear growth of height $\frac{1}{2}$.

The point is, for any $n,m \in \mathbb N_+$ such that $n \neq m$, you have $$\|f_n-f_m\| \ge \frac{1}{2} $$
Indeed, if $n < m$ then $f_n(n+1) = \frac{1}{2} + f(n+1)$, while $f_m(n+1) = f(n+1)$

Hence, we've found a countable family of functions (bounded, increasing and continuous) such that every two are at distance (in $\|\cdot\|_{\infty}$) at least $\frac{1}{2}$. Hence our set cannot be totally bounded.


问:
if $p(x)=ax^2+bx+c$ and $|p(\alpha)|\le 1$ prove $|2aα+b|≤4$
Source: Challenge and Thrill of Pre-College Mathematics

"Let $$p(x)=ax^2+bx+c$$ be a polynomial in $ℝ(x)$ such that $|p(α)|≤1$ for $|α|≤1$. Prove that $|2aα+b|≤4$ for $|α|<1$."

First of all, it is interesting to note that the exercise has multiple questions of the same pattern, and while I was able to do all other such questions, this one seems slightly more trickier. An answer to one of the other questions may be found here, for context/reference: Prove that $|a|+|b|+|c|\le17$ if $p(x)=ax^2+bx+c$ is a real polynomial with $|p|\le1$ for $0\le x\le1$

My attempt: Expressing $a,b,c$ in terms of $p(0.5), p(0)$ and $p(1)$ as in the answer given in the link I just shared. This can be done as follows:

$1) a=2(a+b+c)-4(0.25a+0.5b+c)+2c
     =2p(1)-4p(0.5)+2p(0)$
⇒ $max (a)=2+4+2=8$

Similarly, it can be found that $max(b)=8$, but this in-fact doesn't satisfy our proof.

The problem with my approach is that I considered my proof to assume $|α|≤1$, and not $|α|<1$ as asked by the question.

Could anyone help me on how to proceed with $|α|<1$ ? I can't quite get my finger on it.

Body:
Hint:

Show both $f'(1) = \frac32f(1)+\frac12f(-1)-2f(0)$ and $f'(-1) = -\frac12f(1)-\frac32f(-1)+2f(0)$ are in $[-4, 4]$.  Further, $f'$ is linear.


问:
Find all $4$ subspaces as the span of some basis vectors
We are given a $3\times4$ matrix and it’s row echelon form and asked this question. I just need help interpreting it as I don’t really understand what my professor means. I understand the concept of finding a basis for the subspaces of a matrix. I think the wording is just confusing me, can anyone help me interpret this question please?

答:
The question appears to be asking for a basis of the domain and a basis of the codomain such that subsets of the basis form bases for the subspaces.  Essentially, find bases for the subspaces separately and then put them together into a full basis for the domain or codomain.  For example, $$\begin{bmatrix}1&0&0&0\\0&0&1&0\end{bmatrix}$$ would have $$\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}$$ as a basis for the column space, and
$$\begin{bmatrix}0\\1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\0\\0\\1\end{bmatrix}$$ is a basis for the null space, and the vectors $$\begin{bmatrix}1\\0\\0\\0\end{bmatrix}, \begin{bmatrix}0\\0\\1\\0\end{bmatrix}$$ are orthogonal to the null space and complete a basis for the domain.


问:
Characterization of complex exponential in terms of real identities
It is well known (or at least easy to see) that the complex exponential function $\exp:\mathbb{C}\to \mathbb{C}$ can be charaterized as the unique holomorphic function $f:\mathbb{C}\to \mathbb{C}$ such that
$$
\lim_{h\to 0} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(z+w)=f(z)f(w) \quad \text{for all } z,w\in\mathbb{C}.
$$
In fact we can make a stronger charaterization:
$$
\lim_{h\to 0,\; h\in\mathbb{R}} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(z+w)=f(z)f(w) \quad \text{for all } z,w\in\mathbb{C}.
$$

My question is: What are all the holomorphic functions $f:\mathbb{C}\to \mathbb{C}$ such that
$$
\lim_{h\to 0,\; h\in\mathbb{R}} \frac{f(h)-1}{h}=1 \quad \text{and} \quad f(x+y)=f(x)f(y) \quad \text{for all } x,y\in\mathbb{R}?
$$
Note the (perhaps big) difference: now the condition $f(x+y)=f(x)f(y)$ is given for all real $x$ and $y$. I know that the complex exponential function is an example of such $f$, but are there any others? If not, how to prove that the exponential is the only function satisfying this conditions?

Thank you in advance!

答:
The answer is the same: If $f$ is holomorphic in $\Bbb C$ with $f(x+y)=f(x)f(y)$ for all $x, y\in \Bbb R$, then $f(z+w)=f(z)f(w)$ holds for all $z, w \in \Bbb C$, due to the identity theorem for holomorphic functions.

First fix $y\in \Bbb R$: Then $f(x+y)=f(x)f(y)$ for all $x \in \Bbb R$ implies that $f(z+y)=f(z)f(y)$ for all $z \in \Bbb C$.

Now fix $z \in \Bbb C$: Then $f(z+y)=f(z)f(y)$ for all $y \in \Bbb R$ implies that $f(z+w)=f(z)f(w)$ for all $z, w \in \Bbb C$


问:
What's wrong with this proof that irreps over $\mathbb{C}$ are dimension $1$?
Let $V$ be an irreducible representation of a finite group $G$ over the complex numbers $\mathbb{C}$.

Then $[V, V]_{\mathbb{C}[G]} \cong \mathbb{C}$ since $\mathbb{C}$ is algebraically closed. It follows that $\mathbb{C}[G]$ acts on $V$ by scalars, so that all $\mathbb{C}$-subspaces of $V$ are $G$-invariant. Therefore $V$ must have only $2$ subspaces as a $k$-vector space, so that it must be $1$-dimensional.

On the other hand, $S_3$ has an irreducible $2$ dimensional representation, $\mathbb{C}^2$, which occurs as a sub representation of the canonical representation of $S_3$ on $\mathbb{C}^3$.

答:
I'm not sure why you'd say "it follows $\mathbb C[G]$ acts on $V$ by scalars."

Taking your example of $\mathbb C[S_3]$, you know it is a noncommutative semisimple ring, and you are aware that $M_2(\mathbb C)$ appears in its factoriation. The way $\mathbb C[S_3]$ acts on the simple submodule for that piece is not via a scalar, but via $M_2(\mathbb C)$.


问:
Finding the tangent space of a torus

Problem: Find Basis for Tangent Space of Torus

Suppose I have a torus
$$
T^2 = \left\{(x,y,z)\in\mathbb{R}^3\, : \, \left(x^2 + y^2 + z^2 + R^2 - r^2\right)^2 - 4R^2(x^2+
 y^2) = 0\right\}
$$
How do I find an orthonormal basis for its tangent space $T_x$ and for $T_x^{\perp}$?

My Attempt

In this paper that one should basically first compute $Q$, the transpose of the Jacobian and then compute its QR decomposition to find the basis, so I tried this approach. Since the constraint defining the torus is $q:\mathbb{R}^3\to\mathbb{R}$ we know that the transpose of its Jacobian will be $3\times 1$ and given by
$$
Q =
\begin{pmatrix}
    \partial_x q(x,y,z) \\
    \partial_y q(x,y,z) \\
    \partial_z q(x,y,z)
\end{pmatrix} =
\begin{pmatrix}
    4x(x^2 + y^2 + z^2 + R^2 - r^2) - 8xR^2 \\
    4y(x^2 + y^2 + z^2 + R^2 - r^2) - 8yR^2 \\
    4z(x^2 + y^2 + z^2 + R^2 - r^2)
\end{pmatrix} \in \mathbb{R}^{3\times 1}
$$
Now, however, I don't end up with a matrix, but with a colum vector and I should be taking its QR decomposition to find the basis?

答:
I find the case of a sphere sphere $S^2 = \{(x, y, z) \in \mathbb R^3 : x^2 + y^2 + z^2 - r^2 = 0\}$ easier to illustrate. As explained here, for all $p \in S^2$ we have $$T_p S^2 = \ker df_p,$$ where $f(x, y, z) = x^2 + y^2 + z^2 - r^2$. Explicitly, for $p = (a, b, c)$ we compute that
$$df_p = \begin{pmatrix} 2a & 2b & 2c  \end{pmatrix},$$
so we see that the kernel is
$$\ker df_p = \{v \in \mathbb R^3: \langle a, b, c \rangle^T \cdot v = 0  \}.$$ This set is no other than a plane with normal vector $\langle a, b, c \rangle$, which indeed is the tangent space to $S^2$ at $p$.

Edit: So, finding an ON basis for the tangent space amounts to finding an ON basis of the kernel of $df_p$. As the kernel is perpendicular to the row space of $df_p$, an ON basis of the normal space amounts to an ON basis of the row space of $df_p$. In this case $df_p$ has one row, so the vector $df_p^T$, once normalized, is your desired basis.

Title: Finding the tangent space of a torus


问:
trigonometric representation of a complex number.
Let $z=e^{it}+1$ where $0\leq t\leq \pi$, Find the trigonometric representation of $z^2+z+1$.

(The trigonometric representation should be in the form of : $r(\cos \theta +i \sin \theta)$, where $r,\theta \in \mathbb{R}$ and $r>0$.

What I have done : $$z^2+z+1=e^{2it}+3z=(\cos (t)+i \sin (t)) (-2 i \sin (t)+4 \cos (t)+3).$$
The problem is that the expression in the right bracket is not real always.

答:
The complex number $a+bi$ can be written in trigonometric form as

$$a+bi=\sqrt{a^{2}+b^{2}}(\cos \theta +i\sin \theta ),\qquad\text{with } \tan \theta =\frac{b}{a}.$$

For $z=e^{it}+1=(\cos t+1)+i\sin t$  we have

\begin{eqnarray*}
z^{2}+z+1 &=&\left( e^{it}+1\right) ^{2}+(e^{it}+1)+1= (e^{i2t}+2e^{it}+1)+(e^{it}+1)+1 \\
&=&e^{2it}+3e^{it}+3 \\
&=&((\cos 2t+1)+i\sin 2t)+3\left( (\cos t+1)+i\sin t\right) +3 \\
&=&(\cos 2t+3\cos t+3)+i\left( \sin 2t+3\sin t\right)  \\
&=&u+iv,\qquad u=\cos 2t+3\cos t+3,v=\sin 2t+3\sin t \\
&=&r\left( \cos \theta +i\sin \theta \right) ,
\end{eqnarray*}

where
\begin{eqnarray*}
r &=&\sqrt{u^{2}+v^{2}}=\sqrt{(\cos 2t+3\cos t+3)^{2}+\left( \sin 2t+3\sin
t\right) ^{2}} \\
\tan \theta  &=&\frac{v}{u}.
\end{eqnarray*}

For $0\leq t\leq \pi ,v=\sin 2t+3\sin t\geq 0$ and $u=\cos 2t+3\cos t+3\geq 0
$. So

$$
\begin{equation*}
\theta =\arctan  \frac{v}{u} =\arctan \left( \frac{\sin
2t+3\sin t}{\cos 2t+3\cos t+3}\right) \geq 0,
\end{equation*}
$$

because $w=z^2+z+1=u+iv=\operatorname{Re}(w)+i\operatorname{Im}(w)$ is in the first quadrant.

Plots of $u,v$:

$$u=\cos 2t+3\cos t+3 \text{ (blue) },\quad v=\sin 2t+3\sin t \text{ (red) },\quad 0\leq t\leq \pi.$$

ContentLicense: CC BY-SA 3.0


问:
If the sum of two i.i.d. random variables is normal, must the variables themselves be normal?
It is well known that if two i.i.d. random variables are normally distributed, their sum is also normally distributed.

Is the converse also true? That is, suppose $X$ and $Y$ are two i.i.d. random variables such that $X+Y$ is normal. Is it necessarily the case that $X$ and $Y$ are also normal?

Thoughts: If the pdfs of $X$ and $Y$ are both $f$, the pdf of $X+Y$ is $f*f$, and its Fourier transform is $\hat f^2$. So if $\hat g=\hat f^2$ is a Gaussian function, then $\hat f$ could be any pointwise product of $\sqrt{\hat g}$ and an even function $\mathbb R\mapsto\{-1,1\}$. But it is not clear that any of those correspond to probability distribution functions, which must be nonnegative everywhere.

答:
The OP asked

Suppose $X$ and $Y$ are two i.i.d. random variables such that $X+Y$ is normal. Is it necessarily the case that $X$ and $Y$ are also normal?

The answer is Yes as @saz has already said. A slightly more general result has been known for quite some time.  According to a theorem conjectured by P. Lévy and proved by H. Cramér (see Feller, Chapter XV.8, Theorem 1),

If $X$ and $Y$ are independent random variables and $X+Y$ is normally distributed, then both $X$ and $Y$ are normally distributed.

Note that $X$ and $Y$ need not be identically distributed; just being independent suffices. Feller's book does not include Cramér's proof of this theorem on the grounds that it is based on analytic function theory and thus quite different from Feller's treatment of characteristic functions.


问:
Problem with a continuous and *almost* monotonic function

Problem

Let $f: [a,b] \to \mathbb{R}$ be a continuous function. Suppose that for all x in $[a,b)$ and for all $\delta>0$ there exists a $y \in (x, x+\delta)\cap [a,b]$ such that $f(x)<f(y)$. The goal is to prove that f is monotonically increasing.

Conclusion analysis

$f$ is monotonically increasing on its domain, if and only if $$\forall x\in [a,b]\  \forall \epsilon>0 \ ( f(x+\epsilon)>f(x)).$$

Hypotheses analysis

What this given property says about $f$ is that given any point on the domain, except the boundary point $b$ and given any distance to the right of $x$ there always exists an intermediate point $y$ for which $f$ assumes a value larger than $x$.

Attempt at proof

Let x be a point of the domain, not $b$, and furthermore let $\epsilon>0$ be given. I need to show that $f(x+\epsilon)>f(x)$.
I naturally choose $\delta = \epsilon$ in my hypothesis. I am assured by the function property of the existence of a $y\in (x, x+\epsilon)\cap[a,b]$ for which $f(y)>f(x)$.

I suspect that here I need to use the fact that $f$ is continuous in order to show explicitly that
$f(x+\epsilon) > f(y)$, so that I can conclude the proof.

What I need at this point is a hint. Furhtermore, the problem forced me to consider two seperate cases
with $x\neq b$ and $x=b$; I wonder if the boundary point $x=b$ possess a significant problem down the line.

答:
Assume that $f$ is not increasing. Then there are $u, v \in [a,b]$ such that $u < v$ and $f(u) \ge f(v)$. Since $f$ is continuous, it attains a maximum on $[u,v]$. That is, there exists $\zeta \in [u,v]$ such that $f(z) \le f(\zeta)$ for all $z \in [u,v]$.

$\zeta = v$ is impossible: In that case $f(z) \le f(v) \le f(u)$ for all $z \in[u,v]$ which contradicts the assumption if we take $x = u$ and $\delta = v-u$.

Hence we must have $\zeta \in [u,v)$. But this contradicts the assumption if we take $x = \zeta$ and $\delta = v-\zeta$.


问:
Using AM-GM to find the minimum

Find the minimum value of $\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}}$ if $x$ and $y$ are positive real numbers such that $x > y$.

This is a question from the 22nd Philippine Mathematical Olympiad. I have no idea how to solve this. I am curious as to how it was solved. Can anyone help me, please?

Update: I found a solution here and it looks like it is longer than expected.

答: Hint: put $t= \dfrac{x}{y} \implies t > 1$. Then consider : $f(t) = \dfrac{7t^2 - 2t + 3}{t^2 - 1}, t > 1$. Can you take it from here using calculus tools ?

Title: Using AM-GM to find the minimum


问:
Finding the length of a half of a side of a triangle (can be unequal half)
The Question

Given, $AB$, $AC$ and $BC$. $DE$ is parallel to $BC$. The area ratio between $\triangle ADE$ and $\triangle BDEC$ is also provided. How can I find $AD$?

My Understanding

I know that $\frac{AD}{AB}=\frac{AE}{AC}
=\frac{DE}{BC}$ in this problem.

答:
Let AB=c, BC=a and AC=b. Draw altitude from A, mark its foot as H and it's intersection with DE as G. Let the ratio of areas is:

$\frac{ADE}{BCDE}=k$

We have:

$\frac{ADE}{BCDE}+1=\frac{ADE+BCDE=ABC}{BCDE}=k+1$

$S=ABC=\frac{AH\times BC}2$

$ADE=\frac{DE\times AG}2$

$$\frac{\frac{AH\times BC}2}{\frac{AH\times BC}2-\frac{DE\times AG}2}=k+1$$

$\Rightarrow \frac S{S-DE\times AG}=k+1 \Rightarrow DE\times AG=S\big(1-\frac 1{k+1}\big)=\frac{S\cdot k}{k+1}$

$S=ABC=\sqrt{p(p-a)(p-b)(p-c)}$

$\Rightarrow AH=\frac {2S} a$

Also:

$\frac{AG}{AH}=\frac{DE}{BC}\Rightarrow \frac{DE}{AG}=\frac {BC}{AH}$

Now we have following system of equations:

$\begin{cases}\frac{DE}{AG}=\frac {BC}{AH}=\frac{a^2}{2S}\\DE\times AG=\frac{S\cdot k}{k+1}\end{cases}$

Solving this system gives DE and AG.Now we have:

$$\frac {AD}{c}=\frac{AG}{AH}$$

$$AD=\frac{c\cdot AG}{AH}$$


问:
Proving conjunction of two implications
Let $p, q, r, s$ be propositions. I am supposed to prove the following statement.

$(p \Rightarrow q \ \ \land \ \ r \Rightarrow s) \Rightarrow (p \land r \Rightarrow q \land s) \land (p \lor r \Rightarrow q \lor s)$

Would it be true if I show this statement is a tautology with a truth table and so this implication is always true? If it is not true how can I show this implication or is there any better way for this?

Thanks in advance for any help.

答:
As Shubham Johri mentions in a comment, yes, you can test the validity of the formula through a truth table, but it would be a huge, nightmarish one.

So, when is an implication false? When the antecedent, in this case $(p\rightarrow q)\wedge (r\rightarrow s)$, is true, but the consequent, here $\Big((p\wedge r)\rightarrow(q\wedge s)\Big)\wedge\Big((p\vee r)\rightarrow(q\vee s)\Big)$, is false.

When is a conjunction true and when is it false? When, respectively, both statements are simultaneously true, and when any of the two is false, meaning in our case that $p\rightarrow q$ and $r\rightarrow s$ are true, but either $(p\wedge r)\rightarrow(q\wedge s)$ or $(p\vee r)\rightarrow(q\vee s)$ is false.

If $(p\wedge r)\rightarrow(q\wedge s)$ is false, $p\wedge r$ is true but $q\wedge s$ is false, meaning $p$ and $r$ are true, while $q$ or $s$ is false; either way, we got a contradiction, since $p$, $r$, $p\rightarrow q$ and $r\rightarrow s$ being true imply (by Modus Ponens), that BOTH $q$ and $s$ must be true.

If $(p\vee r)\rightarrow(q\vee s)$ is false, $p\vee r$ is true and $q\vee s$ is false, meaning $p$ or $r$ is true, while both $q$ and $s$ are false. Again, we reach a contradiction regardless of the case, since if $p$ is true, given $p\rightarrow q$ is true one finds $q$ is true; and if $r$ is true, given $r\rightarrow s$ is also true one derives $s$ is true.

Since, by assuming the formula is not a tautology, we reached a contradiction, we must reach the conclusion that the original formula is, in fact, a tautology.

Title: Proving conjunction of two implications


答:
I did it this way:

Calling that if a triangle (and any 2D polygon) is scaled by a factor $\alpha$ then its surface is scaled by a factor $\alpha^2$ then we have that \begin{equation}
ADE=\left(\dfrac{AD}{AB}\right)^2\cdot ABC.
\end{equation}
Also, we can write the surface $ABC$ using the ratio $ADE/BDEC$ given in the following way:
\begin{equation}
ABC=ADE+ADE\cdot\left(\dfrac{BDEC}{ADE}\right)=ADE\cdot\left(1+\dfrac{BDEC}{ADE}\right).
\end{equation}
Joinng both equations we have that
\begin{equation}
ADE=\left(\dfrac{AD}{AB}\right)^2ABC=ADE\left(\dfrac{AD}{AB}\right)^2\left(1+\dfrac{BDEC}{ADE}\right).
\end{equation}
Now, since $ADE\not=0$ (otherwise, the exercise is trivial!) we get that
\begin{equation}
1=\left(\dfrac{AD}{AB}\right)^2\cdot\left(1+\dfrac{BDEC}{ADE}\right).
\end{equation}
Solving for $AD$ we end up with \begin{equation}
AD=\dfrac{AB}{\sqrt{1+\dfrac{BDEC}{ADE}}}.
\end{equation}
Note: Notice that although we divided by $ADE$ in some equations, the final formula does not depend on its value (but in its ratio with $BDEC$, which is given).

Title: Finding the length of a half of a side of a triangle (can be unequal half)

Body:
The Question

Given, $AB$, $AC$ and $BC$. $DE$ is parallel to $BC$. The area ratio between $\triangle ADE$ and $\triangle BDEC$ is also provided. How can I find $AD$?

My Understanding

I know that $\frac{AD}{AB}=\frac{AE}{AC}
=\frac{DE}{BC}$ in this problem.


问:
The (largest) bipartite subgraph
I'm looking for any idea or algorithm that solves the problem of finding bipartite subgraphs in any graph. In general, I'd like to divide graph into the smallest number of bipartite subgraphs. My first thought was to find the largest bipartite subgraph and repeat it until they set of the remaining vertices is empty, but I don't know how to solve this problem in polynomial (or pseudo-polynomial) time.</p>

答:
I assume here that you're looking for induced subgraphs, and I address your "general" problem. It's NP-complete.

A bipartite graph is a 2-colorable graph ; so an induced subgraph that is bipartite is an incomplete (not going through all the vertices) 2-coloration of the graph.
Looking for the smallest decomposition in bipartite graphs, it means looking for a complete coloration of the graph with $2k$ colors, where $k$ is your number of bipartite graphs. Say you have decomposition in $k$ bipartite graphs, then you can easily deduce a $2k$ coloration of your graph, and conversely when you have a $2k$ coloration you can group your colors two by two, and the induced subgraphs will be bipartite.

But deciding whether a graph is $2k$-colorable is NP-complete. So either your subproblem itself is NP-complete, or it is unoptimal for your global problem. Probably both.

Concerning your "local" subproblem, it can be reformulated as finding the biggest set of two colors in the graph. I'd guess that it must be NP-hard ; finding the biggest set of one color (which is the largest anticlique, or independent set) is already strongly NP-hard.

Title: The (largest) bipartite subgraph


问:
Conditional Expectation with discrete Poisson Random Variables
Let $X$, $Y$ be two discrete Poisson Random Variables such that $X \sim \mathcal P(\lambda)$ and $X \sim \mathcal P(\mu)$.

We are asked to evaluate $\mathbb E (X^2+Y^2|X+Y=n)$

We know the sum of $X$ and $Y$ :
$$X+Y \sim \mathcal P(\lambda+\mu)$$
I tried using this methode :

$X^2 + Y^2 = (X+Y)^2 - 2XY$

We get by replacing in the expectation and by the linearity of the expectation :

$$\begin{align*} \mathbb E (X^2+Y^2|X+Y=n)
&= \mathbb E ((X+Y)^2-2XY|X+Y=n)\\
&= \mathbb E ((X+Y)^2|X+Y=n)-2\mathbb E(XY|X+Y=n)\\
&= \mathbb E (n^2|X+Y=n)-2\mathbb E(XY|X+Y=n) \hspace{1cm} \text{since }X+Y=n \\
&= n^2 -2\mathbb E(XY|X+Y=n)
\end{align*}
$$
And here I got stuck calculating the term $\mathbb E(XY|X+Y=n)$

Also this idea
$$\mathbb E (X^2+Y^2|X+Y=n)=\mathbb E (X^2|X+Y=n) +\mathbb E (Y^2|X+Y=n)$$
didn't work quite well. Maybe I still have some problems understanding the concept.

Thank you !

答:
It depends on what you know about Poisson distributions.

If $X$ and $Y$ are independent then the conditional distribution for $X$ given $X+Y=n$ is $\mathrm{Bin}(n, \frac{\lambda}{\lambda+\mu})$, and similarly for $Y$ is $\mathrm{Bin}(n, \frac{\mu}{\lambda+\mu})$

So $\mathbb E[X \mid X+Y=n]=n\frac{\lambda}{\lambda+\mu}$, $\mathrm{Var}(X \mid X+Y=n)= n\frac{\lambda}{\lambda+mu}(1-\frac{\lambda}{\lambda+\mu})$ and  $\mathbb E[X^2 \mid X+Y=n]=n\frac{\lambda}{\lambda+\mu} +(n^2-n)\left(\frac{\lambda}{\lambda+\mu}\right)^2 $ and similarly  $E[Y^2 \mid X+Y=n]=n\frac{\mu}{\lambda+\mu} +(n^2-n)\left(\frac{\mu}{\lambda+\mu}\right)^2$

Thus $E[X^2 +Y^2\mid X+Y=n] = n +(n^2-n)\frac{\lambda^2+\mu^2}{(\lambda+\mu)^2}= n^2 -\frac{2\lambda\mu}{(\lambda+\mu)^2} n(n-1) $


答:
Without using calculus and arithmetic-geometric mean inequality:

Substitute $u=\dfrac xy$ and $u>1$ then,

$$\dfrac {7u^2-2u+3}{u^2-1}=a, a>0$$

$$\implies u^2(7-a)-2u+(3+a)=0$$

$$\implies \Delta =1-(7-a)(3+a)≥0, ~ \text {where} ~0<a≠7 $$

$$\implies a≥ 2+2\sqrt 6$$

$$\implies \text {min} [a]=2+2\sqrt 6<7$$

Finally,

$$u=\dfrac {1}{5-2\sqrt 6}>1$$ which implies $x>y.$

This means, conditions are hold.

Title: Using AM-GM to find the minimum

Body:
Find the minimum value of $\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}}$ if $x$ and $y$ are positive real numbers such that $x > y$.

This is a question from the 22nd Philippine Mathematical Olympiad. I have no idea how to solve this. I am curious as to how it was solved. Can anyone help me, please?

Update: I found a solution here and it looks like it is longer than expected.


问:
Cramer-Rao lower bound for the variance of the unbiased estimator of $\tau(\theta)$
Given the pdf $f(x;\theta)=\frac{1}{\pi[1+(x-\theta)^2]}$ ; $-\inf < x<\inf$, $-\inf < \theta<\inf$

Show that the Cramer-Rao lower bound is 2/n

where n is the sample size.

答:
1- you have the assumption that the samples are independent.

2- calculate the log likelihood ratio for $n$ samples. All likelihoods are multiplicative due to independence.

3- find the second derivative of the log likelihood function.

4- take the minus expectation w.r.t $X$ of what you've found at 3. And take the -1st power.

See http://en.m.wikipedia.org/wiki/Cramér–Rao_bound for more information


问:
Local vs global truncation error
I was reading about local and global truncation error, and, I must be honest, I'm not really getting the idea of the two and what's the difference.

Lets focus on the forward Euler method in particular.

According to the book I'm reading the global error is defined as $$e_i = y(t_i) - y_i, \text{i = 0..N}$$  where, if I understood correctly, $y(t_i)$ is the exact value, whereas $y_i$ is its approximation obtained, in this case, from the forward Euler method. This seems to be the usual error I'm used to, except that here there's no absolute value (or norm in general).

According to my book, the local truncation error satisfies the following:

$$d_i = \frac{y(t_{i+1}) - y(t_i)}{h} - f(t_i, y_(t_i))$$ and also

$$0 = \frac{y_{i + 1} - y_i}{h} - f(t_i, y_i)$$

I don't see why the truncation error satisfies the previous equations. What's the local truncation error and why is it useful? What's the different between the LTE and the global error (which actually for me doesn't seem to be "global")?

Apparently the LTE and the "global" error are not just concepts related to the forward Euler method.

I also had a look at the Wikipedia's article regarding the topic, and it seems to describe the LTE a little bit differently...

答:
Two quite exhaustive answers have already been given.

Local error is the error you get from your truncated approximation of the function of interest and is thus the error you get in a single time step from using your propagator function.

Global error is the accrued error you get from propagating your truncated solution many times and is thus defined for multiple time steps.

I really just want to add a reference that helped me understand it and a nice derivation of it for the Euler method case.By what I understand the procedure depicted in the document is also the general procedure for deriving the global error for a specific method, you try and relate $e_{k+1}$ to $e_{k}$ and in this way derive a bound. This is also a nice video.

P.S. This my first answer on stack and I am also quite new to numerics, please edit or remove if inappropriate.

Title: Local vs global truncation error


问:
Solve for $x$ and $y$ in integers.

In the equation $x^3+y^3=(x+y)^2$, solve for $x$ and $y$ in integers.

So far, I factorized and cancelled out a copy of $(x+y)$ on both sides, leaving me with $x+y=x^2-xy+y^2$. Then, I added $xy+1$ on both sides, and got $(x+1)(y+1)=x^2+y^2+1$. I can't continue from this point.

答:
The simplified expression that you obtained defines an ellipse:

Additionally, the line $y=-x$ also satisfies the relation and so, apart from the solutions already found on the elipse ($(0,0), (1,0), (0,1),(2,1),(1,2)$ and $(2,2)$), you also get all points of the form $(n,-n), n \in \mathbb{Z}$.

Title: Solve for $x$ and $y$ in integers.


问:
Rotation of a wheel in the positive direction and clockwise?
The Question

I’m riding my bike at a constant speed of 20 miles per hour on flat
land when i ride over a dot of fresh yellow paint, thereby getting a
yellow spot on my front wheel, which is 2 feet in diameter. I continue
riding my bike in a straight line at 20 miles per hour. When t seconds
passed since running over the dot, how many feet is the yellow spot on
my tire from the yellow dot i ran over?

My Understanding

I faced some difficulties in understanding an exercise about parameterization (pre-calculus level). I understood the solution except this point; the spot starts at the bottom of the circle, corresponding to an angle of $\frac{3\pi}{2}$. We let the wheel move in the positive direction, so that the wheel rotates clockwise

Is it possible that we can move in the positive direction with a clockwise rotation?

答:
The positive direction is given to be the direction you are traveling. The clockwise vs counterclockwise direction of rotation depends on the side of the bike to the observer. If you are on the right side of the bike – it moving to your right – the wheel will appear to rotate clockwise.

The wheel moves horizontally according to a cycloid curve:
$$\quad x=r(1-\cos \theta)\quad$$ and
$\theta$
changes at a rate of
$$\dfrac{d}{\pi}=\dfrac{20\times5280\space ft/sec}{3600}\bigg/2\pi\space ft=\dfrac{44}{3\pi\space}\dfrac{radians}{sec}.\quad$$
If we start at
$\space \theta=\dfrac{3\pi}{2}\space $ and continue in the clockwise direction, we get
$$d=x=r(1-\cos \theta)\quad\text{where}\quad
\theta=\dfrac{3\pi}{2} -\dfrac{44t}{3\pi}$$


问:
Confusion about length of module over different rings and Macaulay2 code about computing length of module.
I was trying to compute some examples dealing with length of modules and got stuck with this simple example:

Let $R=k[t]/(t^2)$ where $k$ is a field and $J=(x^3)$ be the ideal of the polynomial ring $R[x]$. I want to compute the length $l_{R}(R[x]/J)$ and $l_{R[x]}(R[x]/J)$.

Using the Macaulay2 code, I found the length $l_{R[x]}(R[x]/J) =6$. (I don't know how to use Macaulay2 to compute $l_{R}(R[x]/J)$ though).

This is where I got confused. Is this true that the composition series of $R[x]/J$ over $R[x]$ is $0\subseteq (x^2) \subseteq (x) \subseteq R[x]/J$? If so, $l_{R[x]}(R[x]/J)=3$. Can we have a larger composition series here? If not, is it still true that if $R$ has finite length and $J$ is a monomial ideal in the polynomial ring, the length of $R[x_1,\ldots ,x_n]/J$ over $R[x]$ is the number of monomials that not in $J$ (as the case when $R$ is a field)? And I can't see why we can have a length-6 composition series for $R[x]/J$ over $R$.

Thank you in advance for any help!

答:
In your series ideal $(x)$ is not even maximal: $R[x]/(x)=R$ which is not a field. A composition series of length six is
$$
0 \subset (x^2t) \subset (xt) \subset (x^2) \subset (x^2, t) \subset (x, t) \subset R[x]/J.
$$


问:
Prove that $b^2-4ac$ is NOT a perfect square.
Given that the $3$-digit number $abc= 100a+10b+c$ (not to be confused with $a*b*c$) is a prime, $p$, prove that $b^2-4*a*c$ cannot be a perfect square.

So far, I tried to prove it by contradiction. Let $f(x)=ax^2+bx+c$, so that $f(10)=p$, and both roots of $f(x)$ must be negative integers (since $a,b,c$ are digits). Then, I substituted the values in for $p$, and got $p=a(10-x_1)(10-x_2)$, where $x_1, x_2$ are the roots of $f(x)$. $p$ must divide one of the factors, say $(10-x_1)$, but I cannot continue from this point.

答:
If $b^2 - 4ac$ is a perfect square, considering the quadratic formula or its derivation, we see that the quadratic equation you mentioned, $ax^2 + bx + c$, has two negative rational roots. Hence we can write

$$f(x) = ax^2 + bx + c = (kx + l)(mx +n )$$

for positive integers $k,l,m,n$. We have $f(10) = p = (10k+l)(10m+n)$, so $p$ must have two 2-digit factors. This contradicts the fact that $p$ is prime.


问:
Graph $f(x,y)=\ln(x)-y$
I'm trying to graph $f(x,y)=\ln(x)-y$, however, I am not sure how as all of my tools are refusing to graph it.

Can you please help me?

Thanks

答: You cannot draw it on the $Oxy$-plane. You have two input values $x$ and $y$ and so you need one more axis to represent the value of $f$. Below is the graph drawn via GeoGebra:

Title: Graph $f(x,y)=\ln(x)-y$


答:
Note that there is no way to plot this in 2D because you would need an equation of the form $f(x,y) = 0$, and what you have is $f(x,y) = \ln(x) - y$, which can be thought of as $f(x,y) = z$.  That means you need a 3D plot to visualize it.

There are several good online options for plotting 3D graphs.  Here are two good ones:

1.) Calcplot3D, which produces the following plot.  This one is my personal favorite, but you might want to see here for key commands for camera controls.

2.) Geogebra, which produces the plot

Title: Graph $f(x,y)=\ln(x)-y$

Body:
I'm trying to graph $f(x,y)=\ln(x)-y$, however, I am not sure how as all of my tools are refusing to graph it.

Can you please help me?

Thanks

Body:
Note that there is no way to plot this in 2D because you would need an equation of the form $f(x,y) = 0$, and what you have is $f(x,y) = \ln(x) - y$, which can be thought of as $f(x,y) = z$.  That means you need a 3D plot to visualize it.

There are several good online options for plotting 3D graphs.  Here are two good ones:

1.) <a href="https://c3d.libretexts.org/CalcPlot3D/index.html?type=z;z=ln(x)-y;visible=true;umin=-6;umax=6;vmin=-6;vmax=6;grid=50;format=normal;alpha=-1;constcol=rgb(255,0,0);view=0;contourcolor=red;fixdomain=true&type=window;hsrmode=3;nomidpts=true;anaglyph=-1;center=-6.513098527506768,-7.382907616178462,1.7527757135334454,1;focus=0,0,0,1;up=-0.028634991195109524,0.2547385033929391,0.9665859155648717,1;transparent=false;alpha=140;twoviews=false;unlinkviews=false;axisextension=0.7;xaxislabel=x;yaxislabel=y;zaxislabel=z;edgeson=true;faceson=true;showbox=true;showaxes=true;showticks=true;perspective=true;centerxpercent=0.5;centerypercent=0.5;rotationsteps=30;autospin=true;xygrid=false;yzgrid=false;xzgrid=false;gridsonbox=false;gridplanes=false;gridcolor=rgb(128,128,128);xmin=-6;xmax=6;ymin=-6;ymax=6;zmin=-6;zmax=6;xscale=5;yscale=5;zscale=5;zcmin=-4;zcmax=4;zoom=0.277778;xscalefactor=1;yscalefactor=1;zscalefactor=1" rel="nofollow noreferrer">Calcplot3D, which produces the following plot.  This one is my personal favorite, but you might want to see here for key commands for camera controls.

2.) Geogebra, which produces the plot


问:
Density of gradient of restricted-function in $L^p$.
Let $A \subset Q$ be two regular bounded subset of $\mathbb{R}^3$ and denotes

$$V=\{\nabla (\varphi|_{A}) \ |\ \varphi \in C^\infty_0(Q) \}.$$

It is clear that $V \subset L^p(A)^3$ for any $p \in ]1,+\infty[$ and therefore $\bar{V} \subset L^p(A)^3$ with respect to the norm $||\cdot||_{L^p(A)^3}$.

Can we prove the opposite inclusion and that therefore $V$ is dense in $L^p (A)^3 $ ?

答: This is not true: We have $\nabla \times v=0$ for all $v\in V$. So these functions in $V$ cannot approximate non-curl-free functions.


问:
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?
The population of a certain bacteria can multiply threefold in 24 hours. If there are 500 bacteria now, how many will there be in 96 hours?

I figured out this bacteria $=500(3)^{96/24}$

but then my mate told me it's bacteria $=500(1+3/1)^{96/24 }$

what do I do what is the next step.

答:
Regardless of the ambiguity, that solution doesn't look right because that doesn't seem like the right rate constant. In order to solve an exponential growth problem we normally assume a growth given by
$$ P(t) = P(0) e ^{k t}$$
We can find k by using the given information (there is where the ambiguity lies, I agree that threefold prabably means 3 times the present value like the rest).
If we then use that
$$P(24) = 3 P(0) = P(0) e ^{k 24}$$
we can then find k and the problem will be a little easier, since once we have k we can then find $P(t)$ for any future value of $t$.</p>


问:
Lower bound for number of points of elliptic curve over finite field
I'm asked this kind of question:

Show there exists $q+1$ points on the elliptic curve over $\mathbb{F}_q$ given by $y^2 = x^3-x$ when $q \equiv 3 (mod 4)$

The fact that I'm asked for an approximation shows you that I don't have much tools available to prove it. Basically, I tried Hasse's theorem but I don't get reasonable conclusion (in fact I get a contradiction).

Is there a simple method to produce such an approximation?

Body:
Since $q\equiv 3\bmod 4$ then $-1$ is not a square thus for any $a\in \Bbb{F}_q-\{0,1,-1\}$ exactly one of $\sqrt{a^3-a}, \sqrt{(-a)^3-(-a)}$ is in $\Bbb{F}_q$ which means that

$$2\ \# \{ (a,b)\in \Bbb{F}_q{}^2, a\ne 0,1,-1,b^2=a^3-a\} = 2 (q-3)$$

Adding $(\infty,\infty),(0,0),(1,0),(-1,0)$ we get that
$$\# \{ (a,b)\in \Bbb{F}_q{}^2, b^2=a^3-a\}\cup O = q+1$$


问:
How Many Uniquely Enumerated $4\times4$ Sudoku Grids Exist?
I'm looking to find how many uniquely enumerated $4\times4$ Sudoku grids exist. I am aware that there are other questions with solutions to this question, however I am asking again as there seem to be conflicting answers online. Some say $384$ whilst some say $288$.

I make my solution to be 384, as I see that there are $4!$ ways of enumerating the first (top left) $2\times2$ block, and then there are $2!$ ways of enumerating the remaining cells of each of the first column, second column, first row and second row. Once the first block and the adjacent blocks (blocks two and three, if you will) are fully enumerated, the fourth block (bottom right) can only be enumerated in one way due to the restrictions on number placement caused by the enumeration of blocks 2 (top right) and 3 (bottom left).

From this, I therefore make my total number of possible enumerations to be:
$$4!\times2!\times2!\times2!\times2!=384.$$

It is clear that quite a few others have calculated it in a slightly different way and made their solution to be 288. This doesn't make sense to me as it seems that the total should be divisible by 8, as the any given enumerated grid should have 8 symmetries (rotations and flips) which are all equally valid sudoku enumerations.

Is my calculation incorrect or are some of the answers online incorrect?

Edit: I made a typo initially and wrote 388 instead of 288. I was concerned that 8 did not divide 388, but it does in fact divide 288 - which makes the issue of divisibility by 8 go away.

答:
Not all of your $384$ possibilities actually work.  For example one of your $384$ choices seems to be

$\begin{matrix}
1&2&|&3&? \\
3&?&|&2&? \\
-&-&+&-&-  \\
2&3&|& ? & ?   \\
?&?&|& ? & ?
\end{matrix}$

forcing

$\begin{matrix}
1&2&|&3&4 \\
3&4&|&2&1 \\
-&-&+&-&-  \\
2&3&|& 1/4 & ?   \\
4&1&|& ? & 2/3
\end{matrix}$

but it is impossible to complete the grid.

Your argument about $8$ symmetries might fail if there is a pattern which is rotationally symmetric with a $180^\circ$ turn, which there is:

$\begin{matrix}
1&2&|&3&4 \\
3&4&|&1&2 \\
-&-&+&-&-  \\
2&1&|& 4 & 3   \\
4&3&|& 2 & 1
\end{matrix}$


问:
Prove that $\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{ \frac {36-8a}{a+9}}}\leqslant 3+\sqrt{3}$
For $0\leqslant a \leqslant 3.$ Prove that:
$$f(a)=\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{
\frac {36-8a}{a+9}}}\leqslant 3+\sqrt{3}$$

The text for LHS:  sqrt(2a+(36-8a)/(a+9)+2sqrt(a^2+3))+sqrt((36-8a)/(a+9))

I found that the equality holds when $a=3$ or $a=1.$ I also found that when $3\ge a\ge 1$ then $f'(a)\ge 0$ but I can not prove it and I don't know what to do in the opposite case. Helps me!

答:
Consider that you look for the extrema of function $$f(a)=\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{ \frac {36-8a}{a+9}}}$$ Its ugly first derivative is almost $0$ very close to $a=1$
$$f'(1)=\frac{12}{5 \sqrt{55}}-\frac{27}{10 \sqrt{70}}\sim  0.000904209$$ Similarly
$$f''(1)=\frac{4161}{4400 \sqrt{55}}+\frac{27}{1400 \sqrt{70}}\sim 0.129821 >0$$ confirms that it is a minimum. We also have
$$f(1)=\frac{1}{5} \left(2 \sqrt{55}+\sqrt{70}\right)\sim  4.63980$$

Now, you "just" need to prove that, in the range, $f'(a)$ does not cancel a second time.

Edit

If we make a series expansion around $a=1$, the minimum is
$$a_{min} \sim \frac{27 \left(5687 \sqrt{14}-3871 \sqrt{11}\right)}{67963 \sqrt{11}+1089 \sqrt{14}}\approx 0.993035$$
$$f(a_{min} )\approx 4.639796308297$$ while a full optimization gives
$$a_{min}\approx 0.993066 \qquad \text{and} \qquad f(a_{min} )\approx 4.639796308235$$

In the same spirit, we could show that the first derivative goes though a maximum close to $a=2$ (exactly at $a\approx 2.06292$) and an expansion around $a=2$ shows that it would cancel again around $a\approx 3.23940 >3$ (the exact value would be $a\approx3.24253$).


问:
Is Dividing By a Percentage Valid in the Context of Determining Cost per Acquisition/Return on Investment?
I'm new to the site, and am hoping someone can assist. I'm currently trying to find the return on investment (ROI), and I can't figure out if the formula below is valid in the context it's being applied; please see the first part of the formula below (keeping in mind this is example data):

$90 / 0.3 = 300$

Above, 90 represents the cost per acquisition (cost to obtain a lead), and 0.3 represents the close ratio (the percentage of leads that result in a sale).

The second part of formula continues as follows:

$300 - 100 - 300 = -100$

Directly above, the first 300 value listed represents the revenue of a sold lead, the 100 value represents operation costs to process the sale, and the second 300 value is the value that was generated in the first part of the formula. As you can see, this results in a negative ROI, which doesn't quite make sense to me.

Please let me know if the logic above is valid in the context of generating CPA/ROI figures.

Thank you.

答:
The first calculation is fine.  You are claiming that $30\%$ of the leads result in a sale and computing that it takes $300$ leads to make $90$ sales.

Note:  this was written when the first paragraph talked about numbers of leads and orders, before it was changed to the cost of leads and orders.

The second makes no sense.  The units do not match.  The first $300$ and $100$ are dollars, but the second $300$ is the number of leads.  What you should do is subtract the cost of fulfilling the order ($100$) and the cost of generating an order.  It takes you $\frac 1{0.3}=\frac {10}3$ leads to generate an order.  You should multiply that by the cost of a lead to get the profit on one order.  If a lead costs you $45$, the profit on one order is
$$300-100-45\cdot \frac {10}3=50$$


问:
What's meant by "regularization" of ODEs?
What's meant by "regularization" of ODEs?

Such as, "in order to be solved by conventional ODE solvers such as ode45"?

The context where I encountered this was related to discontinuities of ODEs. And there it seemed that by adding some term or function, one may fix the discontinuous behavior.

答:
To encourage simple solutions and avoid overly complicated solutions.

To make the problem more well posed and robust to noise in the data or instabilities caused by properties in the numerical solver.

You can read more here Regularization (Wikipedia)


问:
Online course in representation theory or differential geometry
Are there any courses in representation theory that are available online? I'm looking for a course including videos, notes as well as assignments. I'd also be interested in a course in differential geometry. I couldn't find anything. The MIT open courseware courses are all without video. </p>

答:
See this link and this one.

In particular:

Introduction to Lie Algebras

Representation theory

Differential geometry


问:
Exercise 11.1.G Vakil FOAG
I am trying to solve Ex 11.1.G from Ravi Vakil's FOAG. It says if $X$ is an affine scheme over $k$, a field and $K|_k$ is an algebraic field extension, then $X$ is of pure dimension $n$ iff $X_K:=X\times_k K$ is of pure dimension $n$.

Here's my attempt at a solution. Let $X=\operatorname{Spec}A$. Then $X_K=\operatorname{Spec}(A\otimes_k K)$. The canonical map $X_K \rightarrow X$, in this case corresponds to the homomorphism of $k$-algebras namely $A\rightarrow A\otimes_k K$; $a\mapsto a\otimes 1$

Since $K|_k$ is an integral extension, so is $A\otimes_k K|_A$. Thus the Going-Up theorem holds true.

Also since $K$ is a free $k$-module, $A\otimes_k K$ is a free $A$-module and faithfully flat. Since $A\hookrightarrow A\otimes_k K$ is a flat ring extension, the Going-Down theorem holds true.

Now say $P$ be a minimal prime ideal of $A\otimes_k K$. Then $p:=P \cap A$ is a minimal prime ideal of $A$ by Going-Down theorem. Conversely if $p$ is a minimal prime of $A$ and $P$ lies over $p$ (since it is an integral extension) then $P$ is a minimal prime ideal of $A\otimes _k K$ by Incomparability.

Assume $X$ is of pure dimension $n$. Then for any minimal prime $p$ of $A$, we have $\operatorname{coht}p=\dim X$. If $P$ is a minimal prime of $A\otimes _k K$, then  $\operatorname{coht} P =\operatorname{coht} P\cap A=\dim X $ by Going-Up Theorem and Incomparability. Thus $X_K$ is of pure dimension $n$ as well.

Now assume $X_K$ is of pure dimension $n$. Then if $p$ is a minimal prime ideal of $A$, then $p=P\cap A$ where $P$ is a minimal prime of $A\otimes_k K$. Again by Going-Up and Incomparability, we get $\dim X_K=\operatorname{coht}P=\operatorname{coht} P\cap A $. Thus $X$ is of pure dimension $n$.

I hope this is a correct solution. I am pretty sure there exists alternative solution along the lines of Vakil's hints or preferably without using flatness since Vakil does not develop it upto this point but I simply can't get it. To be more precise, Vakil has first asked to reduce the problem to $A$ being an integral domain. I am stuck with this part without using the Going-Down theorem for flat ring extensions. Any help/insight/suggestion will be most welcome.

答:
Does this work for the reduction?  If $Q\subset A':=A\otimes_k K$ is a minimal prime, then $PA'\subset Q$ for some minimal prime $P\subset A.$  Thus it suffices to check that $A'/PA'$ is pure of dimension $n$. But $$A'/PA' = (A\otimes_k K)\otimes_A (A/P)= (A\otimes_A A/P)\otimes_k K = A/P\otimes_k K   $$ So we've reduced to $A/P\subset A/P\otimes_k K$, as desired.

The only thing that seems slightly sketchy is the second equality in the centered line...

Edit: If you downvote, please let me know what's wrong with the argument, as I am interested in the correct reduction.

Title: Exercise 11.1.G Vakil FOAG


问:
Derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ without qoutient rule
I'm trying to find derivative of  $\frac{\cos t-\sin t}{\cos t+\sin t}$ in a different way: there is a trick to find derivative of the form $\frac{ax+b}{cx+d}$:
$$\left(\frac{ax+b}{cx+d}\right)'=\frac{ad-bc}{(cx+d)^2}$$
I'm trying to use it for taking derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ too. the result of  is $\frac{-2}{(\cos t+\sin t)^2}$. but I don't know how to use this method for the function

答: Your function is$$\frac{1-u}{1+u}=\frac{2}{1+u}-1$$with $u:=\tan t$, so by the chain rule its derivative is $\frac{-2\sec^2t}{(1+\tan t)^2}=\frac{-2}{(\cos t+\sin t)^2}$.


问:
Why is the inverse image sheaf a sheaf of rings
Given an inclusion of topological spaces $i:X \hookrightarrow Y$ where $(Y,\mathcal{O}_Y)$ is a ringed topological space, I'm trying to understand why $(X,i^{-1}\mathcal{O}_Y)$ is a ringed topological space.

Now I see why $(i^{-1}\mathcal{O}_Y)_x$ for any $x \in X$must be a local ring, given that stalks of the inverse image sheaf are the same as stalks on the sheaf $\mathcal{O}_Y$. But while trying to write the entire proof out I realized I didn't quite understand what the ring structure on the inverse image sheaf looks like (for this inclusion case and in general).

By definition, the inverse image sheaf is the sheaf associated to the presheaf $U \mapsto$ lim $_{W \supset i(U)}\mathcal{O}_Y(W)$, which going by Hartsehorne's construction of the sheafification (page 64, Prop-Def 1.2), is the set of functions

$\{f:U \rightarrow \bigcup_{p \in U} ($lim $_{W \supset U}\mathcal{O}_Y(W))_p\}$

But I don't see explicitly what the ring structure on this looks like. Any help would be appreciated

答:
What is an element of $(\lim_{W\supseteq i(U)}\mathcal{O}_Y(W))_p$?  Well, it is an element of $\lim_{W\supseteq i(U)}\mathcal{O}_Y(W)$ for some neighborhood $U$ of $p$ (modulo the equivalence relation given by restriction to smaller neighborhoods).  And what is such an element?  It is just an element of $\mathcal{O}_Y(W)$ for some $W$ that contains $i(U)$ (again, modulo an equivalence relation that lets us restrict to smaller neighborhoods).  Now suppose you have a second such element, which is an element of $\mathcal{O}_Y(W')$ for some $W'$ that contains $i(U')$ for some other neighborhood $U'$ of $p$.  How do you add or multiply these two elements?  Well, you just restrict them both to $\mathcal{O}_Y(W\cap W')$, which contains $i(U\cap U')$, and then add or multiply them using the ring structure of $\mathcal{O}_Y(W\cap W')$.

In more abstract terms, all we're doing in this construction is taking certain directed colimits of rings of the form $\mathcal{O}_Y(W)$.  The forgetful functor from rings to sets preserves directed colimits, so we could have just taken all our colimits in the category of rings, rather than in the category of sets.  This would make it so everything is obviously a ring, and we'd still have the same underlying sets.


答:
Another approach: we can rewrite your function as
$$
\frac{\cos t - \sin t}{\cos t + \sin t} = \frac{\cos(t)\sin(\pi/4) - \sin(t)\cos(\pi/4)}{\cos(t)\cos(\pi/4) + \sin(t) \sin(\pi/4)}
= \frac{\sin(\pi/4 - t)}{\cos(\pi/4 - t)} = \tan(\pi/4 - t).
$$
By the chain rule, we find the derivative of this function to be
$$
-\sec^2(\pi/4 - t) = -\frac{1}{\cos^2(\pi/4 - t)} = -\frac{1}{[\cos(t)/\sqrt{2} + \sin(t)/\sqrt{2}]^2}
\\= \frac{-2}{(\cos(t)+\sin(t))^2}.
$$

Title: Derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ without qoutient rule

Body:
I'm trying to find derivative of  $\frac{\cos t-\sin t}{\cos t+\sin t}$ in a different way: there is a trick to find derivative of the form $\frac{ax+b}{cx+d}$:
$$\left(\frac{ax+b}{cx+d}\right)'=\frac{ad-bc}{(cx+d)^2}$$
I'm trying to use it for taking derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ too. the result of  is $\frac{-2}{(\cos t+\sin t)^2}$. but I don't know how to use this method for the function


问:
Strategy while solving quadratics by splitting the middle term
While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)

答:
As you know, the factors of $\space 1344\space$ are $\space{1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 24, 28, 32, 42, 48, 56,\\ 64, 84, 96, 112, 168, 192, 224, 336, 448, 672, 1344}$

You should work down the largest factor where  $x<124$ because "it" and its cofactor must add up to $124$.  It happens on the first try here that
$\quad 112+12=124\quad\land\quad  112\times 12=1344\quad$ so there you have it. If you did not find a "pair" by the time the factor became less than the cofactor, then the solution(s) would be rational or irrational, possibly complex, but not integers.


答:
You can also do:
$$\frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \frac{(\cos{t}-\sin{t})(\cos{t}-\sin{t})}{(\cos{t}+\sin{t})(\cos{t}-\sin{t})} = \frac{1-\sin{2t}}{\cos^2{t}-\sin^2{t}} = \frac{1}{\cos{2t}}-\tan{2t} = \sec{2t}-\tan{2t}$$ Take the derivative to get $$2\sec{2t}(\tan{2t}-\sec{2t}) = \frac{-2}{(\cos{t}+\sin{t})^2}$$ after some manipulation.


问:
Show that $\lim_{n \to \infty}(b_1b_2...b_n)^{\frac{1}{n}}=b$
Let $b_n>0$ with the following property: $\lim_{n \to \infty}b_n=b>0$.

Show that $\lim_{n \to \infty}(b_1b_2...b_n)^{\frac{1}{n}}=b$.

I have an idea and at one moment of the proof i'm stuck. I would like to know if my idea holds and how to finish it (if it works), please.

As $(b_n)$ converges to $b$, we have by definition of convergent sequence that:

$\forall \varepsilon>0 \ \exists N>0 \ \forall n\ge N$: $|b_n-b|<\varepsilon \iff |b_n|<\varepsilon+b$.

We want to show that: $\forall \varepsilon>0 \ \exists N'>0 \ \forall n\ge N'$: $|(b_1b_2...b_n)^{\frac{1}{n}}-b|<\varepsilon \iff |b_1b_2...b_n|<(\varepsilon+b)^{n}$.

But,

$|b_1b_2...b_n|=|b_1b_2...b_N...b_n|=|b_1b_2...b_{N-1}|\cdot|b_N...b_n|<|b_1b_2...b_{N-1}|\cdot
(\varepsilon+b)^{n-N+1}$

After this step i'm really stuck and don't know how to continue if it is possible. I tried as well to bound $|b_1b_2...b_{N-1}|$ by $M^{N-1}$ with $M=\max\{x_k:1\le k\le N-1\}$ but as it doesn't tend to $0$ i can't to anything using it...

P.S I know that there are some posts on this problem, but i would like to know if my approach works and if not to understand why (or have corrections). Thanks in advance for help!

答:
By the AM-GM-HM inequality, for $n \in \mathbb{N} :$ $$\frac{\sum b_i}{n} \geq (b_1b_2 \dots b_n)^{\frac{1}{n}} \geq \frac{n}{\sum \frac{1}{b_i}}$$

Prove and use the following Lemma : If $\displaystyle\lim_{n \to \infty} a_n = L$ for a sequence $(a_i)_{i \in \mathbb{N}}$ where each $a_i \in \mathbb{R}$, then $\displaystyle\lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = L$.

Using the above lemma and Sandwich theorem, conclude that $(b_1b_2 \dots b_n)^{\frac{1}{n}} = b$.


问:
Diagonalization of complex, symmetric but defect matrices
Real symmetric matrices (not Hermetian $M^\dagger = M$, but symmetric $M^T = M$) can according to Autonne–Takagi factorization, as described here, be diagonalized. Thus, we can find a unitary matrix $U$ such that $A = U D U^T$, for a diagonal matrix $D$. However, as described here, the matrix
$$
A = \begin{pmatrix}
2i&1\\
1&0
\end{pmatrix}
$$
is defect, in that it only has one eigenvector $(i, 1)^T$, and thus $U$ can't be made up of eigenvectors as it usually is when diagonalizing. However, writing $U^T = (u_1, u_2, ..., u_n)$, for some vectors $u_i$, we have that
$$
U A U^T = D \implies A U^T = A (u_1, ..., u_n) = (Au_1, ..., Au_n) = U^T D = (u_1, ..., u_n) \begin{pmatrix}
d_1 &0 &0  \\
 0& ... &0 \\\
0&0 & d_n
\end{pmatrix}
= (d_1 u_1, ..., d_nu_n).
$$
So it seems that $U$ is made up of the eigenvectors of $A$ anyway. Want went wrong?

答:
It is not true that
$$
UAU^T = D \implies AU^T = U^TD.
$$
Note that in the case that $A$ is defective, $U$ is unitary but not orthogonal. That is, we have $U^{-1} = U^\dagger \neq U^T$.


问:
Solve for $x$ and $y$ in integers.

In the equation $x^3+y^3=(x+y)^2$, solve for $x$ and $y$ in integers.

So far, I factorized and cancelled out a copy of $(x+y)$ on both sides, leaving me with $x+y=x^2-xy+y^2$. Then, I added $xy+1$ on both sides, and got $(x+1)(y+1)=x^2+y^2+1$. I can't continue from this point.

答:
Let's write everything on one side and factor (and not discard infinite families of solutions).
\begin{align*}
x^3 + y^3 - (x+y)^2 &= 0  \\
(x+y)(y^2 - xy + x^2 -y -x) &= 0
\end{align*}

A product is zero when any factor is zero, so, from $x+y = 0$, we find the infinite family of solutions
$$  y = -x  \text{.}  $$
Now we attack $y^2 - xy + x^2 -y -x = 0$.  Note that this is the equation of an ellipse with major axis parallel to the line $y = x$ and  centered at $(1,1)$.  If you sketch that ellipse, you will find that it is in the box $\left[\frac{1}{3}(3 - 2 \sqrt{3}),\frac{1}{3}(3 + 2 \sqrt{3})\right] \times{}$ $\left[\frac{1}{3}(3 - 2 \sqrt{3}),\frac{1}{3}(3 + 2 \sqrt{3})\right]$.  So we check the nine integer points in that box. (We could exclude $(1,1)$ since it is the center, but the automatic/unthinking way to proceed is to check every point in the box).
\begin{align*}
&x  &  &y  &  y^2 - xy + &x^2 -y -x  &  \\ \hline
&0  &  &0  &  0&  &\checkmark  \\
&0  &  &1  &  0&  &\checkmark  \\
&0  &  &2  &  2&  &  \\
&1  &  &0  &  0&  &\checkmark  \\
&1  &  &1  &  -1&  &  \\
&1  &  &2  &  0&  &\checkmark  \\
&2  &  &0  &  2&  &  \\
&2  &  &1  &  0&  &\checkmark  \\
&2  &  &2  &  0&  &\checkmark  \\
\end{align*}

So the full solution set is $\{(x,y) \mid y = -x\} \cup {}$ $\{(0,0)$, $(0,1)$, $(1,0)$, $(1,2)$, $(2,1)$, $(2,2)\}$.

Title: Solve for $x$ and $y$ in integers.


问:
Adjoining element in a finite field $\mathbb F_{p^n}$ to $\mathbb F_p$ to get $\mathbb F_{p^6}$
Given that $p$ is a prime and $n$ is a natural number, I want to know how many $\alpha$ are in $\mathbb F_{p^n}$ such that $\mathbb F_p (\alpha) = \mathbb F_{p^6}$.

I know that $[\mathbb F_{p^6} : \mathbb F_p] = 6$, so $\mathbb F_p (\alpha) = \mathbb F_{p^6}$ would happen if and only if the irreducible polynomial of $\alpha$ with coefficients in $\mathbb F_p$ has a degree of $6$, but not sure how to proceed from here.

答:
To long for a comment, let me know if there are mistakes in my argument.

For $a,b\in \mathbb N$ it holds $\mathbb F_{p^a} \subset \mathbb F_{p^b} \Leftrightarrow a|b$, so if $n$ is not a multiple of $6$ the field $\mathbb F_{p^6}$ is not contained in $\mathbb F_{p^n}$.

If $n$ is a multiple of $6$, fix an algebraic closure $\bar{\mathbb F}$ of $\mathbb F_p$. The elements of $\mathbb F_{p^n}$ are exaclty the roots of $X^{p^n}-X\in \mathbb F[X]$ in $\bar{\mathbb F}$,  and the elements of $\mathbb F_{p^6}$ are exaclty the roots of $X^{p^6}-X$. So the elements $\alpha\in \mathbb F_{p^n}$ generating for $\mathbb F_{p^6}$ are the elements in this field which are not contained in a proper subfield: $\mathbb F_{p}$, $ \mathbb F_{p^2}$ and $\mathbb F_{p^3}$, therefore there are
$$p^6-(p^3-p)-(p^2-p)-p= p^6-p^3-p^2+p$$
generating elements, since $\mathbb F_{p^3}\cap \mathbb F_{p^2}= \mathbb F_{p}$.


问:
Question about uniqueness function series
Please pardon me if this is elementary, but I've looked hard for an answer to this and am very surprised I have yet to find a good one. My question is simple: under what condition(s) does $$\sum_{n=1}^{\infty}f_n(x) = \sum_{n=1}^{\infty}g_n(x)$$ imply that $$f_n(x) = g_n(x)$$?</p>

答:
It's hard to find any reasonable conditions for such an implication. The reason is simply that given any series of the form $\sum_{n=1}^\infty f_n(x)$ that converges, then changing the order of summation, or making small or large changes to $f_1$ only to absorb these in other $f_k(x)$ gives too much freedom for finding infinitely many possibilities for functions $g_k(x)$ that will add up to the same value.

If you place some severe restrictions on the functions though then some (not terribly interesting) implications are possible. For instance, if all functions attain non-negative values, and moreover the inequalities $f_k(x)\le g_k(x)$ hold for all $x$ in the relevant domain, then equality of the series will imply the equality of the functions (in the relevant domain).

But really, this is hopeless. It's like asking for conditions that will assure that two series of real numbers are equal. There are simply too many possibilities. Even with sums of just two real numbers, what conditions are there on $a,b,c,d$ that ensure that if $a+b=c+d$, then $a=c$ and $b=d$???


问:
How do I know what scales to look for for asymptotic expansion? E.g. cubic case?
How do I know what scales to look for for asymptotic expansion? E.g. cubic case?

How do I check for their validity?

Such as in the example here:

http://www.macs.hw.ac.uk/~simonm/ae.pdf

where they find that

$x^3-x^2-(1+\epsilon)x+1=0$

has double root at $x=1$ and it suggests them to try an expansion in powers of $\epsilon^{1/2}$.

But one could want to solve e.g.

$x^3-5x^2+4x+\epsilon=0$

Given here:

http://sycon.rutgers.edu/~speer/528s17/Perturbation-DG.pdf

which has roots $x=0,x=1,x=4$.

答:
In the first equation you have
$$
(x+1)(x-1)^2=ϵx\iff (x-1)^2=\frac{ϵx}{1+x}
$$
Close to $x=1$ this gives approximations for the roots there of
$$
x=1\pm\sqrt{\frac{ϵx}{1+x}}\approx 1\pm\sqrt{\fracϵ2}
$$
For the root close to $-1$ you get similarly
$$
x=-1+\frac{ϵx}{(1-x)^2}\approx -1+\fracϵ4
$$

For the second equation goes the same,
$$
x(x-1)(x-4)=-ϵ
$$
gives approximations close to the roots of the undisturbed equation of
$$
-\fracϵ4,~~1+\fracϵ3,~~ 4-\fracϵ{12}
$$


问:
Using AM-GM to find the minimum

Find the minimum value of $\dfrac{7x^{2} - 2xy + 3y^{2}}{x^{2} - y^{2}}$ if $x$ and $y$ are positive real numbers such that $x > y$.

This is a question from the 22nd Philippine Mathematical Olympiad. I have no idea how to solve this. I am curious as to how it was solved. Can anyone help me, please?

Update: I found a solution here and it looks like it is longer than expected.

答:
Let, $x=a+b$ and $y=a-b$ where $a>b$, we have

$$f(x,y)= \dfrac {2a^2+3b^2}{ab}+2$$

$$\implies \underbrace{\dfrac { 2a^2+3b^2}{ab}≥\dfrac{2\sqrt 6 ab}{ab}}_{\text{ arithmetic-geometric mean inequality}}=2 \sqrt 6$$

$$\implies \text{min} [ f(x,y)]= 2+2 \sqrt 6$$

REMARK.

According to equality condition of arithmetic-geometric mean inequality, we have

$$2a^2=3b^2 \Longrightarrow a= \dfrac {b\sqrt 3}{\sqrt 2}$$

which follows

$$y := b \sqrt {\dfrac {3}{2}}-b>0$$

So, conditions are hold.

Title: Using AM-GM to find the minimum


问:
How do I calculate the Time Complexity of a polynomial within a square root?
So say I have the following polynomial $F(n)=\sqrt{5n^2+7n-13}$ and I want to calculate it's upper and lower bounds ($O(n)$ and $\Omega(n)$), as well as the overall complexity $\Theta(n)$. What is the general process for doing so if there is a square root? I know for a normal polynomial I am proving for example $f=O(g)$ & $f=\Omega(g)$ by using the $c$ and setting $n_0 = 1$ in order to prove the inequalities. However when it comes down to having a square root, I'm at a loss of where/how to start. My initial thoughts are I have to use $g(n)=\sqrt(n^2)$ but at second look that doesn't make a whole lot of sense to me? Can anyone point me in the right direction?

答: Intuitively, when $n$ is very large the first term dominates over the other two, so $F(n)\approx \sqrt{5n^2}=n\sqrt 5$.  We don't care about multiplicative constants, so $F(n) \in \Theta(n)$.  You can make this more precise if you want by dividing the $n$ out of the square root and seeing that the other terms have factors of $\frac 1n$


答: Notice that $\sqrt{5n^2+7n-13}=\sqrt{5}n\sqrt{1+\frac{7}{5n}-\frac{13}{5n^2}}$ and that $\lim\sqrt{1+\frac{7}{5n}-\frac{13}{5n^2}}=1$ as $n\to\infty$, so asymptotically the function grows as $\sqrt{5}n$.

Title: How do I calculate the Time Complexity of a polynomial within a square root?

Body: So say I have the following polynomial $F(n)=\sqrt{5n^2+7n-13}$ and I want to calculate it's upper and lower bounds ($O(n)$ and $\Omega(n)$), as well as the overall complexity $\Theta(n)$. What is the general process for doing so if there is a square root? I know for a normal polynomial I am proving for example $f=O(g)$ & $f=\Omega(g)$ by using the $c$ and setting $n_0 = 1$ in order to prove the inequalities. However when it comes down to having a square root, I'm at a loss of where/how to start. My initial thoughts are I have to use $g(n)=\sqrt(n^2)$ but at second look that doesn't make a whole lot of sense to me? Can anyone point me in the right direction?


问:
Does any theory of infinite quantities provide info for $-1$ raised to infinite power, its absolute value, finite part and a series or integral form?
I've heard about surreal numbers, hyperreal numbers, Hardy fields, nonstandard analysis, cardinal ariththmetic, ordinal arithmetic, games, etc.

My impression is that neither of them can exactly show the result of the following operation: $(-1)^w$, where $w$ is some infinite number from that theory.

The only theory that exactly gives a result is extended real line ${\overline {\mathbb {R} }}$, where, of course, $(-1)^\infty=0$ (because of Cesàro mean).

Am I correct in my skepticism about all these theories?

答:
There are somewhat natural sine and cosine functions on the class $\mathbf{No}$ of surreal numbers. The class of zeroes of $\sin$ is $\pi \ \mathbf{Oz}$ where $\mathbf{Oz}$ is Conway's ring of omnific integers (a class of "integers" for $\mathbf{No}$).

For $a \in \mathbb{No}$, one can define $(-1)^a:=\cos(a \pi)+i\sin(a \pi) \in \mathbf{No}[i]$ (algebraic closure), which yields $(-1)^{a+b}=(-1)^a (-1)^b$ for all $a,b \in \mathbf{No}$. Note that for $z \in \mathbf{Oz}$, we have $(-1)^z=1$ if $z$ is even and $(-1)^z=-1$ if $z$ is odd.

Now I don't claim that this definition is useful or interesting. Is it the right definition of $(-1)^a$? To answer that, one should first say what one expects of such an operation...


问:
What does it mean by "The differential of a map is independent of coordinate charts"?
We know that the differential is represented by the Jacobian matrix $[\partial F^i/\partial x^j(p)]$, so it definitely depends on the choice of coordinates $x^j$. But the statement in question is written in Tu's introduction to smooth manifolds in the discussion of regular level set theorem. Can somebody please explain what that statement mean?

答:
Quick answer:

The differential of a map associate a vector $df[v]$ to any vector $[v]$, and that vector $df[v]$ depends only on $v$, not on the coordinate system.

Detailed discussion:

What is a map? A map $f:M\to \mathbb R^n$ associates to any point  $x\in M$ a vector $f(x)$. The value $f(x)$ depends only on the point, not on local coordinates.

What is the differential of a map? The differential of a map is not a function from $M$ to $\mathbb R^n$. One usual way to define the differential of a function $f:M\to \mathbb R^n$ is the following:

$f$ is differentiable at $x\in M$ if there exists a linear map $L$ so that $$f(x+\epsilon v)=f(x)+\epsilon L(v)+o(\epsilon)$$

In other words, if $f$ is approximable by a linear map at first order. So, now look at $L$. Who is $L$? It is a linear map bur from where to where? The origin vector space is the tangent space of $M$ at point $x$, usually denoted $T_xM$, and the target space is the tanget space at $f(x)$. In the present case, we can canonically identity the tangent space at $f(x)$ with $\mathbb R^n$ (but if you have  $f:M\to N$ you have to take $T_{f(x)}N$). So $$L\in\hom(T_xM,\mathbb R^n)$$.

The map $L$ is called the differential of $f$ at $x$ and it is usually denoted by $d_xf$.

Now, it is a general linear algebra fact that if you have two finite dimensional vector spaces $V,W$ then, for any choices of basis $B_V,B_W$ of $V$ and $W$, you can associate to any $F\in\hom(V,W)$ the matrix of $F$ in those chosen basis. The linear map $F$ exists independently on $B_V,B_W$: it is the matrix that changes when you change coordinates.

Well, for $df$ is the same: the linear map $d_xf:T_xM\to \mathbb R^n$ does not depend on coordinates, the Jacobian matrix --- which describes $d_xf$ as matrix --- does indeed depend on coordinates.


问:
Pattern for all the binary chains divisible by 5
For instance, $x = 101$ is divisible by $5$ because it is the integer 5. Same thing for $x=1111$ is also divisible by 5 as it is the integer 15. However, $x=1100$ is not divisible by $5$ as it is the integer 12.

Is there a pattern the recognise the binary chains divisible by 5?

答:
We can use the happy face of five-divisibility!

Start in state $0$. Follow the appropriate arrows as you read digits from your binary number from left-to-right. If you end up in state $0$ again your number is divisible by $5$ (and if not, the state number gives you the remainder).

How does it work? Well if we're in state $k$ it means the digits we have read so far form the number $n$ with remainder $k \equiv n \mod 5$. If we then read another digit $b$, we effectively move to the new number $n' = 2n + b$. Thus we need to move to state $(2k + b) \bmod 5$, which is exactly what we do in the above graph. Thus if we end up in state $0$ in the end we know there is no remainder, and the number that we read is divisible by 5.

The state diagram above is just this logic graphically displayed. You could have it as a table instead as well:

\begin{array}{ccc}
k & b & 2k + b & (2k + b) \bmod 5\\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 2 & 2 \\
1 & 1 & 3 & 3 \\
2 & 0 & 4 & 4 \\
2 & 1 & 5 & 0 \\
3 & 0 & 6 & 1 \\
3 & 1 & 7 & 2 \\
4 & 0 & 8 & 3 \\
4 & 1 & 9 & 4 \\
\hline
\end{array}

This also makes for a nice mental rule. You start with the number $0$ in your head and look at the digits from left-to-right. For each digit you multiply the number in your head by 2 and add the digit you just read. If the number goes to five or above you subtract five. If you end up with $0$ the number is divisible by 5.

As an example for the binary number $1111101000_2 = 1000$, you go:

```<code>              0 is our starting value
1111101000
^             2*0 + 1 = 1
1111101000
 ^            2*1 + 1 = 3
1111101000
  ^           2*3 + 1 = 7, is >= 5 so we subtract 5
              7 - 5   = 2
1111101000
   ^          2*2 + 1 = 5, is >= 5 so we subtract 5
              5 - 5   = 0
1111101000
    ^         2*0 + 1 = 1
1111101000
     ^        2*1 + 0 = 2
1111101000
      ^       2*2 + 1 = 5, is >= 5 so we subtract 5
              5 - 5   = 0
1111101000
       ^      2*0 + 0 = 0
1111101000
        ^     2*0 + 0 = 0
1111101000
         ^    2*0 + 0 = 0
</code>```

Our remainder is $0$, thus we are divisible by five!

Body:
We can use the happy face of five-divisibility!

Start in state $0$. Follow the appropriate arrows as you read digits from your binary number from left-to-right. If you end up in state $0$ again your number is divisible by $5$ (and if not, the state number gives you the remainder).

How does it work? Well if we're in state $k$ it means the digits we have read so far form the number $n$ with remainder $k \equiv n \mod 5$. If we then read another digit $b$, we effectively move to the new number $n' = 2n + b$. Thus we need to move to state $(2k + b) \bmod 5$, which is exactly what we do in the above graph. Thus if we end up in state $0$ in the end we know there is no remainder, and the number that we read is divisible by 5.

The state diagram above is just this logic graphically displayed. You could have it as a table instead as well:

\begin{array}{ccc}
k & b & 2k + b & (2k + b) \bmod 5\\
\hline
0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 2 & 2 \\
1 & 1 & 3 & 3 \\
2 & 0 & 4 & 4 \\
2 & 1 & 5 & 0 \\
3 & 0 & 6 & 1 \\
3 & 1 & 7 & 2 \\
4 & 0 & 8 & 3 \\
4 & 1 & 9 & 4 \\
\hline
\end{array}

This also makes for a nice mental rule. You start with the number $0$ in your head and look at the digits from left-to-right. For each digit you multiply the number in your head by 2 and add the digit you just read. If the number goes to five or above you subtract five. If you end up with $0$ the number is divisible by 5.

As an example for the binary number $1111101000_2 = 1000$, you go:

              0 is our starting value
1111101000
^             2*0 + 1 = 1
1111101000
 ^            2*1 + 1 = 3
1111101000
  ^           2*3 + 1 = 7, is >= 5 so we subtract 5
              7 - 5   = 2
1111101000
   ^          2*2 + 1 = 5, is >= 5 so we subtract 5
              5 - 5   = 0
1111101000
    ^         2*0 + 1 = 1
1111101000
     ^        2*1 + 0 = 2
1111101000
      ^       2*2 + 1 = 5, is >= 5 so we subtract 5
              5 - 5   = 0
1111101000
       ^      2*0 + 0 = 0
1111101000
        ^     2*0 + 0 = 0
1111101000
         ^    2*0 + 0 = 0

Our remainder is $0$, thus we are divisible by five!


问:
Line integral over vector fields
Evaluate the line integral $\oint_
CH\cdot n\mathrm{d}s$
through the sum of the integrals over the sides
of :
$$∮
_
H · n\mathrm{d} =
∫
_{1}
H · n\mathrm{d} +
∫
_{2}
H · n\mathrm{d} +
∫
_{3}H · n  +
∫
_{4}
H · n\mathrm{d}$$
where 1, 2, 3 and 4 are the four sides of .
enter image description here

The link has the diagram needed. Please help. I have used a variant of green's theorem and used $\mathrm{d}s = \mathrm{d}x~\hat{\mathrm{i}} + \mathrm{d}y~\hat{\mathrm{j}}$ which then allows me to use $n\mathrm{d}s  = \mathrm{d}y~\hat{\mathrm{i}} - \mathrm{d}x~\hat{\mathrm{j}}$. After this working out the equation of each line allows me to substitute and do the integration with respect to $\mathrm{d}x$. I have done this and got 4 as my answer but when I do the double integral to check my answers don't match. $H = x~\hat{\mathrm{i}}+y~\hat{\mathrm{j}}$

答:
You have made a mistake in your line integral somewhere as the sum should be zero and not $4$.

Please note that if the vector field is $\vec H = (x,y)$,

$Q_x = P_y = 0$ where $P = x, Q = y$.

So the line integral over a closed curve will be zero.

Line integral $\displaystyle = \oint_C (Q_x - P_y) \ dA = 0$

Line integral over $S2$ in anticlockwise direction,

$r(t) = (1,2) + (0-1, 1-2)t = (1-t,2-t), 0 \leq t \leq 1$

Line integral over $S2 = \displaystyle \int_0^1 (1-t,2-t) \cdot (-1,-1) \ dt = - 2$

Line integral over $S4$ in anticlockwise direction,

$r(t) = (1,0) + (2-1, 1-0)t = (1+t,t), 0 \leq t \leq 1$

Line integral over $S4 = \displaystyle \int_0^1 (1+t,t) \cdot (1,1) \ dt = 2$

So the sum over $S2$ and $S4$ is zero. Also line integral along $S1$ and $S3$ is zero. Please do the working and check.

Title: Line integral over vector fields


问:
what does the professor mean by "the convergence to the limit point is fast?"
I am considering the following facility location problem:

Let $S$ be a subset of interval $[0,1]$, which denotes the set of positions of agents. With a little abuse of notation, $S$ also denotes the set of all agents. For instance, $S=\{1, \frac{1}{2}, \frac{1}{3}, \cdots\}$ indicates that there is an agent in $1, \frac{1}{2}, \frac{1}{3}, \cdots$, respectively. Now we need to locate a facility on the real line(to serve all the agents). Suppose the $y\in \mathbb{R}$ is the location of the facility, then the cost of agent $x\in S$ is definte as $cost(x, y):=|x-y|$. The total cost is the sum of all agents' cost, i.e., $\sum_{x\in S} |x-y|$.

In a discussion with a professor by email, he made the following three observations and asked me to check their correctness:

If $y$ is not a limit point of $S$ then the total cost is infinite.

The total cost is finite iff $y$ is the only limit point of $S$ and the convergent to the limit point is fast.

Suppose there are $m$ facilities, define the cost of an agent to be the distance to the closest facility. Then the total cost is bounded iff each facility is in a limit point of $S$ and the convergence to each facility is fast.

I understood his observations except the "convergence to the limit point is fast". And I proposed $S=\{1, \frac{1}{2}, \frac{1}{3},\cdots\}$ as an counter example to observation $2$ since the only limit point here is $0$ but the total cost $\lim_{n=1}^{\infty}\frac{1}{n}$ is unbounded. He argued that it is not a counter example because the convergence is slow.

I've learnt the concepts of the rate of convergence and the order of convergence, but what does he mean by the convergence is fast or slow? I think there must be some missing prerequisite knowledge I should know first.

答:
$\newcommand{\d}[1]{\mathtt{\large{\underline{#1}}}}$I will cover, for the sake of others who may not have understood what you could from the Professor's e-mail, the entire trail and logic of reasoning of the Professor. Therefore, while OP will know some of the material here, it is important everybody can read through the entire mail.

The setting

The basic idea is that of serving agents. You have agents, and they all require a certain facility. For example , people live in a certain locality and you want to place a grocery store to satisfy their food requirements. The way you place the store would be based on how far you are from each of them. Each person will associate a cost , which is the amount of effort they must expend to reach the store, and you want to satisfy everybody to the best of your ability. This is modeled in the following way mathematically.

We have a subset $S \subset [0,1]$. There is meant to be an agent or customer at $S$ i.e. a person who needs to be served. There is a single facility or grocery store to satisfy everybody, located at a $y \in [0,1]$. Then we talk about the cost on the entire locality of having the store at this place.

The notion used here is the notion of the $l_1$ distance. So for each point $x \in S$, the cost of having the facility at $y$ is $|y-x|$, which you can think of as the distance to travel from $y$ to $x$. Then we sum over $x \in S$ to get the function $f(y) = \sum_{x \in S} |y-x|$.

$f$ is then the cost to the locality of having the facility at $y$. The reason why $f$ can be interpreted as such , I have already expounded on.

Naturally, the question to be asked is : does $\min_{y \in [0,1]} f(y)$ exist? Is it unique?  Can we find it(or one of the minima) in a certain way (maybe using an algorithm)? Of course you wish to answer these questions so you can place the locality in a place obtained by using such an algorithm.

The questions addressed of $\color{blue}{\text{existence}}$, $\color{red}{\mathrm{uniqueness}}$ and  $\color{green}{\mathrm{construction}}$ are all part of the general notion of well-formulation or well-posedness of a problem in the field of optimal control theory. Indeed , the specific question addressed by the Professor is a very minimal example in the word of optimal control.

On the well-definition of $f$

First of all, $f$ is defined as the sum over all $x\in S$ of non-negative quantities. It can be noticed that if any reasonable notion of summation allowed $S$ to be uncountable , then $f$ is obsolete, in the sense that it is either zero or infinite everywhere. Thus, such a situation is not interesting. You can see here for the details on that.

We therefore restrict ourselves to $S$ being either finite or countable. However, it seems that all assertions have been made for the countable case. With this, let us prove the first assertion generalized :

If $S$ is countable and $y$ is not a limit point of $S$ then $f(y) = \infty$. On the other hand, if $S$ is finite then $f(y)< \infty$ for all $y \in [0,1]$.

Proof : The easy case first. If $|S| = m< \infty$ then $|x-y| \leq 1$ for all $x \in S, y \in [0,1]$ so $f(y) \leq m < \infty$ for any $y \in [0,1]$.

The harder case now : if $|S| = \infty$ and $y$ is not a limit point of $S$, then in particular there exists an $\epsilon>0$ and points such that $|x - y| > \epsilon$ for all $x \in S$ (using the non-finiteness of $S$ here). But then $f(y)\geq \sum_{i=1}^n |x_i - y| \geq n \epsilon$ (ordering the points in $S$ like $x_1,x_2,...$ arbitrarily) for all $n$ so $f(y) = \infty$.

This means that $f$ doesn't make sense whenever $y$ is not a limit point of $S$. But what if $S$ had no limit points? The answer to that question is addressed by the following result :

Every infinite subset of a compact subset of $\mathbb R$ has a limit point.

This is a consequence of the Bolzano-Weierstrass theorem. Thus, we can rest assured that $S$ will certainly have limit points, and $f$ can possibly be finite at  one of these limit points, so we at least have a problem to work with. We may then seek to look at the behaviour of these points.

Looking at the limit points

The problem with more than one limit point, is that if points are close to one limit point they are not close to the other, so that makes $f$ infinite at both these limit points.  However, even if $y$ is a limit point of $S$, the rate at which points in $S$ approach $y$ may not be good enough for summation. It is illustrative to explain this with examples.

Let $S = \{1,\frac 12, \frac 13 , ...\}$ (not including $0$). Then $0$ is the only limit point of $S$ (easy to see, it is a monotonically decreasing sequence and then one can use the Archimedean axiom), so we can look at $f(0) = \sum_{n \in \mathbb N} \frac 1n$. This is the famous harmonic series, which is known to be infinite. Thus $f(0) = \infty$. In particular, for $S = \{1,\frac 12,...\}$, the problem of finding a facility is obsolete!

We may consider a slight modification : let $S = \{1 , \frac 14 , \frac 19 ,...\}$ denote the squares of the reciprocals of the natural numbers. Once again, $0$ is the only limit point. However, note that $f(0) = \frac 1{n^2} < \infty$ (a well-known result). Thus, the problem is well-defined, and furthermore we have that $0$ must be the unique minimizer, since $f$ is infinite at all other points.

For some situation involving multiple limit points, let $$S = \left\{\frac 14 , \frac 19, \frac 1{16},...\right\} \cup \left\{1 - \frac 1{8} , 1 - \frac 1{27} , 1 - \frac 1{64} ,...\right\}$$, so it's the squares of the reciprocals except $1$, along with the cubes of the natural numbers except $1$, subtracted from $1$. The deliberate construction of the sequence tells us that $0,1$ are the only limit points. However, one can see that $f(1) = f(0)=\infty$, because although the first part of the definition of $S$ converges to $0$, it goes far away from $1$, so $f(1) = \infty$. Similar the second part converges to $1$ so goes far away from $0$ so $f(0) = \infty$. So the problem becomes obsolete.

We can now prove using the examples, the following result for $S$ countable:

For $S$ countable, there is at most point $y \in [0,1]$ such that $f(y) < \infty$. That point, if it exists, is the unique limit point of $S$.

Proof : Suppose that $f(y_1)<\infty$ and $f(y_2) < \infty$ for $y_1 \neq y_2$. Then they are limit points from the explanation in the previous section, so there exist subsequences $x_1,x_2,...$ and $z_1,z_2,...$ of $S$ such that $|x_i - y_1|< \frac{|y_1 - y_2|}{2}$ and $|z_i - y_2| < \frac{|y_1 - y_2|}{2}$ for all $i$. By the triangle inequality, $|x_i - y_2| > \frac{|y_1 - y_2|}{2}$ and $|z_i - y_1|> \frac {|y_1-y_2|}2$ for all $i$. It follows that $f(y_1) = f(y_2) = \infty$.

Which means that $f(y)<\infty$ at possibly only one point, which must a limit point as well. There cannot be any other limit point by the previous argument so $y$ is the unique limit point of $S$.

Now, given $S$ we can assume that $y$ is the unique limit point of $S$ and talk about $f(y) = \sum_{x \in S} |x-y|$. However, as I illustrated in the first two examples above, the finiteness of this sum depends upon how fast $|x-y|$ approaches $0$ as $x$ approaches $y$.

The theory of convergence of series will dictate how fast it needs to be. Tests like the ratio test, comparison test etc. are the basic tools in this direction.

The notions of rate and order are important, because these dictate the long-term behaviour of how $|x-y|$ looks like for $x$ close to $y$, and thus determine the summability of $f(y)$. For example, if $S = \{\frac 2n\}$ then the rate of convergence to the limit point $0$ is $O(\frac 1n)$ which is too slow. However, take $S = \{\frac 1{2n^3} + \frac 1{2n^2}\}$ then the convergence to $0$ is $O(\frac 1{n^2})$ which is fast enough for convergence.

The example proposed by the OP is not a counterexample, but rather an example! Indeed, the  shrinking of the harmonic series is too slow to allow convergence to happen, as Professor's point number $2$ asserts.

Finally, to address $S$ finite :

If $S$ is finite then $f(y)$ is a continuous function of $y$.

Proof : $f(y) = \sum_{x \in S} |x-y|$ is now a finite sum of continuous functions, hence continuous.

Thus $f$ attains a minimum on $[0,1]$, so the problem is well-defined! You can try to see when the minimum is unique (for example, $S$ consisting of one point?) by experimentation : it is not unique all the time!

On the construction part : if $S$ is infinite, then the question is of the location of the limit point of $S$, if it is unique, and if the rate of convergence is fast enough, which are all found analytically from a symbolic expression of $S$ (I actually don't know how $S$ is specified in finite terms if it is an infinite set , so I won't speak of the algorithmics). In the case of $S$ finite, the additional factor helping us is that $f$ is convex, whence we can apply a gradient descent or so to get a minimum.

We see that the problem is a bit obsolete, in the case of $S$ countable (for $S$ finite, it is a tractable problem, even if the solution is not unique). We allow more flexibility for $S$ countable, by changing the $f$.

A changed $f$ and the third point

We change our setup now. We have $m$ competing grocery stores , or facilities, and each person prefers the grocery store closest to him or her. That is , for a facility located at $y$, we look at all the $x$ such that $x$ is closer to $y$ than to any other facility, call that subset $S'$, and then $f(y) = \sum_{x \in S'} |x-y|$. So the cost of a facility doesn't take everyone into account now, but because of competition only those customers which are likely to be closest to it.

To be rigorous : Let $\{y_1,...,y_m\}$ be the locations of the facilities. For $y \in \{y_1,...,y_m\}$ let $S'_y = \{x \in [0,1] : |x-y| \leq |x-y_i| \text{ for all} i = 1,2,...,m\}$, and then define :
$$
f(y) = \sum_{x \in S'_y} |x-y|
$$

With this in mind, let us look at example $3$ that we wrote earlier. There, the cost was infinite.Now, with this new metric , this is not the case.

Indeed, $1$ and $0$ are the two limit points, but the sums for both are finite, because we aren't counting the points closer to $0$ in the sum for $1$, and vice-versa. In that case, we have under the new $f$ :
$$
f(0) = \frac 14 + \frac 19 + \frac 1{16} + ... , f(1) = \frac 1{8} + \frac 1{27} + ...
$$

and $f(1)<f(0)<\infty$ (To see $f(1)<f(0)$ we can use a pointwise minimization that $\frac 18<\frac 14 , \frac 1{27}<\frac 19$ etc. and sum up). So $1$ is a better facility than $0$ under the cost structure now imposed!

We can instantly see how the change of $f$ has improved the quality of the problem we seek to solve. This is innovative mathematics in a nutshell : changing the parameters of the problem to produce and give meaning to new unexplored situations.

With this cost structure, we can talk about the total cost $F(y_1,...,y_m) = \sum_{i=1}^m \sum_{x \in S'_{y_m}} |x-y_m|$ where $F : [0,1]^m \to [0,\infty]$. So you place each of the facilities, and total up the costs for each agent. You want to minimize $F$ , naturally, so you can place your facilities there. The question is : when is $F$ finite?

The answer to this question is given wrongly by the Professor. The correct statement is this:

Let $F$ be given as previously. Then $F$ is finite if and only if for each $i$ we have either $S'_{y_i}$ is finite or $y_i$ is a limit point of $S'_{y_i}$ (and hence of $S$) and the decay $|x-y_i|$ is fast enough.

Proof : Let any $y_i$ not be a limit point of $S$. It is then not a limit point of $S'_{y_i}$ either. Suppose that $S'_{y_i}$ is finite, then the resulting sum $\sum_{x \in S'_{y_i}} |x-y_i|$ is finite even if $y_i$ is not a limit point of $S$!

However, if $S'_{y_i}$ is infinite, then we can use the $\epsilon$ method we used earlier to get a contradiction. Thus $y_i$ must be a limit point of $S'_{y_i}$.

Now the statement is near obvious, a finite sum of quantities is finite if and only if each one is finite, so $F$ is finite if and only if each summand is finite.

Let me give an example to clarify this. Let $S = \{\frac 14, \frac 19,\frac 16,...\} \cup \{0.85,0.95,0.99\}$. Let's say we want two facilities. I can place one facility at $0$ to accommodate for the first sequence. To acommodate the finitely many leftover points, I can put the second facility anywhere close to them to retain finiteness of $S'$ of this point! For example, $F(0,0.75) < \infty$ although $0.75$ is not a limit point of $S$. In fact, $F(0,a)<\infty$ for any $a$, since any other $a$ is not a limit point so the set $S'$ for it will be finite and hence sumability will follow.

For finite $S$, things are clear anyway. Construction and uniqueness arguments are skipped, these are similar to the previous situation.

Conclusion

In conclusion, we have seen two problems in optimal control and studied their well-definition and the existence of solutions to such problems. This serves as a primer for subsequent theory in optimal control, which includes :

Convex analysis and optimization (which is interesting because the individual cost functions were convex in the first part).

Using theorems to assert existence (we had existence here from the theorem that a continuous function on a closed and bounded interval attains its minimum : generalizations exist, but the more common tool seen is the Banach fixed point theorem and cousins)

Innovating our notion of distances to allow for flourishing of existing theory (the second problem was much richer than the first, and generalized the first problem in fact. We can also attempt to perform an $l_2$ optimization by taking $(\sum_{x \in S} (y-x)^2)^{\frac 12}$ in place of the $l_1$ sum, which leads to very interesting mathematics because there's an explicit formula for a minimizer in this case).


问:
How many different Shidoku Boards are there?
```<code>A Shidoku board is a 4 x 4 grid of numbers where each of the numbers 1-4
appears exactly once in each row, column, and in each of the four 2 x 2 sub-
grids. How many different shidoku boards are there?"
</code>```

For the first row there are $4!$ different possibilities. Second row there are $4$ different possibilities. Now for the third row... the bottom-right sub-board row depends on the bottom-left sub-board top row.

Ex:

1 2 3 4

3 4 2 1

4 1 x x

The first $x$ couldn't be 3 or 2 since that would mess up the column and it can't be 4 or 1 since those numbers are already used.

This is the point where I am hitting my mental wall and am not progressing in the problem.

答:
This is just a slight variant of 6005's (old) answer. The number of Shidoku boards is $4!\times2\times2$ times the number of boards of the form

```<code> a b c d
 c d ? ?
 b ? d ?
 d ? ? ?
</code>```

(That is, the numbers $\{1,2,3,4\}$ can be assigned to the letters {a,b,c,d} in any of $4!$ ways, after which the third and fourth columns can be swapped or not, as can the third and fourth rows.) The main new feature here is the presence of a "d" in the lower right hand quadrant, whose position is forced. Now the "a" in the lower right hand quadrant has $3$ positions it can go in. And for each of these, it's easy to see that the positions of the "b" and "c" are forced, at which point the rest of the unspecified entries (in the upper right and lower left quadrants) are also forced. So the total number of Shidoku boards is $4!\times2\times2\times3=288$.

Body:
This is just a slight variant of 6005's (old) answer. The number of Shidoku boards is $4!\times2\times2$ times the number of boards of the form

 a b c d
 c d ? ?
 b ? d ?
 d ? ? ?

(That is, the numbers $\{1,2,3,4\}$ can be assigned to the letters {a,b,c,d} in any of $4!$ ways, after which the third and fourth columns can be swapped or not, as can the third and fourth rows.) The main new feature here is the presence of a "d" in the lower right hand quadrant, whose position is forced. Now the "a" in the lower right hand quadrant has $3$ positions it can go in. And for each of these, it's easy to see that the positions of the "b" and "c" are forced, at which point the rest of the unspecified entries (in the upper right and lower left quadrants) are also forced. So the total number of Shidoku boards is $4!\times2\times2\times3=288$.


问:
Find determinant for a matrix with $x$ on the diagonal
Find determinant for a matrix with $x$ on the diagonal and $-1$ at the end of the first row and $-1$ at first of the last row other elements are $0$.

```<code>{ x  0 0 -1
  0  x 0  0
  .  0 x
  .  0 0  x
  -1 0 .  .  x
}
</code>```

$$
\pmatrix{x & 0 & \cdots & 0 & -1\\
0 & x & \ddots & \vdots & 0\\
\vdots & \ddots & \ddots & 0 &\vdots\\
0& \cdots & 0 & x&0\\
-1&0&\cdots & 0 & x}
$$

I know if i add 1/xR1 to R2 the determinat will be $(x^{n-1})(x-1/x)$. I asked my professor she said what if $x$ is zero. Now i don't know what to do.

答: If $x=0$, then the determinant will be zero ( cause there is 1 row with all elements are zeros )


问:
If $(a^{n}+n ) \mid (b^{n}+n)$ for all $n$, then $ a=b$
I happened to receive this from my friend.

Let $a,b \in \mathbb{N}$, such that $a^{n}+n \: \bigl|\: b^{n}+n$ for all $n \in \mathbb{N}$. Prove that $a=b$. How do we proceed?

答:
Chandru1 asks how we might proceed, so in that spirit let me offer an idea that provides partial results and connects this problem to a long-standing one recently characterized as a "frustrating question."

Let's aim to show that under the conditions assumed of $a$ and $b$, $a$ necessarily divides $b$.  This additional relation suffices to show that $a = b$.  (There does not seem to be a simple demonstration of this, but it is much easier than the original problem so I'll let it go for now as an "exercise.")

Suppose $b < a^2$.  Then

$\left( \frac{b}{a} \right) ^n - \frac{b^n + n}{a^n+n} = \frac{n (b^n - a^n)}{a^n (a^n+n)}$.

The right hand side, in the limit of large $n$, approaches zero from above.  The left hand side is the difference between $\left( \frac{b}{a} \right) ^n$ and an integer.  We conclude that eventually the fractional part of $\left( \frac{b}{a} \right) ^n$ approaches zero.  Results of Pisot, Vijayaraghavan and Andre Weil then imply that $\frac{b}{a}$ must be integral.  (See Akayama et al., Powers of rationals modulo 1 and rational base number systems, http://perso.telecom-paristech.fr/~jsaka/PUB/Files/RBNS-rev.pdf .)  The intuition is that the fractional parts of powers of non-integers ought to fill the interval [0, 1) in a fairly "random" way.  Indeed, numerical experiments verify this for rational numbers (but not for all irrationals!): see http://mathworld.wolfram.com/PowerFractionalParts.html .  So, convergence of the fractional part to zero--which is assured by the sequence of divisibility conditions in the problem--implies the ratio $b / a$ is not behaving like a proper fraction: it must actually be an integer.  That gives us enough leverage to show the equality $a = b$.

I suspect a similar approach should work for $b \ge a^2$, but I haven't found it.  Indeed some of the papers in the literature note changes in the behavior of powers of $b / a$ when $b$ exceeds $a^2$, so we should be cautious.

Finally, note that elementary methods of number theory show that all primes dividing $a$ must also divide $b$.  That, however, doesn't seem to get us very far.

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问:
Range of: $\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$
Range of:
$$\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$$

Any help will be appreciated.

答: HINT: where are the functions defined?, where is a root defined? what can you already say without knowing anything about what is in the roots? Given that you cant do complex</p>