问:
Showing $d$ is a metric
The metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by

$$d(x, y) = \left\{
  \begin{array}{lr}
      ||x||+||y|| & \mbox{if } x \neq y \\
      0 &  \mbox{if } x = y \\
      \end{array}
\right.$$
is a variation of the SNCF metric. Let $x = (a_1, b_1)$ and $y = (a_2, b_2)$. Prove that $d$ is a metric.

My attempt:

Axiom 1.

Suppose $d(x, y) = 0$. Then $||x|| + ||y|| = \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} = 0$.

This is a sum of roots so $a_1^2 + b_1^2 = 0$ and $a_2^2 + b_2^2 = 0$. The sum of squares can only be $0$ if the terms are $0$. So $a_1 = 0$, $b_1 = 0$ and $a_2 = 0$, $b_2 = 0$. Therefore $(a_1, b_1) = (a_2, b_2)$.

Going the other way, suppose $x = y$. From the very definition of the metric $d(x, y) = 0$.

Axiom 2.

Notice that

$$d(y, x) = \left\{
  \begin{array}{lr}
      ||y||+||x|| & \mbox{if } y \neq x \\
      0 &  \mbox{if } y = x \\
      \end{array}
\right.$$

You can see that in the top piece of the proposed metric, $||x|| + ||y|| = ||y|| + ||x||$ by commutativity of addition. On the other hand the bottom piece is $0$ which is present in both cases. Hence $d(x, y) = d(y, x)$.

Axiom 3.

Case 1. $x \neq y \neq z$ and $x \neq z$.

$||x|| \leq ||x|| + ||z||$ and $||y|| \leq ||y|| + ||z||$

Adding these: $||x|| + ||y|| \leq ||x|| + ||z|| + ||z|| + ||y|| \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 2. $x \neq y$ and $y = z$.

Since $y = z$ we know $d(z, y) = 0$ and $||y|| = ||z||$.

$||x|| + ||y|| = ||x|| + ||z|| + 0 \implies d(x, y) = d(x, z) + d(z, y) \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 3. $x \neq y$, $y \neq z$ and $x = z$.

Since $x = z$ we know $d(x, z) = 0$ and $||x|| = ||z||$.

$||x|| + ||y|| = 0 + ||z|| + ||y|| \implies d(x, y) = d(x, z) + d(z, y) \implies d(x, y) \leq d(x, z) + d(z, y)$

Case 4. $x = y$.

Since $x = y$ we know $d(x, y) = 0$ and $||x|| = ||y||$.

If $y = z$ then $x = y = z$ and so $d(x, y) = d(x, z) = d(y, z) = 0$ $\implies d(x, y) \leq d(x, z) + d(z, y)$.

If $y \neq z$ and $x = z$ then $d(x, y) = d(x, z) = 0$. But $d(x, y) = 0 \leq 0 + ||y|| + ||z|| = d(x, z) + d(y, z)$

If $y \neq z$ and $x \neq z$ then $||x|| + ||z|| + ||z|| + ||y|| > 0 \implies d(x, z) + d(z, y) \geq d(x, y)$.

So in any case the triangle inequality holds.

QED.

Is this proof correct? Did I go through all the cases for the triangle inequality or did I miss something there? Also is there a simpler way to tackle the triangle inequality axiom or is it necessary to do it in cases as I have done. Any assistance is much appreciated.

答:
Your argument for Axiom $1$ isn’t right, because $d(x,y)=0$ does not imply that $\|x\|+\|y\|=0$. What is true is that either $x=y$, which is fine, or $\|x\|+\|y\|=0$. And since $\|z\|\ge 0$ for all $z\in\Bbb R^2$, $\|x\|+\|y\|=0$ if and only if $\|x\|=\|y\|=0$, which in turn is true if and only if $x=\langle 0,0\rangle=y$. Thus, in all cases we find that $x=y$.

It’s actually a little easier to prove the contrapositive: if $x\ne y$, then $d(x,y)\ne 0$. If $x\ne y$, then without loss of generality we may assume that $x\ne\langle 0,0\rangle$, in which case $\|x\|+\|y\|\ge\|x\|>0$.

For Axiom $3$ you’ve covered the cases, but the argument could safely be shortened considerably:

If $z=x$, then $d(x,y)=d(z,y)=d(x,z)+d(z,y)$. Similarly, if $z=y$, then $d(x,y)=d(x,z)=d(x,z)+d(z,y)$. And if $x\ne z\ne y$, then

$$\begin{align*}d(x,z)+d(z,y)&=\|x\|+\|z\|+\|z\|+\|y\|\\
&\ge\|x\|+\|y\|\\
&\color{red}\ge d(x,y)
\end{align*}$$

since $\|z\|\ge 0$. (Note that the red inequality is true whether or not $x=y$, since $\|x\|+\|y\|$ is always non-negative.)

It would be a good exercise to try to figure out what topology on $\Bbb R^2$ is generated by this metric; I’ve left a brief description in the spoiler-protected box.

 If $x\ne\langle 0,0\rangle$, then $B(x,\|x\|)=\{x\}$, so $\{x\}$ is open for every $x\in\Bbb R^2\setminus\{\langle 0,0\rangle\}$. On the other hand, for each $r>0$ we have $$B(\langle 0,0\rangle,r)=\left\{x\in\Bbb R^2:\|x\|<r\right\}\,,$$ the usual open ball of radius $r$ centred at the origin. Thus, the topology is discrete everywhere except at the origin, which has its usual nbhds.

Title: Showing $d$ is a metric


问:
Value of a hamiltonian path along a function closely related to the hamiltonian function
Consider $T^*M$ with the canonical symplectic structure.

Let $H:T^*M\rightarrow \mathbb{R}$ be an hamiltonian function and $h:\mathbb{R}\rightarrow \mathbb{R}$ a smooth function. Let $\gamma(t)$ be such that $\dot \gamma(t)=X_{h\circ H}(\gamma(t))$, where $X_{h\circ H}$ is the Hamiltonian vector field of $h\circ H$. Is it true that the quantity $2h'(H(\gamma(t)))H(\gamma(t))$ will be constant ?  I know that $h\circ H(\gamma(t))$ is constant by definition of the hamiltonian vector field and flow , but I am not sure about what happens to the previous one. I have tried to compute it but I didn't get anywhere useful.  One thing it led me to is to wonder wether $\omega(X_{h\circ H}, X_h)=0$ or not ?

Any help is appreciated.
Thanks in advance.

答:
Your expectations are correct on any symplectic manifold $(X, \omega)$. Roughly speaking, the level sets of $h \circ H$ are "the same" as those of $H$ and as such, $X_{h \circ H}$ and $X_{H}$ are colinear at every point.

A more rigorous argument (which is however insufficient to establish the colinearity statement I just mentioned) is to prove that $\omega(X_{h \circ H}, X_H) = 0$, which is the same as showing that the Poisson bracket $\{h \circ H, H\}$ vanishes. But the map $\{ - , F \} : C^{\infty}(X) \to C^{\infty}(X)$ is a derivation, so it satisfies the chain rule $\{h \circ H, F\} = (h'\circ H) \cdot \{H, F\}$ (check this identity in a Darboux chart if you wish). Since $\{H, H\} =0$, we get indeed $\{h \circ H, H\} = 0$.

It follows that $H$ is constant along the integral curves of $h \circ H$, which in turn implies that $h' \circ H$ is constant along those curves. The product $(h' \circ H) \cdot H$ is therefore also constant along those curves.

ContentLicense: CC BY-SA 4.0


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
I spent ages fighting with this stuff. Let me spare you some grief.  It gets taught horribly almost everywhere. I have had assignments and textbooks that contained utter gibberish pretending to be coherent questions.

First of all, there is no such thing as a value of a Dirac delta.  I cannot stress that enough.  There is no such thing as a value of a Dirac delta. (I never call it a function, because it isn't one). A function is defined by its values.  A Dirac delta is defined by its integrals.

If you have any doubt about Dirac delta not being a function, then consider: what would be the values of $5\delta(x)$?  How could you tell the difference between that and $\delta(x)$?  Answer: by its alleged values, you can't.  Values do not give anything like enough information.  Values are enough to define a function.  Therefore, the Dirac delta is NOT a function.

Now, there IS a convoluted way in which you can sort of say that "$\delta(x) = 0$ for $x \neq 0$", but it is highly misleading.  What they are saying is that there is a generalized distribution, the $0$ distribution, whose integrals all equal $0$.  If you take the integral of $\delta(x)$ over any interval that does not include $x = 0$, then you get the number $0$.  Two generalized distributions are equal if they have the same values of their integrals.  So you can say their claim, but it is like saying "dogs meow" when by "dogs" you mean "fish" and by "meow" you mean "swim."  It is incredibly misleading.

What I find ironic is that you can define a perfectly ordinary function:$$I = \int_{a}^{b}\delta(x-c) dx$$
This is a function of three variables: $a,b,$ and $c$.  It is not a function of $x$.  Such a function $I$ can be defined for any function $f(x)$, but there are functions $I$ which do not come from integrating any function. In this example, $I(a,b,c) = 1$ if $a<c<b$ and $0$ otherwise.  I don't know why they don't just teach it that way.

But remember: calling a Dirac delta a function is almost exactly like calling $\sqrt{2}$ a fraction.  If you ask various questions about $\sqrt{2}$ while insisting it is a fraction, you might be able to get away with it, but if you ask what the numerator of $\sqrt{2}$ is, it becomes obvious that something is very wrong.  $\sqrt{2}$ is the limit of many sequences of fractions; the Dirac delta is the limit of many sequences of functions.

Consult Lighthill's little book on distributions if you want more details.  Good luck!


问:
Showing the existence of $\phi$ in the proof of the Implicit function Theorem
The Implicit Function Theorem I am working with:
Let $f \in C^1$ from an open set in $\mathbb{R}^m \times \mathbb{R}^n$ to $\mathbb{R}^m$. Let $(y_0, x_0)$ be a point in this open set such that $f(y_0, x_0) = 0$, and
\begin{align*}
L =  \left( \frac{\partial f_i}{\partial y_j} (y_0, x_0) \right)_{i,j = 1, \dots, m}
\end{align*}
is nonsingular. Then there exist open sets $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ such that $(y_0, x_0) \in V \times U$, and to every $x \in U$, there exists a unique $y = \phi (x)$ in $V$ such that $f(\phi (x), x) = 0$, and $\phi \in C^1$ on $U$.

The proof begins by defining a function $F: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^m \times \mathbb{R}^n$ as $F(y,x) = (f(y,x), x)$. Then the Inverse Function Theorem is applied to $F$. We have a neighborhoods $U' \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ so that $F$ has a $C^1$ inverse on $W = F(V \times U')$, $U = \{x \in U': (0,x) \in W\}$

My question is: how can we show that there exists a $C^1$ function $\Phi: W \rightarrow \mathbb{R}^m$ so that $F^{-1}(y,x)) = (\Phi (y,x),x))$ on $W$? My approach has been to find the derivative of $F$ and then try to invert it by the Inverse Function Theorem, but I don't know how this shows $\Phi$ exists.

Furthermore, if we let $\phi(x) = \Phi(0,x)$, how can we show that $\phi$ then satisfies the conclusion of the theorem?

答:
Forget about all the open neighbourhoods. If you have an invertible mapping $F:\Bbb{R}^m\times\Bbb{R}^n\to \Bbb{R}^m\times\Bbb{R}^n$ then we can consider the following two functions:
\begin{align}
\alpha:= \pi_{\Bbb{R}^m}\circ F^{-1}\quad \text{and}\quad \beta:=
\pi_{\Bbb{R}^n}\circ F^{-1},
\end{align}
where the $\pi$'s are the standard projections.
In other words, we can express $F^{-1}$ in the form $F^{-1}(u,v)=(\alpha(u,v),\beta(u,v))$. In your case $\Phi$ is $\alpha$ and $\beta=\pi_{\Bbb{R}^n}$ because of the additional assumption that $F(y,x)$ has the special form $(f(y,x),x)$.


问:
Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$
Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$

I know intuitively that it is unbounded but how can I show that it is unbounded. In particular I want to show that it is monotone increasing and then show there cannot be an upper bound. However I can't seem to figure out what I should do to show that it is monotone increasing.

答: It's clear that $a_i>1 \implies a_n>a_{n-1}+\frac1n \implies a_i>1+\frac12+\frac13+\dotsb+\frac1i$ but clearly the RHS is not bounded and we're done


问:
Integral of the inverse of a "simple" sum
$$\Large \int_{0}^{\infty} \left (\frac{1}{ \sum_{k=0}^{\infty} x^k} \right)~ dx$$

It seems like this converges to $\approx \frac{1}{2}$ but when I try to calculate this I get really stuck, and for no reason!

$$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$

And so the integral becomes:

$$\large \int_{0}^{\infty} (1-x)~ dx = \text{Does not converge}$$

Even if I use the "regular" geometric progression sum $$ \frac{ 1 \cdot (x^n - 1)}{x-1} $$

Where $n \to \infty$ (number of "elements" in the progression) it seems to actually converge to $0$.

Why these $2$ tries failed? I don't get it... what is the correct way to find the value?

答:
This is not exactly the most sensible approach to the problem but it does offer another way to get the solution and I'm a sucker for special functions. We have
$$
I=\int_0^1\lim_{n\to\infty}\frac{1-x}{1-x^n}\,\mathrm dx+\int_1^\infty\lim_{n\to\infty}\frac{1-x}{1-x^n}\,\mathrm dx.
$$
For $n\geq 0$ the integrand of the first integral is nonnegative and bounded above by one, which is integrable on $[0,1]$. Likewise, for $n\geq2$ the integrand of the second integral is nonnegative and bounded above by $(1+x+x^2)^{-1}$ which again is integrable on $[1,\infty)$; hence, by argument of dominated convergence
$$
I=\lim_{n\to\infty}\int_0^1\frac{1-x}{1-x^n}\,\mathrm dx+\lim_{n\to\infty}\int_1^\infty\frac{1-x}{1-x^n}\,\mathrm dx.
$$
Substituting $t=x^n$ and $t=x^{-n}$ into the first and second integral, respectively, we have from the integral representation of the harmonic numbers:
$$
I=\lim_{n\to\infty}\frac{1}{n}\left(H_{-1/n}-H_{1/n-1}+H_{2/n-1}-H_{-2/n}\right),
$$
which upon application of the functional identities and recurrence identities for $H_z$ simplifies to
$$
I=\lim_{n\to\infty}\frac{\pi}{n}(\cot(\pi/n)-\cot(2\pi/n))=\frac{1}{2}.
$$


问:
When can I define a map from a group presentation to $\mathbb{Z}$?
I'm trying to wrap my head around group presentations and I came up with the question:

When can I define maps out of a presentation?

So, for example, given a presentation $G=\langle S\mid R\rangle$ for some generators $S$ and relations $R$, when can I define a homomorphism $\phi:G\to\mathbb{Z}$?

I'm trying to map generators to generators but I also think relations need to coincide as well so I'm not sure.

答: Let $H$ be any group. Suppose $f:S\to H$ is a function which you can extend to $S\cup S^{-1}$ by letting $f(s^{-1})=f(s)^{-1}$. If for every $r=s_1s_2...s_n\in R$ we have $f(s_1)f(s_2)...f(s_n)=e_H$ (intuitively, all elements of $R$ belong to the "kernel" of $f$) then $f$ can be uniquely extended to a homomorphism $\hat{f}:G\to H$. This is the fundamental theorem of group presentations.


问:
Augmenting paths of two matchings
Given a graph $G(V, E)$ and two matchings $M \subset E$ and $M' \subseteq E$ with $|M'| > |M|$, how can one prove that  $M\oplus M'$ ($\oplus$ denotes the symmetric difference) must contain at least $|M'|-|M|$ augmenting paths of $M$? I know that there have to exist some $M$-augmenting paths since $M$ is not a maximum matching, but I don't really know how to prove the above.

答:
If $M$ and $M'$ have $k$ edges in common, then $M \oplus M'$ has $|M|-k$ edges from $M$, and $|M'|-k$ edges from $M'$.

In particular, there are still $|M'|-|M|$ more edges from $M'$ than from $M$ in the symmetric difference.

This excess of $|M'| - |M|$ simply cannot be accomplished by fewer than $|M'|-|M|$ augmenting paths. To see this, you should convince yourself that:

Any component of $M \oplus M'$ that's not an augmenting path, there are equally many edges from $M$ and from $M'$, so it doesn't help at all.

Any component of $M \oplus M'$ that is an $M$-augmenting path has only one more edge from $M'$ than from $M$.

We could also have components that are $M'$-augmenting paths, which are even less useful: they have more edges from $M$ than from $M'$.

Therefore, to get to a difference of $|M'|-|M|$ in the number of edges, we need at least $|M'|-|M|$ or more $M$-augmenting paths.

Title: Augmenting paths of two matchings


问:
Is this proof of $p(z)=0 \iff p(\bar{z})=0$ correct?
So, I want to show $p(z)=0 \iff p(\bar{z})=0$. $$p(z)=0\\ \overline{p(z)}=\bar{0}\\p(\bar{z})=0$$

In a previous exercise I have shown that $\overline{p(z)}=p(\overline{z})$, so felt that it was okay to use the conjugate on both sides.

答: Your proof is correct given that $P(z)$ is a polynomial with real coefficients. If the coefficients are complex $\overline{P(z)} = P(\bar{z})$ doesn't hold. This shows that complex roots occur in conjugate pairs.


问:
Stationary point of $ x \mapsto \frac{1}{2} \| A x - b \|^2 + \frac{\lambda}{2} \| D x \|^2 $
Given full rank matrix $A \in \mathbb{R}^{n \times m}$, matrix $D \in \mathbb{R}^{p \times m}$, vector $b \in \mathbb{R}^n$ and scalar $\lambda > 0$, let scalar field $f: \mathbb{R}^n \to \mathbb{R} $ be defined as follows

$$ f(x)= \frac{1}{2} \| A x - b\|^2 + \frac{\lambda}{2} \| D x \|^2 $$

Is $f$ convex?

Calculate and classify the stationary point of $f$.

I already prove that $f(x)$ is convex using the fact that $\| \cdot \|^2$ is convex. But I cannot calculate the critical points. Is better to procede using the vector form or using every entry if $f(x)$?

答:
Take the derivative and set it equal to $0$

$$\frac{1}{2}\nabla\left(x^TA^TAx-x^TA^Tb-b^TAx+b^Tb + \lambda x^TD^TDx\right)
$$

$$= x^T(A^TA+\lambda D^TD)-b^TA = 0$$

which means

$$x = (A^TA+\lambda D^TD)^{-1}A^Tb$$


问:
Is $\mathbb{R}$ a Borel set?
Is the set of all real numbers, $\mathbb{R}$, a Borel set? If yes, how can this be proved? I know that all closed subsets of $\mathbb{R}$ are Borel sets, but can this fact be used to prove that $\mathbb{R}$, a Borel set? Also, I think I can show that $\mathbb{R}$ is a $\sigma$-algebra on itself (if that is meaningful), but I don't know where to go from there.

答:
Part of the definition of a sigma algebra on a set $X$ is that $X$ itself belongs to it. In particular, $\mathbb{R}$ belongs to the Borel sigma algebra on $\mathbb{R}$.

Also, by the definition of a topology, any topological space is both open and closed in itself. So $\mathbb{R}$ is also closed in itself.

Title: Is $\mathbb{R}$ a Borel set?


答:
Your idea is exactly correct. I must make a nitpick, though. Your "proof" is just a sequence of equations, while a proof should consist of complete sentences which form a coherent logical argument. For example, here we are trying to show a biconditional statement, so your proof should clearly explain why $p(z) = 0$ implies $p(\overline{z}) = 0$, and vice versa.

Here's how I would write this proof:

Let $z \in \mathbb{C}$ be arbitrary. First, suppose $p(z) = 0$. Then $\overline{p(z)} = \overline{0} = 0$, and we already know $\overline{p(z)} = p(\overline{z})$, so we conclude that $p(\overline{z}) = 0$, as desired. For the reverse direction, suppose $p(\overline{z}) = 0$. Then $\overline{p(\overline{z})} = \overline{0} = 0$, and we know that $\overline{p(\overline{z})} = p(\overline{\overline{z}}) = p(z)$, so we conclude that $p(z) = 0$. We've now shown both directions of implication, so we conclude that $p(z) = 0 \iff p(\overline{z}) = 0$.

This might seem overly wordy, and could certainly be made shorter, but (I presume) at this level, a major goal of whatever course you're taking is to get practice writing clear, precise proofs. Most importantly, proofs should always consist of complete sentences! I'll say it again: when you write a proof, you should write entirely in complete sentences. This rule has very few exceptions.

Title: Is this proof of $p(z)=0 \iff p(\bar{z})=0$ correct?

Body:
So, I want to show $p(z)=0 \iff p(\bar{z})=0$. $$p(z)=0\\ \overline{p(z)}=\bar{0}\\p(\bar{z})=0$$

In a previous exercise I have shown that $\overline{p(z)}=p(\overline{z})$, so felt that it was okay to use the conjugate on both sides.


问:
How to get rid of the ceiling function on one side of an equation
Given this function,

$\left\lceil \log _{b-1} \left(\lceil \frac{1}{b} \rceil\right) \right\rceil=1$

How do I get rid of the ceiling function so I can obtain the value of $B$ ?

答: By definition, $x\le \lceil x\rceil<x+1$; therefore $\lceil y\rceil = N$ is equivalent to $N-1< y\le N$.


问:
Inequality on the family of intersecting antichain
Let $\{A_1, A_2,..., A_m\}$ be an intersecting antichain of subsets of $[n]$ s.t. $|A_i| \leq n/2$ for each $i$. Prove that
$$\sum_{i=1}^m \binom{n-1}{|A_i|-1}^{-1} \leq 1.$$

I know it originally appeared in Sperner Systems Consisting of Pairs of Complementary Subsets [Bollobas, 1972], but the proof in the paper is somehow confusing to me. Could anyone give me more explanations on that? Thanks a lot!

答:
I think I can answer it myself now.

For $A \in \mathcal{F}$ and $\pi$, a permutation of $[n]$. We let $w(\pi, A) = 1/|A|$ if $\pi(A)$ is cyclically consecutive and let $w(\pi, A) = 0$ otherwise. We do double counting on $\sum_{\pi, A} w(\pi, A)$. First we fix $A$ and we have
$$\sum_{\pi, A} w(\pi, A) = \sum_{A \in \mathcal{F}} = \sum_A n|A|!(n-|A|)!/|A|.$$
Then we fix $\pi$ and by the Lemma in the paper we have
$$\sum_{\pi, A} w(\pi, A) = \sum_\pi \sum_{A \in \mathcal{F}} w(\pi, A) \leq \sum_\pi 1 = n!,$$
completing the proof.


答: If you are starting from real analysis without the definition of 'a topology' you can use as a definition of an open set in $\mathbb{R}$, or more generally in a metric space, that a set $O$ is open if for all $x\in O$ there is a ball $B_r(x)\subseteq O$ for some $r>0$. For the empty set this is vacuously true (there are no $x$ to quantify over, so you can't find one to disprove the universal) therefore $\emptyset$ is open which means $\mathbb{R}$ is closed.

Title: Is $\mathbb{R}$ a Borel set?

Body: Is the set of all real numbers, $\mathbb{R}$, a Borel set? If yes, how can this be proved? I know that all closed subsets of $\mathbb{R}$ are Borel sets, but can this fact be used to prove that $\mathbb{R}$, a Borel set? Also, I think I can show that $\mathbb{R}$ is a $\sigma$-algebra on itself (if that is meaningful), but I don't know where to go from there.


问:
Complete/Connected space
The metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by

$$d(x, y) = \left\{
  \begin{array}{lr}
      ||x||+||y|| & \mbox{if } x \neq y \\
      0 &  \mbox{if } x = y \\
      \end{array}
\right.$$
is a variation of the SNCF metric. Given that $d$ is a metric, show that if $X = \mathbb{R}^2$, then $(X, d)$ is a complete but not connected metric space.

My attempt:

Complete:

A Metric Space is said to be complete if every Cauchy Sequence converges in the space.

Let $\{x_n\}$ be a Cauchy sequence in $d$. We need to show that for all $\epsilon > 0$, there is an $N \in \mathbb{N}$ so that $d(x_n, x) < \epsilon$ (for some $x \in X$) for all $n \geq N$.

Not very sure how to proceed here.

Not connected:

$(X, d)$ is disconnected if it can be written as the union of two non-empty separated sets. We need to find $\emptyset \neq A, \emptyset \neq B \subseteq X$ so that $X = A \cup B$ where $\overline{A} \cap B = A \cap \overline{B} = \emptyset$.

I cannot find any such subsets here.

Any assistance is much appreciated.

答:
The most straightforward approach is to figure out what topology on $\Bbb R^2$ is generated by $d$.

HINT: If $x\ne\langle 0,0\rangle$, and $0<r\le\|x\|$, what is $B(x,r)$? And what is $B(\langle 0,0\rangle,r)$? You should try to figure this out on your own, but I left a description, without proof, in a spoiler box in my answer to your previous question about this metric.

Once you have that, showing that the space is not connected is very easy. To show that it is complete, you need to figure out what the Cauchy sequences are.

HINT: Every Cauchy sequence in $\left\langle\Bbb R^2,d\right\rangle$ either converges to the origin or is eventually constant.

Title: Complete/Connected space


问:
Graph clique problem
I'm not sure to what degree this is a graph problem, and algorithms question, or what, but I'll give the setup:

I have a simple undirected graph given in the form (for example) $E=\{\{v_1,v_2\},\{v_0,v_1\},\{v_0,v_2\},\{v_0,v_3\},\dots\}$ where $V=\bigcup E$ and $E$ is a set of pairs, and I wish to find a representation $K=\{\{v_0,v_1,v_2\},\{v_0,v_3\},\dots\}$ such that $E=\{x:|x|=2\wedge\exists y\in K\,x\subseteq y\}$ and $|K|$ is minimal. What can be said about this problem: How hard is it? Can it be stated in terms of more common graph problems? If possible, give an algorithm for finding such a $K$ ($K$ is not unique).

It is easily observed that $K$ represents a set of cliques in the graph, so I want to relate it to the maximal clique problem, but my goal is not necessarily to find a maximal clique, just a minimal covering set of cliques.

Edit: An alternative minimality condition is to minimize $|K|+\alpha\sum_{x\in K}|x|$, which more accurately reflects my goal of using this format as a compression system for storing such undirected graphs. In my application, $\alpha=2/7$.

答: This is known as (Edge) Clique Cover. It is NP-hard and hard to approximate. Here's a paper (of which I am a coauthor) that contains some experiments. The variant where the sum of clique sizes is maximized is also NP-hard. In my experience, choosing either objective does not make a big difference, so unless you're looking for optimal solutions, the heuristics from either paper might be good for your objective, which is a mixture of these two.</p>

ContentLicense: CC BY-SA 3.0

Title: Graph clique problem


答:
Alternatively, first we show $a_n > 1, \forall n \ge 2$ by induction. $a_2 = 1\cdot(1+\frac{1}{2})= \frac{3}{2} > 1$. Assume $a_n > 1, n \ge 2 \implies a_{n+1} =a_n^2++\dfrac{a_n}{n+1}> 1+\dfrac{1}{n+1} > 1\implies a_{n+1} > 1\implies a_n > 1, \forall n \ge 2$. Thus using the recursive formula we have: $a_n = a_{n-1}^2+\dfrac{a_{n-1}}{n}> a_{n-1}^2 \implies a_n > a_{n-1}^2$. Apply this new inequality repeatedly on the way to the $2$nd term:
$a_n > a_{n-1}^2> (a_{n-2})^{2^2}> (a_{n-3})^{2^3}> \cdots> (a_2)^{2^{n-2}}= \left(\frac{3}{2}\right)^{2^{n-2}}$. This shows $a_n$ is unbounded.

Body:
Show that the sequence $a_n = a_{n-1}(a_{n-1} + \frac{1}{n})$ is unbounded when $a_1 = 1$

I know intuitively that it is unbounded but how can I show that it is unbounded. In particular I want to show that it is monotone increasing and then show there cannot be an upper bound. However I can't seem to figure out what I should do to show that it is monotone increasing.


问:
If every non-trivial proper subgroup of $G$ is cyclic then is $G$ cyclic?
I know that the fundamental Fundamental Theorem of Cyclic Groups stats that if $G$ is cyclic then so is every subgroup. So if I am given a group $G$, then by demonstrating that every non-trivial proper subgroup is cyclic can I conclude that $G$ is cyclic?

I am inclined to say no. This is because the fundamental theorem is not an if and only if statement, but I cannot seem to come up with a counter example.

答: The Klein four-group is a counterexample, as is the symmetric group on three elements.</p>


问:
Non-homogeneous Differential Equation
Trying to solve a non-homogeneous differential equation, whether it is linear, Bernoulli, Euler, you solve the related homogeneous equation and then you look for a particular solution depending on the "class" of the non-homogeneous term.

Just to make an example, let's say we have this equation

$$y'' - y' - 2y = 2x e^x$$

we solve $$y'' -2y' +y =0$$ and we get solutions of the form $y_{\text{hom}}(x) = C_1 e^{2x} + C_2 e^{-x}$.

Then, since the known term is of the form $p(x)(x) e^{\mu x}$, and $\mu$ is not a solution to the characteristic polynomial, we should find the solutions through those of the form $q(x)e^{\mu x}$. If, on the other hand, $\mu$ were a solution of the characteristic polynomial, we should have looked for solutions through those of the form $q(x) x^q e^{\mu x}$ where $q$ is the multiplicity of the root $\mu$. (where $q(x)$ is the general polynomial of the same degree as $p(x)$.

I have a whole list of these "classes" of known terms, like $p(x)$, $p(x)e^{\mu x}$, $e^{\alpha x}(a \sin(\beta x) + a' \cos(\beta x))$, and the corresponding general solutions (like I said before, also with the multiplicity involved).

My question is, where do these come from? I have no idea how such classes were found or anything; my book and my professors just said these were given and we should just learn them.

Edit: from the comment I see that something's going on, unfortunately I don't have enough background to know how to work with Laurent series. I would appreciate any explanation. Thanks.

答:
The possible right-hand-sides for this method are, themselves, solutions of linear homogeneous differential equations with constant coefficients.  User8268 provided another description of this class in a comment.

If your equation is $\mathsf Ly=f$, where $\mathsf L$ is a linear differential operator with constant coefficients, and $\mathsf Mf=0$ where $\mathsf M$ is another such operator, then your desired solution is also a solution of $\mathsf{ML}y=0$, and knowledge of solutions of such (homogeneous differential equations with constant coefficients) can then be used to show what the form of $y$ must be.

Title: Non-homogeneous Differential Equation


问:
Determine counterclockwise moving
in my app, I let user touch and move to draw an arc. After drawing, I got a set of points. Is there any way to determine that user draw the arc counterclockwise or reverse counter clock wise?</p>

答:
If it is indeed a circle you need only find an expression in a form that defines an angle $\theta$ between the initial point and later points.

Once you've got that angle check if $\tan \frac{1}{4}\theta$ is increasing. If it is, it is moving counter clockwise.

Note this only works for angles between $0$ and $2 \pi$

Title: Determine counterclockwise moving


问:
Norm attaining functionals on a Banach space
Do every linear functional on a Banach space attains its norm? In case of finite-dimension, the argument follows via compactness of the unit ball. What about in infinite-dimensional case? Is there any any general version of it?

Any help will be appreciated!

答:
Since the accepted answer did not give a counter example, it might be good to exhibit one here.

Consider the linear functional on $c_0$ given by
$$f(x) =\sum_{n=1}^\infty x_n/2^n.$$

It is easy to show that $\|f\|=1$, but yet there is no unit vector $x$ in $c_0$ such that $|f(x)|=1$.

One would like to take $x=(1,1,1,\ldots)$, but unfortunately this in not a vector in $c_0$.


问:
How to introduce conjunctions into a conditional statements so as to get the CNF and DNF?
I have the following conditional statement: ((p → q) → p) → p

Knowing that CNF is a conjunction of disjunctions (and that DNF is a disjunction of conjunctions), we will obviously have to introduce conjunctions and disjunctions.

Knowing that p → q can be alternatively expressed as ¬p ∨ q, it seems (to me) fairly easy how we could convert all of these conditionals into disjunctions.

However, I'm completely clueless as to how to introduce conjunctions. How could we do this?

Thanks in advance!

答:
Almost.   You are thinking of Implication Equivalence.   There is also Implication's Negation Equivalence.

$$\begin{align}\text{I.E.}~&&\phi\to\psi~~\iff \lnot \phi\vee\psi\tag{note: disjunction}\\\text{I.N.E.}~&&\lnot(\phi\to\psi)\iff\phi\land\lnot\psi\tag{note: conjunction}\end{align}$$

So using these, step by step::$$\begin{align}1.&&((p\to q)\to p)\to p\tag{Premise}\\2.&&\lnot((p\to q)\to p)\lor p\tag{by I.E. on 1}\\3.&&((p\to q)\land\lnot p)\lor p\tag{by I.N.E on 2}\\4.&&((\lnot p\lor q)\land\lnot p)\lor p\tag{by I.E. on 3}\end{align}$$

Then we may distribute one way, aiming for a DNF::

$$\begin{align}5.&&(\lnot p\land\lnot p)\lor (q\land\lnot p)\lor p\tag{Distribute $\land/\lor$ on 4}\end{align}$$

Or we may distribute the other way, aiming for a CNF::

$$\begin{align}6.&&(\lnot p\lor q\lor p)\land(\lnot p\lor p)\tag{Distribute $\lor/\land$ on 4}\end{align}$$

Of course, neither of these are their final form.   You will still need to reduce them by using a few other Equivalence Rules.


问:
Can I define such an average norm of an infinite-dimensional vector?
Suppoes that there is an infinite-dimensional vector $x = \{x_1,x_2,\dots,x_{\infty}\}$. Can I define such a norm
$$
\|x\| =  \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^{N} x_k,
$$
which implies that the limit must exist.

I believe that it satisfies the definition of a norm, but I have never seen it in any contexts before. Is this norm properly defined?

答: This average norm is not well-defined! For a non-zero vector the norm may also be zero, which contradicts the definition of a norm. For example, the norm of $x_k = \frac{1}{k}$ is zero but it is a non-zero vector.


问:
Skorokhod Embedding theorem proof
It occurs in Durrett's proof of Skorokhod embedding that he needs the following.

Suppose that we have a Brownian motion $B_t$ in $1$-d that starts at $0$. (To be clear, it is not necessarily constructed on the Wiener space, but I do require it to be everywhere continuous.  With that assumption, then we have that $T_{b, a}=\inf\{t\geq 0 : B_t \notin (a, b)\}$ is a stopping time wrt the standard right continuous (RC) filtration.)  If $U, V$ are RVs with $U\geq0$ and $V\leq0$ then supposedly the function $T_{U, V}$ is measurable. (with respect to the original sigma algebra for which $U, V$ are RVs.)  Notice that this does not make it a stopping time.

I need help with that.

He also then uses the optional stopping theorem for continuous RC martingales with RC filtrations.  This theorem implies that if $T$ is the stopping time and $X_t$ is the martingale, then $X_T$ is measurable wrt the sigma algebra of the stopping time $T$ and also for all $t\geq0$ we have $X_T=E(X_t\mid \mathfrak{F}_T)$ on $\{T<t\}$ a.s.

Durrett only says that by conditioning on the values of $(U, V)$, we can apply this to our situation above to conclude that (By now he has already started assuming the measurability of $T_{U, V}$.)

$$E(T_{U, V})=E\lbrace E(T_{U, V}\mid (U, V))\rbrace=E(-UV).$$

He has already shown that $E(T_{b,a})=-ab$.  Where I object is it seems like his manipulation is "treating $U, V$ as constant" just because they are the subject of being "conditioned upon."  I don't understand this, especially since $U, V$ are allowed to take values in continua.

(For those with the book, see p. 384 and the surrounding.)

答:
Let $X\in\mathcal X$ and $Y\in\mathcal Y$ be independent random elements on some joint probability space.

You can rewrite Fubini's theorem as: for every measurable function $f:\mathcal X\times \mathcal Y\to\mathbb R$ $$\mathbb E(f(X,Y)|X=x) =\mathbb  Ef(x,Y).$$

So for your case as $B_{T_{U,V}}$ is a function of the two random elements $B_t\in C^0(\mathbb R)$ and $(U,V)\in\mathbb R^2$ we have

$$\mathbb E\left(\left. B_{T_{U,V}} \right|U=u,V=v\right) = \mathbb E\left(B_{T_{u.v}} \right). $$

Title: Skorokhod Embedding theorem proof


问:
Convex set in a Vector Field
I'm studying a specific reaction-diffusion equation, and to prove well-posedness I need to find a convex set such that the vector field points inwards at the border.

The vector field is:
$ F(u_1,u_2,u_3)=(\alpha(u_{3}^{\gamma}-u_{1}^{\alpha} u_{2}^{\beta}),\beta(u_{3}^{\gamma}-u_{1}^{\alpha} u_{2}^{\beta}), \gamma(-u_{3}^{\gamma}+u_{1}^{\alpha} u_{2}^{\beta}))$,
with $u_1,u_2,u_3,\alpha,\beta, \gamma>0$

What are the techniques or approaches to find convex sets for this type of vector field?
Any source is appreciated.

Thanks.

答: You basically need to find a set such that the vector field $F$ is the inward pointing normal vector.  To achieve this let $G$ be the function $G(x,y,z)=\langle -F(x,y,z), (x,y,z) \rangle$.  I don't want to do the algebra but it should work out that $DG_w$, the differential of $G$ at $w$, has full rank and each $w$ such that $G(w)=0$ (you need to show this).  If that is the case then you can use the submersion theorem to state that the set $G^{-1}(0)$ is a manifold and, moreover, the inward $F$ is by definition the inward pointing vector to that manifold.  If they didn't ask to prove all that then you can probably get away by just saying the set $G^{-1}(0)$ is the one you want.  Convexity can be shown just by saying that if $G(w)=0=G(z)$ then linear combinations of w and z are also zero after applying G to it.

Title: Convex set in a Vector Field


问:
General rule of thumb for deciding functions of two variables are continuous or have limits
Is there any particular strategy to determine whether for sure a function of two variables or perhaps three is continuous or necessarily has a limit.? For example the function $f(x,y)=\frac{x^3y^2}{x^4+y^4}$ if $ \textbf{x}\neq 0$ and $0$ if $\textbf{x}=0$ is continuous, but what is a good strategy to figure this out efficiently?

Generally the way I approach this is to try and find two sequences $\{(x_n,y_n)\}$,$\{(w_n,z_n)\}$ approaching $(0,0)$. Then if I cannot find the function applied to these sequences converging to different limits I try to prove the result by finding an upper bound on $f$. What is an easier/ more efficient strategy for doing this?

答:
Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, $f:X\to Y$ a function, $E\subseteq X$ s.t. $x_{0}\in X$ is an adherent point of $E$ and $L\in Y$. We say that $f$ approaches $L$ as $x$ approaches $x_{0}$ along $E$ iff for all $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} > 0$ such that for every $x\in E$ we have that
\begin{align*}
d_{X}(x,x_{0}) < \delta_{\varepsilon} \Rightarrow d_{Y}(f(x),L) < \varepsilon
\end{align*}

Based on such context, there is an additional result which states when $f$ has limit $L$. More precisely, the following two statements are logically equivalent:

The limit of $f$ is $L$ when $x$ approaches $x_{0}$ along $E$.

Whenever $x_{n}$ is a sequence of points in $E$ which converges to $x_{0}$, the sequence $f(x_{n})$ converges to $L$.

Based on such result, you can test whether or not a function between metric spaces converges.

Hopefully this helps!


问:
Prove that each $f_n(x) = \frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}$ is Borel measurable
How can I prove that for each $n \in \mathbb{N}, f_n(x) = \frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}$ is Borel measurable where $X\subset \mathbb{R}$ and $f: X \mapsto \mathbb{R}$ is a real-valued function? I was thinking of first proving that each $f_n(x)$ is continuous, but I am not sure if that is even true. Are there any other ways of proving this statement?

答: I assume $f$ is measurable in this case.  If so then since the composition of measurable functions is measurable, and since $x+\frac{1}{n}$ is measurable, so is $f(x+\frac{1}{n})$.  The rest follows from the fact that linear combinations of measurable functions are measurable.


问:
Does the zero ideal contain the product of prime ideals?
It is simple to show that non-trivial ideals contain the product of non-zero prime ideals, but is it the case for the zero ideal as well?</p>

答: Take $F$ a field, and let $R=F\times F$. Then $P=F\times\{0\}$ is a prime ideal (in fact, it is maximal), as $R/(F\times\{0\})\cong F$ is a field. Symmetrically, $Q=\{0\}\times F$ if a prime ideal of $R$. Now notice that $PQ$ is the zero ideal.


答:
If all you know is that $f$ is a real-valued, this can't possibly be the case because it implies every function is measurable. For given any function $g : \mathbb{R} \to \mathbb{R}$, we could define a function $f : \mathbb{R} \to \mathbb{R}$ by $f(x) = 0$ if $x \in [0, 1)$, $f(x) = g(x) + f(x - 1)$ if $x \geq 1$, and $f(x) = f(x + 1) - g(x + 1)$ if $x < 0$. Then $f_1(x) = g(x)$.

But not every function $\mathbb{R} \to \mathbb{R}$ is measurable. For example, the characteristic function of any non-measurable set will be non-measurable.

On the other hand, if we know $f$ is measurable, this follows from a few facts:

Continuous functions are measurable when the measure spaces have the Borel $\sigma$-algebra. Hence, $-$ and multiplication by a constant are measurable functions.

The composition of measurable functions is measurable.

If $f, g : X \to \mathbb{R}$ are measurable, then $f \times g : X \to \mathbb{R}^2$ defined by $(f \times g)(x) = (f(x), g(x))$ is measurable.

So we have $f(x + 1/n)$ is measurable, because it is the composition of measurable $f$ and measurable function $x \mapsto x + 1/n$. Then the map $x \mapsto (f(x + 1/n), f(x))$ is measurable. Compose this with subtraction to get the measurable map $x \mapsto f(x + 1/n) - f(x)$. Compose this with multiplication by constant $n$ to get that the map $f_n$ is measurable.

Note that result (3.) can actually be generalised as follows:

If $A, B$ are measure spaces and $A \times B$ has the product $\sigma$-algebra, then $A \times B$ is the product of $A$ and $B$ in the category of $\sigma$-algebras. That is, given maps $f : X \to A$ and $g : X \to B$, the map $x \mapsto (f(x), g(x)) : X \to A \times B$ is measurable iff $f$ and $g$ are.

If $A$ and $B$ are topological spaces, then the product $\sigma$-algebra of the Borel algebras of $A$ and $B$ is exactly the Borel $\sigma$-algebra of $A \times B$.

Title: Prove that each $f_n(x) = \frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}$ is Borel measurable

Body: How can I prove that for each $n \in \mathbb{N}, f_n(x) = \frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}$ is Borel measurable where $X\subset \mathbb{R}$ and $f: X \mapsto \mathbb{R}$ is a real-valued function? I was thinking of first proving that each $f_n(x)$ is continuous, but I am not sure if that is even true. Are there any other ways of proving this statement?


问:
How to count the number of subsets of $X$ of size $k$ that are disjoint from $A$?
Apologies if my formatting is incorrect, this is my first post.

I'm currently taking Discrete Mathematics, but I'm struggling to understand more complicated uses of PIE. The only examples we've covered in class are things such as giving out cookies using binomial coefficients, but we've never done anything like this problem, although this problem is labeled as 1.6 in our book, we've got no examples similar, unless I'm not interpreting the problem correctly. Any help?

Let
$X = \{1, 2, 3, \dots, n\}$ and $A = \{1, 2, 3, \dots, k\}$
for some $n \ge k \ge 0$.

We say that a subset $B$ of $X$ is disjoint from $A$ if $A \cap B = \emptyset$. Using PIE, count the number of subsets of $X$ of size $k$ that are disjoint from $A$.

答:
For PIE, the $k$ properties to be avoided here are $i\in B$, where $i=1$ to $k$.  For a given count of $r$ properties, there are $\binom{k}{r}$ choices of properties and then $\binom{n-r}{k-r}$ ways to complete the $k$-subset, so PIE yields
$$\sum_{r=0}^k (-1)^r \binom{k}{r} \binom{n-r}{k-r}
=\binom{n}{k}-k \binom{n-1}{k-1}+\binom{k}{2} \binom{n-2}{k-2}-\binom{k}{3} \binom{n-3}{k-3}+\dots+(-1)^k \binom{n-k}{0}$$


问:
Get 2 unknowns from a equation with answer of equation
I want to do a simple math operation, but because I'm bad at math I don't know how can I actually do it

I have this Equation

out = (var1 - 128) + 64 * (var2 - 1)

in this Equation both var1 and var2 cant go upper then 255

also var1 is always higher then 198

A example of use

803 = (227 - 128) + 64 * (12 - 1)

I want to know its possible to get var1 & var2 with out?

sorry if my question is too bad or I cant tell what I mean right (I also love to do it With programming language

--EDIT--
I can do this in this way, and its almost fast, but Im not sure if its a good way for it our not

```<code>$var1 = 198
$var2 = 1
$Answer = 0
While $Answer = 0
    $ActualLen = ($var1 - 128) + 64 * ($var2 - 1)
    if $ActualLen = 1600 Then
        ConsoleWrite("CStrlen= " & $var1 & @CRLF & "ExtraStrlen= " & $var2 & @CRLF)
        $Answer = 1
    EndIf
    If $var2 = 255 Then
        $var1 += 1
        $var2 = 1
    Else
        $var2 += 1
    EndIf
WEnd
</code>```

答:
As others have stated, normally one linear constraint on two variables won't do.  In this case, there is one and only one way that the problem is solvable: if var2 is required to be an integer.

Let $a$ denote var1, $b$ denote var2, $c$ denote out.

Reason that the equation is solvable:

In the computation of $c$, $b$ is multiplied by the scalar, $64$.

This means that as $b$ changes value from $n$ to $n+1$ or $n-1$, the value of $64 \times b$ changes value from $64n$ to $64n+64$ or $64n - 64.$

Although there may be other bounds on $b$ (i.e. var2) that is irrelevant.

What is critical is that you have specified that $198 < a \leq 255$, where $a$ is var1.

Since $(255 - 198) < 64,$ once $c$ (i.e. out) is known, there is one and only one integer value for $b$ such that $64b$ will result in $a$ being in the range $198 < a \leq 255.$

It is irrelevant whether $a$ (i.e. var1) is required to be an integer.  All that is necessary is that (to summarize)

var2 is required to be an integer

var2 has the scalar of 64 applied to it

var1 has no scalar applied to it (or you can say that the scalar is 1).

the width of the range (i.e. $255 - 198$) of allowable values of var1 is less than 64

This begs the question: given $c$ (i.e. out), how do you calculate $a$ and $b$ (i.e. var1 and var2).

The simplest approach is as follows:

Since $c = (a - 128) + 64(b-1)$, you have that

$c - 64b + 64 = a - 128 \implies$

$$c + 192 - 64b = a.\tag1$$

Therefore, programatically, create a guess for $b$ (i.e. $g$) as follows.

Let $g$ represent the computation from the LHS of equation (1), above, under the assumption that $b = 0.$

If the result is in the range $198 < g \leq 255$, then, you are done.  Simply set $a = g$.

Suppose that that is not the case.
Then, you simply have to find the right integer value for $b$, so that the result of the LHS computation in equation (1) above will be in the required range of $198 < a \leq 255.$

As your experience/sophistication with math grows, you will find (for yourself) a shortcut to the algorithm that I am going to give you, that you can easily write the code for.

If $g \leq 198$, then, subtract 1 from $b$, and recompute $g$.  Do this in steps, one at a time.  Sooner or later, $g$ will be in range, and you will then have solved the problem, because you will have identified the appropriate value for $b$, and because you can then set $a = g.$

If $g > 255$, then follow the algorithm in the above paragraph, in reverse.  Simple add 1 to $b$, recompute $g$, and check if it is in range.  Sooner or later, it will be.

One final note.  It may be strange that I am guaranteeing that an appropriate value for $a$ will be found, even though the width of the range of allowable values for $a$ is slightly less than $64$, and the integer value of $64b$ will change by $64$ as $b$ is increased or decreased by $1$.

I am assuming that you will only be applying this algorithm to legitimate values for $c$ (i.e. out), where $c$ was initially computed by unknown but legitimate values for $a,b$.

As a simple example of when the above algorithm will not work, consider $c$ (i.e. out) $= 66$.

Then, with $b = 0$, your initial value for $g$ is $258$ which is too high.  So, you set $b = +1$ and recompute $g$ and now get $g = 194$, which is too low.  So neither $b = 0$ nor $b = +1$ provides a satisfactory value for $g$, so you can never set $a = g$.

The reason that this happened is that your initial value for $c$ (i.e. out) was illegitimate.  That means that $c=66$ could not have been computed by any legitimate values for $a$ and $b$.


问:
Suppose that characteristic $F$ is $p$. If $K/F$ is separable then $K = F(K^{p})$ where $K^{p} = \{ x^{p} \, |\, x\in K\}$.
I am having difficulty finishing this problem.  So far I have this:  </p>

<p>Want to show $K \subset F(K^{p})$.  Since $K/F$ is separable then $K/F$ is algebraic.  In particular, $\alpha\in K$ is separable over $F$ hence separable over $F(\alpha^{p})$.  So $\alpha\in F(\alpha^{p})$ by an exercise already done.  But then I do not know how to conclude $\alpha\in F(K^{p})$. Any suggestions?  Or is my approach completely wrong? </p>

答: Let $L=F(K^p)$. Clearly, we have $F\subseteq L\subseteq K$. It is also clear that for every $\alpha\in K$, we have $\alpha^p\in L$. This shows that the extension $K/L$ is purely inseparable (see characterization 2 here). But $K/F$ separable implies that $K/L$ is separable (see the second bullet here) and because $K/L$ is both separable and purely inseparable, we have $K=L$.</p>


问:
if $$ is a finite abelian group $()$ divides the exponent of $$, for each $\in $
Define the exponent of a group as $\sup o(a)$ ,where $a \in G$. Show that if $G$ is a finite abelian group $o(a)$ divides the exponent of $G$, for each $\in $

Here $o(a)$ denotes the order of an element $a$?
How can I show this? Can someone please show the right approach to attempt this question.

Thanks.

答:
By the fundamental theorem of finitely generated groups, we can write $G = \bigoplus\limits_{i = 1}^n \bigoplus\limits_{j = 1}^{m_i} \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z}$ with inclusion maps $k_{ij} : \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z} \to G$, such that the $p_i$ are distinct primes and for each $i$, the $e_{ij}$ are sorted in increasing order.

Let $M = \prod\limits_{i = 1}^n p_i^{e_{i m_j}}$. Consider the map $f: G \to G$ defined by multiplication by $M$. It's easy to see that $f = 0$. Thus, we see that for all $a \in A$, $o(a) | M$. And in particular, for all $a \in A$, $o(a) \leq M$.

Now note that for $x = \sum\limits_{i = 1}^n k_{i m_j}(1)$, we have $o(x) = M$.

Then we see that $M$ is the exponent of $G$ and that for all $a \in G$, $o(a)$ is a factor of $M$.


问:
Is it possible to isolate $ac/bc/bd$ from the equation $x=ab+cd$?
Here is a tricky algebra problem I couldn't figure out, I am wondering if anyone else could, or if it's even possible. $$ x = ab + cd$$ The goal is to completely isolate ac (or $bc$ or $bd$ or $ad$ it doesn't matter)

Everything I try results in $a$ or $c$ being stuck somewhere on both sides.

答:
It can't be done.

More precisely, from the equation
$$
x=ab+cd
$$
it's not possible to solve for $ac$ in terms of $b,d,x$.

For example, suppose $b=d=x=1$.

Then the equation reduces to
$$
a+c=1
$$
so for any choice of $a$ we get
$$
ac=a(1-a)
$$
which varies with $a$.


问:
Prove indexed set and Cartesian product
I have a problem as follows:

And this is my work:

Can some one suggest a way to do the opposite direction:

Starting with: Conversely, suppose $P\ \in\bigcup{\ (A}_i\times\ S)$ for some $i\ \in I$.

Thank you so much!

答:
Suppose $P \in \bigcup\limits_{i \in I} (A_i \times S)$. Then take some $i \in I$ such that $P \in A_i \times S$. Then we can write $P = (a, s)$ for $a \in A_i$ and $s \in S$. Then $a \in \bigcup\limits_{i \in I} A_i$. Then $P = (a, s) \in (\bigcup\limits_{i \in I} A_i) \times S$.

Thus, we see that $\bigcup\limits_{i \in I} (A_i \times S) \subseteq (\bigcup\limits_{i \in I} A_i) \times S$.

Title: Prove indexed set and Cartesian product


问:
Showing partial derivatives are constant on connected domain, means $f$ is constant
I know this question has been asked before, I just want to know how I can use this approach to successfully complete the proof.

Let $f: A \rightarrow \mathbb{R}$, where $A\subseteq \mathbb{R}^n$ is an open, $\text{connected}$ set.  Prove that, if $\frac{\partial f}{\partial x_i}(\textbf{x}) = 0$ for all $i \in \{1,\cdots,n\}$ and $\textbf{x} \in A$, then $f$ is constant.

Attempt:Choose $r>0$ such that $B_r(x) \subset A$. Let $\textbf{h}$ be a point in $\mathbb{R}^n$ such that $||\textbf{h}||<r$. By the mean value theorem there is a $\theta \in (0,1)$ with $f(\textbf{x}+\textbf{h})-f(\textbf{x})=\langle \nabla f(\textbf{x}+\theta \textbf{h}),\textbf{h}\rangle$. But $\frac{\partial f}{\partial x_i}(x)=0$ for each $i$. Thus $f(\textbf{x}+\textbf{h})=f(\textbf{x})$. Since this holds for all $\textbf{x}+\textbf{h} \in B_r(x)$, $f$ is constant on $B_r(x)$. So $f(y)=c$ for all $y \in B_r(x)$. Now I know $B_r(x) \subset f^{-1}\{c\}$ but I cannot figure out how to show that $B_r(x)$ must also be closed and equal to all of $A$ by connectedness Any ideas?

Edit: Now I am noticing some issues with my attempt based on the mean value theorem I am using, are there any ways to resolve these issues?

答:
Let $U$ be the connected open set in question, and suppose we have $x \in U$. We wish to show $V = \{y \in U | f(y) = f(x)\}$ is open.

Indeed, consider such a $y$ and take $r$ such that $B_r(y) \subseteq U$. Then consider $z \in B_r(y)$. We can get from $y$ to $z$ by taking a path where each segment of the path is parallel to an axis. Thus, along each segment, $f$ must be constant (since its partial derivative in the relevant direction is 0). Then we see that $f(y) = f(z) = f(x)$. Thus, $B_r(y) \subseteq V$. Then $V$ is indeed open.

But it is also the case that $W = \{y \in U | f(y) \neq f(x)\}$ is open, by continuity. And we know that $V \cup W = U$ and $V$ has an element - namely, $x$. And clearly, $V \cap W = \emptyset$. Therefore, it must be the case that $V = U$.

In that case, for all $y \in U$, $f(y) = f(x)$.


问:
Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$
I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.

I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theorem, by parametrizing the sphere and by algebraic manipulating the function but I haven't been able to come up with the solution.

Could anybody help me?

Thank you in advance.

答:
Integrand's concern about the case for $ \ xyz \ \ne \ 0 \ $ turns out to be unnecessary.  If we solve the original Lagrange equations for $ \ \lambda \ $ , we obtain

$$ \left\{ \begin{array}{c}
2·(x^3-yz)  \ = \ \lambda x \\ 2·(y^3-4xz)  \ = \ \lambda y \\
2·(z^3- xy) \ = \ \lambda z
\end{array} \right. \ \ \Rightarrow \ \ \lambda \  =  \ \frac{2·(x^3-yz)}{x} \ =  \ \frac{2·(y^3-xz)}{y} \ =  \ \frac{2·(z^3-xy)}{z} \ \ . $$

Cross-multiplying the first ratio-pair produces

$$ x^3y \ - \  y^2z \ \ = \ \ xy^3  -  x^2z \ \ \Rightarrow \ \ x^2z \ -  \ y^2z \ \ = \ \ xy^3 - x^3y $$ $$ \Rightarrow \ \ z·(x^2-y^2) \ \ = \ \ xy·(y^2 - x^2) \ \ \Rightarrow \ \  (z \ + \ xy) \ · \ (x^2-y^2) \ \ = \ \ 0  \ \ ;   $$

similarly, we have from the other ratio-pairs,
$$ (y \ + \ xz) \ · \ (x^2-z^2) \ \ = \ \ 0  \ \ \ , \ \ \ (x \ + \ yz) \ · \ (y^2-z^2) \ \ = \ \ 0  \ \ .   $$

[That each of these equations looks like a "cyclic-permutation" re-labeling of the others is due to the same symmetry being found in the expression for $ \ f(x,y,z) \ $ and the symmetry of the spherical constraint.]

The "difference-of-squares" factors would tell us such things as $ \ y = \pm x \ , $ but since we are also restricted to the first octant, we have just $ \ x = y = z \ $ as one case.  This gives the $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ results.

The remaining case comes from the factors such as $ \ z + xy \ = \ 0 \ \Rightarrow \  z \ = \ -xy \ . $  Again because we are in the first octant, including those portions of the coordinate planes, this requires that $ \ z = 0 \ $ and either $ \ x = 0 \ \ \text{or} \ \ y=0 \ . $  Together with the related results from the other two equations, we obtain the three critical points which are the permutations of $ \ (3 \ , \ 0 \ , \ 0 ) \ $  and the three permutations of $ \ (3\sqrt{2} \ , \ 3\sqrt{2} \ , \ 0 ) \ . $  So it does not appear that any critical points have been overlooked and the conclusion in Integrand's 19 May post is confirmed.

(Incidentally, the origin is a rather interesting saddle point.  The function $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ has an inflection point at $ \ x = 0 \ $ , so this change in concavity occurs along each of the lines passing between opposing octants:
$$ -x = -y = -z \ \rightarrow \ x = y = z \ \ , \ \ x = -y = -z \ \rightarrow \ -x  = y = z \ \ , $$
$$ -x = y = -z \ \rightarrow \ x = -y = z \ \ \text{and} \ \ -x = -y = z \ \rightarrow \ x  = y = -z \ \ . ) $$


问:
Cauchy sequences proof
Let $A \subset l^{\infty}$ be a subset such that $A = \{\{a_n\}$ (for $n\geq 1) \subset \mathbb{R} : a_n \neq 0$ for finitely many $n\}$ (That is, it consists of sequences of reals with finitely many non-zero elements). Let $A$ be equipped with the metric $d_{\infty}$; that is, $d_{\infty}(a, b) = \sup_{n \geq 1} |a_n - b_n|$, where we have $a = a_n$ and $b = b_n$ being points in $A$. Let $\{a^i\} \subset A$ for $i \geq 1$ given by $a^i = (1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{i}, 0, 0, ...)$ be a sequence. Prove that $\{a^i\}$ is Cauchy and deduce that the metric space $(A, d_{\infty})$ is not complete.

I know that after proving that it is Cauchy, I will have found a Cauchy sequence that does not converge making the space not complete. But I am struggling to show that it is Cauchy. I have spent hours thinking about this problem but to no avail. Any assistance with a proof is much appreciated.

Edit: Something I just realized: Knowing that $l^{\infty}$ is complete and $A$ is a subspace then $A$ is closed if and only if $A$ is complete. Would it suffice to show that $A$ is not closed?

Edit 2: Someone mentioned this to me:

Note that $a_1$ = $(1, 0, 0, ...)$, $a_2$ = $(1, \frac{1}{2}, 0, 0, ...)$ and so on. Then

$$\lim_{n \rightarrow \infty} ||a_n - a|| = \lim_{n \rightarrow \infty} \sqrt{\sum_{j = n + 1}^{\infty} \frac{1}{j^2}} = 0$$

So $a_n \rightarrow a$ in $l^{\infty}$ and $\{a_n\}$ is Cauchy. Is this valid? If so, could somebody elaborate?

答:
We have $|a^i - a^j| \leq \frac{1}{\min(i, j)}$. Consider some $\epsilon > 0$. Take $N > \frac{1}{\epsilon}$, which exists by the Archimedean property. Then for all $i, j \geq N$, we have $|a^i - a^j| \leq \frac{1}{\min(i, j)} \leq \frac{1}{N} < \epsilon$.

Thus, the sequence is Cauchy.

Title: Cauchy sequences proof


问:
An special application about Jensen formula
Let:$f: \mathbb{C} \rightarrow \mathbb{C}$ be entire non-constant, and let us set:

$T(r)=\frac{1}{2\pi}\int_{0}^{2\pi}log_+|f(re^{i\phi})|d\phi$,
here $\log_+{s}=max(logs,0)$, show that:

$T(r)\rightarrow\infty$ as $r\rightarrow\infty$

I have two tries:

1.Use Jensen formula, however, the trunction to logarithm makes it be scarcely possible

2. Use Laurent series, take T(r) to be $T(r)=\frac{1}{\pi}\int_{0}^{2\pi}\log_+|f(re^{i\phi})|^2d\phi$, this also false.

Thanks for your advice!

答:
One needs to use the Poisson-Jensen formula (Jensen formula is for $w=0$):

$\log |f(w)|- \sum_{|a_m|<R}\log |\frac{R(w-a_m)}{R^2-\bar a_m w}|=\frac{1}{2\pi}\int_0^{2\pi}\log |f(Re^{i\theta})|\frac{R^2-|w|^2}{|R-w|^2}d\theta, |w| <R$ and as usual $a_m$ are the zeroes of $f$ counted with multiplicity (where we assume wlog $w \ne a_1,..,a_m,..$ otherwise we need obvious modifications).

Poisson Jensen implies that (as we subtract negative terms on LHS):

$\log |f(w)| \le \frac{1}{2\pi}\int_0^{2\pi}\log |f(Re^{i\theta})|\frac{R^2-|w|^2}{|R-w|^2}d\theta \le \frac{1}{2\pi}\int_0^{2\pi}\log_+ |f(Re^{i\theta})|\frac{R^2-|w|^2}{|R-w|^2}d\theta$

Since $\frac{R^2-|w|^2}{|R-w|^2} \le \frac{R+|w|}{R-|w|}$ we get that (using crucially the fact that $\log_+ \ge 0$ here):

$\log |f(w)| \le \frac{1}{2\pi}\frac{R+|w|}{R-|w|}\int_0^{2\pi}\log_+ |f(Re^{i\theta})|d\theta=\frac{R+|w|}{R-|w|}T(R)$

Letting $M_f(r)=\max_{|z|=r}|f(w)|$ and appying the above with $r=R/2$ we get that:

$\log M_f(R/2) \le 3T(R)$ and now by Liouville $\log M_f(R/2) \to \infty, R \to \infty$ so we are done!

Note that if $f=e^g$ (in other words if $f$ has no zeroes), then $\int_0^{2\pi}\log |f(Re^{i\theta})|d\theta=2\pi \log |f(0)|$ so we do need $\log_+$ to ensure that the integrals go to infinity; more generally $\int_0^{2\pi}\log |f(Re^{i\theta})|d\theta$ counts the number of zeroes, while $\int_0^{2\pi}\log_+ |f(Re^{i\theta})|d\theta$ measures the growth rate of the maximum modulus of $f$ as we saw above

(the opposite inequality $T(R)=\frac{1}{2\pi}\int_0^{2\pi}\log_+ |f(Re^{i\theta})|d\theta \le \log M_f(R)$ being trivial)


问:
Uncertainty Principle For a Generic State
Firstly, I know this is a physics problem, but this forum is so much more active, and I'm sure someone here could help me. The problem is the following:

Given a spin-1/2 particle, and the generic state:

$$ |\psi\rangle = e^{i\phi}cos(\theta)|z_+\rangle +
> e^{-i\phi}\sin(\theta)|z_{-}\rangle $$

find $\Delta S_x$, $\Delta S_y$ and $\langle S_z \rangle$ to show $\Delta S_x \Delta S_y \ge \frac{\hbar}{2} |\langle S_z \rangle|$

I think this should be pretty straightforward. Given I know the uncertainty of an opperator is given by

$$
\Delta A =\sqrt{\langle A^2\rangle - \langle A \rangle^2 }
$$

I also know that

\begin{align*}
    S_z= \frac{\hbar}{2}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad S_x= \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\qquad S_y= \frac{\hbar}{2}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\\
   \implies S_z^2=S_x^2=S_z^2= \frac{\hbar^2}{4}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\end{align*}

So after a bunch of calculations I get:

\begin{align*}
\langle S_z\rangle&=\frac{\hbar}{2}\cos(2\theta)\\
\langle S_x\rangle&=\hbar\sin(\theta)\cos(\theta)\cos(2\phi), \qquad  \langle S_x^2\rangle=\frac{\hbar^2}{4}\\
\langle S_y\rangle&=-\hbar\sin(\theta)\cos(\theta)\sin(2\phi), \qquad   \langle S_y^2\rangle=\frac{\hbar^2}{4}
\end{align*}

Finaly:

\begin{align*}
    \Delta S_x = \sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\cos(2\phi))^2}\\
    \Delta S_y = \sqrt{\frac{\hbar^2}{4} -(-\hbar \sin(\theta)\cos(\theta)\sin(2\phi))^2}\\
    \frac{\hbar}{2}|\langle S_z\rangle|= \frac{\hbar^2}{4}|\cos(2\theta)|
\end{align*}

Somehow I simply can't prove what is asked from this. Either it is a simple step I am not seeing, or I have some calculation mistake. I would appreciate any help

答:
All you have left is to show that $$\sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\cos(2\phi))^2}\sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\sin(2\phi))^2} \ge \frac{\hbar^2}{4}|\cos(2\theta)|$$

You can multiply by $\frac{4}{\hbar^2}$ to make it $$\sqrt{1 -(2\sin(\theta)\cos(\theta)\cos(2\phi))^2}\sqrt{1 - (2\sin(\theta)\cos(\theta)\sin(2\phi))^2} \ge |\cos(2\theta)|$$

Or equivalently, since $2\sin(\theta) \cos(\theta) = \sin(2\theta)$, $$\sqrt{1 -(\sin(2\theta)\cos(2\phi))^2}\sqrt{1 - (\sin(2\theta)\sin(2\phi))^2} \ge |\cos(2\theta)|$$

Square both sides, and use that $|\cos(2\theta)| = \sqrt{1-\sin^2(2\theta)}$ to make it $$\left(1-\left(\sin\left(2\theta\right)\cos\left(2\phi\right)\right)^{2}\right)\left(1-\left(\sin\left(2\theta\right)\sin\left(2\phi\right)\right)^{2}\right) \ge 1-\sin^{2}\left(2\theta\right)$$

The LHS simplifies to (after using that $\sin^2(2\phi) + \cos^2(2\phi) = 1$) $$1-\sin^{2}(2\theta)+\left(\sin(2\theta)\sin(2\phi)\right)^{2}\left(\sin(2\theta)\cos(2\phi)\right)^{2}$$

which must be greater than or equal to $1-\sin^2(2\theta)$, since the last term is always nonnegative.


问:
How to get rid of the ceiling function on one side of an equation
Given this function,

$\left\lceil \log _{b-1} \left(\lceil \frac{1}{b} \rceil\right) \right\rceil=1$

How do I get rid of the ceiling function so I can obtain the value of $B$ ?

答: If $b \lt 0$ the ceiling of $\frac 1b$ is at most $0$ and the log is undefined.  If $b \ge 1$ the ceiling of $\frac 1b$ is $1$ and the log is $0$, which fails, so we have $0 \lt b \lt 1$.  In that case $b-1 \lt 0$.  The definition I know for logs to a base different from $e$ is $\log_a(x)=\frac {\log x}{\log a}$.  If $b-1 \lt 0, \log_{b-1}x$ is undefined and there is no answer.


问:
An experiment consists of selecting a card at random from a 52-card deck. What is the probability that a club or a king is drawn?
what I did is P(a Club or a king)= (13+4)/52=0.327
But the answer is wrong.
Can someone explain me where I did wrong and what would be the correct way to solve?</p>

答: You counted the king of clubs twice, once as a king and once as a club.  There are only $13+4-1$ cards that are either a king or club, where $1$ is the number of cards that are both.  You may have seen $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ and you have lost the third term.


答:
Let $n>m$, if $a_n$ denotes a term of the sequence $\{a^n\}_n$,

$$\begin{align*}
&a_n-a_m\\
&=\left(1, \frac{1}{2}, \frac{1}{3}, \cdots, \frac{1}{m}, \frac{1}{m+1}, \cdots, \frac{1}{n}, 0,0,0, \cdots\right)-\left(1, \frac{1}{2}, \frac{1}{3}, \cdots, \frac{1}{m}, 0,0,0, \cdots\right)\\
&=\left(\underbrace{0,0,\cdots,0}_{m\text{ zeroes}},\frac{1}{m+1}, \cdots, \frac{1}{n}, 0,0,0, \cdots\right)\,.
\end{align*}$$

So $d_\infty(a^n, a^m)=\frac{1}{m+1} \rightarrow 0$ as $n, m \rightarrow \infty$.

$$\begin{align*}
&a_n-a\\
&=\left(1, \frac{1}{2}, \frac{1}{3}, \cdots, \frac{1}{n}, 0,0,0, \cdots\right)-\left(1, \frac{1}{2}, \frac{1}{3}, \cdots, \frac{1}{n},\frac{1}{n+1},\cdots\right)\\
&=\left(\underbrace{0,0,\cdots,0}_{n\text{ zeroes}},-\frac{1}{n+1}, -\frac{1}{n+2}, \cdots\right),,
\end{align*}$$

so $d_\infty(a^n,a)=\frac{1}{n+1} \rightarrow 0$ as $n \rightarrow \infty$.

But $a=\left(1, \frac{1}{2}, \frac{1}{3}, \cdots, \frac{1}{n},\frac{1}{n+1},\cdots\right) \notin A$, hence $A$ isn't complete.

Title: Cauchy sequences proof

Body:
Let $A \subset l^{\infty}$ be a subset such that $A = \{\{a_n\}$ (for $n\geq 1) \subset \mathbb{R} : a_n \neq 0$ for finitely many $n\}$ (That is, it consists of sequences of reals with finitely many non-zero elements). Let $A$ be equipped with the metric $d_{\infty}$; that is, $d_{\infty}(a, b) = \sup_{n \geq 1} |a_n - b_n|$, where we have $a = a_n$ and $b = b_n$ being points in $A$. Let $\{a^i\} \subset A$ for $i \geq 1$ given by $a^i = (1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{i}, 0, 0, ...)$ be a sequence. Prove that $\{a^i\}$ is Cauchy and deduce that the metric space $(A, d_{\infty})$ is not complete.

I know that after proving that it is Cauchy, I will have found a Cauchy sequence that does not converge making the space not complete. But I am struggling to show that it is Cauchy. I have spent hours thinking about this problem but to no avail. Any assistance with a proof is much appreciated.

Edit: Something I just realized: Knowing that $l^{\infty}$ is complete and $A$ is a subspace then $A$ is closed if and only if $A$ is complete. Would it suffice to show that $A$ is not closed?

Edit 2: Someone mentioned this to me:

Note that $a_1$ = $(1, 0, 0, ...)$, $a_2$ = $(1, \frac{1}{2}, 0, 0, ...)$ and so on. Then

$$\lim_{n \rightarrow \infty} ||a_n - a|| = \lim_{n \rightarrow \infty} \sqrt{\sum_{j = n + 1}^{\infty} \frac{1}{j^2}} = 0$$

So $a_n \rightarrow a$ in $l^{\infty}$ and $\{a_n\}$ is Cauchy. Is this valid? If so, could somebody elaborate?


问:
Extrema of $(1+\sin x)(1+\cos x)$

Find the extrema of $(1+\sin x)(1+\cos x)$ without using calculus.

I was able to figure out the minima by observing that each of the brackets range from $0$ to $2$. Therefore the minima has to be $0$ when either one of the brackets is zero.

However I couldnt figure out the maxima. I tried expanding it to complete the square but it didnt quite work out well.

Any hint is appreciated!

答: Hint: $(1+\sin x)(1+\cos x)=1+\sin x+\cos x+\frac 12\sin 2x=1+\sqrt 2\sin(x+\frac {\pi}{4})+\frac 12\sin 2x$

Title: Extrema of $(1+\sin x)(1+\cos x)$


问:
REVISITED $^2$: Does a solution in $\mathbb{R}^n$ imply a solution in $\mathbb{Q}^n$?
Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear
equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution
in $\mathbb{Q}^n$?

Where do I start?

I feel I can use this to help:

$$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$

Here is a thought that just crossed my mind, but I'm not sure if this is legal:

Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . .

答:
I'll try to restate correctly what user1938185 wanted to say. Because $\Bbb Q$ is a field, solving any linear system cannot force your variables to take values outside$~\Bbb Q$. By comparison consider a set of linear equations in a real vector space (each one has a hyperplane of solutions, you are trying to find the intersection of those solution sets). I hope you would be mighty surprised if you were told that no solution exists (the intersection is empty) but that a generalised solution can be constructed by taking complex coordinates instread of real ones. Indeed this can never happen (although such a thing may happen for polynomial equations; indeed the complex numbers are created by "inventing" a solution to the quadratic equation $x^2=-1$).

If you look at the operations you have learned to solve linear equations (forming linear combinations of equations, Gaussian elimination, substitution, whatever), you will see that they only involve arithmetic operations (including division after checking that it is not by zero) which can never cause you to go outside of a field$~F$ that contains the coefficients of the original system (in your question that would be $F=\Bbb Q$). Either you and up with a plain contradiction of the type $0x=1$, which won't have a solution either in whatever field larger than$~F$ you may invent, or you found a unique solution using arithmetic operations (which will then only involve values in$~F$), or you get multiple solutions, where some variables can be chosen freely and the other ones can be expressed in terms of them by arithmetic operations. Only in the final situation can it occur that you also have solutions where variables take values outside$~F$ (the freely chosen ones can certainly be taken that way), but you don't need to do this: if all freely chosen variables are taken to have values in$~F$, then all variables will take variables in$~F$.

To summarize, if you have a system of linear equations with coefficients in a subfield$~F$ of a strictly larger field$~K$, then the existence of solutions with values in$~K$ implies the existence of solutions with values in$~F$. In particular if there is a unique solution with values in$~K$, it actually has values in$~F$. You can apply this in amongst others with $F=\Bbb Q$, $K=\Bbb R$, or with $F=\Bbb R$, $K=\Bbb C$.


问:
Proving this inequality for an algorithm
I want to prove that for all $a \ge 2$,

$$ a \cdot {2a-1 \choose a} \le 2 \cdot \sum_{i=0}^{a-1} i \cdot {2a-1 \choose i} $$

I am thinking the best way might be induction, but I have had no luck going from k to (k+1). Any hints?

答:
Hint

Compare the RHS scalar (including the factor of $2$) applied against the specific term that involves
$\displaystyle \binom{2a - 1}{a-1} = \binom{2a - 1}{a}.$


问:
Find the radius of convergence for the power series
Find the radius of convergence:

a) $\sum_{n=0}^\infty (-1+(1)^)z^n$

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

These problems are different from what I encountered before. From what I remember, I use the ratio test. Does this method still apply here?

答:
a)   $R=\frac{1}{\lim_{n \rightarrow \infty}|e^{in}|^\frac{1}{n}}$ and $|e^{in}|=|\cos n+i\sin n|=1$

b)   $R=\frac{1}{\lim_{n \rightarrow \infty}|e^{-n^2}|^\frac{1}{n}}=\frac{1}{\lim_{n \rightarrow \infty}e^{-n}}=\infty$


答:
$(1 + \sin x)(1+ \cos x)=1+\sin x + \cos x + \sin x \cos x.$

Note that $\sin x \cos x = \frac 12 \sin 2x.$

Note also that $\sin x + \cos x = \sqrt 2 \left ( \frac {\sqrt 2}{2} \sin x + \frac {\sqrt 2}{2} \cos x \right) = \sqrt 2 \left ( \cos \frac {\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x \right )= \sqrt 2 \sin (x+ \frac{\pi}{4}).$

Both of these factors reach their maximum when $x=\frac{\pi}{4}$, so the maximum is $\frac 32 + \sqrt 2$.

Title: Extrema of $(1+\sin x)(1+\cos x)$

Body:
Find the extrema of $(1+\sin x)(1+\cos x)$ without using calculus.

I was able to figure out the minima by observing that each of the brackets range from $0$ to $2$. Therefore the minima has to be $0$ when either one of the brackets is zero.

However I couldnt figure out the maxima. I tried expanding it to complete the square but it didnt quite work out well.

Any hint is appreciated!


问:
Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005)
Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.

Source : Vietnamese TST 2005

I'm sure that the only solutions are the zero function, the identity, and the negative of the identity.  First, when $x=y=z=0$ we have $f(0)=3f(0)^3$ so $f(0)=0$ is the only integer solution.  Next if $y=-x$ and $z=0$ then $f(-x)=-f(x)$ so $f$ is odd.  Then $x=y=1$ and $z=-1$ give $f(1)^3=f(1)$ so $f(1)=0,1,-1$.

Since $f$ is a soln iff $-f$ is, we can take $f(1)=0,1$.  Then $f(2)=f(1^3+1^3+0^3)=0,2$.  Then $f(3)=f(1^3+1^3+1^3)=0,3$.  Also $f(4^3-3^3-3^3)=f(10)=f(1^3+1^3+2^3)$, so $f(4)=0,4$.  Then $f(5^3-4^3-4^3)=f(-3)$ so $f(5)=0,5$.  Then $f(6)=f(2^3-1^3-1^3)=0,6$.  Then, $f(7)=f(2^3-1^3+0^3)=0,7$.  Then $f(8)=f(2^3+0^3+0^3)=0,8$.  Then $f(9)=f(2^3+1^3+0^3)=0,9$ and $f(10)=f(2^3+1^3+1^3)=0,10$.  Etc...

There should be a smart way to make induction, but I haven't found that.  May be a nice identity.  This is the closest to the claim that $n^3$ is the sum of five cubes of integers with absolute values $<|n|$:
$$n^3=6^3+(n-2)^3+(n-3)^3-(n-6)^3-(n-11)^3+(n-12)^3.$$
At least the identity above gives me $f(11)=0,11$ and $f(12)=0,12$.

答:
Using hgmath's comment, here is a way to write $n^3$ as a sum of five cubes of integers with absolute values $<|n|$ for $n=2k$ and $k\ge 8$.  We write
$$(2k)^3=(2k-4)^3+(k+7)^3-(k-9)^3-10^3-2^3.$$
For this to work, we need to have $f(0)$, $f(1)$, $f(2)$, ..., $f(15)$.  I already got $f(0)$ until $f(12)$.

For $f(13)$ and $f(15)$ we use hgmath's comment (note that hgmath's comment is good for $k\geq 4$, but I already found until $f(12)$).  For $f(14)$, we can use
$14^3=12^3+10^3+2^3+2^3$.


问:
largest singular value of real symmetric matrix
I am trying to find a proof of the following fact: Let $M$ be a real symmetric matrix, then the largest singular value satisfies:
$$
\sigma_1(M) = \lim_{k \to \infty} \left[\text{Trace}(M^{2k})\right]^{\frac{1}{2k}}
$$

答:
$\newcommand{\diag}{\mathrm{diag}}$
$\newcommand{\tr}{\mathrm{Trace}}$
Since $M$ is symmetric, $M^2 = M'M$, which can then be decomposed as
\begin{align*}
M'M = O\diag(\sigma_1^2(M), \ldots, \sigma_n^2(M))O',
\end{align*}
where $O$ is order $n$ orthogonal matrix. It then follows that $M^{2k} = (M'M)^k =
O\diag(\sigma_1^{2k}, \ldots, \sigma_n^{2k})O'$, whence $\tr(M^{2k}) = \sigma_1^{2k} + \cdots + \sigma_n^{2k}$. Hence by squeeze principle, it is easy to verify that
\begin{align*}
\lim_{k \to \infty}(\sigma_1^{2k} + \cdots + \sigma_n^{2k})^{1/2k} = \sigma_1.
\end{align*}


问:
Proving $(A+B)^{-1} = A^{-1} + B^{-1}$ when there exists $J$ so that $J^2 = -I$
I'm having trouble proving this biconditional statement:

Prove that there exist $A, B \in \mathcal{M}_{n\times n}(\mathbb{R}^n)$ such that $(A+B)^{-1} = A^{-1} + B^{-1}$ if and only if there exists $J \in \mathcal{M}_{n \times n}(\mathbb{R}^n)$ such that $J^2 = -I_n$.

Here, $I_n$ is the $n \times n$ identity matrix. So far, none of my ideas have panned out and I don't feel like I have enough information to solve this problem. Any hints would be appreciated!

答:
First, write $B = XA.$ (this is always possible, since $B$ is invertible). Then, you get

$$(A+XA)^{-1} = A^{-1} + A^{-1}X^{-1}.$$ Factoring out the $A^{-1}$ from both sides, we get

$$(I+X)^{-1}= (I + X^{-1}).$$  Multiply this through by $I+X$ to get:

$$I = (I+X^{-1})(I+X),$$ whence it follows that $$X = (I+X)^2,$$ so the minimal polynomial of $X$ is $$X^2+X+I=0,$$ or $$\left(X+\frac12I\right)^2 = -\frac{3}{4} I,$$ so

$$\left(\frac{2\sqrt{3}}{3}(X+\frac12I)\right)^2 = -I.$$

Every step in the above is an equivalence, so proves both directions.


问:
prove that any set $S$ subset of {1,2,,...100} with |S|=26 contains at least one odd number or y is divisible by x, where y different from x
prove that any set $S \subset\{1,2,,...100\}$ with $|S|=26$ contains at least one odd number or it contains two numbers $x$ different from $y$ for which $x\,|\,y$.

I know I should use pigeonhole principle, but I am having trouble to come up with an idea of what to consider as pigeonholes. I was thinking to consider two cases, one where I have at least one odd number but not necessarily $x\,|\,y$, and the opposite.

答: If it has no odd number, then every number is even, so divide them all by $2.$ Now you have $26$ numbers between $1$ and $50.$ Can you finish the argument?


问:
Limit point of the specific set is in a set of the open cover.
I am currently working on the exercise 2.26 of Baby Rudin.

It is mentioned the limit point $z$ must be in some set $G_n$. I don't get why should the limit point be covered as well, I tried proving that the set is closed but I am not able to get anywhere. Thanks for helping.

答: This is because the $G_n$ are a cover of the space $X$. Since $z$ is an element of $X$, it must be in one of the covering elements.


问:
Point Spectrum of an operator from orthogonal decomposition
Suppose I have an operator $ H $ that is self-adjoint and densely defined on  $ L^2 (\mathbb{R} , \mathbb{C}) \times L^2 (\mathbb{R} , \mathbb{C}) $.

Suppose further that $ L^2 (\mathbb{R} , \mathbb{C}) $ is an invariant subspace of $ H $.

Now, suppose that the spectrum of $ H $ consists purely of the point spectrum. I am thinking about the Quantum Harmonic Oscillator, for example.

Suppose further that the restriction of $ H $ onto $  L^2 (\mathbb{R} , \mathbb{C}) \times 0 $ and $  0 \times L^2 (\mathbb{R} , \mathbb{C})  $  has purely simple eigenvalues.

Is it true that the spectrum of $ H $ will be the union of the point spectrum of the restriction? If so, how could one prove it?

Proof Attempt

Let $ \lambda \in \sigma(H)= \sigma_p (H) $. Then,there is a $ u=(u_1,u_2) \in L^2 (\mathbb{R} , \mathbb{C})\times L^2 (\mathbb{R} , \mathbb{C})  $ such that
\begin{equation}\label{key}
 Hu= \lambda u.
\end{equation}
Now,
\begin{equation}
 H = \begin{pmatrix}
  H_{11} & 0 \\
  0 &  H_{22}.
 \end{pmatrix}
\end{equation}
for $ H_{11} $, $ H_{22} $ self adjoint operators on $ L^2 (\mathbb{R} , \mathbb{C})  $.

Then,
\begin{align}
 H_{11} u_1 &= \lambda u_1, \\
 H_{22} u_2 &= \lambda u_2.
\end{align}

and so certainly $ \sigma(H) \subset \sigma(H_{11}) \cup \sigma(H_{22}) $.

For the other direction, is it correct to simply take $(u,0)$ and $(0,u)$?

I feel like some form of what I am trying to prove- a statement about reconstructing the spectrum of an operator from its restrictions on invariant subspaces which are orthogonally decomposed- should be a standard proposition or theorem in a Functional analysis textbook, but I can't seem to find it.

答:
Your argument is fine. You don't need to restrict the type of the elements of the spectrum, though.

From
$$
H-\lambda I=\begin{bmatrix} H_{11}-\lambda I&0\\0& H_{22}-\lambda I\end{bmatrix}
$$
it is easy to see (proof below) that $H-\lambda I$
is invertible if and only if both $H_{11}-\lambda I$ and $H_{22}-\lambda I$ are. So $$
\rho(H)=\rho(H_{11})\cap \rho(H_{22}),$$
and taking complements you get $$\sigma(H)=\sigma(H_{11})\cup\sigma(H_{22}).$$

We want to show that $\begin{bmatrix} X&0\\0& Y\end{bmatrix}$ is invertible if and only if $X$ and $Y$ are.

If $AX=XA=I$ and $BY=YB=I$, then $\begin{bmatrix} A&0\\0& B\end{bmatrix}$ is an inverse. Conversely, if
$$
\begin{bmatrix} X&0\\0& Y\end{bmatrix}\begin{bmatrix} A&B\\ C& D\end{bmatrix}=\begin{bmatrix} A&B\\ C& D\end{bmatrix}\begin{bmatrix} X&0\\0& Y\end{bmatrix}=\begin{bmatrix} I&0\\0& I\end{bmatrix},
$$
looking at the diagonal coordinates we get the equalities
$$
XA=AX=I, \qquad YD=DY=I.
$$
So $X,Y$ are invertible.


问:
How/why $\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}$ turn into (negative fraction) $-\frac{x+a}{x²a²}$?
I have that the solution of $\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)$ is $-\frac{x+a}{x²a²}$, but when I try to solve it I get (positive) $\frac{x+a}{x²a²}$.

The way I've been getting it is by making the denominators equals $\left(\frac{1}{x²(x-a)}\frac{a²}{a²}\right)-\left(\frac{1}{a²(x-a)}\frac{x²}{x²}\right)$ which gives $\frac{a²-x²}{x²a²(x-a)}$ that by factoring the nominator is $\frac{(a-x)(a+x)}{x²a²(x-a)}$, allowing me to cancel $\frac{(a-x)}{(a-x)}$, finishing as $\frac{x+a}{x²a²}$.

Where it should've become negative?

OBS: I don't know if matters but it's all part of a limits problem that goes like $\lim_{x\to a}\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)=-\frac{x+a}{x²a²} = \frac{-2}{a³}$, but my problem is just where it became negative as it's suppose to become.

答:
The correct cancellation is
$$
\frac{a-x}{x-a}=-1
$$
since $a-x=(-1)(x-a)$.


答:
We can also use the inequality
\begin{gather*}
AM\geq GM\\
So,\ \\
\sqrt{( 1+\sin x)( 1+\cos x)} \leq \frac{1+\sin x+1+\cos x}{2}\\
Now,\ the\ function\ on\ the\ RHS\ is\\
1+\frac{\sin x+\cos x}{2} =1+\frac{1}{\sqrt{2}}\sin\left( x+\frac{\pi }{4}\right)\\
The\ maxima\ of\ this\ function\ is\ 1+\frac{1}{\sqrt{2}} ,\\
because\ maximum\ value\ of\ \sin\left( x+\frac{\pi }{4}\right) =1\\
\sqrt{( 1+\sin x)( 1+\cos x)} \leq 1+\frac{1}{\sqrt{2}}\\
( 1+\sin x)( 1+\cos x) \leq \left( 1+\frac{1}{\sqrt{2}}\right)^{2} =\frac{3+2\sqrt{2}}{2} =\frac{3}{2} +\sqrt{2}
\end{gather*}

Title: Extrema of $(1+\sin x)(1+\cos x)$


Title: How/why $\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}$ turn into (negative fraction) $-\frac{x+a}{x²a²}$?

Body:
I have that the solution of $\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)$ is $-\frac{x+a}{x²a²}$, but when I try to solve it I get (positive) $\frac{x+a}{x²a²}$.

The way I've been getting it is by making the denominators equals $\left(\frac{1}{x²(x-a)}\frac{a²}{a²}\right)-\left(\frac{1}{a²(x-a)}\frac{x²}{x²}\right)$ which gives $\frac{a²-x²}{x²a²(x-a)}$ that by factoring the nominator is $\frac{(a-x)(a+x)}{x²a²(x-a)}$, allowing me to cancel $\frac{(a-x)}{(a-x)}$, finishing as $\frac{x+a}{x²a²}$.

Where it should've become negative?

OBS: I don't know if matters but it's all part of a limits problem that goes like $\lim_{x\to a}\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)=-\frac{x+a}{x²a²} = \frac{-2}{a³}$, but my problem is just where it became negative as it's suppose to become.

Body:
$$\frac{1}{x^2(x - a)} - \frac{1}{a^2(x - a)}$$
$$\frac{1}{(x-a)}\left(\frac{1}{x^2} - \frac{1}{a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{a^2 - x^2}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x)(a - x)}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x) -1(x - a)}{x^2a^2}\right)$$
$$\frac{-(a + x)}{x^2a^2}$$


Title: Proving $(A+B)^{-1} = A^{-1} + B^{-1}$ when there exists $J$ so that $J^2 = -I$

Body:
I'm having trouble proving this biconditional statement:

Prove that there exist $A, B \in \mathcal{M}_{n\times n}(\mathbb{R}^n)$ such that $(A+B)^{-1} = A^{-1} + B^{-1}$ if and only if there exists $J \in \mathcal{M}_{n \times n}(\mathbb{R}^n)$ such that $J^2 = -I_n$.

Here, $I_n$ is the $n \times n$ identity matrix. So far, none of my ideas have panned out and I don't feel like I have enough information to solve this problem. Any hints would be appreciated!

Body:
$(\Longrightarrow)$

Since
$(A+B)(A^{-1}+B^{-1})= I + AB^{-1}+BA^{-1} + I=(A+B)(A+B)^{-1}=I$

$AB^{-1}+BA^{-1}=AB^{-1}+(AB^{-1})^{-1}=-I$ and $ (AB^{-1})^2+I=-AB^{-1}$

and if we let $X=AB^{-1}$, then we have $X^2+X+I=O$.

We can do a change of variables $X=Y+C$

$(Y+C)^2+(Y+C)+I=Y^2 +(2C+I)Y+C^2+C+I$

and by letting $C=-\frac{1}{2}I$, then we get $Y^2+0Y+\frac{3}{4}I=Y^2+\frac{3}{4}I=O$

now $\left(\frac{2}{\sqrt{3}}Y\right)^2+I=Z^2+I=0$ and $Z=\frac{2}{\sqrt{3}}\left(X+\frac{1}{2}I\right)$ satisfies the properties of $J$.

$(\Longleftarrow)$

Given $J$ s.t. $J^2=-I$, then let $Z=-\frac{1}{2}I+\frac{\sqrt{3}}{2}J$

then $\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)\left(-\frac{1}{2}I-\frac{\sqrt{3}}{2}J\right)= \frac{1}{4}I^2-\frac{3}{4}J^2= \frac{1}{4}I+\frac{3}{4}I=I$ and $Z$ is invertible.

$Z$ also satisfies $Z^2 + Z +I =O$

$\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)^2 + \left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)+ I=\frac{1}{4}I^2-\frac{\sqrt{3}}{2}J+\frac{3}{4}J^2-\frac{1}{2}I+\frac{\sqrt{3}}{2}J+I$

$=\frac{1}{4}I-\frac{3}{4}I-\frac{1}{2}I+I=O$

Given an invertible matrix $B$, let $ZB=A$, and then $Z=AB^{-1}$ where $A$ is invertible because $Z$ and $B$ are inverible.

$(A^{-1}+B^{-1})=(A+B)^{-1}$ because $Z+Z^{-1}=AB^{-1}+BA^{-1} = -\frac{1}{2}I + \frac{\sqrt{3}}{2}J -\frac{1}{2}I -\frac{\sqrt{3}}{2}J =-I$

thus $I+AB^{-1}+BA^{-1}+I=(A+B)(A^{-1}+B^{-1})=I$.


问:
Find the measure of $\angle G$ in a triangle
In the following figure, $\Delta$ADB, $\Delta$PCB and $\Delta$EFG are right triangles.
$PB=AE$,$AC=CB$.

Question:  What is the value of $\angle G$?

I figured out $\angle A = \angle CPB$ ,  $EF=AC$. What should I do next?

答:
$CP$ is orthogonal bisector of segment $AB$. Hence
$$AP = BP = AE$$ which means that triangle $AEP$ is isosceles. Therefore
$$\angle \, AEP = \angle \, APE = \alpha \,\,\, \text{and} \,\,\, \angle \, PAE = 180^{\circ} - 2\alpha$$
Since $\angle \, DPE = 90^{\circ} - \angle \, AEP = 90^{\circ} - \alpha$,
$$\angle \, APB = \angle \, APE - \angle \, DPE = 2\alpha - 90^{\circ}$$
But $ABP$ is isosceles ($AP = BP$), so
$$\angle \, PAG = \angle  \, PAC = 90^{\circ} - \frac{1}{2}(\angle \, APB ) = 135^{\circ} - \alpha$$
From here,
$$\angle \, EAG = \angle \, PAG - \angle \, PAE = 135^{\circ} - \alpha - 180^{\circ} + 2\alpha = \alpha - 45^{\circ}$$
Now, in triangle $AEG$ we know the angle at vertex $A$, which is $\angle \, EAG = \alpha - 45^{\circ}$ and we know the external angle at vertex $E$, which is $\angle \, AEP = \alpha$. Theretore,
$$\angle \, G = \angle \, AGE = \angle \, AEP - \angle \, EAG = \alpha - \alpha + 45^{\circ} = 45^{\circ}$$


问:
Evaluating $\int t \exp(2t) \sin(2t) \mathrm dt$
Can someone show me how to evaluate

$$\int te^{2t}\sin 2t\,dt$$

without using a complex analysis approach? Thanks,

David

答:
$\int e^{t}\sin (t)\,dt$ is known.

By (1), $\int e^{2t}\sin (2t)\,dt$ is known.

$\int te^{2t}\sin (2t)\,dt$ is known with integration by parts: Let $u=t$ and $dv = e^{2t}\sin (2t)\,dt$, where $dv$ is known by (2).


问:
Is there a polynomial accepting all positive and only positive values?
Please help to prove that there is no polynomial
$p\colon \mathbb{R}^n \to (0,+\infty)$.

答:
Although it may be counterintuitive, yes, there is; for example,
$$f(x,y) = (1-xy)^2+x^2,$$
which I found here: https://math.stackexchange.com/a/3822, which in turn comes from a comment in math overflow.

You can see that $f(x,y)>0$ everywhere, and we cannot have $f(x,y)=0$. On the other hand, for any $\epsilon > 0$, you can find $x,y$ such that $f(x,y) = \epsilon$. (Specifically, $f(\sqrt{\epsilon}, \frac1{\sqrt{\epsilon}}) = \epsilon$.)


问:
Adding powers of $i$
I've been struggling with figuring out how to add powers of $i$.

For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...

It has to do something with division by 4, since the "power cycle" of $i$ repeats every fourth power.

Thank you for any clues.

答:
Observing that $i^{3}+i^{4}+\ldots +i^{50}$ is a geometric progression with ratio $i$, first term $i^3$ and $50-3+1=48$ terms, we have

$i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}=i^{3}\times
\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}$

$=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0$

Edit: "arithmetic" corrected to "geometric"

Title: Adding powers of $i$

LastEditorDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Which of these variables affect y more?
I have two equations:

1: $\displaystyle y=\frac{0.0029}{\sqrt{k}}$

2: $\displaystyle y=\frac{0.0060}{\sqrt{m}}$

Which variable has a greater effect on $y$? i.e., does doubling $m$ have a greater effect on $y$ than doubling $k$? How do I describe/prove that?

Thanks!

(I asked a similar question previously, here. However I feel that the square root nature of this question makes it an entirely different question, and so I've posted a new question.)

答:
If you double $k$ you have

$$y'=\frac{0.0029}{\sqrt{2\cdot k}}=\frac{1}{\sqrt{2}}\cdot y$$

Simillary if you double $m$ you also get

$$y'=\frac{0.0060}{\sqrt{2\cdot m}}=\frac{1}{\sqrt{2}}\cdot y$$

So your value will always change by a constant factor that is $\displaystyle \frac{1}{\sqrt{2}}$. How much the "effect" is depends on how big $y$ was before.

Title: Which of these variables affect y more?


问:
Subgroups of $G$ if there is a homomorphism from $G$ to $C_2$
Is there always a subgroup of index 2 in $G$ if there is a homomorphism $f: G \rightarrow C_2$? My inkling is that this is true because in this case $G$ must have a normal subgroup of index 2, and the result follows from Lagrange's theorem. Could someone please help me and provide a solution?

答:
Need not have a subgroup of index 2.

For counter-example: It is well known that $A_5$ does not have a subgroup of index 2.
But, there is a trivial homomorphism from $A_5$ to $C_2$.
Even if it is not onto, which is not mandatory in the question.


答:
We have similar right $\triangle PCB \cong \triangle AFE$ since $PB=AE$. Hence let $CB=x=EF=AC$. Also let $PC=y=AF$.

Then $CF=AF-AC=y-x$. Complete the rectangle $PCFQ$. We get $PQ=CF=y-x$. Also $QE=QF-EF=y-x$. Hence $\triangle PQE$ is right isosceles. Using similarity or since $PQ \parallel CG$, $$\angle G = 45^{\circ}$$

Title: Find the measure of $\angle G$ in a triangle

Body:
In the following figure, $\Delta$ADB, $\Delta$PCB and $\Delta$EFG are right triangles.
$PB=AE$,$AC=CB$.

Question:  What is the value of $\angle G$?

I figured out $\angle A = \angle CPB$ ,  $EF=AC$. What should I do next?


问:
Is there a higher order chain rule such that $\frac{d^2 r}{ds^2}= \frac{d^2 r}{dx^2} \frac{d^2 x}{ds^2} $
I don't have much more to add the question is right there in the title. I just came across this in a long calculation that I'm working on. It would make everything easier if this were true.</p>

答:
You can use that $(fg)' = f'g + fg'$ in order to find $\frac{d^2r}{ds^2}$. Starting off from $$\frac{dr}{ds}=\frac{dr}{dx} \frac{dx}{ds}$$

Differentiate wrt $s$ to get

$$\frac{d^2r}{ds^2}=\frac{dx}{ds}\left(\frac{d}{ds}\frac{dr}{dx}\right) + \frac{dr}{dx} \frac{d^2x}{ds^2}$$

Use the chain rule to get $$\frac{d^2r}{ds^2}=\frac{dx}{ds}\left(\frac{d^2r}{dx^2} \frac{dx}{ds}\right) + \frac{dr}{dx} \frac{d^2x}{ds^2}$$

Or equivalently, $$\frac{d^2r}{ds^2}=\left(\frac{dx}{ds}\right)^2\frac{d^2r}{dx^2}+\frac{dr}{dx}\frac{d^2x}{ds^2}$$

For higher order derivatives, you can use Faa di Bruno's formula.


问:
Given $G = (V,E)$, a planar, connected graph with cycles, Prove: $|E| \leq \frac{s}{s-2}(|V|-2)$. $s$ is the length of smallest cycle

Given $G = (V,E)$, a planar, connected graph with cycles, where the smallest simple cycle is of length $s$. Prove: $|E| \leq \frac{s}{s-2}(|V|-2)$.

The first thing I thought about was Euler's Formula where $v - e + f = 2$. But I really could not connect $v$, $e$ or $f$ to the fact that we have a cycle with minimum length $s$.

Any direction will be appreciated, thanks!

答:
This is a question of Grimaldi, Discrete and Combinatorial Mathematics, chapter 11 section 4 exercise 23,a.

You may read the last few pages of this section, if you still do not see I can give you the hint.

Good luck

The responce for the last comment:

Each edge is in the boundary of at most 2 faces. Then if you consider the sum of lengths of all faces in G, or regions, the total lenght is always less equal to 2$e$. And obviously, since $s$ is the smallest lenght, the sum of lengths is greater and equal to $sr$. Finally we have this: $2e \geq \sum deg(r_i) \geq sr$ where $deg(r_i)$ is the number of edges in the boundary of face.


问:
Pontryagin Duality and Injective Direct Limits
Let $(G_n)_{n \in \Bbb{N}}$ be an injective direct limit of discrete groups. How does pontryagin duality play with injective direct limits? More precisely, what is $\widehat{\lim G_n}$ isomorphic to? Is it $\lim \widehat{G_n}$? If it helps, we may assume the $G_n$ are abelian groups.

EDIT: I suspect maybe they aren't isomorphic...I don't think that the direct limit of compact spaces is necessarily compact. Isn't $\Bbb{R}$ isomorphic to the injective direct limit of the compact intervals $[-n,n]$? If so, $\lim \widehat{G_n}$ probably won't always be compact, but $\widehat{\lim G_n}$ is always compact.

答: Since Pontryagin duality is a contravariant equivalence of categories, it turns colimits into limits.  So, if $G$ is the direct limit of a system $(G_n)$, the Pontryagin dual $\widehat{G}$ is the inverse limit of the dual system $(\widehat{G_n})$ (which is an inverse system, not a direct system, since the arrows got reversed) in the category of compact abelian groups.  Inverse limits in the category of compact abelian groups are computed by just taking the inverse limit as sets and giving it the inverse limit topology (compact Hausdorff spaces are closed under inverse limits of topological spaces) and group structure.


答:
There is no complex analysis when you write with $x=2t$
$$\int t\,e^{2t}\sin 2t\,dt=\frac 14\Im \int x\,e^{x}e^{ix}\,dx=\frac 14\Im \int x\,e^{(1+i)x}\,dx$$ One single integration by parts gives the result.

Title: Evaluating $\int t \exp(2t) \sin(2t) \mathrm dt$

Body:
Can someone show me how to evaluate

$$\int te^{2t}\sin 2t\,dt$$

without using a complex analysis approach? Thanks,

David


答:
There are elementary proofs, as exhibited by the other answer, but the identity in question can also be viewed as a special instance of Gelfand's formula. Since $M$ is real symmetric, we have $\sigma_1(M)=\rho(M)$ and $\operatorname{tr}(M^{2k})= \|M^k\|_F^2$, where $\|\cdot\|_F$ denotes Frobenius norm. Therefore, by Gelfand's formula,
$$
\sigma_1(M)=\rho(M)
=\lim_{k\to\infty}\|M^k\|_F^{1/k}
=\lim_{k\to\infty}\left(\|M^k\|_F^2\right)^{1/2k}
=\lim_{k\to\infty}\left(\operatorname{tr}(M^{2k})\right)^{1/2k}.
$$

Title: largest singular value of real symmetric matrix

Body:
I am trying to find a proof of the following fact: Let $M$ be a real symmetric matrix, then the largest singular value satisfies:
$$
\sigma_1(M) = \lim_{k \to \infty} \left[\text{Trace}(M^{2k})\right]^{\frac{1}{2k}}
$$