问:
Countability of $\mathbb{Z} \times \mathbb{Z}$ using isomorphism with $\Bbb Z[i]$ - how to make this rigorous?

Visualise the set $\mathbb{Z}\times \mathbb{Z}$  as points on the complex plane,  and take origin as the center of a circle of radius $r$. With $r$ going from $0$ to $\infty$ - as you encounter points on the circle, write them down in counterclockwise order. This way you have a bijective mapping, thus $\mathbb{Z}\times \mathbb{Z}$ is countable.

This proof about the countability of $\mathbb{Z}\times \mathbb{Z}$ doesn't look rigorous enough to me. It was proposed by one of my colleagues. The proposal is to write down elements of $Z[i]$ as we encounter them in some sort of a spiral order.

I need help making this more rigorous.

Is there an explicit bijection for this? It'd be great if someone could help me construct the same.

Perhaps we must show that every such circle has only a countable number of such points and that only countably many such circles, i.e. values of $r$ matter. Thanks!

答:
Adding to strawberry-sunshine's answer: An explicit bijection is hard because it is hard (a) to give an explicit enumeration of all radii that contain a number from $\mathbb Z [i]$ and (b) given one such radius to enumerate all numbers in $\mathbb Z [i]$ with that radius with an explicit function.

As the other answer pointed out, for any radius $r$ there is a finite number of associated numbers in $\mathbb Z [i]$, say S. We can order these formally counterclockwise by first ordering all the numbers $S_+$ from the upper half plane (plus the nonnegative real number halfline) by setting $S_0 = \emptyset$, $S_n=\{a_1+ib_1,\ldots,a_n+ib_n\}$ and

$$a_{n+1} = \max \{a\in \mathbb R: \exists b\in \mathbb R: a+bi\in S_+\setminus S_n\}, \quad b_{n+1} = \sqrt{r^2-a_{n+1}^2} $$

and after that, if $r\neq 0$, set $a_{|S_+|+n} = a_{|S_+|-n+1}$ and $b_{|S_+|+n} = -b_{|S_+|-n+1}$. So by induction, you can order all those numbers counterclockwise, but a convenient definition without using recursion escapes me.

However, this is all unneededly complicated. Also the line

Visualise the set $\mathbb Z \times \mathbb Z$ as points on the complex plane

basically already is a bijection, namely $f:\mathbb Z \times \mathbb Z\rightarrow \mathbb Z[i], (a,b)\mapsto a+bi$. If you are looking for formal proofs always be suspicious when someone uses terms like "visualize", as that is not a valid concept in formal proofs. The given bijection can easily be seen as bijective by confirming the function $g:\mathbb Z[i]\rightarrow \mathbb Z \times \mathbb Z, z\mapsto (\Re(z),\Im(z))$ is the inverse (which we can check by checking that $g\circ f$ and $f \circ g$ are the identities on $\mathbb Z \times \mathbb Z$ and $\mathbb Z[i]$ respectively). Therefore, the enumerating as done above with the counterclockwise ordering is unneededly contrived, albeit a good thought exercise nonetheless.

Finally a way I would prefer is simply using that if $|A|=|C|$ and $|B|=|D|$ (i.e. bijections between these sets $A, B, C, D$ exist), then $|A\times B|=|C\times D|$ (if the given bijections are $f$ and $g$ respectively, then the new bijection is given by $(a,c)\mapsto (f(a),g(c))$). Then set $A=B=\mathbb N$ and $C=D=\mathbb Z$ and the countability of $\mathbb Z \times \mathbb Z$ follows from the countability of $\mathbb N \times \mathbb N$.


问:
A vector represents a force with its magnitude and direction but could it also represent the time it was applied for?
I'm an A2 igsce math student and I'm taking mechanics for the first time in math this session.

I know that what I'm going to say is wrong judging from my teacher's reaction when I asked him about this but unfortunately he didn't explain to me why it's wrong.

So, when we draw two forces on an object, we usually assume that they are applied at the same time but do we also assume that they were applied for a certain duration of time, like one second maybe?

I mean, how are we supposed to get the magnitude of the resultant of these two forces meanwhile we don't know the duration? To make my point clearer, for example, couldn't there be more than one force in the exact direction of the resultant force that would make our object/point reach the final speed it is supposed to reach after these two forces are applied for a certain time? — thinking about it now the examiner never mentions for how long two forces should be applied, but anyway.

But each force would need a certain amount of time, like a weaker force will need more time while a stronger one will take less time and so on. And all these forces are in the same direction of the resultant force of course.

答:
At this point in your course it appears that all forces that are treated are constant in time and infinitely present like gravity, a steady wind, or friction (there the force is not constant, but the cause of it, or friction coefficient, is).

You might be thinking more like of temporary forces like a rocket engine (or more mundane, a car engine) or changing forces like the magnetism of an electro-magnet under alternating current.

A force only provides acceleration (actually, changes in impulse) as long as it is "switched on", two forces can only be added in the intended sense if they are "switched on" at the same time. The confusion of your teacher could be because the forces discussed so far are "constantly on".

ContentLicense: CC BY-SA 4.0


问:
Proof Solver Geometry
Are there any programs that can solve Geometry Problems?

An example of such a problem would be:
The centroid of a triangle always divides its medians into two sections with a 1:2 ratio.

While that problem is very easy to solve by hand, I am wondering whether it can be automated in any way.
Thanks in advance!

答:
With GeoGebra Discovery you can even ask GeoGebra for the relation of the two segments that the centroid produces in the median.

The Relation tool "guesses" (but rigorously), and it is 2:1.

First, only numerically:

Then, the symbolic relation is shown:

Finally, if you do not need to "guess" the relation and you know it already, you can use directly the Prove command in GeoGebra (see the bottom of the picture: d=Prove(2k==j) and the answer is true...

The proof is algebraic geometry based, so it is not readable.

Title: Proof Solver Geometry


问:
Generalized Nakayama's lemma over a non-commutative ring
Let $R$ be a ring, $J(R)$ its Jacobson radical, $M$ is a finite $R$-module. The following statement is usually called Nakayama's lemma: if $IM=M$ then $M=0$. This is true over any ring (commutative or not).

Over a commutative ring the generalized version of Nakayama's lemma holds: if $IM=M$ for some ideal $I$ and finite $M$ then there is $x \in I$ such that $(1-x)M=0$.

The only proof I know is based on determinant trick, which doesn't generalize to non-commutative situation. I doubt that this is true over non-commutative ring. What are the simplest examples of a non-commutative ring and finite module over it that generalized Nakayama's lemma is false?

答:
I don't know about the simplest example, but here's a low-dimensional one. Take $R$ to be the path algebra of the quiver with vertices 1,2 and arrows $a\colon 1\to 2$ and $b\colon 2\to 2$, modulo the relation $b^2$. Thus $R$ has basis $e_1,e_2,a,b,ba$. Take $M$ to be the module $Re_1/Rba$, so having basis $e_1,a$. Take $I$ to be the two-sided ideal generated by $e_1$, so having basis $e_1,a,ba$. Now $IM=M$, since $e_1^2=e_1$ and $ae_1=a$, but $(1-x)M\neq0$ for all $x\in I$, since $(1-x)a=a$.

EDIT: Actually one can remove the $b$ to get a smaller example: Take $A$ to be the path algebra of the quiver $1\xrightarrow{a}2$, so having basis $e_1,e_2,a$ such that $e_1,e_2$ are orthogonal idempotents, and $a=e_2ae_1$. Take $I=M=Ae_1$, a projective left ideal. Then $I$ has basis $e_1,a$ and $I=IA$ is actually a two-sided ideal. Now $I^2=I$, but $(1-x)a=a$ for all $x\in I$, so $(1-x)I$ is nonzero.


问:
How to redistribute probabilities when one outcome's probabilities changes
Let's say that one scenario has 3 possible outcomes A, B and C, for each outcome there's a 33% chance of it happening, however, if suddenly A's chances changed to 80%, how would I redistribute the chances among the remaining outcomes?

My first thought would be to split the remaining evenly between B and C, however this would not suffice if the initial chances for B and C weren't the same.

With that in mind is there a way to redistribute the chances correctly even if B / C weren't initially the same?

Notes:

*For this specific question only one outcome will change with a predefined chance.

*There can be n number of outcomes with x chances of each occurring initially.

答:
Consider this example: we have three possible outcomes $A,B,C$, with probability $0.5,0.3,0.2$ respectively. Note that $\mathbb P(B):\mathbb P(C)=3:2$, so we can say that $\mathbb P(B)=3k,\mathbb P(C)=2k$ where $k$ is a constant. If $\mathbb P(A)$ is changed to $0.8:$ \begin{align*}
\mathbb P(A)+\mathbb P(B)+\mathbb P(C)&=1\longrightarrow 0.8+3k+2k=1\longrightarrow k=0.04
\end{align*}
So we have the new probabilities: $$\mathbb P(B)=3\times 0.04=0.12,\:\mathbb P(C)=2\times 0.04=0.08$$


问:
Calculate $\int_D x^2(y+1)\,\mathrm{d}x\,\mathrm{d}y$ where $D=\left\{(x,y)\in \mathbb R^2 : |x|+|y|\le 1\right\}$
I have to calculate
$$\int_D x^2(y+1)\,\mathrm{d}x\,\mathrm{d}y\,,$$

where $$D=\left\{(x,y)\in \mathbb R^2 : |x|+|y|\le 1\right\}$$

Now I went on to try to write this $D$ as a set of the form $$ E = \left\{(x,y) \in \mathbb{R^2}: a < x < b \,,\, \alpha(x)<y<\beta(x)\right\} \,,$$ since in these cases we know how to calculate the integral. But there are many possible solutions to the inequation in D, namely:

and I wouldn't know which to take as borders of the integral because I don't think they all give the same result.

Moreover I tried to use symmetry. The set in my case looks like this

so maybe we could assume it is symmetric and the Integral is equal to zero. But I am very unsure. This topic was explained very briefly in the lecture therefore I am having a lot of problems.

Thank you in advance

答:
Values at a finite number of points have no effect on the integral. It is not even necessary to cosnider $x>0, x<0$ etc.

What we have is $\int_{-1}^{1}\int_{-|1-|x|}^{1-|x|} x^{2}(y+1)dy dx$. Note that $x^{2}$ can  be pulled out of the inside integral and the integral of $y$ is $0$ becasue it is an odd function. Hence, we are left with $\int_{-1}^{1} 2x^{2}(1-|x|)dx$. Once  again, note that this is twice the integral from $0$ to $1$. I will let you finish.


问:
Are all integer fractions rational?
Any repeating decimal can be written as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers. But is the reverse true. Will any fraction $\frac{a}{b}$ where $a$ and $b$ are integers produce a quotient with a repeating decimal?</p>

答:
This question is confused.  The DEFINITION of "rational number" is a number that is a quotient of two integers.

It can be shown that every repeating decimal is rational and every rational number is a repeating decimal.  But: DO NOT think that "repeating (or terminating) decimal" is the definition of rational number.  No mathematician does that.

The answer is "yes" because of the pigeonhole principle.  Say, for example you divide $5$ by $82$:

$$
\begin{array}{cccccccccccccc}
& & 0 & . & 6 & 0 & 9 & 7 & 5 &   \\[10pt]
82 & ) & 5 & . & 0 & 0 & 0 & 0 & 0 & 0 \\
   &   & 4 &   & 9 & 2 \\  \\
   &   &   &   &   & 8 & 0 \\
   &   &   &   &   &   & 0 \\  \\
   &   &   &   &   & 8 & 0 & 0 \\
   &   &   &   &   & 7 & 3 & 8 \\  \\
   &   &   &   &   &   & 6 & 2 & 0 \\
   &   &   &   &   &   & 5 & 7 & 4 \\  \\
   &   &   &   &   &   &   & 4 & 6 & 0 \\
   &   &   &   &   &   &   & 4 & 1 & 0 \\  \\
   &   &   &   &   &   &   &   & 5 & 0 & 0 & \longleftarrow\text{ return to starting point}\\
\end{array}
$$
(We started by asking how many times $82$ goes into $500$ and we said $6$.  Now we are again faced with the question of how many times $82$ goes into $500$, and the answer is still $6$.)

How do you know that you will eventually return to the starting point, and therefore start repeating?  The answer is: the pigeonhole principle.  Thare are only 82 possible remainders: $0,1,2,3,\ldots, 81$.  If you ever get $0$, the decimal expansion terminates.  But if you never get $0$, you can't go more than $81$ steps without getting a remainder that you've seen before, since after $81$ steps you've used them all up.

ContentLicense: CC BY-SA 3.0

Title: Are all integer fractions rational?


问:
proving limits exist
Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.

it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is  $\frac{n^3}2≥6$.

答:
For $n \neq 0$,$$ \frac{ 4n^3+3n}{n^3-6}=  \frac{ 4+\frac{3}{n^2}}{1-\frac{6}{n^3}}.$$

You should be using the properties of the limits of sequences, like here, or Rudin's PMA Theorem $3.3$. You should prove the properties first, then use them. This then reduces your epsilon-delta task significantly for questions like this one.

Title: proving limits exist


问:
Let $G$ be a group with $|a_i|=n_i$ for all $1\le i \le m$. Show that $|x|=n_1n_2\ldots n_m$ where $x$ defined as follows
Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that
$|x| = n_1n_2\ldots n_m$.

Does it could be proved using induction on $m$? If yes, what about the base case? If I let $P(m)$ be the statement, then the base case is: $P(1)$ true
since $|x| = |a_1| = n_1$, isn't it?

答:
Here I got so far. Please give me a correction if I make some mistakes.

Let $a_i \in G, |a_i| = n_i$ where $1\le i \le m$. Given that $\gcd(n_i,n_j) = 1$ and
$a_ia_j = a_ja_i$ for all $i$ and $j$.
Let $e_G$ be the identity element in $G$.

Let $P(m)$ be the statement:

The order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$.

Base Case:
Consider $P(2)$. That is, the order of
$a_1a_2$ is $n_1n_2$. We'll show it now.
Note that $|a_1| = n_1$ and $|a_2|=n_2$.
Let $|a_1a_2| = s$.

To show that $s=n_1n_2$:

$s \mid n_1n_2$:
Notice that
\begin{align*}
(a_1a_2)^{n_1n_2} &= \overset{n_1n_2}{\overbrace{abab \ldots ab}} \\
&= \overset{n_1n_2}{\overbrace{a_1a_1\ldots a_1}}
\overset{n_1n_2}{\overbrace{a_2a_2\ldots a_2}} \\
&= a_1^{n_1n_2}a_2^{n_1n_2} \\
&= (a_1^{n_1})^{n_2}((a_2)^{n_2})^{n_1} \\
&= (e_G)(e_G) \\
&= e_G
\end{align*}
Hence, $s \mid n_1n_2$, as desired.

$n_1n_2 \mid s$:
Consider that
\begin{align*}
e_G = (a_1a_2)^s &= (a_1a_2)^{sn_1} \\
&= a_1^{sn_1}a_2^{sn_1} \\
&= (a_1^{n_1})^sa_2^{sn_1} \\
&= (e_G)a_2^{sn_1} \\
&= a_2^{sn_1}.
\end{align*}
Hence, $n_2 \mid sn_1$. Since
$\gcd(n_1,n_2)=1$, then we must have
$n_2 \mid s$. Similarly, consider
\begin{align*}
e_G = (a_1a_2)^s &= (a_1a_2)^{sn_2} \\
&= a_1^{sn_2}a_2^{sn_2} \\
&= a_1^{sn_2}(a_2^{n_2})^s \\
&= a_1^{sn_2}(e_G) \\
&= a_1^{sn_2}.
\end{align*}
Hence, $n_1 \mid sn_2$. Since
$\gcd(n_1,n_2) = 1$, then we must have
$n_1 \mid s$.
Thus, we have $n_1n_2 \mid s$, as desired.

Hence, $s = n_1n_2. \Box$.

Thus, the order of $a_1a_2$ is $n_1n_2$
i.e. $P(2)$ is true.

Inductive Step:
Let $x=a_1a_2\ldots a_k$ and $n=n_1n_2\cdots n_k$.
Assume that $P(k)$ is true. That is, the order of $x$ is $n$.

To show $P(k+1)$ is also true (that is, the order of $xa_{k+1}$ is $nn_{k+1}$.):
Notice that
\begin{align*}
a_1a_2\ldots a_{k+1} &= a_1a_2\ldots a_ka_{k+1} \\
&= (a_1a_2\ldots a_k)a_{k+1} \\
&= xa_{k+1}.
\end{align*}
Now, we apply the same method as in the base case above. Here we go:

Let $|xa_{k+1}| = t$. To show that
$t = nn_{k+1}$:

$t \mid nn_{k+1}$:
Consider that
\begin{align*}
(xa_{k+1})^{nn_{k+1}} &=
\overset{nn_{k+1}}{\overbrace{xa_{k+1}xa_{k+1}\ldots xa_{k+1}}} \\
&= \overset{nn_{k+1}}{\overbrace{xx\ldots x}}\overset{nn_{k+1}}{\overbrace{a_{k+1}a_{k+1}\ldots a_{k+1}}} \\
&= (x^n)^{n_{k+1}}(a_{k+1}^{n_{k+1}})^n
\\
&= (e_G)(e_G) \\
&= e_G.
\end{align*}
Hence, $t \mid nn_{k+1}$, as desired.

$nn_{k+1} \mid t$:
Consider that
\begin{align*}
e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn} \\
&= x^{tn}a_{k+1}^{tn} \\
&= (x^n)^ta_{k+1}^{tn} \\
&= (e_G)a_{k+1}^{tn} \\
&= a_{k+1}^{tn}.
\end{align*}
Hence, $n_{k+1} \mid tn$.
But, since $\gcd(n,n_{k+1}) = 1$, then we must have $n_{k+1} \mid t$.
Similarly, consider that
\begin{align*}
e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn_{k+1}} \\
&= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\
&= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\
&= x^{tn_{k+1}}(a_{k+1}^{n_{k+1}})^t \\
&= x^{tn_{k+1}}(e_G) \\
&= x^{tn_{k+1}}.
\end{align*}
Hence, $n \mid tn_{k+1}$. Since
$\gcd(n,n_{k+1}) = 1$, then we must have
$n \mid t$.
Thus, we have $nn_{k+1} \mid t$, as desired.

Hence, $nn_{k+1} = t. \Box$.

Thus, the order of $xa_{k+1}$ is
$nn_{k+1}$ i.e. $P(k+1)$ holds.

Therefore, the order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$ and hence proved. $\Box$


问:
Proving that $\gcd\left(\frac a {\gcd(a,b)},\frac b {\gcd(a,b)}\right) =1$
I need to show that the $$\gcd\left(\frac a {\gcd(a,b)},\frac b {\gcd(a,b)}\right) =1$$

I'm not sure exactly how to approach this .. I tried using the fact that
$\gcd(a,b)=d$ so $d=ma+nb$ but I didn't get too far

Could anyone suggest where to start?

答:
There is a very simple proof, Let
$$\gcd(a,b) = d = am+bn$$
Now let’s multiply both side with ${1}/{d}$
$$\frac{1}{d}\cdot d = \frac{1}{d} (am+bn)$$
$$1= \frac{a}{d}\cdot m+\frac{b}{d}\cdot n$$
And this is just the definition of the linear combination, So
$$\gcd\left(\frac{a}{b},\frac{b}{d}\right)=1$$


问:
"NONCOMMUTATIVE Algebra with a view towards Algebraic Geometry"?
Is there a noncommutative algebra book that is similar to Eisenbud's "Commutative Algebra with a view towards Algebraic Geometry" in the sense that fundamental and geometrically motivated notions are covered in a systematic manner ? I was trying to look one online but found myself not knowing where to even start.

Thanks!

答:
I am not sure you will find a single textbook with all informations. As suggested in comment, one answer could be "look up any book on $D$-modules" (for example these lectures by Bernstein are very good).

Here are some other pointers that could be useful :

These lectures by Ginzburg on non-commutative geometry, that contains a lot of important constructions. I really learned a lot just by reading the first chapters.

These lectures by Gaitsgory on geometric representation theory. It includes the study of category $\mathcal O$ for complex semisimple Lie algebras $\mathfrak g$, and the study of $\mathscr D$-modules, an important example of non-commutative algebras appearing in algebraic geometry.

I also think that books on Hochschild cohomology are relevant : these lectures by Pieter Belmans are geometry-oriented, and this book by Sarah Witherspoon is more algebraic but still could be a good reference.


答:
You need to integrate over the square region as you have sketched. Equation of the lines of the square are $x+y = -1, y-x = 1, x-y=1, x+y=1$.

So,

$-1-x \leq y \leq 1+x, -1 \leq x \leq 0$

$x-1 \leq y \leq 1-x, 0 \leq x \leq 1$

Here is one way to set it up -

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

But please also note that $x^2y$ is an odd function wrt. $y$ and so due to symmetry above and below $x-$axis, the integral will cancel out and be zero. So you can just choose to integrate $x^2 \ dy \ dx$ which comes to $\frac{1}{3}$.

Title: Calculate $\int_D x^2(y+1)\,\mathrm{d}x\,\mathrm{d}y$ where $D=\left\{(x,y)\in \mathbb R^2 : |x|+|y|\le 1\right\}$

Body:
I have to calculate
$$\int_D x^2(y+1)\,\mathrm{d}x\,\mathrm{d}y\,,$$

where $$D=\left\{(x,y)\in \mathbb R^2 : |x|+|y|\le 1\right\}$$

Now I went on to try to write this $D$ as a set of the form $$ E = \left\{(x,y) \in \mathbb{R^2}: a < x < b \,,\, \alpha(x)<y<\beta(x)\right\} \,,$$ since in these cases we know how to calculate the integral. But there are many possible solutions to the inequation in D, namely:

and I wouldn't know which to take as borders of the integral because I don't think they all give the same result.

Moreover I tried to use symmetry. The set in my case looks like this

so maybe we could assume it is symmetric and the Integral is equal to zero. But I am very unsure. This topic was explained very briefly in the lecture therefore I am having a lot of problems.

Thank you in advance


问:
General question about proving that a function is differentiable
This is a general question about a concept so let's assume $f(x)$ is some function.

Let $f(x)=\left\{
  \begin{array}{l l}
    some-function & \quad \text{if $x \neq 0$}\\
    some-other-function & \quad \text{if $x=0$}
  \end{array} \right.$

I know that if I prove that $f'(x)$ exists on all of the domain of the function, then it's differentiable.
But I also know that if the function is not continuous, it can't be differentiable.

What I don't understand is, how the fact that I show that the derivative exists for every $x\in domain$, makes sure that the function is continuous?

For a more concrete example,

Let $g(x)=\left\{
  \begin{array}{l l}
    \frac{sinx}{x} & \quad \text{if $x \neq 0$}\\
    1 & \quad \text{if $x=0$}
  \end{array} \right.$

It's obvious that the function is differentiable at $\frac{sinx}{x}$ and at $1$, but why does it imply that the function is necessarily continuous?
Is it enough to say that since it's differentiable when $x\neq 0$ and when $x=0$ then it's differentiable?

答:
The fact that a function is defined “by cases” has no relevance, except perhaps on how you decide whether it is continuous or differentiable.

The theorem “a function that's differentiable at a point $c$ is continuous at $c$” holds whatever is the definition of the function.

In your case, the derivative at $0$ is, if it exists,
$$
\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}\frac{f(x)-1}{x}\mathrel{\color{red}{=}}\lim_{x\to0}\frac{\dfrac{\sin x}{x}-1}{x}=\lim_{x\to0}\frac{\sin x-x}{x^2}=0
$$
where the “how you decide” mentioned above comes into place: since we're computing the limit for $x\to0$, the value of $f(x)$ when we perform the step denoted with the red equal sign can be taken as $f(x)=\frac{\sin x}{x}$.

Thus the function $f$ is differentiable at $0$ and, by general theorem, continuous at $0$.

Of course, the continuity at zero can be checked separately, but it's not necessary as we proved differentiability directly.


问:
Gradient of $∇x: A^2x$
Let $x ∈ \mathbb R^n$ and $A ∈ \mathbb R^{n×n}$

What is the gradient $∇x:A^2x$?

I see a rules that often used when differentiating multivariate functions:

For all  $A∈\mathbb R^{m×n}$ ,  $∇x:Ax=A^⊤$,

So I process the $x: A^2x$ to $∇x: A(Ax)$ and use previous rules

It becomes  $A  A^⊤$

Is it correct?
If not, what should I do with it?

Title: Gradient of $∇x: A^2x$

Body: Since $\nabla_xAx=A^T$, $\nabla_xA^2x=(A^2)^T=(A^T)^2$.


问:
Why does this function series converges uniformly?
$f_n (x):= \dfrac{\sin (2n\pi x)}{2n\pi}.$

I am working with a function series: $\sum_{n=1}^{\infty} f_n(x).$

My mathematics book says that this function series converges uniformly in every closed interval that contains no integer.

But I cannot understand why:

$$\bigg| \sum_{n=1}^{\infty} f_n(x) \bigg| \leqq \sum_{n=1}^{\infty} \dfrac{1}{2n\pi}$$ but the right side of  this inequality doesn't converge.

Furthermore, I don't understand why my book says
"in every closed interval that contains no integer."

Even if $x$ is integer, or $x\in \mathbb{Z} $,
$\quad \sum_{n=1}^{\infty} f_n(x)=0, \quad$ so I think this series converges uniformly.

答: Accorging to Theorem 7.2.2, part (1),  page 112 of Fourier Series by Edwards this statement is false.</p>


问:
The PDF of $X$ is $\frac{1+\theta x}{2}$ for $x\in [-1,1], \theta \in[-1,1]$. Is the Method of Moments estimator of $\theta$ consistent?
We know that the method of moments estimator, $\hat{\theta}_n(\textbf{X})$, is consistent if

$$\lim_{n\to \infty}\mathbb{P}(|\hat{\theta}_n(\textbf{X})-\theta|>\epsilon)=0$$.

We know that $\mathbb{E}(X;\theta)=\frac{1}{2}\int_{-1}^1(x+\theta x^2) dx=\frac{\theta}{3}$, so $\hat{\theta}_n(\textbf{X})=3\bar{X}$.

Therefore $\lim_{n\to \infty}\mathbb{P}(|\hat{\theta}_n(\textbf{X})-\theta|>\epsilon)= \lim _{n\to \infty}\mathbb{P}(|3\bar{X}-\theta|>\epsilon) $.

At this point I'm not sure how to proceed? I've tried using Chebyshev's inequality but this doesn't lead anywhere. I know that if a sequence in random variables converges in distribution to a constant then that is equivalent to  a sequence in random variables converging in probability to that constant. Now $\lim_{n\to \infty}\mathbb{P}(\hat{\theta}_n(\textbf{X})\leq x)= \lim_{n\to \infty}\mathbb{P}(3\bar{X}\leq x)$ and I don't think this equals $\mathbb{P}(\theta\leq x)$? So would I be right in saying the method of moments estimator is not consistent?

答:
You can easy use a necessary and sufficient condition for convergence in $L^2$ that is sufficient condition for consistency:

$$\mathbb{E}[\hat{\theta}]=\mathbb{E}[3\overline{X}_n]=3\cdot \frac{\theta}{3}=\theta$$

$$\mathbb{V}[\hat{\theta}]=\mathbb{V}[3\overline{X}_n]=\frac{9\sigma^2}{n}=\frac{3-\theta^2}{n}$$

Now being

$$\mathbb{E}[\hat{\theta}]=\theta$$

and

$$\lim\limits_{n \to \infty}\mathbb{V}[\hat{\theta}]=0$$

the estimator is consistent

Of course applying Chebishev's inequality the result is the same

$$\lim\limits_{n\to \infty}\mathbb{P}\Big\{\Big|3\overline{X}_n-\theta\Big|<\epsilon\Big\}\geq 1-\frac{3-\theta^2}{n \epsilon^2}=1$$


答:
I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.

For $n > 9$, $0 < (n^3 - 64n),$ which is a lower bound for the denominator.

For $n > 9$, $(n-8) > 1 \implies$ that the numerator is

less than $3(n+8)(n-8) = 3(n^2 - 64).$

Therefore, the fraction is less than

$$\frac{3(n^2 - 64)}{n^3 - 64n} = \frac{3}{n}.$$

Therefore, choose $n$ such that $n > 9$ and

$\frac{3}{n} < \epsilon \implies \frac{3}{\epsilon} < n.$

Title: proving limits exist

Body:
Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.

it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is  $\frac{n^3}2≥6$.


问:
Continuous, Non-analytic function with maximum on the interior of a compact set?
Is there an example of a complex function on a compact set K that is continuous, non-analytic, and achieves its maximum on int(K)?

Since the max mod principle says that if the function is analytic, then the maximum is on the boundary. I'm looking for a non-analytic function where the maximum is not on the boundary, but the function is continuous.

答:
For any compact set $K \subset \Bbb C$ with nonempty interior and $a \in \operatorname{int}(K)$, the function $f(z) = -|z-a|$ is such an example.

There is nothing special about the complex numbers here, the same example works on any metric space $(X, d)$ with $f(x) = -d(x, a)$ for some $a \in \operatorname{int}(K)$.


问:
Wave equation with piecewise (discontinuous) initial condition
I have the equation $u_{tt}-u_{xx}=0$, with initial conditions $u(0,x)=f(x)=0$ and $u_t(0,x)=g(x)=\begin{cases} 1, & \mbox{if } x>0 \\ 0, & \mbox{if } x \leq 0 \end{cases}$. I have to solve for $x,t\in \mathbb R$.

First, I tried to find the solution for $t>0$. However, when $x+t>0$ and $x-t<0$, $g$ isn't continuous, and d'Alembert's formula yields the (incorrect) solution $u(t,x)=\frac{1}{2}(x+t)$. I've also tried solving for $t>0$ and $x>0$, but I don't think I can find any initial condition $u(t,0)=h(t)$. Am I missing something or should I consider a different approach? Thanks in advance.

答:
$f$ is continuous, and $g$ has discontinuity of the first kind in zero, so we can use d'Alembert's formula: $u(t,x) = \frac{1}{2} \int_{x-t}^{x+t} g(z) dz = \frac{1}{2} (\int_{x-t}^{0} g(z)dz+\int_{0}^{x+t} g(z)dz)$. In order to calculate this integral we consider three sets:
$$K = \{(t,x)|t>0,x-t<0,x+t<0\}\\
M = \{(t,x)|t>0,x-t<0,x+t>0\}\\
P = \{(t,x)|t>0,x-t>0,x+t>0\}$$

And we obtain
$u(t,x)=\left\{\begin{matrix}
0, & (t,x) \in K, \\
\frac{x+t}{2}, & (t,x) \in M,  \\
t, & (t,x) \in N,
\end{matrix}\right.$

It is easy to prove that $u$ belongs to the class $C([0,\infty)\times\mathbb{R})$. But this functions is not continuously differentiable.

Also, we can prove that this problem has unique solution from class $C([0,\infty)\times\mathbb{R})$.


问:
If $\psi=x\sin(\frac{1}{x})$, and $f$ is integrable, is then $f\circ \psi$ also integrable?
Let
$$
\psi(x)=
\left\{
\begin{array}{cll}
x \sin\Big(\dfrac{1}{x}\Big) & \text{if} & x\in (0,1],\, \\
0 & \text{if} & x=0,
\end{array}
\right.
$$
and let $f:[-1,1]\rightarrow \mathbb{R}$ be Riemann integrable.

How can I show that $f\circ \psi$ is Riemann integrable?

I have several theorems in my book that could work if $\psi$ were a $C^1$ diffeomorphism or a homeomorphism with inverse satisfying Lipschitz condition. But $\psi$ is clearly neither of those. Any help with a zero set argument or reverting to definitions?

答:
Note. If $f$ is Riemann integrable and $\varphi$ continuous, then their composition $f\circ\varphi$ is NOT necessarily Riemann integrable. (See here.)

Sketch of proof.

Claim 1. If $f$ is Riemann integrable and $\varphi$ is differentiable and $\varphi'(x)\ge \eta>0$, then $f\circ\varphi$ is Riemann integrable.

Proof. Let $P=\{a=t_0<\cdots<t_n=b\}$ be a partition of $[a,b]$, then
$$
U(f\circ\varphi,P)-L(f\circ\varphi,P)=
\sum_{i=1}^n\big(M_i(f\circ\varphi)-m_i(f\circ\varphi)\big)(t_i-t_{i-1})
$$
where
$$
m_i(f\circ\varphi)=\inf_{t\in [t_{i-1},t_i]}(f\circ\varphi)(t)=
\inf_{x\in [\varphi(t_{i-1}),\varphi(t_i)]}f(t) \quad\text{and}\quad
M_i(f\circ\varphi)=\sup_{t\in [t_{i-1},t_i]}(f\circ\varphi)(t)=
\sup_{x\in [\varphi(t_{i-1}),\varphi(t_i)]}f(t)
$$
since $\varphi$ is increasing and hence
$$
U(f\circ\varphi,P)-L(f\circ\varphi,P)=
\sum_{i=1}^n\big(M_i(f\circ\varphi)-m_i(f\circ\varphi)\big)(t_i-t_{i-1})
\\=
\sum_{i=1}^n\big(M_i(f\circ\varphi)-m_i(f\circ\varphi)\big)
\big(\varphi(t_{i})-\varphi(t_{i-1})\big)
\frac{t_i-t_{i-1}}{\varphi(t_{i})-\varphi(t_{i-1})}
\\=
\sum_{i=1}^n\big(M_i(f\circ\varphi)-m_i(f\circ\varphi)\big)
\big(\varphi(t_{i})-\varphi(t_{i-1})\big)
\frac{1}{\varphi'(\tau_{i})}\le \frac{1}{\eta}\sum_{i=1}^n\big(M_i(f\circ\varphi)-m_i(f\circ\varphi)\big)=\frac{1}{\eta}\big(U(f,\tilde P)-L(f,\tilde P)\big)
$$
where $\tau_i\in(t_{i-1},t_i)$ and $\,\tilde P=\{\varphi(a)=\tilde t_0<\cdots<\tilde t_n=\varphi(b)\}$.

Clearly, for suitable choice of $\tilde P$, and subsequently of $P$, the $\frac{1}{\eta}\big(U(f,\tilde P)-L(f,\tilde P)\big)$ can become smaller than any given $\varepsilon$, and hence $f\circ\varphi$ is Riemann integrable.

Claim 2. If $g:[a,b]\to\mathbb R$ is bounded and for every $\delta_1,\delta_2>0$, the function $g$ is Riemann integrable on $[a+\delta_1,b-\delta_2]$. Then $g$ is Riemann integrable on $[a,b]$.

Proof. Simple application of the integrability criterion, which was used above.

Therefore, due to Claim 1, if $\varphi(x)=x\sin(1/x)$, then $\varphi'(x)\ge \eta>0$ or
$\varphi'(x)\le -\eta<0$, for some $\eta>0$ is any interval of the form
$$
\Big[\frac{1}{\frac{\pi}{2}+k\pi}+\delta,\frac{1}{-\frac{\pi}{2}+k\pi}-\delta\Big], \quad \delta>0,\,\,k\in\mathbb Z^+
$$
and hence, $f\circ \varphi$ is integrable in $\big[\frac{1}{\frac{\pi}{2}+k\pi}+\delta,\frac{1}{-\frac{\pi}{2}+k\pi}-\delta\big]$. Due to Claim 2,
$f\circ \varphi$ is integrable in $J_k=\big[\frac{1}{\frac{\pi}{2}+k\pi},\frac{1}{-\frac{\pi}{2}+k\pi}\big]$, and consequently in $I_n=\big[\frac{1}{\frac{\pi}{2}+k\pi},1\big]$, for all $n\in\mathbb N$, which is a finite union of $J_k$'s.
Applying Claim 2 once more we obtain that $f\circ \varphi$ is integrable in $[0,1]$.


答:
A function $f$ is continuous at $x_0$ if and only if
$$\lim_{x\to x_0}f(x)=f(x_0).$$
Which is really the same as
$$\lim_{x\to x_0}(f(x)-f(x_0))=0.$$
Now you probably recognize the expression on the left side. It appears in the difference quotient:
$$\frac{f(x)-f(x_0)}{x-x_0}.$$
A function is differentiable at $x_0$ if the limit of this quotient as $x\to x_0$ exists. The denominator is guaranteed to converge to $0$. But for the limit of the quotient to exist, the numerator then also has to go to $0$, otherwise the whole thing diverges. And we just noticed before that the numerator going to $0$ is exactly what's needed for continuity.

Note that none of this has anything to do with how the function has been written down, wether it's piecewise or not, doesn't matter ($\sqrt{x^2}$ and $\vert x\vert$ have the exact same properties, since they are equal, but $\vert x\vert$ is defined piecewise, and $\sqrt{x^2}$ isn't). How we represent a function can at most influence how convenient our calculations end up being. But the fact that differentiability implies continuity isn't based on such a representation.

Title: General question about proving that a function is differentiable

Body:
This is a general question about a concept so let's assume $f(x)$ is some function.

Let $f(x)=\left\{
  \begin{array}{l l}
    some-function & \quad \text{if $x \neq 0$}\\
    some-other-function & \quad \text{if $x=0$}
  \end{array} \right.$

I know that if I prove that $f'(x)$ exists on all of the domain of the function, then it's differentiable.
But I also know that if the function is not continuous, it can't be differentiable.

What I don't understand is, how the fact that I show that the derivative exists for every $x\in domain$, makes sure that the function is continuous?

For a more concrete example,

Let $g(x)=\left\{
  \begin{array}{l l}
    \frac{sinx}{x} & \quad \text{if $x \neq 0$}\\
    1 & \quad \text{if $x=0$}
  \end{array} \right.$

It's obvious that the function is differentiable at $\frac{sinx}{x}$ and at $1$, but why does it imply that the function is necessarily continuous?
Is it enough to say that since it's differentiable when $x\neq 0$ and when $x=0$ then it's differentiable?


问:
Sufficiency and MLE of $\theta$ where $f(x;\theta)=\dfrac{1}{2}e^{|x-\theta|},-\infty<\theta<\infty$

Let $X_1,X_2,...,X_n$ be a random sample from the density.
$f(x;\theta)=\dfrac{1}{2}e^{|x-\theta|},-\infty<\theta<\infty$.

Discuss the sufficiency and MlE of $\theta$ for this density

I can write this density function as

$ f(x;\theta) = \begin{cases} \dfrac{1}{2}e^{-(x-\theta)} ,  \theta < x < \infty \\
      \dfrac{1}{2}e^{(x-\theta)}  ,  -\infty < x < \theta \end{cases}$

I am skipping the whole proof as I am preparing for time based exam. I just want to know
for the first part,we have $\theta<x_{(1)}$ and for second part  $\theta>x_{(n)}$

So $x_{(1)}$ and $x_{(n)}$ are jointly sufficient. Am I right ?

and for MLE   $\theta$ satisfies $ x_{(n)}<\theta<x_{(1)}$ which seems absurd. Mle doesnt exist in this case?

答:
To find MLE for $\theta$ an easy approach is the usual one

$$L(\theta)\propto e^{-\Sigma_i |X_i-\theta|}$$

$$l(\theta)=-\Sigma_i |X_i-\theta|$$

To be maximized...

Maximizing $-\Sigma_i |X_i-\theta|$ is the same as minimizing

$$ \bbox[5px,border:2px solid black]
{
\Sigma_i |X_i-\theta|
\qquad (1)
}
$$

It is well known that $\hat{\theta}$ which minimize (1) is the sample median

(the proof is easy and I can show it to you if necessary)


问:
Independent Spaces
What does it mean for spaces to be independent?

答:
Given a vector space $V$, we say that a collection of subspaces $\{W_1, \dots, W_k\}$ is independent if for every $1 \le i \le k$,

$$W_i \cap \sum_{j \ne i} W_j = \{0\}$$

In other words, no nonzero vector in $W_i$ can be a linear combination of vectors from the other subspaces.

Title: Independent Spaces


问:
Generalisation of an expression.
I have been solving matrices questions lately and this pattern keeps showing up:

Given matrix $ P = \begin{bmatrix}
1 & 0 & 0\\
4 & 1 & 0\\
16 & 4 & 1\\
\end{bmatrix}$ and $I$ be identity matrix of order 3. If $[Q] = [q_{ij}]$ is a matrix such that $P^{50}-Q= I$ then $\frac{q_{31}+q_{32}}{q_{21}} = ?$

While solving I found this pattern for a general matrix of the type:
$$L = \begin{bmatrix}
1 & 0 & 0\\
a & 1 & 0\\
b & a & 1\\
\end{bmatrix}$$
As

$$L^n = \begin{bmatrix}
1 & 0 & 0\\
a×n & 1 & 0\\
K & a×n & 1\\
\end{bmatrix}$$

Where $K = (L_{31})_{{n-1}}+a^2×(n-1) + b$

If I have to figure out $L_{50}$ using this algorithm, I need to find $L_{49}, L_{48}$ and so on which basically defeats the purpose. How do you reiterate the value of $L_{n-1}$ so that I can use directly the value of $b$ to calculate $L_{31}$ of any power of matrix?

Title: Generalisation of an expression.

Body:
Note that $L=I+N$ with $N^3=0$. Therefore, $L^n = I+n N + \frac{n(n-1)}{2}N^2$. Since
$$
N^2 = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
a^2 & 0 & 0\\
\end{bmatrix}
$$
we get
$$
L^n = \begin{bmatrix}
1 & 0 & 0\\
n a & 1 & 0\\
nb +\frac{n(n-1)}{2}a^2& n a & 1\\
\end{bmatrix}
$$


问:
Integral trick explanation or link
We used the following integral trick in our lecture, I really can't understand why it works. I would appreciate an explanation or even just a name so i can search for it :)

So apparently:

$$\int_{-\infty}^\infty\boldsymbol{x}\cdot f(x)dx=\int_{-\infty}^\infty\boldsymbol{(x-x_{a})}f(x)dx+\boldsymbol{x}_{\boldsymbol{a}}\cdot\int_{-\infty}^\infty f(x)dx$$

Where $f(x)$ is some function of $x$ and $x_a$ is .. not sure about that either, probably any constant..

答:
This is just because
$$xf(x)=(x-x_a)f(x)+x_af(x)$$
$x_a$ is a constant, so you can pull it outside of the integral.

Title: Integral trick explanation or link


问:
Deciding if these integrals Converge or Diverge (comparison test).
I have $4$ improper integrals to decide if they converge or diverge:

1)$\int_1^\infty\frac {|\sin x|}{x^2}dx$.

2)$\int_{-\infty}^\infty \frac {1} {\sqrt{1+4x^2}}dx$

3)$\int_1^\infty \frac {1}{x\sqrt {x^2-1}}dx$

4)$\int_1^{10}\frac {1}{\sqrt{x^3-1}}dx$

What I did and my answers:

For (1), I said that ($\sin x\approx x$) at ($x\to \infty$), and since its in positive I can look at it like $\frac {1}{x}$, and that diverges.

For (2), I said that $\left(\frac {1}{ \sqrt {1+4x^2}}\approx \frac {1}{\sqrt{4x^2}}\approx \frac {1}{x}\right),   $  (at infinity) and that also diverges, so I don't need to check the other part.
For (3), $\left(\approx \frac {1}{x^2}\right)$ and that converges at infinity so my integral converges.

For (4), $\left(\frac {1}{\sqrt{x^3-1}}\approx\frac {1}{x^{1.5}}\right)$, and that converges at infinity.

 I feel like I must have made some mistakes, I would love it if someone can approve/correct my work, appreciate all the help.

答:
Converges because $\frac{\left| \sin x\right| }{x^2}\le \frac{1}{x^2}$. It's about $0.775$

Correct

$x=1$ is a problem, too.  $\int_1^{\infty } \frac{1}{x \sqrt{x^2-1}} \, dx=\left[\arctan\left(\sqrt{x^2-1}\right)\right]_1^{\infty}=\frac{\pi}{2}$.  Converges for direct calculation

$\int_1^{10} \frac{1}{\sqrt{x^3-1}} \, dx$ converges, but not for the reason you wrote, because  $\infty$ is not involved.

The integral is improper at $x=1$

Set $$\frac{1}{\sqrt{x^3-1}}=u\to x=\sqrt[3]{\frac{1}{u^2}+1}\to dx=-\frac{2du}{3 \left(\frac{1}{u^2}+1\right)^{2/3} u^3}$$
The integral becomes
$$\int_{\frac{1}{3 \sqrt{111}}}^{\infty } \frac{2}{3 \left(\frac{1}{u^2}+1\right)^{2/3} u^2} \, du$$
which converges because $\frac{1}{\left(\frac{1}{u^2}+1\right)^{2/3} u^2}\sim \frac{1}{u^2}$ as $u\to \infty$


问:
Proving an equality involving a product $\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$
How can I prove that:

$$\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$$

When $n\in\mathbb{N}$ such that $n\ge2$.

I think that we need to prove that all terms are monotonically increasing functions. But how do I prove that?

答: Well, plug $n=2$ in and see that the inequality holds. Then observe that every factor is monotonically increasing for $n \ge 2$, note that $3n^3-3n=3n(n^2-1)$.


答:
Each factor is ultimately increasing. Find their minima:

For $3 n^2 - 2$ it's minimum is at $n = 0$, it is positive for $n > \sqrt{2} / \sqrt{3}$, for $n \ge 2$ it is at least $10$.

For $3 n^2 - 1$ the the minimum is also at $n = 0$, for $n \ge 2$ it is at least $11$.

For $3 n^2 - 3 n + 1$ the minimum is at $n = 1/2$, for $n \ge 2$ it is at least $7$.

Pulling all together, for $n \ge 2$ your expression is at least:
$\begin{align*}
  \frac{10 \cdot 11 \cdot 7}{2}
    &= 385
\end{align*}$

Title: Proving an equality involving a product $\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$

Body:
How can I prove that:

$$\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$$

When $n\in\mathbb{N}$ such that $n\ge2$.

I think that we need to prove that all terms are monotonically increasing functions. But how do I prove that?


问:
The unfair subway
This is problem 24, "The Unfair Subway", in Mosteller's Fifty Challenging Problems in Probability with Solutions

Marvin gets off work at random times between 3 and 5 P.M. His mother lives uptown, his girl friend downtown. He takes the first subway that comes in either direction and eats dinner with the one he is first delivered to. His mother complains that he never comes to see her, but he says she has a 50-50 chance. He has had dinner with her twice in the last 20 working days. Explain.

The accompanying solution says that it's because the uptown train always arrives one minute after the downtown train, which in turn arrives nine minutes after the uptown train, in this time span. So there's a nine-to-one chance that Marvin will get on the downtown train and not the uptown one.

Huh? Then what happened to the "50-50 chance" part of the problem?

The problem seemed to be posed as a probabilistic inference problem, i.e. one where the goal is to calculate:
$$\binom{20}{2} (0.5)^2 (1-0.5)^{18} \approx 0.00018$$
but it turns out it was a statistical inference problem (one based on maximum likelihood estimates at that) that contradicts information in the problem itself.

So my question is: is this a valid problem in probability? Am I missing something that would make this a valid problem?

答:
There's a better solution to this. Suppose there are ten equidistant stops, and one train that shuttles back and forth. Then if Marvin's station is the second, there is a 90% chance that the train is uptown of him when he arrives, so he will go downtown.

In fact, this solution is so much better than Mosteller's (IMHO) that I'm tempted to think he heard it from a friend a long time ago and garbled it.

Title: The unfair subway


问:
Does $f, f' \in L^1([0, \infty))$ imply that $\lim_{x \to \infty} xf(x) = 0$?
Does $\int_0^\infty |f(x)| \, dx$ and $\int_0^\infty |f'(x)| \, dx$ being finite imply that $\lim_{x \to \infty} xf(x) = 0$? </p>

<p>(Context: I am working through an analytic number theory textbook. In a lemma, the book assumes the conditions on $f$ and $f'$. The term $xf(x)$ appears from an integration by parts, but is tossed out, which must mean that $\lim_{x\to\infty} xf(x) = 0$. However, I've tried to prove this and was not successful. I also cannot think of any counterexamples.)</p>

答:
Here's a construction of a counterexample. The idea is based on a simple construction for series that is given on p. 350 of the centenary edition of G. H. Hardy's A Course of Pure Mathematics. The idea in Hardy's example is to construct a sequence $\{a_n\}$ of positive numbers such that $\sum a_n < \infty$ but $na_n \not\to 0$ as $n\to\infty$; to do this he simply puts $a_n = 1/n^2$ unless $n$ is a perfect square, in which case he puts $a_n = 1/n$.

To adapt this idea, choose a smooth function $\rho$, with $0\leq \rho \leq 1$, that is supported in $(-2,2)$ and that satisfies $\rho(x)=1$ for all $x \in (-1,1)$. Let $M$ be a bound for $|\rho'|$. Put
$$
\begin{align*}
\eta(x) & = \sum_{n = 2}^\infty \rho(x - n^2).
\end{align*}
$$
Basically, $\eta$ is identically $1$ in a neighborhood of each squared integer, and vanishes at each $x$ for which $(x-2,x+2)$ contains no squared integer. In fact, since the supports of $\rho(x-n^2)$ are disjoint, we have $0\leq \eta\leq 1$ and $|\eta'|\leq M$.

Finally, define
$$
f(x) = {\eta(x)\over 1 + x}.
$$
for $x \geq 0$. Obviously $f(x)\geq 0$ for all $x$. Also, $f$ is integrable, since
$$
\begin{align}
\int_0^\infty f(x)\,dx &\leq \sum_{n = 2}^\infty \int_{n^2 - 2}^{n^2 + 2}{dx\over 1 + x}  \leq \sum_{n = 2}^\infty{4\over n^2 - 1}.
\end{align}
$$

Now
$$
\begin{align}
f'(x) & = {\eta'(x)\over 1+x} + O\left({1\over (1+x)^2}\right)
\end{align}
$$
as $x \to \infty$. Since $|\eta'|\leq M$, we get
$$
\begin{align*}
\int_0^\infty{|\eta'(x)|\over 1+x}\,dx & = \sum_{n = 2}^\infty \int_{n^2-2}^{n^2+2}{|\eta'(x)|\over 1+x}\,dx  \leq M\sum_{n = 2}^\infty {4\over n^2-1},
\end{align*}
$$
and it follows that $f'(x)$ is integrable over $(0,\infty)$ as well.

On the other hand, $xf(x) \not\to0$ as $x\to\infty$, since
$$
n^2f(n^2) = {n^2\over 1 + n^2} \to 1
$$
as $n\to\infty$. Thus $f$ provides a counterexample.


问:
Calculating the residues of $f(z)=\frac{ze^{iaz}}{z^4+4}$
I have $f(z)=\frac{ze^{iaz}}{z^4+4}$. The poles of $f$ are {$\pm1\pm i$}.

How do I calculate $2\pi i \sum \operatorname{Res}\limits_{z = \pm 1 + i} f(z)$ ? (I'm only looking at positive $i$'s.)

I've tried using the property $\operatorname{Res}f(z) = \frac1{g'(p)}$ and that gives me $\frac1 4 e^{ia(\pm1+i)}$. But then I realised the property only works for one pole. How do I do this?

答: Consider, for instance, the residue at $1+i$. Since$$z^4+4=(z-1-i)(1-1+i)(z+1-i)(z+1+i),$$define$$g(z)=\frac{ze^{iaz}}{(z-1+i)(z+1-i)(z+1+i)},$$and you have $f(z)=\frac{g(z)}{z-1-i}$. Therefore,\begin{align}\operatorname{res}_{z=1+i}\bigl(f(z)\bigr)&=\operatorname{res}_{z=1+i}\left(\frac{g(z)}{z-1-i}\right)\\&=g(1+i)\\&=-\frac{ie^{(-1+i)a}}8.\end{align}


问:
Given a directional derivative to find a point
Let $ f(x,y,z)$ be differentiable, and assume that

$$ f(x, y, x^2 + y) = 3x - y $$

(for all values of $x,y$).

Also, given the direct-derivative of the point $A=(0, 12, 12)$ and the direction vector is $(1, 0, 1)$ is equal to $3$, I need to find the gradient of $f$ at $A$.

I have no idea how to even start, because $f$ is not given explicitly, only the derivative — but again, not explicitly for $f(x,y,z)$, but for this weird combination $f(x, y, x^2 + y)$.

What should I do?

答:
It is possible that the directional derivative is being treated as a linear operator (i.e. the Gateux derivative). I'm assuming this is true, so I won't normalise (not least because it's a lot simpler!). Whether you should or not depends on the definition given in your course.

Let $\nabla f: \mathbb R^3 → \mathbb R^3$ denote the gradient of $f$, and let

$$g(x,y) = (x,y,x^2+y).$$

In particular, we have the derivative (Jacobean?)

$$Dg(x,y) = \begin{pmatrix}1&0\\0&1\\2x&1\end{pmatrix},$$

for our point $A = g(0,12)$, we'll just need

$$ Dg(0,12) = \begin{pmatrix}1&0\\\ 0&1 \\\ 0&1 \end{pmatrix}.$$

Firstly, with my assumption, your directional derivative equation can be written as

$$ \nabla f(A) \cdot (1,0,1) = 3 $$

(this is one definition of directional derivative.)

As for the others, simply differentiate your first equation with respect to $x$ and $y$. The chain rule in matrix form says

$$
   \begin{align} (3,-1)
=
   \nabla_{(x,y)}
   \left( 3x - y \right)
&=
   \nabla \big( f(x,y,x^2+y)\big)
\\\\&=
   \nabla \big( f\circ g(x,y)\big)
\\\\&=
\nabla f\big(g(x,y)\big)Dg(x,y):
\end{align}$$

i.e., $\nabla f$ evaluated at $g(x,y)$, (a row vector) multiplied by the matrix $Dg(x,y)$ on the right.

To get our point $A$, simply fix $(x,y) = (0,12)$, so that the previous equation becomes

$$ (3,-1) = \nabla f(A)\, Dg(0,12) = \nabla f (A)\begin{pmatrix}1&0\\\ 0&1 \\\ 0&1 \end{pmatrix}.$$

Now since, from before,

$$ 3 = \nabla f(A) \cdot (1,0,1) = \nabla f(A) \begin{pmatrix}1\\0\\1\end{pmatrix}, $$

This gives us a matrix equation:

$$(3,-1,3) = \nabla f \begin{pmatrix}1&0&1\\\ 0&1&0 \\\ 0&1&1 \end{pmatrix}.$$

Multiplying on the right by the inverse of the matrix on the right hand side will finish it off. Since the matrix is close to the identity, you can use two steps of row reduction to solve it, i.e. writing the simultaneous equations in matrix form,

$$\left(\begin{array}{ccc|r}1&0&0&3\\\ 0&1&1&-1 \\\ 1&0&1&3 \end{array}\right) → \left(\begin{array}{ccc|r}1&0&0&3\\\ 0&1&1&-1 \\\ 0&0&1&0 \end{array}\right) → \left(\begin{array}{ccc|r}1&0&0&3\\\ 0&1&1&-1 \\\ 1&0&1&3 \end{array}\right) → \left(\begin{array}{ccc|r}1&0&0&3\\\ 0&1&0&-1 \\\ 0&0&1&0 \end{array}\right).$$

The last matrix reads $\nabla f (A) = (3,-1,0)$. The normalised version of this simply gives you a harder matrix to invert, which other answers have done.


问:
Inner product space-scalar multiplication
After reading i am still a bit confused:

(because it's about 2 different fields $\mathbb {C}$ and $\mathbb {R}$ in many explanations i have read is very confusing bcause of the different fields
so i wanted to ask just to be sure i understood)

If the inner product space is over $\mathbb {C}$
$$\langle a,λb\rangle=\overlineλ \langle a,b\rangle$$
and if it is over $\Bbb R$,
$$\langle a,λb\rangle=λ \langle a,b\rangle$$
But in both cases:
$$\langle λa,b\rangle=λ \langle a,b\rangle?$$

答: We say an inner product is linear in its leftmost argument and antilinear in its rightmost one. (In some contexts, such as physicists' use of the bra-ket notation for quantum mechanics, you'll see this convention reversed.) This one-of-each property is not ordinary bilinearity, which is linear in both arguments; instead, this axiom is called sesquilinearity. But for an inner product space over a self-conjugate scalar field such as $\Bbb R$, $\bar{\lambda}=\lambda$, so for brevity we drop the conjugate for the antilinear argument.


Title: Calculating the residues of $f(z)=\frac{ze^{iaz}}{z^4+4}$

Body:
I have $f(z)=\frac{ze^{iaz}}{z^4+4}$. The poles of $f$ are {$\pm1\pm i$}.

How do I calculate $2\pi i \sum \operatorname{Res}\limits_{z = \pm 1 + i} f(z)$ ? (I'm only looking at positive $i$'s.)

I've tried using the property $\operatorname{Res}f(z) = \frac1{g'(p)}$ and that gives me $\frac1 4 e^{ia(\pm1+i)}$. But then I realised the property only works for one pole. How do I do this?

Body: By l'Hôpital's rule, the residue is$$\begin{align}\lim_{z\to\pm1+i}\frac{z-(\pm1+i)}{(z^3+4z^{-1})e^{-iaz}}&=\lim_{z\to\pm1+i}\frac{1}{(-iaz^3-4iaz^{-1}+3z^2-4z^{-2})e^{-iaz}}\\&=\frac{e^{-a\pm ia}}{-ia(\pm1+i)^3-4ia(\pm1+i)^{-1}+3(\pm1+i)^2-4(\pm1+i)^{-2}},\end{align}$$which I leave you to simplify further using e.g. $(\pm1+i)^{-1}=(\pm1-i)/2$.


问:
Categorical description of matrix similarity
I am wondering if there is a nice categorical description of matrix similarity, i.e. the equivalence relation on matrices given by $A\sim B \iff A=QBQ^{-1}$, for some invertible $Q$.

In particular, I am considering the matrices as linear maps from a vector space into itself, which in turn forms a category $\mathrm{End}(V)$ with one object $V$ and arrows linear maps. We can then form the functor category $\mathrm{End}(V)^{\mathbb2},$ where $\mathbb2$ is the category with $2$ objects and $1$ arrow between them, i.e. this is the 2-category of linear maps and pairs of maps between them making the following commute:

\begin{array}{ccccccccc} V & \xrightarrow{A} & V \\
\downarrow &   &  \downarrow \\
V & \xrightarrow{B} & V \end{array}

So similarity between $A$ and $B$ is stronger than isomorphism in this category, as $(C,D):A\rightarrow B$ is an isomorphism if and only if $C$ and $D$ are linear isomorphisms, whereas similarity would require $C=D$. Do similar matrices in this category obey some characterising or otherwise interesting property, or is isomorphism the strongest we can get?

答:
$\DeclareMathOperator\rank{rank}$If you allow $C\neq D$, an isomorphism $A\cong B$ in the category you describe is equivalent to $\rank(A)=\rank(B)$. You can of course define a category where as morphisms you only allow commutative squares with $C=D$, then an isomorphism $A\cong B$ is indeed equivalent to $A$ and $B$ being similar.

Another way you might want to look at this is in the setting of $K[X]$-modules, where $K$ is the underlying field. A $K[X]$-module amounts to a $K$-vector space $V$ together with a $K$-linear map $f\colon V\to V$ that describes the action of $X$, that is $f(v) = X\cdot v$. In this setting, matrices $A$ and $B$ yield $K[X]$-modules and those are isomorphic if and only if $A$ and $B$ are similar.


问:
Does Monty Hall apply here?
Alice picks a card from a deck of cards. Bob guesses what her card is. Say Bob guesses the King of Hearts. Alice says her card is a heart, and gives Bob the option to keep his current guess or to change his guess. Is he more likely to guess correctly if he changes his guess?

I think this is Monty with 52 doors. When Alice reveals that her card is a heart, she is opening 39 doors (corresponding to the other 3 suits) to reveal goats. Bob is therefore more likely to win by changing his guess, since the probability he guessed correctly at the outset is $\dfrac{1}{52}$ so the probability one of the other 12 hearts is Alice's card is $\dfrac{51}{52}$. Each of these 12 cards is equally likely to be Alice's card, so the probability Bob wins if he switches his guess to another heart is $\dfrac{1}{12}\cdot\dfrac{51}{52}$.

Is this correct? Or does Monty Hall not apply here? If it does not apply, why does it not apply?

答:
I think Monty Hall dose not apply here, since there is a major difference between your question and Monty Hall: Monty Hall's choice will not release any information about whether the car is behind the door of your current choice (since he cannot open the door you have chosen), so when Monty Hall opens the door, it increases the probability of other doors that he do not open but remains the probability of your choice unchanged. However, in your question, once Alice say her card is a heart (she is opening 39 doors), the probability of any heart cards (including Bob's current choice) increase.

In a word, Monty Hall's choice dose not release information about your current choice, whlie Alice's words dose release information about whether Bob's current choice is right.

Title: Does Monty Hall apply here?


问:
why $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle ?$.
Prove that  every Hilbert space is reflexive.

I found the answer here

I have some doubt in  given below  answer marked in red box

My thinking : I think this  sentence ($f_{\varphi} \in \mathcal H^*$. By the RRT, there exists a unique $y_{f_{\varphi}}$ such that for all $f \in \mathcal H^*$, we have $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.)  is WRONG

why that given above sentence is wrong ?

Ans:  According to RRT  it should be like this $f_{\varphi}(y)=(y,y_{f_{\varphi}})$

My doubt :According to RRT :If $T$ is a bounded linear functional on a Hilbert space $H$ then
there exists some $g \in H$ such that for every $f \in H$ we have
$T(f) =\langle f, g \rangle $.

Here $\langle f, g \rangle $  contain $f$

If  $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.then where $y_{f_{\varphi}}$ is gone

and $\langle f, f_{\varphi}\rangle$   doesn't contain $y_{f_{\varphi}}$

Therefore  $f(y_{f_{\varphi}}) \neq \langle f, f_{\varphi}\rangle$

So  im not getting why  $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle ?$.

答:
Yes im right.
It should be  like this

Now, $f_{\varphi} \in \mathcal H^*$. By the RRT, there exists a unique $y_{f}$ such that for all $f \in \mathcal H^*$, we  have

$$\ f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle$$ for all $y_{f_\varphi} \in \mathcal H$

Using the RRT two times  we have  $||y_f||=||f_{\varphi}||$  and $||f||=||y_{f_\varphi}||$

Therefore $f(y_{f_{\varphi}}) = \langle f, f_{\varphi}\rangle$.


答:
If you just plug $n+2$ in the expression and simply develop each term you get:

$$f(n+2)=\frac 12(3n^2+12n+11)(3n^2+12n+10)(19+33n+18n^2+3n^3)$$

There are $+$ signs everywhere, so for $n\ge 0$ we have $f(n+2)>0$

Note: With the same method, you can have the same result for $f(n+1)$.


问:
Convergence of $\int_0^n f(x)\, dx - [\frac{1}{2} f(0) + f(1) +\dots + f(n-1) + \frac{1}{2} f(n)]$
If $f(x)$ is increasing, strictly concave, and twice continuously differentiable for $x > 0$, does the difference
$$
d_n = \int_0^n f(x)\, dx - [\tfrac{1}{2} f(0) + f(1) + f(2) +\dots + f(n-1) + \tfrac{1}{2} f(n)]
$$
converge to a finite positive limit when $n\to +\infty$?

The error form provided by the trapezoidal rule is not helpful since it just asserts that $d_n = -n f''(\xi_n)/12\geqslant 0$, for some $0 < \xi_n < n$, or that
$$
d_n\leqslant -\tfrac{1}{12}[f''(\xi_1) + f''(\xi_2) +\dots + f''(\xi_n)]
$$
with  $k-1 < \xi_k < k$. Since $f(x)$ is increasing and concave, then $f''(x)$ is negative, bounded, and increasing for $x > 0$, but then again, how to assure the converge of $\sum_{k=1}^n f''(\xi_n)$ when $n\to +\infty$?

答:
As you found out, the answer has to lie in the conflict field of "increasing" and "concave". Let's try a solution without referring to derivatives.

Concave: implies that for $x\in[a,b]$ one has that the function graph is above the secant,
$$
f(x)\ge \frac{(x-a)f(b) + (b-x)f(a)}{b-a}
\implies \int_n^{n+1} f(x)\, dx\ge \frac12 (f(n) + f(n+1)).
$$

Switching the points around, that is, exchanging $x$ and $b$ and again isolating $f(x)$ gives for $a < b < x$
$$
(x-a)f(b)\ge (b-a)f(x) + (x-b)f(a)\\
f(x)\le \frac{(x-a)f(b) + (b-x)f(a)}{b-a},
$$
which implies
\begin{align}
\int_{n+1}^{n+2} f(x)\, dx
&\le \int_{n+1}^{n+2}[(x-n)f(n+1) + (n+1-x)f(n)]\, dx\\
&=\frac12(3f(n+1)-f(n)),
\end{align}
so that
$$
0\le \int_n^{n+1} f(x)\, dx - \frac1 2(f(n)+f(n+1))\le \frac12 [2f(n)-f(n+1)-f(n-1)].
$$

Summing up from $m$ till $n$ we get
\begin{align}
0 &\le d_n-d_m\\
&= \int_m^n f(x)\, dx -\Bigl(\frac12 f(m)+ f(m+1) +\dots + f(n-1) +\frac12 f(n)\Bigr)\\
&\le \frac12 (f(m)-f(m-1)) -\frac12 (f(n) - f(n-1)). \tag{*}
\end{align}

From concavity we saw that $2f(n) - f(n+1) - f(n-1)\ge 0$. It immediately follows that the sequence $v_n = f(n+1) - f(n)$ is monotonically falling. Because $f$ is increasing, this sequence stays non-negative, so it has a non-negative limit $v$. This means that the sequence $v_n$ is a Cauchy sequence, and (*) then implies that also the sequence of the $d_n$ is Cauchy. Thus it has a limit, which can not be negative.


问:
$f$ differentiable. Show there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$
let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be differentiable for all $x \in \mathbb{R}$, and let $L\in\mathbb{R}$ so that $f(0)=L$ and $\underset{x\rightarrow\infty}{\lim}f\left(x\right)=L$.

prove there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$.
I get that I'm supposed to use Rolle's theorem, but I can't find 2 x values so $f(x_1)=f(x_2)$.

答:
Let $t=\arctan x$ and $g(t)=f(\tan t)$. Define
$$ g(\frac\pi2)=\lim_{x\to\infty}f(x)=L $$
Then $g(t)$ is defined in $[0,\frac\pi2]$ and satisfies the condition of Rolle's Theorem. Therefore there is $c\in(0,\frac\pi2)$ such that
$$ g'(c_1)=0. $$
Note
$$ g'(t)=f'(\tan t)\sec^2t $$
and hence $g'(c_1)=0$ becomes
$$ f'(\tan c_1)= 0. $$
Let $c=\tan c_1\in(0,\infty)$ and then $f'(c)=0$.


问:
How many subgraphs can be formed from a graph having $e$ sides?
A couple of days ago I was asked this question, and with no prior experience of Graph Theory I attempted to come up with a solution ,

$H(\alpha, \beta)$ is a subgraph of $G(V, E)$ if and only if $\alpha \subseteq V$ and $\beta \subseteq E$.

$e$ as the edge count of graph $G$ represents the cardinality of the edge set of $G$, i.e. $E$.

Since a graph must have at least one vertex but can contain $0$ edges, $\phi$ can represent an edge set as well. Hence, the total number of subgraphs of $G$ is the cardinality of $P(E)$ where $P$ is the power set of a given set.

$$|P(E)| = 2^e$$
$$\therefore \text{the number of subgraphs is } 2^e$$

If this solution is not correct, please let me know of any pointers or where in the concept I went wrong here, I'm soon going to start Graph Theory and I haven't found any solution to this current problem anywhere (Web search and Discrete mathematics and its applications by Rosen) yet.

答: This isn't right, you are effectively choosing subgraphs by choosing which edges to include. But that doesn't say which vertices you are including in each subgraph (apart from the ones attached to the chosen edges). E.g. if an edge is not chosen, do we include the 2 vertices joined by that edge, perhaps just one of them, or perhaps neither? Also, it depends on whether or not the subgraphs have to "look" different from one another (isomorphically distinct). E.g. if you have a graph on 3 vertices with 3 edges (a triangle) there are 3 ways to make a subgraph with 2 vertices and 1 edge. But they are all the same graph, so should we count them as distinct subgraphs? The job of counting numbers of graphs with a certain property is generally a hard one. Hope this helps :)</p>


问:
Divergence of $\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$ for $x>1$
How can we show that the series
$$\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$$
diverges for $x>1$ ?

The book gives the following hint:
consider
$$\sum_{k=1}^\infty\sum_{n=M_{k-1}+1}^{M_k}\frac{1}{(\ln M_k)^x}$$
where $\ln M_k=k$ and note that $M_k-M_{k-1}=e^{-1}(e-1)M_k$; hence show that the series diverges.

But I really can't figure out what this hint means.

答: Or; as the following limit: $$\lim_{n\to\infty} n^1\times\frac{1}{\ln^x(n)}=\infty,~~x>1$$ so the Comparison Test tells us the series diverges.</p>


问:
Show that the sequence $x_{n+1}=\sqrt{2-x_n}$ converges
I want to show that the sequence defined by $x_{n+1}=\sqrt{2-x_n}$ with $x_0=\frac{2}{3}$ converges. To be honest, I have no idea how to prove that because this sequence is not even monotonic but it is bounded. Thanks in advance.

答:
It may help to know what this converges to (if it converges). The limit would have to fulfil
$$ x=\sqrt{2-x},$$
i.e., be a solution of $x^2+x-2=0$. So $x=\frac{-1\pm \sqrt 9}2$, but only the positive solution (that is $1$) makes sense.

So we may try to show that the distance to $1$ decreases. Let $y_n=x_n^2-1$. Then

$$y_{n+1}=x_{n+1}^2-1=(2-x_n)-1=1-x_n=-\frac{y_n}{1+x_n} $$
and as certainly $x_n>0$, we conclude that $|y_{n+1}|<|y_n|$, i.e., $|y_n|$ is strictly decreasing. This also implies that $|y_n|\le |y_0|=\frac59$, so $x_n\ge\sqrt{1-\frac59}=\frac23$, and we find the fetter inequality $$|y_{n+1}|\le \frac{3}{5}|y_n|$$
so that $y_n\to 0$ at least geometrically.


问:
On the Besov space
I want to prove this:

$$
C^{-1}\lambda^{s-\frac dp}\lVert u\rVert\smash{\dot B}_{p,r}^s
\,\le \, \lVert u(\lambda\cdot)\rVert_{\smash{\dot B}_{p,r}^s} \,\le\,
C\lambda^{s-\frac dp}\lVert u\rVert\smash{\dot B}_{p,r}^s.
$$

(original screenshot)

I proved it for $\lambda = 2^k$ for every $k$ integer!

Can you help me to conclude..

Thank You..

答: You can get the proof on from this lecture (Proposition 1.5.4) .</p>

Title: On the Besov space


答:
If $\lim_{n\to \infty}{x_n}=l$ then $\lim_{n\to \infty}{x_{n+1}}=l$ so if $\{x_n\}$ converges then
$$\lim_{n\to \infty}{x_n}=\lim_{n\to \infty}{x_{n+1}}=l \Rightarrow l=\sqrt{2-l} \Rightarrow l=-2 \ or\ 1$$
So $x_n\ge0 $ then $l=1.$
It is easy check $\lim_{n\to \infty}{x_n}=l=1$ by definition.

Title: Show that the sequence $x_{n+1}=\sqrt{2-x_n}$ converges

Body: I want to show that the sequence defined by $x_{n+1}=\sqrt{2-x_n}$ with $x_0=\frac{2}{3}$ converges. To be honest, I have no idea how to prove that because this sequence is not even monotonic but it is bounded. Thanks in advance.


问:
How to solve for this inverse question?
```<code>The time taken to complete a school project is inversely proportional to the number of students involved.
A team of 5 students can complete the project in 6 weeks.
b) Suppose that 4 weeks into the project, one of the students left the team.
   The remaining students decide to continue with the project by themselves.
   Calculate the number of days needed to complete the project in this case.
</code>```

My working:
$$y=\frac{k}{x}$$
From the question, $$k=210$$
So for 5 students, it would take 42 days to complete the project.

So for 4 students, it would take 52.5 days to complete the project.

From the question, (into week 4), so it would be 21 days completed for 5 students.

The remaining days would be:

```<code>21 days + Days of 4 students
</code>```

The difference between 1 Student is :
$$52.5-42=10.5 days $$

How to go on from here?? I'm confused to find the days to complete the project since the number of days depend on the number of students involve..

答:
I interpreted "4 weeks into the project, one of the students left the team"

to mean that one student left after $4$ weeks.

After $4$ weeks with $5$ students, the project is $\frac23$ complete.

To complete the remaining $\frac13$, it will take $4$ students $\frac{52.5}3$ days.

Title: How to solve for this inverse question?


问:
confused about a statement of pure extension in Rotman's abstract algebra textbook

I don't know how to prove the statement in the red rectangles. Namely, How to prove any field extension $K/k$ with $[K:k]\leq 2$ is a pure extension.

答: I assume we are not in characteristic $2$ and $[K:k]=2$. Let $v\in K\setminus k$ be any element. Then $K=k(v)$. As the extension is of degree $2$ and we are not in characteristic $2$ it is Galois. Let $\sigma$ be the non-trivial $k$-automorphism of $K$. Define $u=v-\sigma(v)$. Since we are not in characterstic $2$ we have $\sigma(u)\ne u$, hence $u\in K\setminus k$ and therefore $K=k(u)$. Furthermore we have $\sigma(u^2)=u^2$, so $u^2\in k$.


问:
Proving an equality involving a product $\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$
How can I prove that:

$$\frac{(3n^2-2)(3n^2-1)(1-3n+3n^3)}{2}\ge1$$

When $n\in\mathbb{N}$ such that $n\ge2$.

I think that we need to prove that all terms are monotonically increasing functions. But how do I prove that?

Body:
Even $$3n^{3}-3n+1=3n(n^{2}-1)+1\geq 9n+1$$ for $$n\geq 2$$ thus
Note that $$\frac{9n+1}{2}\geq 4.5\geq 1$$.
Note that for $$n\geq 2$$ $$3n^{2}-2\geq 1$$ and $$3n^{2}-1\geq 1$$.
Thus $$\frac{(3n^{3}-3n+1)(3n^{2}-1)(3n^{2}-2)}{2}\geq \frac{9n+1}{2}*1*1\geq 4.5\geq 1$$


问:
Does this algorithm terminate in finite time?
I am trying to determine whether the following algorithm terminates:

```<code>int n;
int s;
s=3n;

while s>0
{
  if s is even
  {
    s=floor(n/4);
  }
  else
  {
    s=2s;
  }
}
</code>```

So far, I have tried to see whether I can come up with a pattern for how the algorithm behaves for arbitrary n. But I haven't had any luck so far. I'm pretty stumped right now. Any ideas or suggestions as to how I show that this algorithm terminates?

答: Hint:  for what $s$ will the loop terminate the next time through?  You need $s=0$ at the end of the loop. How about trying it for $n=1$ through $8$?</p>


问:
Implicit function theorem - derivatives
Let us have two function $n(\alpha)$ and $s(\alpha)$ and a set of two implicit equations $F_1(\alpha, n(\alpha), s(\alpha))=0$ and $F_2(\alpha, n(\alpha), s(\alpha))=0$. In this paper I've been reading they state that one can use the implicit function theorem to this set of implicit equations and gain
$
\frac{\partial n}{\partial \alpha}=-\frac{det A}{det B}
$
where
$A=\begin{bmatrix}
 \frac{\partial F_1}{\partial s} & \frac{\partial F_1}{\partial \alpha} \\
 \frac{\partial F_2}{\partial s} & \frac{\partial F_2}{\partial \alpha}
 \end{bmatrix}
$
$B = \begin{bmatrix}
 \frac{\partial F_1}{\partial s} & \frac{\partial F_1}{ \partial n} \\
 \frac{\partial F_2}{\partial s} & \frac{\partial F_2}{\partial n}
 \end{bmatrix}$

I cannot comprehend how they gained that partial derivative of n. Would be grateful for any advice.

答:
Let's replace $F_1$ and $F_2$ with $F$ and $G$. Differentiating each implicit equation, and using $F_1(u, v, w)$ to denote the derivative of $F$ with respect to its first argument, etc.) we get
\begin{align}
\newcommand{\al}{\alpha}
0 &= F_1(\al, n(\al), s(\al)) + F_2(\al, n(\al), s(\al))n'(\al) + F_3(\al, n(\al), s(\al))s'(\al)\\
0 &= G_1(\al, n(\al), s(\al)) + G_2(\al, n(\al), s(\al))n'(\al) + G_3(\al, n(\al), s(\al))s'(\al)
\end{align}
where I've used $n'(\al)$ to denote the thing you've called
$$
\frac{\partial n}{\partial \alpha},
$$
but which should probably be
$$
\frac{dn}{d\al}
$$
because $n$ appears to be a function of a single variable. Rewriting those two equations a little, and getting rid of all the arguments, we have

\begin{align}
0 &= F_1  + F_2 n' + F_3 s'\\
0 &= G_1  + G_2 n' + G_3 s'
\end{align}
which we can rewrite in matrix form as
$$
\pmatrix{0\\0} =
\pmatrix{
F_1 & F_2 & F_3 \\
G_1 & G_2 & G_3
}
\pmatrix{1\\n'\\s'}
$$
Adding a row to make this $3 \times 3$, we get
$$
\pmatrix{0\\0\\1} =
\pmatrix{
F_1 & F_2 & F_3 \\
G_1 & G_2 & G_3 \\
1 & 0 & 0
}
\pmatrix{1\\n'\\s'}
$$
and calling the $3 \times 3$ matrix $M$, we can solve
$$
M^{-1} \pmatrix{0\\0\\1} =
\pmatrix{1\\n'\\s'}
$$
And we see that all we really need is the second entry of
$$
M^{-1}\pmatrix{0\\0\\1}.
$$
The authors have applied Cramer's rule to deduce that it is the ratio of determinants that they wrote down (I assume).

Reasonable questions are: "How did you know that adding [1 0 0] gave you an invertible matrix?" and "How did you decided to do that, anyhow?"

For the first: I didn't know that. But the authors don't seem to worry that $\det B$ might be zero, so I figure we're just doing some handwaving in some kind of generic case for the time being.

For the second: A ratio of determinants reminded me of Cramer's rule, so having written things in $3 \times 2$ matrix form, I just wanted to add the simplest possible thing to get a $3 \times 3$ form.

I'll be honest...I didn't actually go through Cramer's rule and check that this all works out. But I'll bet it does...

Title: Implicit function theorem - derivatives


问:
$G=\langle a,b \rangle$, show $a^{p^3} \in Center(G)$
Let G be a finite group of order $p^5$ where p is a prime and $G=\langle a,b \rangle$

where $|a|=p^4, b^p=a^p, bab^{-1}=a^{1+2p^3}$

Show that $a^{p^3} \in Center(G)$

For a it is trivial $aa^{p^3}a^{-1}=a^{p^3}$

For b the solution says $ba^{p^3}b^{-1}=ba^pa^pa^pb^{-1}=a^{p^3}$

My questions is if the solution for b is correct? For me it is incorrect that $a^{p^3}=a^pa^pa^p$. I would say that $a^pa^pa^p=a^{3p} \neq a^{p^3}$ or am I missing something here

Body:
It's enough to show $b.a^{p^3}=a^{p^3}.b$ because any element of $G=\langle a,b \rangle$ is in the form $g=a^{n_1}b^{m_1}...a^{n_k}b^{m_k},\ k\in \mathbb{N}$ and if $b.a^{p^3}=a^{p^3}.b$ then $g.a^{p^3}=a^{p^3}.g$. So
$$b.a^{p^3}=b.(a^p)^{p^2}=b.(b^p)^{p^2}=b.b^{p^3}=b^{p^3}.b=a^{p^3}.b$$


问:
Area under conformal Mapping $ e^z $
The question is as follows:

Consider the region

$\Omega = {x+iy: -1 \le x \le 2, -\pi/3 \le y \le \pi/3 } $

in the complex plane. The transformation $x + iy → e^{x+iy}$ maps the region Ω onto the region S ⊂ ℂ (the set of all complex numbers). Then the area of the region S is equal to.

I managed to find the relationship between (x,y) in input plane to (u,v) in the output plane as:

$y = tan^-1(v/u)$

$x = 1/2 ln(v^2 + u^2)$

Using this and the conditions in the question, I got the area as the red shaded region shown below

The two circles have radii $e^4$ and $e^{-2}$.

The answer should come out to be $2\pi/3 (e^4 - e^{-2})$

But the answer shown in solution paper is $\pi/3 (e^4 - e^{-2})$

My question is, is it not necessary to consider the left half of U-V plane for this?

答:
$u = e^x cosy \ge 0$

So only the region in the right half plane matters

Title: Area under conformal Mapping $ e^z $


问:
How to calculate $\int_{C}{}\frac{\bar{z}}{z+i} dz$ if $C$ is a circle: $|z+i|=3$?
The task:
$$
\text{Calculate } \int_{C}{}f(z)dz\text{, where } f(z)=\frac{\bar{z}}{z+i}\text{, and } C \text{ is a circle } |z+i|=3\text{.}
$$
Finding the circle's center and radius:
$$
|z+i|=|x+yi+i|=|x+(y+1)i|=3
\\
x^2+(y+1)^2=3^2
$$
Parametrizing the circle:
$$
z(t)=i+3e^{-2\pi i t}
$$

Now I need to calculate this integral:
$$
\int_{0}^{1}{f(z(t))z'(t)dt}=
2\pi\int_{0}^{1}{
\frac{1-3ie^{-2\pi i t}}{e^{4 \pi i t}}dt
}
$$

Unfortunately I calculated this integral, and it's equal to $0$. Is this correct? I don't think so. Where did I go wrong? Maybe I made a mistake when calculating the integral - what would be the best way to calculate it?

答:
Observe that
$$
\frac{\overline{z}}{z+i}
=\frac{\overline{z}\cdot \overline{(z+i)}}{(z+i)\cdot \overline{(z+i)}}
=\frac{\overline{z}\cdot \overline{(z+i)}}{|z+i|^2}
=\frac{1}{9}\cdot \overline{z}\cdot \overline{(z+i)}
$$

Your parametrization of the path should be
$$
z(t)=\color{red}{-i}+3e^{2\pi i t},\quad [0,1]
$$

If your integral is $I$, then
\begin{align}
9I
&=\int_C \overline{z}\cdot \overline{(z+i)}\,dz\\
&=\int_0^1 (i+3e^{-2\pi i t})\cdot 3e^{-2\pi i t}\cdot 6\pi i\cdot  e^{2\pi it}
\,dt\\
&=\int_0^1 (i+3e^{-2\pi i t})\cdot 18\pi i\,dt=-18\pi
\end{align}

and thus $I=-2\pi$.


问:
Semisimple perverse sheaf?
Let $\mathcal F$ be a complex of constructible sheaves on a stratified algebraic variety $X$ of dimension $d$.

I read (*) that if $\mathcal F$ is perverse and $\mathscr H^j(\mathcal F) = 0$ for $j \neq -d$ (here $\mathscr H^*$ are the ordinary cohomology sheaves associated to a complex of sheaves), then $\mathcal F$ is a semisimple perverse sheaf.

However, if $X$ is smooth with a single strata, and $\mathcal F = \mathcal L[-1]$ where $\mathcal L$ is a not semisimple local system (for example, $X = \mathbb C^*$, $\mathcal L$ of rank $2$ associated to the endomorphism $e_1 \mapsto e_1, e_2 \mapsto e_1 + e_2$ ). It seems to me that $\mathcal F$ is still concentrated in a single degree by definition, and is also not a semisimple object of $\mathscr P(X)$ by definition. Where is my mistake ?

(*) Reference : Göttsche, Lothar and Soergel, Wolfgang - "Perverse sheaves
and the cohomology of Hilbert schemes of smooth algebraic surfaces",  (see the proof of proposition $3$ for the claim)

答:
the paper you mentioned does not make this claim. in that paper you have a sheaf $F$ on $X$,$G$ is a finite group acting on $X$,and $Y$ is $X/G$. now if you look at $\pi^{*}(L)[d]$ where $L$ is a local system on $X$ then the fact about vanishing of the cohomology of$\pi_{*}L$ (and self-duality) just means that $\pi_{*}L$ is perverse.

semisimplicity deduced from the relation between sheaves on $Y$ and $G$ invariant sheaves on $X$,and the fact that $G$ is finite and hence all its representations are semisimple, and the relation between local systems on $X$ and representations of the fundamental group.

in fact you start with a representation of fundamental group $V$. then you can associate to this a representation $V\otimes F[G]$ of $G$ and then this representation is semisimple because $G$ is finite.

Title: Semisimple perverse sheaf?


问:
Hydraulic radius of a complex shape
I'm working on a project involving thermoacoustics, and one of the important parameters is known as the hydraulic radius. If you have a pipe with some odd geometry, the hydraulic radius is its cross-sectional area of flow (i.e.: of the area not occupied by the shape) divided by its wetted perimeter.

The project I'm working on uses a "stack" for a certain component of the piping system. This comprises a series of plates with some thickness t and spacing x that fit within a tube. One of the plates is bisected widthwise by a diameter of the tube. If the thickness, number of plates, and tube diameter are known, what is the hydraulic radius of this shape?

The white area inside the red circle is the cross-sectional flow area
The perimeter of that white area is the wetted perimeter
The ratio of these two values is the value in question

答:
If a given region has a low point of y, then the angle formed between the lower-left and upper-left of the region and the downward line are $\theta_1$ and $\theta_2$ respectively.

$Perimeter(ABCD)$ is $|\overline{AD}| + |\overline{BC}| + Len(AB) + Len(CD)$

$$ \begin{align}
|\overline{AD}| = & 2R \sin(\theta_1) \\
|\overline{BC}| = & 2R \sin(\theta_2) \\
Len(AB) = & \frac{\theta_2 - \theta_1}{2\pi} \cdot 2\pi R = R(\theta_2 - \theta_1)\\
Len(BC) = &  R(\theta_2 - \theta_1)\\
\end{align}
$$
Hence, $Perimeter(ABCD) = 2R[\sin(\theta_1)+\sin(\theta_2)+\theta_2-\theta_1 ]$

$Area(ABCD)$ can be computed as $Area(OBMC) - Area(OAMD)$.

Given $y_1=|\overline{ME}|$ for arbitrary $y_1$, and your $t = |\overline{EF}|, y_2=y_1+t$:
$$ \begin{align}
& Area(OAMD) = \pi R^2 \frac{\theta_1}{\pi} = \theta_1 R^2 \\
& Area(OAED) = R^2\sin(\theta_1)\cos(\theta_1)\\
& \implies Area(AEDM) = R^2[ \theta_1 - \sin(\theta_1)\cos(\theta_1)] \\
\end{align}
$$

So the area of the region is the difference of the two:

$$ Area(ABCD) = Area(BCM) - Area(ADM) = \\
R^2[( \theta_2 - \sin(\theta_2)\cos(\theta_2) ) -
( \theta_1 - \sin(\theta_1)\cos(\theta_1) ) ]
$$

$\theta_1$ can be calculated as: $$\theta_1 = \cos^{-1}\left(\frac{R-y}{R}\right), y=|\overline{ME}| $$

Note: this is just the area and perimeter for one of the white areas.  I don't know if it's valid to combine all of the regions for hydraulic radius.

Title: Hydraulic radius of a complex shape


问:
Why $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$?
I passed through this answer here and I don't understand the following. We are at the moment:

If $u^2+v^2 \neq 0$, then $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$. Now by Cauchy-Riemann, we also get that $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial y}=0$. Hence, $f'(z) = 0$ which gives us that $f$ is constant.

Why we get "$\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$? I don't see this...

答: There is a comment in the answer to the question you linked that answers your question. The original question was about proving that for a holomorphic function $f$, $|f|$ constant implies $f$ constant. Now if $|f(z)|$ is constant, so is $|f(iz)|$, and running the same argument that showed $\frac{\partial v}{\partial y}=0$ on this rotated function gives $\frac{\partial v}{\partial x}=0$.


问:
Is the empty tuple denoted by () a empty function f: ∅- > ∅?

I'm studying a book analysis 1 by T.Tao and I have three questions.

what i wrote as Title is the first one and the rests are these.

1.Does the empty tuple denoted by () mean a empty function f: Ø → Ø ?

Is an ordered n-tuple of sets (X_i)1≤i≤n always needed when defining a n-fold Cartesian product?

If the answer to the number 2 question is true then I need an ordered tuple of sets (X_i)1≤i≤0 to define a Cartesian product ⫪(X_i)1≤i≤0.

but what is this ordered tuple of sets (X_i)1≤i≤0 ?

I asked the author on his blog but he replied all other questions except mine ㅜㅜ

Are these silly questions?

anyway please help..

答:
The definition of "empty tuple" is in Example 3.5.11 of the text. The empty tuple is the only element in the "empty Cartesian product $\prod_{1\le i\le 0} X_i$.

It is not that the n-tuple is "needed" but that it is "by definition" an element of the Cartesian product. See Definition of 3.5.7.

No.


问:
Synthetic Axiomatization - Meaning
I was reading an answer about the construction of the real numbers and there was a mathematical therm that I could not understand.

On that answer they said that the real numbers can be constructed through synthetic axiomatization, what is the meaning of this expression in mathematics?

Thank you for your time

答: Whatever you read was wrong. You cannot construct something via any kind of axiomatization. What you can do is to stipulate that the real numbers are a collection of objects together with some operations on them such that as a whole they satisfy some axiomatization. Of course, when you do so you cannot assume that there is such a thing at all! That is why it is a non-trivial fact that there is in fact a model of the standard axiomatization of the reals. See also this post regarding the two common kinds of axiomatizations of the real numbers (the standard one is the second-order one).

Title: Synthetic Axiomatization - Meaning


问:
kernel and image of a map help
The question states:

a) Suppose that $U = V = M_{2,2} (\mathbb R)$ is the vector space of $2 × 2$ matrices over $\mathbb R $ Let $\phi$ be defined by $\phi(A) = A - A^T $ for each $A \in M_{2,2} \mathbb R$. Determine
$\ker \phi $ and state its dimension.

b) Determine $\operatorname{im}\phi$ and its dimension.

So my thoughts are we use $\ker \phi  = \left \{ f \in U \mid \phi(f) = 0 \right \}$ and then we follow all the steps to show that  $$\ker \phi = \left \langle \begin{pmatrix}
1  & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\right \rangle
$$

and we can conclude that $\dim \ker \phi = 3 $ and so by Rank and Nullity Thm, we say $\dim\operatorname{im}\phi = 1$.

My question is that is this correct, this process to get that answer, and I'm just wondering what $\operatorname{im}\phi$ would be?

Thank you in advance.

Edit: How I got to the kernel:

$$f = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} $$
then $$f^T = \begin{pmatrix}
a & c\\
b & d
\end{pmatrix}$$

then $$f-f^T = \begin{pmatrix}
0 & c-b\\
c-b & 0
\end{pmatrix}$$

So putting into a system of eqs.

$$\left \{ b-c = 0 \right. \\
\left \{ c-b = 0 \right.$$

Which implies that b=c

So $$f = \begin{pmatrix}
a & b\\
c & d
\end{pmatrix} = a\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix} +b \begin{pmatrix}
0 & 1\\
 0& 0
\end{pmatrix} + d \begin{pmatrix}
0 & 0\\
 0& 1
\end{pmatrix} $$

Which allows me to come to the conclusion that :

$$ker \phi = \left \langle \begin{pmatrix}
1  & 0\\
0 & 0
\end{pmatrix}, \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\right \rangle
$$

Thank you for telling me to edit

I just realised that for kernel 2nd part, it should be $$\begin{pmatrix}
0 &1 \\
1 & 0
\end{pmatrix}$$

答:
The first part is correct.  It is easy to see that the kernel is the set of symmetric matrices, which for dimension $n$ has dimension $n(n+1)/2$.

Since the image is one dimensional, it suffices to find one nonzero element of it, and that will form a basis.  For instance, we have that $\begin{pmatrix}1&1\\0&1\end{pmatrix}\mapsto \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, and the latter matrix spans the image.  Notice it is antisymmetric.

Title: kernel and image of a map help


问:
Partial Differentiation-Prove $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$
If $u$ and $v$ are functions of $x$ and $y$ defined by $ x=u+e^{-v}sinu, y=v+e^{-v}cosu$ , then prove that
$$\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$$

My Attempt:

$\frac{\partial x}{\partial v}=0+(sinu)e^{-v}(-1)=-e^{-v}sinu$

and then I reciprocated it. Then I did similar with $\frac{\partial y}{\partial u}$.

Is my method correct??
Can we reciprocate in partial differentiation.?

答: It is incorrect to take reciprocal of a dx/dv to get dv/dx. Take partial differentiation of first equation with respect to y and second equation with respect to x. Since x and y are not dependent on each other dx/dy and dy/dx is equal to 0. Eventually you will prove the result asked in the question i.e. du/dy = dv/dx.</p>


问:
Why doesn't the combined area of the two small squares equal the bigger square?
So, I was trying to solve this problem from a textbook. The Problem is: "A piece of wire 60 inches long is cut into two pieces and then each piece is bent into the shape of a square. If the sum of the areas of the two squares is 117 square inches, find the length of each piece of wire."

And this is the solution.

"A piece of wire 60 inches is cut into two pieces and then each piece was bent into the shape of a square"

$4x + 4y = 60$

$x + y = 15$

$y = (15-x)$:

"the sum of the areas of the two square is 117 square inches"

$x^2 + (15-x)^2 = 117$

$(15-x)(15-x)+x^2=
x^2 + 225 - 15x - 15x + x^2 = 117$

combine like terms

$2x^2 - 30x + 225 - 117 = 0$

$2x^2 - 30x + 108 = 0$

simplify divide by 2

$x^2 - 15x + 54 = 0$.

Factors to
$x = 6$ and
$x = 9$
therefore
4(6) = 24" is one piece of wire
and
4(9)= 36" is the other

But something doesn't make sense here, If we bend the original 60-inch wire into a square. Then it's area would be 400 square inches.

So why do the two smaller squares area equal, 117 square inches and not 400?

答: The area of a square formed by the full piece of wire would be $(60/4)^2=15^2=225$ square inches, not $400$. As for why the total area of two squares isn't $225$ square inches, let's consider what happens when we cut an arbitrary length off the wire, say $x$ inches, where $0<x<60.$ Now we have two lengths of wire--one is $x$ inches, and the other is $60-x$ inches. Bending both into squares gives us one area of $$\left(\frac{x}{4}\right)^2=\frac{x^2}{16}$$ and another of $$\left(\frac{60-x}{4}\right)^2=\frac{(60-x)^2}{16}=\frac{3600-120x+x^2}{16},$$ for a total area of $$\frac{3600-120x+2x^2}{16}=\frac{3600}{16}+\frac{2x^2-120x}{16}=225+\frac{x(x-60)}8.$$ But remember that $0<x<60,$ so $x(x-60)<0,$ and so the total area is less than $225.$


问:
Computing the distance between vector and its projection on a random subspace
Let $V\in\mathbb{R}^{n\times m}$, where $n>m$, be a random matrix of standard normal gaussians.
Given an arbitrary vector $y\in\mathbb{R}^n$, I need to understand the distance $||y-p_V(y)||^2$, where $p_V$ is the projection onto the range of $V$, meaning $p_V = V(V^TV)^{-1}V^T$ as a matrix.
Through a slightly different lens, I would like to understand the distance of $y$ from $R(V)$.

Given that $V$ is random, this makes the above distance a random variable whose distribution (or at least expected value) is important to my research. This feels like a not-so-novel problem but I couldn't find a solution on my own.

How can I approach this? Thanks in advance!

答:
Up to formalizing why $R(V)$ is a uniformly sampled subspace (which should follow from the fact that standard Gaussians are spherically uniform), I think I got it:

By rotation we can assume that $R(V)$ is $\mathbb{R}^m$, and change the question to be "sampling a vector $y$ uniformly from the sphere $S^{n-1}$, what is $\sum_{i=1}^m y_i^2$?".
By this answer, we get that, if $B\sim\text{Beta}\left(\frac{n}{2}, \frac{n}{2}\right)$ then:
$$
y_i^2 = 4B^2-4B+1 \implies \mathbb{E}y_i^2 = 4\cdot \frac{n+1}{4n} - 4\cdot\frac{1}{2}+1 = \frac{1}{n}
$$
Where the computation of the expected value of $B^2$ relies on this answer.

Therefore we get an error of $1-\frac{m}{n}$.

I would appreciate help with first part if anyone has a direction.


问:
Mathematical Modelling Homework
Considering a situation in which two motorists, person A and B, share the same driving route but own different sized vehicles. Person A fills up the vehicle’s tank at a station along normal route for $US"x_1"$ per litre.
On the other hand, person B drives an extra $"d"$ kilometres out of his normal route to fill up his vehicle’s tank for US $"x_2"$ per litre where $x_2 < x_1$ .

1) Choose suitable parameters for Person A and B's vehicles. Justify your choices.
For this question, What are the parameters?

Are the parameters the choices of the vehicle?:

1- speed - maximmumm speed and acceleration

2- Weight

3- engine output - power and torque

4- Fuel consumption - relationship between fuel and the distance

2) On a spreadsheet, choose several sets of values for $x_1$ , $x_2$ and $d$ to investigate further.

3) Define a set of variables that would be relevant in the above situation.

答: Hint: Under 1), which of the parameters are useful in this calculation?  The general idea is that you will use some fuel driving to the cheaper gas station.  What is the tradeoff that makes it worthwhile?</p>

Title: Mathematical Modelling Homework


答:
If you bend the $60$-inch wire into a square, the area should be $\left( \frac{60}4\right)^2=225$

In general, we have

$$\left( p+q\right)^2 \ne p^2+q^2$$

Hence $p=\frac{24}4$ and $q=\frac{36}{4}$

We have $(p+q)^2 = p^2+2pq+2^2$. Hence there should be a difference of $2pq=2\left( \frac{24}{4}\right)\left(\frac{36}{4} \right)=108.$

Geometrically: Don't forget the rectangle region.

Title: Why doesn't the combined area of the two small squares equal the bigger square?

Body:
So, I was trying to solve this problem from a textbook. The Problem is: "A piece of wire 60 inches long is cut into two pieces and then each piece is bent into the shape of a square. If the sum of the areas of the two squares is 117 square inches, find the length of each piece of wire."

And this is the solution.

"A piece of wire 60 inches is cut into two pieces and then each piece was bent into the shape of a square"

$4x + 4y = 60$

$x + y = 15$

$y = (15-x)$:

"the sum of the areas of the two square is 117 square inches"

$x^2 + (15-x)^2 = 117$

$(15-x)(15-x)+x^2=
x^2 + 225 - 15x - 15x + x^2 = 117$

combine like terms

$2x^2 - 30x + 225 - 117 = 0$

$2x^2 - 30x + 108 = 0$

simplify divide by 2

$x^2 - 15x + 54 = 0$.

Factors to
$x = 6$ and
$x = 9$
therefore
4(6) = 24" is one piece of wire
and
4(9)= 36" is the other

But something doesn't make sense here, If we bend the original 60-inch wire into a square. Then it's area would be 400 square inches.

So why do the two smaller squares area equal, 117 square inches and not 400?


问:
Cut-off function multiplied by Lipschitz continuous function in the Schwartz space
In "An SPDE Model for systemic risk with endogeneous contagion" they claim

Let

$\varphi_\lambda\in C_c^\infty(\mathbb{R}, [0,1])$ a standard cut-off function, which is equal to one inside an interval $[-\lambda, \lambda]$ for $\lambda\in\mathbb{R}$,

$\psi\in\text{Lip}(\mathbb{R})$ (i.e. a Lipschitz continuous function on $\mathbb{R}$),

then $\Gamma_\lambda:=\varphi_\lambda \psi\in\mathscr{S}(\mathbb{R})$ (which is the Schwartz space).

How do I show, that this holds, i.e. the function $\Gamma_\lambda$ is smooth, and fullfills the property

$\forall \alpha,\beta\in\mathbb{N}_0:\;\sup_{x\in\mathbb{R}}|x^\alpha D^\beta(\Gamma_\lambda(x))|<\infty.$

$\psi$ as a Lipschitz continuous function doesn't have to be smooth.

答:
I'm confused too; here's my attempt at a counterexample. Suppose $\psi = |x-1|$, which is Lipschitz, and $\phi_\lambda$ is some bump function supported on $[-\lambda,\lambda]$ where $\lambda > 1$. Then
\begin{align*}
\frac{d^2}{dx^2}(\psi \phi_\lambda) &= \phi_\lambda \frac{d}{dx} \frac{x-1}{|x-1|}  ) + \text{other terms} \\
\end{align*}
But $\frac{d}{dx} \frac{x-1}{|x-1|} = \delta_{\{1\}}$ is the Dirac delta at $1$, which is not Schwartz.

The paper does note that the authors

"are mainly interested in the case $\psi = \text{Id}$, other applications may call for a general Lipschitz function, so we allow for this in the ensuing argument"

Perhaps this will be of use, although maybe there's an error in my counterexample


问:
Matrix rank and related questions
Suppose $A_{m\times m},B_{m\times n}$ are matrices and suppose det$(A)=1$ and rank$(B)=r$. Why rank$(C)=r$, where $C=AB$?

Also why rank$(AB)\leq$min$($rank$(A)$,rank$(B))$, the equality comes when atleast one has full rank. In general what is the proof of rank$(AB)\leq$min$($rank$(A)$,rank$(B))$. Can anyone help me to prove this arguments?

答: The rank of a matrix is invariant under multiplication by an invertible matrix (see here or here for example). For your second question, take a look here.</p>

Title: Matrix rank and related questions


答:
Think of matrices as linear transformations. If you have a map of vector spaces $f:V\to W$ and a $n$-dimensional subspace $U$ of $V$, then the image $f(U)$ has dimension at most $n$. Given two matrices $A,B$, they represent linear maps $A:K^m\to K^m, B:K^n\to K^m$. Now by applying the the remark above to the images of these maps we see that the dimension of the image $(AB)(K^n)$ can be at most the rank of $A$ and $B$, hence $$\operatorname{rank}(AB)\leq\min(\operatorname{rank}A,\operatorname{rank}B)$$
Now assume that $A$ has determinant $1$, in this case the corresponding map $K^m\to K^m$ is an isomorphism, so we get $\dim \operatorname{im}(AB)=\dim\operatorname{im} B$, and therefore $\operatorname{rank}(AB)=\operatorname{rank}B$.

Title: Matrix rank and related questions

Body:
Suppose $A_{m\times m},B_{m\times n}$ are matrices and suppose det$(A)=1$ and rank$(B)=r$. Why rank$(C)=r$, where $C=AB$?

Also why rank$(AB)\leq$min$($rank$(A)$,rank$(B))$, the equality comes when atleast one has full rank. In general what is the proof of rank$(AB)\leq$min$($rank$(A)$,rank$(B))$. Can anyone help me to prove this arguments?


问:
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}<a+b<-2$

Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$

I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.

Please give some idea/hint for the other part.

答:
Hint:

Let $x=a+b$ then $b=x-a$ so we get an quadratic equation on $a$: $$3a^2(x+2)-3a(x^2+2x)+x^3+11=0$$ Since $a$ is real it discriminant is non negative, so we have $$-3(x+2)(x^3-6x^2+44)\geq 0$$

 Notice that $x\mapsto x^3$ and $x\mapsto -6x^2$ are increasing for $x \leq 0$.

 Now if $x\leq -{7\over 3}$ then we get $$\Big(-{7\over 3}\Big)^3-6 \Big(-{7\over 3}\Big)^2 +44\geq 0$$ A contradicition.


Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.

Body: If you want to use induction you assume that $(-1)^n-10^n=11k$  Then you want to prove $(-1)^{(n+1)}-10^{(n+1)}=11m$.  So $(-1)^{(n+1)}-10^{(n+1)}=-1(-1)^n-10(10^n)=-1(11k+10^n)-10(10^n)=-11k-11(10^n)=11m$


问:
Question based on Triangle Inequality $\displaystyle |x+y|\leq |x|+|y|$
If $x,y,z\in \mathbb{R}-\left\{0\right\}$. Then prove that  $\displaystyle 1\leq \frac{|x+y|}{|x|+|y|}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3$

My Try:: Using Triangle Inequality

$\displaystyle | x+y | \leq | x |+| y |\Rightarrow \frac{| x+y|}{| x |+| y |}\le 1$

Similarly $\displaystyle | y+z | \leq | y |+| z |\Rightarrow \frac{| y+z|}{| y |+| z |}\le 1$

Similarly $\displaystyle | z+x | \leq | z |+| x | \Rightarrow \frac{| z+x|}{| z |+| x |}\le 1$

Now Add all three equations::

$\displaystyle \frac{| x+y|}{| x |+| y |}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3$

But I did not Understand How can I calculate Lower value of the Given expression.

答:
Case 1: $x, y, z$ are all greater than $0$.

Then the expression simplifies to 3 as we can get rid of the absolute signs.

Case 2: Two of $x, y, z$ are greater than $0$.

Without loss of generality let $x, y$ be greater than $0$. Then the expression reduces to:

$$1 + \frac{|x+z|}{x+|z|} + \frac{|y+z|}{y+|z|}$$

The above expression reduces to $1$ when we set $x=y=-z$ and in every other situation the value of the expression is greater than $1$. Thus, $1$ is the minimum value that we have found so far.

Case 3: Only one of $x, y, z$ is greater than $0$.

Without loss of generality suppose that $z$ is greater than $0$. Then using the strategy outlined for case 4 we can flip this case to that of case 2 an show that the minimum is again 1.

Case 4: $x, y, z$ are all less than $0$.

Just as in case 1, the expression simplifies to 3. We can see this as follows:

Define X=-x, Y=-y and Z=-z. By definiton, $X, Y, Z$ are all greater than 0. Thus,

$$\frac{\mid x+y\mid}{\mid x \mid+\mid y \mid}+\frac{\mid y+z\mid}{\mid y \mid+\mid z \mid}+\frac{\mid z+x\mid}{\mid z \mid+\mid x \mid}$$

is equivalent to:

$$\frac{\mid -X+-Y\mid}{\mid -X \mid+\mid -Y \mid}+\frac{\mid -Y+-Z\mid}{\mid -Y \mid+\mid -Z \mid}+\frac{\mid -Z+-X\mid}{\mid -Z \mid+\mid -X \mid}$$

Factor out the '-1's and we are back to case 1.


问:
Complex Analysis: If f is an odd function, find an expression for f.
Complex Analysis

Suppose that $f$ is a holomorphic function in $\mathbb{C}\setminus\{0\}$ and satisfies $$\vert f(z) \vert \leq \vert z \vert ^{2} +\frac{1}{\vert z \vert ^{2}}$$

If f is an odd function, find an expression for f.

I thought about calculating the Laurent Serie of f, but I can't calculate Laurent's series explicitly (there is no way to calculate the coefficients), Dai thought about estimating, using the invariance of the circle to know which one cancels out. I cannot develop this question

答:
One can immediately see that for all $z\in\Bbb C^\times:=\Bbb C\setminus\{0\},$ we have $$\left|z^2f(z)\right|=|z|^2|f(z)|\le|z|^4+1.$$

As a result, the Laurent series for $g(z):=z^2f(z)$ cannot have any terms of power greater than $4,$ for if there were any, then for sufficiently large $|z|,$ the above inequality would fail to hold. Moreover, we see that $g$ is bounded in the punctured unit disk, so its Laurent series has no terms of negative power, and so is a polynomial of degree no greater than $4.$

Since $f$ is odd, then $g(-z)=(-z)^2f(-z)=z^2f(-z)=-z^2f(z)=-g(z),$ and so $g$ is odd. Hence, all terms of the Laurent series for $g$ have odd power, and so $$g(z)=az^3+bz$$ for some $a,b\in\Bbb C,$ whence $$f(z)=az+\frac{b}{z}.$$

Added: To see why all terms of the Laurent series of $g$ must have odd power, suppose $h$ is an odd function that is holomorphic in $\Bbb C^\times.$ Note that for all $z\in\Bbb C^\times,$ we have $$0=h(z)+h(-z).\tag{$\heartsuit$}$$ Now, write $$h(z)=\sum_{n\in\Bbb Z}b_nz^n,$$ and consider out the expansion for the right-hand side of $(\heartsuit).$