问:
Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}?
Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?

Obviously the set $L$ depends on where $C$ is, which is why I've written $L$ as a function of $C.$

If $C$ is the midpoint of $AB$, or on the perpendicular bisector of $AB$, it is not clear to me that $L= \{$ the perpendicular bisector of $AB\},$ because maybe $L$ contains more points. [Edit: Karan Elangovan has answered this part of the question.]

答:
Geometric Construction for $\boldsymbol{C}$ off the Axes

Here is a geometric construction for the points of the locus.

We are given the points $A=(-1,0)$, $B=(1,0)$, and $C$. For any $\lambda\in\mathbb{R}$, define
$$
\begin{align}
P&=\frac{A+C}2+\lambda\rho(C-A)\tag{1a}\\
Q&=\frac{B+C}2+\lambda\rho(B-C)\tag{1b}
\end{align}
$$
where $\rho(v)=v\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is $v$ rotated by $\frac\pi2$ counter-clockwise.

Then define two circles: one centered at $P$ passing through $A$ and $C$, and another centered at $Q$ passing through $B$ and $C$. Note that we are simply constructing two similar isosceles triangles: $\triangle APC$ and $\triangle BQC$.

By construction, both circles pass through $C$. Let the other point of  intersection of these circles be $D$.

By construction, $\angle APC=2\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BQC$. Thus, by the Inscribed Angle Theorem,
$$
\angle ADC=\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BDC\tag2
$$

The construction above works because $A$ is outside the circle containing $B$ and $B$ is outside the circle containing $A$. This construction also works when $A$ is inside the circle containing $B$ and $B$ is inside the circle containing $A$

Again, $\angle APC=2\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BQC$. Thus, by the Inscribed Angle Theorem,
$$
\angle ADC=\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BDC\tag3
$$

However, when only one of $A$ or $B$ is inside the circle containing the other, we have the following situation

Note that by the Inscribed Angle Theorem, $\angle BDC=\frac12\angle BQC$; however, $\angle ADC=\pi-\frac12\angle APC$.

$\angle ADC$ becomes supplementary to half its associated central angle, $\angle APC$, when $D$ passes through $A$; that is when $A$ changes whether it is inside or outside of the circle containing $B$. This happens when
$$
\lambda=\lambda_A\equiv\frac{(A-C)\cdot(B-A)}{2\rho(A-C)\cdot(B-C)}\tag4
$$

$\angle BDC$ becomes supplementary to half its associated central angle, $\angle BQC$, when $D$ passes through $B$; that is when $B$ changes whether it is inside or outside of the circle containing $A$. This happens when
$$
\lambda=\lambda_B\equiv\frac{(C-B)\cdot(B-A)}{2\rho(A-C)\cdot(B-C)}\tag5
$$

Which angle, $\angle ADC$ or $\angle BDC$, is supplementary to half its associated central angle switches when $D$ passes through $C$; that is, when the circles are internally tangent. This does not happen when the circles are externally tangent since then neither $A$ nor $B$ is inside the circle containing the other.

If we set
$$
\begin{align}
\mu&=\frac{(A-C)\cdot(B-C)}{\rho(A-C)\cdot(B-C)}\tag{6a}\\[3pt]
\sigma&=\operatorname{sign}(\rho(A-C)\cdot(B-C))\tag{6b}
\end{align}
$$
then the circles are internally tangent when
$$
\lambda=\frac{\mu-\sigma\sqrt{\mu^2+1}}2\tag7
$$
and externally tangent when
$$
\lambda=\frac{\mu+\sigma\sqrt{\mu^2+1}}2\tag8
$$
Here is an animation of $D$ (red point) traversing the locus of points (blue curve) where $C$ lies on the bisector of $\angle ADC$. The red curve is where one or the other inscribed angles is supplementary to half of its associated central angle.

Locus for $\boldsymbol{C}$ on the Axes

If $C$ is on the $y$-axis, then the locus is all of the $y$-axis except $C$.

If $C$ is on the $x$-axis, then the locus is all of the $x$-axis with the segment from $(-1,0)$ to $(1,0)$ and $C$ removed.

If $C=(0,0)$ then the locus is the union of the $x$-axis and $y$-axis with the segment of the $x$-axis from $(-1,0)$ to $(1,0)$ removed.


问:
Conditional probability
A girl goes to school by bus every day. If it doesn't rain, probability that she will be late for bus is 1/5. If it rains probability that she will be late is 2/3. Probability that it is raining is 1/4.

Girl forgot to pick up a bus. Define the probability that it was raining.

In this task I have done following using my logic, but I am not sure is it correct:
P(late for bus and rained that day) = P(rains)/P(late for school and is raining )= 1/4/2/3 = 3/8.

I am not sure if it is correct, so I plead for your help.

答: Given that it rains, the probability she is late is 2/3 is P(L|R) not P(R|L), so your answer must be corrected.</p>

ContentLicense: CC BY-SA 3.0

Title: Conditional probability

ContentLicense: CC BY-SA 4.0


问:
Problem on deriving $E(x)$ of a geometric distribution
I want to derive the result that for a geometric distribution
$$X \sim Geo(p)$$
The expected value is equal to
$$E(x)=\frac{1}{p}$$

I have started off by using the definition of $E(x)$ being the mean

$$E(x)=\sum xP_x$$

And we can write $\sum xP_x$ as

$$\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p $$

Multiplying $\sum xP_x$ by $q$ and then taking the difference of the two expressions

$$\left[\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p \right] $$

$$-\left[ q\sum xP_x= qp + 2q^2p + 3q^3p + 4q^4p......+ (n-1)q^{n-1}p + nq^np \right]$$
$$\text{_________________________________________________________________}$$
$$\sum xP_x \left(1-q \right)=p+qp+q^2p.....+q^{n-1}p +nq^np$$
$$\text{_________________________________________________________________}$$

From here we can see that the right hand side excluding the last term forms the sum of a geometric progression hence we can use the formula for the sum of first $n$ terms of a geometric progression which is in this case $$\frac{p(1-q^n)}{1-q}$$
$$\because p=1-q$$ $$\therefore \frac{p(1-q^n)}{1-q} \iff 1-q^n$$

Now the expression simplifies to
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^np$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n(1-q)$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n - nq^{n+1}$$

At this point it is apparent I have made a mistake somewhere as even by simplifying it further

$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n - nq)$$
$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n(1 - q))$$
$$\sum xP_x = \frac{1-q^n(1 + np)}{\left(1-q \right)}$$

And I have no clue on how to proceed, where have I messed up or have I taken a wrong method entirely?

Body:
The geometric distribution has two forms, the expectation you seek is for the form with probability mass function: $f_x(x)=p(1-p)^{x-1}, x=1,2,3,4,....$
By definition, this expectation is computed as: $$\mathbb{E}[x]=\sum_{x\geq1}xp(1-p)^{x-1}$$ We can then simplify this with the steps below:
$$\mathbb{E}[x]=\sum_{x=1}^{\infty}xp(1-p)^{x-1}$$
$$\mathbb{E}[x]=\sum_{x=1}^{\infty}\dfrac{p}{(1-p)}\times x(1-p)^x$$
$$\mathbb{E}[x]=\dfrac{p}{(1-p)}\sum_{x=1}^{\infty}x(1-p)^x$$
But note that, $$\sum_{x=1}^{\infty}x(1-p)^x=(1-p)+2(1-p)^2+3(1-p)^3+4(1-p)^4+....$$
$$\sum_{x=1}^{\infty}x(1-p)^x=(1-p)[1+2(1-p)+3(1-p)^2+4(1-p)^3+.....]$$
$$\sum_{x=1}^{\infty}x(1-p)^x=(1-p)[1+(1-p)+(1-p)^2+(1-p)^3+...+(1-p)+2(1-p)^2+3(1-p)^3+...]$$
Observe that $1+(1-p)+(1-p)^2+(1-p)^3+...$ is a geometric series with common ratio $(1-p)$ and first term 1. Hence the sum to infinity is given by $\dfrac{a}{1-r}=\dfrac{1}{p}$. Also, $(1-p)+2(1-p)^2+3(1-p)^3+...=\sum_{x=1}^{\infty}x(1-p)^x$ hence, we have that:$$\sum_{x=1}^{\infty}x(1-p)^x=(1-p)[\dfrac{1}{p}+\sum_{x=1}^{\infty}x(1-p)^x]$$
Thus, $$\sum_{x=1}^{\infty}x(1-p)^x-(1-p)\sum_{x=1}^{\infty}x(1-p)^x=\dfrac{1-p}{p}$$
$$[1-(1-p)]\sum_{x=1}^{\infty}x(1-p)^x=\dfrac{1-p}{p}$$
$$p\sum_{x=1}^{\infty}x(1-p)^x=\dfrac{1-p}{p}$$
$$\sum_{x=1}^{\infty}x(1-p)^x=\dfrac{1-p}{p^2}$$
Now recall that $$\mathbb{E}[x]=\dfrac{p}{(1-p)}\sum_{x=1}^{\infty}x(1-p)^x$$
Hence, $$\mathbb{E}[x]=\dfrac{p}{(1-p)}\times \dfrac{1-p}{p^2}$$
$$\therefore \mathbb{E}[x]=\dfrac{1}{p}$$


问:
Basics in Complex Analysis
I am learning complex analysis and having doubts in the basics as follows:-

I read that the derivative of a complex valued function $f(z)$
behaves exactly in the similar fashion as if it were a real valued
function $f(x)$.I would like to know why that happens. If i guess
that it is because the definition of differentiability in both the
worlds, the complex as well as the real is $'identically'$ similar .
If there's more to it ,please help me here.

The integral of a complex-valued function is defined along a contour as the sum of the product of the function value at $z$ times the displacement $dz$ .What does this represent if it's not area under the curve ? $$\int_\gamma f(z)\,dz .$$

Following from above ,in complex integrations , we essentially multiply two complex numbers (i.e. $f(z)$ and $dz$) , simplify them to a single number and then integrate. How is this action inside an integral is justifiable ?

In Cauchy's theorem of analytic functions in a simple closed curve , is it necessary to have
function $f(z)$ being $'extending ...smoothly'$ towards the contour. Is it necessary as well as suffficient if the function is only analytic on and inside the contour and outside the contour it can behave in a rather funny way.Can that happen ?

If there's some text that people here can share on complex analysis , that would be really helpful.

Thanks for helping.

答:
Is this one question or four questions?

(1). The derivative of a complex function is not at all (other than superficially) like that of a real function.  For example, if a complex function has one derivative, it has infinitely many derivatives!

(2). A contour integral could represent various physical entities (for example work) depending on the context.  Green's theorem and Stokes' theorem relate a contour integral to an "area" or "volume" integral.  In any case, the integral needs no physical analogue.

(3).  The term $dz$ is a differential, similar to the role of differential in a real integral.

(4).  In Cauchy's theorem, $f(z)$ is holomorphic (it has a derivative).

Please read "Chapter zero" of Stein and Shakarchi, Complex Analysis for more motivation.

Title: Basics in Complex Analysis


问:
Cutting ropes such as to create two sets with equal number of blue and red stones
Given a rope with a Start and an End, and a number of Blue and Red stones that can slide in this rope such as to create a file of stones.

For a given sequence of blue and red stones, what’s the minimum number of cuts required to ensure that the resulting rope sections can be combined to form two separate sets with equal number of Blue and red stones?

Does the number of cuts depend on the sequence of colours?

What’s the number of cuts required for stones with n different colours?

答:
Assume the number of blue stones, $b$, and red stones, $r$, are even.

Form a circle with the rope with the stones equally placed on this circle. Draw a diagonal in an arbitrary position and rotate this diameter. There may initially be more red stones on one side of the diameter than the other but this imbalance will be reversed when the diagonal has rotated $180^o$. So for some diagonal there are equal numbers of red stones on each side of the diameter and therefore also equal numbers of blue stones.

Cut the rope at the points indicated by the diameter and put together the  section(s) of rope containing the start and end points.

At most $2$ cuts are required. (It will be less than $2$ if the chosen diameter indicates the start and end point.)


问:
The gist of the epsilon delta definition of limit
I am currently re-reading and re-learning with basic concepts in calculus/analysis. I have tried to prove certain limits exist using the epsilon delta definition. Many of these examples are fairly easy because they are made so to guide new students like me to choose the right $\delta$. The gist of these problems is that illustred through this example:

Prove $\displaystyle\lim_{x\to5}|3x+2|=17$.

Let $\epsilon>0, \epsilon$ is arbitrary, pick $\delta=\epsilon/3\implies3\delta=\epsilon$. Then $\forall x\in\mathbb{R}$, then within $0<\color{red}{|x-5|}<\delta$, we have $|3x+2-17|<|3x-15|=3\color{red}{|x-5|}$. This makes $|f(x)-L|$ equal to $|x-a|$, and the $3$, let call it a constant C, they make it easy to choose the value of $\epsilon$ because now you can come to the conclusion that $3|x-5|<3\delta\implies3|x-5|<\epsilon/3\implies|x-5|<\epsilon$.

Stewart does provide a harder problem where such a $C$ is not provided, like $\displaystyle\lim_{x\to3}x^2=9$. I will ask the question on this problem in another thread, but here I wish to ask you something.

I have been spoonfed like many students in Calc 1 with a list of limit law to calculate limits. This means, for example, when x approaches $3$, $x^{2}$ approaches $9$. The term approach seems to be forgotten since for simple limit of polynomial, we can just plug the limit in and it pops out the value.

My point is with the definition of epsilon delta, the gist seems to lie in the concept of neighbourhood or nearness. I have seen the title of a book by Dugac and it contains the word "voisinage" in French. This makes me think that we are interested in the neighbour of the limit point, rather than the exact limit point.

I have also read that the definition of limit is independent from the concept of function being defined. A classic example is that $$\displaystyle\lim_{x\to0}\dfrac{\sin(x)}{x}=1$$, despite the fact that $x$ is not defined at $x=0$. But the limit still exists. There are many proofs to this, most useful is trigonometric proof.

I have looked into cases when limit doesn't exist, and this site suggests that:

https://www.mathwarehouse.com/calculus/limits/how-to-determine-when-limits-do-not-exist.php

The one-sided limits are not equal

The function doesn't approach a finite value (see Basic Definition of Limit).

The function doesn't approach a particular value (oscillation).

The x - value is approaching the endpoint of a closed interval

Of all these, the ones with one-sided limit makes me wonder. Since in one dimension, we can easily define left and right, in 3 dimensions, this may not be very useful, am I correct?

So the point of epsilon and delta is to generalize the concept of neighbourhood or nearness of the limit point, because this can be generalized to higher dimensions?

Sorry, I haven't studied multivariate calculus so my experience lies entirely within function of one variable.

答:
"the point of epsilon and delta is to generalize the concept of neighborhood or nearness of the limit point, because this can be generalized to higher dimensions"

I would say it's not to generalize it but to define it. Usually the delta pertains to $x$, the epsilon pertains to $f(x)$.

In higher dimensions one defines something called an open ball. Well, that is a neighborhood in higher dimensions.


问:
Counterexample to inverse Leibniz alternating series test
The alternating series test is a sufficient condition for the convergence of a numerical series. I am searching for a counterexample for its inverse: i.e. a series (alternating, of course) which converges, but for which the hypothesis of the theorem are false.

In particular, if one writes the series as $\sum (-1)^n a_n$, then $a_n$ should not be monotonically decreasing (since it must be infinitesimal, for the series to converge).

答:
Put:
$$
b_n = \begin{cases}
n^{-2} &: n \text{ odd} \\
2^{-n} &: n \text{ even}
\end{cases}
$$

$b_n$ is not monotonically decreasing. Still, $\sum (-1)^n b_n$ converges.


问:
How to classify discrete groups of motions
I'm studying chapter 5 of Artin's Algebra, and it talks about the classification of the discrete groups of motions(in the plane), and that there are only 17 types. I thought about what it means for two discrete groups to be equal, and it didn't seem like group isomorphism was the right criteria to classify these groups. For example, I can't think of a way to prove that all of these groups are non-isomorphic. So what is the correct criteria? It is just group isomorphism? If it is not, why wouldn't Artin include that in his book?</p>

答:
Yes, the classification is up to isomorphism of abstract groups.
All $17$ groups can be written down by a presentation and an explicit semidirect product. This makes it easy to see that the groups are non-isomorphic. For example, the first group is $\Bbb Z^2$, where we have only translations, and the last group is $D_6\ltimes \Bbb Z^2$, where we have the point group $D_6$, the dihedral group.

References:

What are the algebraic structures of the wallpaper groups?

Proof that there exist only 17 Wallpaper Groups (Tilings of the plane)


问:
Pushouts in category of topological spaces modulo homotopy
Let $HTOP$ denote the category whose objects are topological spaces  and whose morphisms are equivalence classes of continuous maps modulo homotopy equivalence. I m wondering what are the properties of this category. Does it have products, coproducts, pushouts ? I tried searching Google but it seems I don't know the right keywords

答:
This category is often called simply “the homotopy category.” It has effectively no limits and colimits, other than products and coproducts which are constructed as in TOP. It was the first natural example of a non-concrete category, as proven by Freyd in “homotopy is not concrete.”

More commonly studied is its full subcategory spanned by the CW complexes, which is equivalent to the category of all spaces localized at the weak homotopy equivalences. This continues to lack any good categorical properties. It doesn’t even have a generating set, though the category of pointed connected CW complexes modulo homotopy famously does—namely, the spheres, by Whitehead’s theorem.

These various homotopy categories are the ur-examples of homotopy categories of model categories, which were invented in large part to get around the non-existence of categorical constructions in homotopy categories by allowing the construction of homotopically meaningful limits and colimits. For instance, a homotopy pushout of $A\leftarrow C\to B$ represents maps from $A,B$, not which are equal on $C$ like a pushout in Top, nor which are equal on $C$ up to homotopy like a (non-existent) pushout in the homotopy category, but maps from $A$ and $B$ together with a chosen homotopy between them on $C$. This is the double mapping cylinder. The more recent theory of $\infty$-categories is also designed in large part to make homotopy limits and colimits behave more like ordinary limits and colimits.


问:
A conditional probability problem on coupon collection
Suppose that there are $n$ types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of $n$ types. Suppose one continues collecting until a complete set of at least one of each type is obtained.
(a) Find the probability that there is exactly one type $i$ coupon in the final collection.

Hint : Condition on T, the number of types that are collected before the first type $i$ appears.

Now, because of each new coupon obtained is independent of past selections and is equally likely to be any of $n$ types, I feel that $T$ is uniformly distributed between $0$ and $n-1$. But, I am not able to prove it.

答:
What is the probability that $T=0$? The first coupon must be $i$, hence it's $1/n$. What about $T=1$? Then we can get any positive number of any single other type, of which there are $n-1$, followed by coupon $i$. $(n-1)/n\times (1+1/n+1/n^2+\cdots)\times 1/n=(n-1)/n\times 1/(1-1/n)\times 1/n=1/n$. That's a good sign...

So can we come up with a simple argument along the lines you suggest? Well, let $X_j$ be the first time a $j$ is collected. Then all the $X_j$ will end up in some particular order (there is 0 probability they are infinity, so we don't bother with that). We want to know where $X_i$ falls. But note that the probability of any one specific order like $4,8,2,1,5,6,3,7$ is equal to that of any other order since we can just relabel all the coupons and nothing can change; there is nothing special about any one labelling. Therefore, the order is a uniform random permutation of $1,\ldots,n$. Then it follows that the location of $X_i$ is uniformly distributed in the order as you require.

I'll leave you to think about how to use this for the rest of the question.


问:
How to find the $\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$
im struggeling to find $$\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$$

Ive tried the following:
$$\lim_{x \to x_0} ax^{bx} = \lim_{x \to x_0} e^{ax^{bx}} =  \lim_{x \to x_0} e^{bx \ln(ax)} =  e^{\lim_{x \to x_0} bx \ln(ax)}$$
wich leads me to
$$ = e^{\lim_{x \to 0} \frac{3\ln(2-e^{\arcsin^2(\sqrt{x})})}{x}}$$
Is this the right way to go? If yes how to get rid of the division by x?
Thanks for any help!

答:
Hint: Consider substituting $y=\arcsin(\sqrt x)$. Then $x=\sin^2y$, and $y\to0$ as $x\to0$. So the limit becomes

$$\lim_{x\to0}\left(2-\exp\left(\arcsin^2\left(\sqrt x\right)\right)\right)^{\frac3x}=\lim_{y\to0}\left(2-\exp\left(y^2\right)\right)^{\frac3{\sin^2y}}$$

The log-exp manipulation you use then converts the limit to

$$\exp\left(3\lim_{y\to0}\frac{\ln\left(2-\exp\left(y^2\right)\right)}{\sin^2y}\right)$$

Can you take it from here?


问:
Population Growth Question
I am not sure how I am supposed to do this, but here is the problem and my attempt.

A farmer wants to produce chickens for the community. Let $P(t)$ represent the population of chickens at time $t$ in months. Assume the population of chickens reproduce at a rate proportional to its size. The farmer cultivates a small number of chickens for a trial run. In three months, the population of chickens increased from 75 to 300. The farmer wants to begin with an initial population of $P(0)$ chickens, and after a year, have $8000 + P(0)$ chickens, so that 8000 can be harvested, and $P(0)$ saved to start breeding again for the following year. What initial population $P(0)$ of chickens does the farmer need to start with?

Attempt:

First, since $P(t)$ represents the population at time $t$ in months, then the population growth equation can be written as: $$P(t) = P_0e^{kt}$$

Then I tried to find the relative growth rate. Since in three months, the population went from 75 to 300, then that means $$300 = 75e^{3k}$$

Or in other words, $k = \frac{\ln{4}}{3}$.

I am not exactly sure how to proceed from here, or if I am on the right track. I would like some assistance, or advice.

答:
Excellent so far!

You have $e^{3k}=4$ and you want $$Pe^{12k}=P+8000.$$
Note that $e^{12k}=4^4$.

Therefore  $256P=P+8000$ and $P=32$ are required.

Title: Population Growth Question


答:
Considering the change of variable $\sqrt{x}= \sin y$, the limit can written in the form $e^L$, where $\displaystyle L = \lim_{y \to 0} \dfrac{3 \ln (2-e^{y^2})}{\sin ^2 y}$. Applying L'Hopital's rule and taking out the factors that do not contribute to the indetermination , we get
$$
\lim_{y \to 0} \dfrac{3 \ln (2-e^{y^2})}{\sin ^2 y}= 3 \lim_{y\to 0} \dfrac{-2y e^{y^2}}{(2-e^{y^2}) \cdot 2 \cos y \sin y}=-3 \lim_{y \to 0} \frac{y}{\sin y} = -3
$$

and so, the final answer is that your limit is $e^{-3}$.

Title: How to find the $\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$

Body:
im struggeling to find $$\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$$

Ive tried the following:
$$\lim_{x \to x_0} ax^{bx} = \lim_{x \to x_0} e^{ax^{bx}} =  \lim_{x \to x_0} e^{bx \ln(ax)} =  e^{\lim_{x \to x_0} bx \ln(ax)}$$
wich leads me to
$$ = e^{\lim_{x \to 0} \frac{3\ln(2-e^{\arcsin^2(\sqrt{x})})}{x}}$$
Is this the right way to go? If yes how to get rid of the division by x?
Thanks for any help!


答:
Note that $k$ is the same in both scenarios because the chickens are the "same" type of chickens.

So, formulate $P(12) - P(0) = 8000 \implies P(0) \left( e^{12 k} - 1 \right) = 8000$, substituting $k = \frac{\ln 4}{3}$ and solve for $P(0)$.

Title: Population Growth Question

Body:
I am not sure how I am supposed to do this, but here is the problem and my attempt.

A farmer wants to produce chickens for the community. Let $P(t)$ represent the population of chickens at time $t$ in months. Assume the population of chickens reproduce at a rate proportional to its size. The farmer cultivates a small number of chickens for a trial run. In three months, the population of chickens increased from 75 to 300. The farmer wants to begin with an initial population of $P(0)$ chickens, and after a year, have $8000 + P(0)$ chickens, so that 8000 can be harvested, and $P(0)$ saved to start breeding again for the following year. What initial population $P(0)$ of chickens does the farmer need to start with?

Attempt:

First, since $P(t)$ represents the population at time $t$ in months, then the population growth equation can be written as: $$P(t) = P_0e^{kt}$$

Then I tried to find the relative growth rate. Since in three months, the population went from 75 to 300, then that means $$300 = 75e^{3k}$$

Or in other words, $k = \frac{\ln{4}}{3}$.

I am not exactly sure how to proceed from here, or if I am on the right track. I would like some assistance, or advice.


问:
Getting useful distributions out of distributions with unknown probabilities
Suppose I have biased coins that all land heads with probability $p\in(0,1)$. Then I can simulate a fair coin by flipping a pair of biased coins until I get heads-tails or tails-heads. I associate to each respectively heads and tails. Since the probability of either is $p(1-p)$, I have extracted a fair coin out of a pair of biased ones.

This is a well-known trick that is often introduced to undergraduates in probability. What I want to know is: Are there other ways to extract "fair" distributions out of "biased" ones?

Two specific examples of what I am asking:

I have a countable infinite sequence of i.i.d. Poisson random variables $X_i$ with mean $\lambda\in\mathbb R_{>0}$. Can I simulate a Poisson random variable $X$ with mean $1$?

I have an uncountable collection of i.i.d. normal random variables $X_t$ with mean $\mu\in\mathbb R$ and variance $\sigma^2\in\mathbb R_{>0}$. Can I simulate a normal random variable $X$ with mean $0$ and variance $1$?

I am not sure if this is important, but I do not mind if the process to generate a "fair" distribution takes infinite time, so long as it converges in some reasonable sense.

答:
The trick of transforming random variables to "fair" ones works for any independent pair of absolutely continuous random variables X and Y, even if X and Y are dependent on each other or the two variables are not identically distributed, as long as the two variables are statistically indifferent (Montes Gutiérrez 2014, De Schuymer et al. 2003); equivalently, their probabilistic index (Acion et al. 2006) is 1/2 or their net benefit is 0. This means in our case that P(X < Y) = P(X > Y). In particular, two independent normal random variables X and Y are statistically indifferent if and only if they have the same mean (they remain so even if their standard deviations differ) (Montes Gutiérrez 2014).

In this case, if you have a "black box" to generate i.i.d. variates from an unknown distribution, you can proceed as follows (Morina et al. 2019):

Draw two variates from the "black box", call them X and Y.

If X is less than Y, output 1. If X is greater than Y, output 0. Otherwise, go to step 1.

Because X and Y are i.i.d., they are statistically indifferent, so that this method will output 1 or 0 with equal probability (see the appendix in my Note on Randomness Extraction).

This method generates i.i.d. random bits, so along with a method to generate uniform integers as well as a method to sample from the Poisson distribution with those integers, that solves example 1.

REFERENCES:

Montes Gutiérrez, I., "Comparison of alternatives under uncertainty and imprecision", doctoral thesis, Universidad de Oviedo, 2014.

De Schuymer, Bart, Hans De Meyer, and Bernard De Baets. "A fuzzy approach to stochastic dominance of random variables", in International Fuzzy Systems Association World Congress 2003.

Morina, G., Łatuszyński, K., et al., "From the Bernoulli Factory to a Dice Enterprise via Perfect Sampling of Markov Chains", arXiv:1912.09229 [math.PR], 2019.

Acion, Laura, John J. Peterson, Scott Temple, and Stephan Arndt. "Probabilistic index: an intuitive non‐parametric approach to measuring the size of treatment effects." Statistics in medicine 25, no. 4 (2006): 591-602.


问:
Complex function with infinitely many zeros on bounded domain
I found a similar post that discussed complex functions that have infinitely many complex zeros here:
Examples of complex functions with infinitely many complex zeros

I was wondering, if one of these functions is in a bounded domain (and its infinitely many zeros are also in the bounded domain) does f have to be identically $0$? Or can it be a non-constant function still?

答:
If you only consider a complex function, without holomorphicity, anything can happen. Simply consider the indicator function $1_{U\setminus\mathbb{Q}}(z)$ where $U$ is the domain.

If $f:U\to\mathbb{C}$ has infinitely many zeros in the bounded domain $U$ and can be extended as an entire function, then yes, $f$ must be zero, since zeros of non-trivial holomorphic functions are isolated.

If $f:U\to\mathbb{C}$ is holomorphic but not necessarily the restriction of an entire function, then Conrad's comment gives an example. In this case, see also this question: Can a non-constant analytic function have infinitely many zeros on a closed disk?


问:
Consider a matrix with integer entries such that $a_{ii}=1$ and $a_{ij}=0$ for $i>j$
Consider a matrix with integer entries such that $a_{ii}=1$ and $a_{ij}=0$ for $i>j$

Then

$A^{-1}$ exists and it has integers entries.

$A^{-1}$ exists and it has some entries which are not integers.

$A^{-1}$ exists and it is a polynomial function in $A$ with integers coefficient.

$A^{-1}$ is a not a power of $A$ unless $A=I$

As $\det A=1$ so $A$ inverse exist and $A^{-1}=\dfrac{\operatorname{Adj} A}{\det A}$ clearly $A^{-1}$ will have all integer entries, $2$ is false,$c_nA^n+\dots +I=0\Rightarrow A^{-1}=-c_nA^n+\dots+ c_{n-1}A$ as $\det A=c_0=1$ so $3$ is true, say $A^{k-1}=A^{-1}$ then $A^k-I=0$ so minpoly must divide $x^k-1$ and as its charpoly is $(x-1)^n=0$ so minpoly must divide $(x-1)^n$ too but that forces minpoly is $(x-1)$ and hence $A=I$. is my all logics are  okay?

答:
Your proofs are correct, apart from some minor issues: the line
$$c_nA^n+\dots +I=0\Rightarrow A^{-1}=-c_nA^n+\dots+ c_{n-1}A$$
should read
$$A^n + c_{n-1}A^{n-1} + \ldots + c_1A + c_0I = 0
\ \Rightarrow\ A^{-1} = (-1)^{n+1}(A^{n-1} + c_{n-1}A^{n-2} + \ldots + c_1I)$$
or
$$c_nA^n + c_{n-1}A^{n-1} + \ldots + c_1A + I = 0
\ \Rightarrow\ A^{-1} = -(c_nA^{n-1} + c_{n-1}A^{n-2} + \ldots + c_1I),$$
depending on whether you force the leading term or the constant term to $1$. Also, you need to explain why the coefficients $c_0,c_1,\ldots,c_{n-1}$ are integers.</p>


问:
Is the picture isomorphic?
enter image description here
Me and my frined were having a descussion i think this is not isomorphic.
What do you think and could you provide some more info about it?</p>

答:
Let $G=(V,E)$ and $G‘=(V’,E‘)$ be two graphs. They are isomorphic if there is a bijection $\varphi: V \to V‘$ such that $(v_x, v_y) \in E \iff (\varphi(v_x), \varphi(v_y)) \in E‘$.

This means that, if we can find a bijective vertex mapping between $V(G)$ and $V(G‘)$, such that adjacency is preserved, then the given graphs are isomorphic. Can you go from here for the provided graphs?

Edit: The number of vertices, the number of edges, the degree sequence. and many more other properties must be the same in $G$ and $G‘$ in order for them to be isomorphic. These are necessary conditions. Take the number of vertices and edges, this follows directly from the defined isomorphism map. But these conditions are not sufficient, meaning you can have graphs having the same number of vertices and edges without them being isomorphic. We need a so called certificate in order to prove isomorphism (have a look at canonical labelling). However the complexity of the graph isomorphism problem is high (actually a precise classification is still missing, but we do not have efficient algorithms for general graphs).

Meaning, finding certificates such as canonical labellings is not faster for a small number of graphs than finding the isomorphism map directly. Since your provided problem is very easy, you should construct the isomorphism map directly.

For example: let $\varphi(v_5)=u_5$. Define the other mappings such that the definition provided is satisfied. (and of course, if some necessary condition fails, you can already state that the two graphs are not isomorphic).

Title: Is the picture isomorphic?


问:
reflexive, symmetric and not transitive relation
If a relation on a set $R$ is reflexive, symmetric but not transitive, does it imply that all the classes $[x] = \{ y\mid xRy\}$ are mutually exclusive, in the sense that for no $[x]$ and $[y]$, they both share an element, and so their intersection is always empty?

答:
If we have a set $X$ and $\mathcal{U}$ is any cover of it, so $\bigcup \mathcal{U} = X$, then we have such a relation in $xRy$ iff $\exists O \in \mathcal{U}: \{x,y\} \subseteq O$. It’s trivially symmetric and reflexive. Gerw’s answer is a special case.

The “class” of $x$ is then just $\bigcup\{O \in \mathcal{U}\mid x \in O\}$, which is usually denoted $\textrm{St}(x,\mathcal{U})$, the star of $x$ wrt the cover. If the cover is disjoint, we exactly get the equivalence relation with those elements as classes, otherwise we get some non-transitivity and other classes too.


Title: Population Growth Question

Body:
Looks good so far!  This means that you have $$P(t) = P_{0}e^{\frac{t}{3}\ln 4}$$
or after rewriting:  $$P(t) = P_{0}e^{\ln 4^{t/3}} = P_{0}(4)^{t/3}.$$  Now it just remains to find the initial population $P_{0}$ so that $$8000 +P_{0} = P_{0}(4)^{12/3}.$$  Solving this equation for $P_{0}:$
\begin{align}
8000 +P_{0} &= P_{0}(4)^{12/3}\\
8000 &= P_{0}\left(4^{1/3} - 1\right)\\
P_{0} &=\frac{8000}{4^{12/3} - 1}.\\
\end{align}
Simplifying this gives us $$P_{0} = \frac{8000}{4^{4} - 1} = \frac{1600}{51} = 31.37...\approx 32.$$


问:
Uniform convergance for $f_n(x)=x^n-x^{2n}$
the function $f_n(x)=x^n-x^{2n}$ converge to $f(x)=0$ in $(-1,1]$. Intuativly the function does not converge uniformally in (-1,1]. How can I prove it?
I tried using the definition $\lim \limits_{n\to\infty}\sup \limits_{ x\in (-1,1]}|f_n(x)-f(x)|$ function is continial fractional on $[-1,1]$ and $x=0,(\frac 1 2 )^{\frac 1 n}$ are the roots of the derivative. I found that the second derivative is negative in the second point. then $\sup=1/4$ and the function does not converge uniformally?</p>

答:
Choose an arbitrarily large odd value of $n$.  There exists some $0<x<1$ such that $x^n>\dfrac 12$.

Then $$\begin{array}{rl}f_n(-x) &= (-x)^n - (-x)^{2n}
\\ &= -\left(x^n + x^2n\right) \\ &\leq -\frac 34 \end{array}$$

So $f_n$ does not converge uniformly on $(-1,1]$.


问:
How to rearrange the following trigonometric integral
How do you rearrange the following;

$\frac{1}{R_0}\int_{0}^{\pi/2}\frac{1-sin^3(\alpha)}{cos^2(\alpha)} d\alpha$

using t=$\tan\frac{\alpha}{2}$

to obtain

$\frac{1}{R_0}\int_{0}^{1}[\frac{2}{(1+t)^2}+4\frac{t}{(1+t^2)^2}]dt$

I do not understand how this rearrangement was done

The final answer to the integral is $\frac{2}{R_0}$

Many thanks

答: The substitution you are talking about is known as the Weierstrass substitution, info on the link. It is a common method that you can derive using half angle formulae etc.</p>


问:
Bounded linear functional separating a hyperplane and a point
I want to find examples of normed linear space, a convex subset $C$ of $X$ and a point $x_0\in X\setminus C$ such that for all bounded linear functionals $f:X\to \mathbb{K}$, Re$f(x_0)<\sup\{\text{Re}f(x):x\in C\}$.

If I start with a proper dense subspace $C$ of $X$, then there exists a sequence $(x_n)$ in $C$ such that $x_n\to x_0$. Thus for all bounded linear functional $f$ on $X$, Re$f(x_n)\to \text{Re}f(x_0)$. Thus Re$f(x_0)\leq \sup\{\text{Re}f(x):x\in C\}$. But how to show the strict inequality?

答:
First of all, as stated the problem fails trivially, since we have the zero functional (which is always bounded and it gives equality, not strict inequality). So suppose that we exclude the zero functional from the question. If the convex set is a subspace, the question has no interest:

If $Y$ is a proper, closed subspace of $X$ and $x_0\in X\setminus Y$, then there exists a functional $f\in X^*$ so that $f=0$ on $Y$ and $f(x_0)>0$, so $\text{Re}(f(x_0))=f(x_0)>0=\sup\{f(x): x\in Y\}$.

If $Y$ is a subspace of $X$ that is not closed, nor dense, then $\bar{Y}$ is a proper closed subspace of $X$ and the previous case yields a non-zero functional that fails to satisfy $\text{Re}(f(x_0))<\sup\text{Re}(f(x))$.

If $Y$ is a dense subspace of $X$, then any bounded functional $f$ that is zero on $Y$ is identically zero on $X$, bceause of continuity. Now suppose that $f$ is a non-zero functional on $Y$, so $f(y_0)\neq0$. Consider the element $y_1=\overline{f(y_0)}\cdot y_0\in Y$. Then $f(y_1)=|f(y_0)|^2>0$. But the elements $y_n=n\cdot y_1$ belong in $Y$ and $f(y_n)=n|f(y_0)|^2$, so $\sup\{\text{Re}(f(y)): y\in Y\}=\infty$ so for any $x\in X$ we have that $\text{Re}(f(x_0))<\sup\{\text{Re}(f(y)): y\in Y\}$.


问:
Proving a subset is a vector subspace using Subspace criterion
Given
$$V =\ {f :\mathbb R\to \mathbb R \mid f(x)=a+bx+cx^2\mbox{ where }a, b, c \in \mathbb R}$$
and
$$W = \{f :\mathbb R\to \mathbb R \mid f(x)=α+βx^2\mbox{ where }α, β \in \mathbb R\}$$
how do I show, using the Subspace criterion, that the subset $W \subseteq V$ is a vector subspace of $V$?

答:
Well, let $f=a+bx^2$ and $g=c+dx^2$ be elements of $W$.

Then $f+g = (a+bx^2) + (c+dx^2) = (a+c) + (b+d)x^2$ lies in $W$.

Moreover, for each scalar $t$, we have that

$tf = t(a+bx^2) = (ta) + (tb)x^2$ lies in $W$ too.


问:
Generic polynomial and irreduciblity.
Let $p= x^2+ ax+b \in R[x] $ where $R= \mathbb{F}[a,b], a$ and $b$ are indeterminates over $F$.

It is mentioned that $p$ is a generic polynomial and it is irreducible, Since specialising the variables(a=0 and b=t) gives $\tilde p= x^2-t $ which is irreducible over $F[t][x]$.

My questions are

What is the meaning of generic polynomial?
In general, what is the meaning of the generic element?.

how does the actual process of specialising variables work?

why $\tilde p $ is irreducible?

Kindly help me with this. Thanks in advance

答:
-I can't comment due to less points!

a generic polynomial refers usually to a polynomial whose coefficients are indeterminates.

You can use Gauss Lemma, Eisenstein's Criteria and Reduction Mod n method to find if, a given polynomial is irreducible or not.

we'll try with Reduction Mod n with example.. ${P= x^2-15x+7}$, we can take mod7 but it can't tell any thing about irreducibility over Z. so we we'll try mod 5.

P(x) mod5 = ${x^2+2}$ = ${x^2+b}$,  now making a=0 it is a good choice to test irreducibility.

Now if it factor then we're left with (x+t1)(x+t2), so we need ${t1 \cdot t2 =2 =b}$, now,the perfect square in the ${Z_5} $ are {0,1,4}, but 2 doesn't appear.

Also [a,b] are field elements, hence it doesn't have zero devisors.

Hence if it is irrecucible over ${Z_n}$ then it is irreducible over ${Z}$

And Due to Gauss Lemma, if it is irreducible over Z, it is irreducible over Q.

Title: Generic polynomial and irreduciblity.


问:
Does the series $w(x) = \sum_{n=1}^{\infty}w_n(x)$ converge uniformly?
Is the following interpretation correct?

In my textbook, I stumbled upon the following theorem;

Suppose that $\{w_n\}$ is a sequence of continuous functions and $w_n : \Omega \rightarrow \mathbb{C}$. If the series
$$w(x) = \sum_{n=1}^{\infty}w_n(x)$$
converges uniformly, then $w(x)$ is continuous on $\Omega$.

I have two questions as regards this theorem;

A) If I know for sure that $w(x)$ is not completely continuous on $\Omega$ (referring to if there are any jump- or essential discontinuities) and $\{w_n\}$ is a sequence of continuous functions and $w_n : \Omega \rightarrow \mathbb{C}$, do I know if the sum $w(x)$ does not converge uniformly?

B) How would an outline for a proof look like?

Many thanks in advance.

答:
A) Yes: if the sum is not continuous, then the series is not niformly convergent. Essentially  this is the contrapositive statement of the theorem.

B) See the Wikipedia article for the theorem: click here.


问:
The gist of the epsilon delta definition of limit
I am currently re-reading and re-learning with basic concepts in calculus/analysis. I have tried to prove certain limits exist using the epsilon delta definition. Many of these examples are fairly easy because they are made so to guide new students like me to choose the right $\delta$. The gist of these problems is that illustred through this example:

Prove $\displaystyle\lim_{x\to5}|3x+2|=17$.

Let $\epsilon>0, \epsilon$ is arbitrary, pick $\delta=\epsilon/3\implies3\delta=\epsilon$. Then $\forall x\in\mathbb{R}$, then within $0<\color{red}{|x-5|}<\delta$, we have $|3x+2-17|<|3x-15|=3\color{red}{|x-5|}$. This makes $|f(x)-L|$ equal to $|x-a|$, and the $3$, let call it a constant C, they make it easy to choose the value of $\epsilon$ because now you can come to the conclusion that $3|x-5|<3\delta\implies3|x-5|<\epsilon/3\implies|x-5|<\epsilon$.

Stewart does provide a harder problem where such a $C$ is not provided, like $\displaystyle\lim_{x\to3}x^2=9$. I will ask the question on this problem in another thread, but here I wish to ask you something.

I have been spoonfed like many students in Calc 1 with a list of limit law to calculate limits. This means, for example, when x approaches $3$, $x^{2}$ approaches $9$. The term approach seems to be forgotten since for simple limit of polynomial, we can just plug the limit in and it pops out the value.

My point is with the definition of epsilon delta, the gist seems to lie in the concept of neighbourhood or nearness. I have seen the title of a book by Dugac and it contains the word "voisinage" in French. This makes me think that we are interested in the neighbour of the limit point, rather than the exact limit point.

I have also read that the definition of limit is independent from the concept of function being defined. A classic example is that $$\displaystyle\lim_{x\to0}\dfrac{\sin(x)}{x}=1$$, despite the fact that $x$ is not defined at $x=0$. But the limit still exists. There are many proofs to this, most useful is trigonometric proof.

I have looked into cases when limit doesn't exist, and this site suggests that:

https://www.mathwarehouse.com/calculus/limits/how-to-determine-when-limits-do-not-exist.php

The one-sided limits are not equal

The function doesn't approach a finite value (see Basic Definition of Limit).

The function doesn't approach a particular value (oscillation).

The x - value is approaching the endpoint of a closed interval

Of all these, the ones with one-sided limit makes me wonder. Since in one dimension, we can easily define left and right, in 3 dimensions, this may not be very useful, am I correct?

So the point of epsilon and delta is to generalize the concept of neighbourhood or nearness of the limit point, because this can be generalized to higher dimensions?

Sorry, I haven't studied multivariate calculus so my experience lies entirely within function of one variable.

答:
Limits are tricky in this sense. It is true that the limit
$$
\lim_{x \to 3}x^2
$$
can be evaluated by simply plugging in $x=3$, but the reason for this is not as simple as you might think. I will explain this in more detail in a moment, but first let's consider what limits are in general. As you have already correctly noted, limits tell us about the behaviour of a function as you approach a value, but not when you actually 'get there'. The example you gave is perfect. When we write
$$
\lim_{x \to 0}\frac{\sin x}{x} = 1
$$
what we mean is that as $x$ approaches $0$, $\frac{\sin x}{x}$ approaches $1$. In more detail, $\frac{\sin x}{x}$ gets arbitrarily close to $1$. Pick a tiny number, say $0.000001$. Provided that $x$ is sufficiently close to $0$, the difference between $\frac{\sin x}{x}$ and $1$ will be smaller than that tiny number. This helps us state the precise definition of limit:

We say that the limit as $x$ approaches $0$ of $\frac{\sin x}{x}$ equals $1$ because we can make $\frac{\sin x}{x}$ arbitrarily close to $1$ by requiring that $x$ is sufficiently close to, but unequal to, $0$.

Hopefully, this is the gist of the definition of a limit that you were looking for. And the precise definition definition with epsilon's and delta's is saying essentially the same thing. It just clarifies our wordy description even further.
Note that the last section of our wordy definition was

...requiring that $x$ is sufficiently close to, but unequal to, $0$.

As we mentioned earlier, limits tell us about what happens we get close to, but never actually reach a certain value. In a way, the limit tells us what we anticipate when we get close to a value, rather what actually happens there. The function $\sin x / x$ is undefined when $x=0$. It makes no sense to actually ask what happens when we 'get to zero'. But we can make meaningful statements about what happens when we get close to zero. Let's go back to your example,
$$
\lim_{x \to 3}x^2 \, .
$$
If limits are all about what happen when we get close to $3$, then why can we simply plug in $x=3$? This is because the function $x^2$ (and all polynomial functions in general) are continuous. A function is said to be continuous at the point $x=a$ if
$$
\lim_{x \to a}f(x) = f(a) \, .
$$
In other words, the anticipated value is the same as the value actually obtained by plugging in $x=a$. Many functions do not have this property. In the earlier example,
$$
\frac{\sin x}{x}
$$
is not even defined at $x=0$, and so it doesn't even make sense to ask whether it is continuous there. Nevertheless, it can be shown that all polynomial functions are continuous everywhere, and so evaluating the limit
$$
\lim_{x \to a}P(x)
$$
where $P$ is an arbitrary polynomial, simply amounts to plugging in $x=a$. This is what is so confusing to newcomers to calculus. Although limits tell you about the behaviour of a function as you get close to a point, a lot of the time you can take a shortcut and just plug in the value being approached. This is because, with a little work, we can show that the 'elementary functions' (polynomials, the exponential function, sine and cosine, etc.) are continuous at the point $x=a$ if they are defined there. But don't let that mislead you. When we say that $\lim_{x \to a}f(x) = L$, we are still talking about the behaviour of a function as you get close to, but unequal to $a$; however, in some instances it is possible to work out this limit simply by plugging, provided that you know beforehand that the function is continuous.


答: You dont need anything involved. $A=I-N$ with $N$ an integer matrix satisfying $N^n=0$. Then by the geometric series $A^{-1}=I+N+N^2+\cdots+N^{n-1}$. Of course $N=I-A$ so any polynomial in $N$ is also a polynomial in $A$.</p>

Body:
Consider a matrix with integer entries such that $a_{ii}=1$ and $a_{ij}=0$ for $i>j$

Then

$A^{-1}$ exists and it has integers entries.

$A^{-1}$ exists and it has some entries which are not integers.

$A^{-1}$ exists and it is a polynomial function in $A$ with integers coefficient.

$A^{-1}$ is a not a power of $A$ unless $A=I$

As $\det A=1$ so $A$ inverse exist and $A^{-1}=\dfrac{\operatorname{Adj} A}{\det A}$ clearly $A^{-1}$ will have all integer entries, $2$ is false,$c_nA^n+\dots +I=0\Rightarrow A^{-1}=-c_nA^n+\dots+ c_{n-1}A$ as $\det A=c_0=1$ so $3$ is true, say $A^{k-1}=A^{-1}$ then $A^k-I=0$ so minpoly must divide $x^k-1$ and as its charpoly is $(x-1)^n=0$ so minpoly must divide $(x-1)^n$ too but that forces minpoly is $(x-1)$ and hence $A=I$. is my all logics are  okay?


问:
Proof of $\tilde{K}^*_G((G/H)_+\bigwedge X)\cong \tilde{K}^*_H(X)$
I have been reading the paper "A generalization of the Atiyah-Segal completion theorem" by Adams, Haeberly, Jackowski and May, where I found the Following claim:

Let $G$ be finite group, $H\subset G$ a subgroup. Let $X$ be a compact pointed space, with $G$ acting on it. Then there is an isomorphism $$\tilde{K}^*_G((G/H)_+\bigwedge X)\cong \tilde{K}^*_H(X)  $$

I know a similiar statement to be true from Segals paper on equivariant $K$-theory, namely that $K_G(G/H\times X)\cong K_H(X)$, but that is only in degree zero.

I suspect the more general version can be proven by noticing that $(/)_+⋀\cong (G/H\times X)/(G/H)$ and using the LES of $K$-theory, but it did not work out so far.

Can anyone give me reference where this more general version is proven, or provide some hints how to deduce it from the version I know to be true?

答:
It suffices to prove this degree 0 since $\tilde{K}^{-q}_G((G/H)_+\bigwedge X):=\tilde{K}_G((G/H)_+\bigwedge (X\bigwedge S^q))$.

The crucial observation is that $G/H\to G/H\times X$ is an equivariant retract, so that the LES for the pair $(G/H\times X,\,G/H)$ falls apart into short exact sequences.

Then the "obvious" map $\tilde{K}_G((G/H)_+\bigwedge X)\to\tilde{K}_H(X)$ can be seen to be an isomorphism from the snake lemma and the already known isomorphisms $K_G(G/H\times X)\to K_H(X)$ and $K_G(G/H)\to K_H(*)$.


答:
You may integrate directly
$$\int_{0}^{\pi/2}\frac{1-\sin^3a}{\cos^2a} da
= \int_{0}^{\pi/2}\frac{(1-\sin a)( 1+\sin a+\sin^2a)}{1-\sin^2a} da\\
=\int_{0}^{\pi/2}\left( \frac1{1+\sin a}+\sin a\right) da
 = \int_{0}^{\pi/2}\left( \frac1{1+\cos a}+\cos a\right) da\\
 = \int_{0}^{\pi/2}\left( \frac12\sec^2\frac a2+\cos a\right) da
 = \left( \tan \frac a2+ \sin a \right)_0^{\frac\pi2}=2
$$

Title: How to rearrange the following trigonometric integral

Body:
How do you rearrange the following;

$\frac{1}{R_0}\int_{0}^{\pi/2}\frac{1-sin^3(\alpha)}{cos^2(\alpha)} d\alpha$

using t=$\tan\frac{\alpha}{2}$

to obtain

$\frac{1}{R_0}\int_{0}^{1}[\frac{2}{(1+t)^2}+4\frac{t}{(1+t^2)^2}]dt$

I do not understand how this rearrangement was done

The final answer to the integral is $\frac{2}{R_0}$

Many thanks


答:
Let me put this as an answer to bring all the comments together and make things clear:

As a concrete example $f(z)=\sin \frac{\pi}{1-z}$ is the easiest example of a (nontrivial) holomorphic function on the unit disc that has infinitely many zeroes $z_n=1-1/n$. However $\sum (1-(1-1/n))=\infty$ and that by a general result forces $f$ to be unbounded which is clear since it has an essential singularity at $1$.

If the zeroes would be say $1-1/n^2$ or any $z_n$ with $\sum (1-|z_n|) < \infty$ one can actually construct a bounded function on the unit disc with zeroes precisely at $z_n$ (Blaschke products) and that is one of the simplest theorem of the type $f$ satisfies some growth condition (here bounded on the unit disc), then its zeroes satisfy some growth restrictions (here $\sum (1-|z_n|) < \infty$) and those are sufficient.

Going even more general, we know that given any sequence in the plane $z_n$ with no (finite) accumulation point, one can construct an entire function $f$ with zeroes precisely at $z_n$ and then there is a factorization theorem that tells us that all such functions are of a specific form (though one needs growth restrictions like finite order for the theorem to be useful as otherwise there are too many choices) so one can ask same about any domain the plane, bounded or not and indeed the analogous result is true though it requires some subtle topological arguments in full generality, while the factorization theorem is even less useful outside specific cases like the unit disc and bounded functions say.

So given any domain (open connected) $G$ and any sequence $z_n$ with no accumulation point in $G$, there is $f$ holomoprhic on $G$ with zeroes at precisely $z_n$

Title: Complex function with infinitely many zeros on bounded domain

Body:
I found a similar post that discussed complex functions that have infinitely many complex zeros here:
Examples of complex functions with infinitely many complex zeros

I was wondering, if one of these functions is in a bounded domain (and its infinitely many zeros are also in the bounded domain) does f have to be identically $0$? Or can it be a non-constant function still?


问:
Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$
I got this question in my maths paper

Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
  and find the sum if it exists.

I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.

答:
Let me add a more general answer.

Note that
$$
{1 \over {\left( {x + 1} \right)\left( {x + 2} \right) \cdots \left( {x + m} \right)}}
 = {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }}
 = {{\Gamma \left( {x + 1} \right)} \over {\Gamma \left( {x + 1 + m} \right)}}
 = x^{\,\underline {\, - m\,} }
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.

In virtue of the definition through the Gamma function, the above is valid for $x,m \in \mathbb C$.

Then we apply the Indefinite Sum concept, by which
$$
\eqalign{
  & \Delta _{\,x} \;x^{\,\underline {\,q\,} }
  = \left( {x + 1} \right)^{\,\underline {\,q\,} }  - x^{\,\underline {\,q\,} }
  = qx^{\,\underline {\,q - 1\,} } \quad  \Rightarrow   \cr
  &  \Rightarrow \quad \Delta _{\,x} ^{\left( { - 1} \right)} \;x^{\,\underline {\,q\,} }
  = \sum\nolimits_x {x^{\,\underline {\,q\,} } }  = \left\{ {\matrix{
   {{1 \over {q + 1}}\;x^{\,\underline {\,q + 1\,} }  + c} & { - 1 \ne q}  \cr
   {\psi (x + 1) + c} & { - 1 = q}  \cr
 } } \right. \cr}
$$
which is to recall that it is valid for $x,q,c \in \mathbb C$.

So in particular, for $q=-m$ and $m \ne 1$, the Indefinite sum becomes
$$
\sum\nolimits_x {x^{\,\underline {\, - m\,} } } \quad \left| {\,1 \ne m} \right.
 = {1 \over { - m + 1}}x^{\,\underline {\, - m + 1\,} }  + c
$$
which means that for the definite sum we get
$$
\eqalign{
  & \sum\limits_{x = a}^b {x^{\,\underline {\, - m\,} } }
  = \sum\nolimits_{x = a}^{b + 1} {x^{\,\underline {\, - m\,} } } \quad \left| \matrix{
  \,m,a,b \in C \hfill \cr
  \;1 \ne m \hfill \cr}  \right. =   \cr
  &  = {1 \over { - m + 1}}\left( {\left( {b + 1} \right)^{\,\underline {\, - m + 1\,} }
  - a^{\,\underline {\, - m + 1\,} } } \right) =   \cr
  &  = {1 \over {m - 1}}\left( {a^{\,\underline {\, - \left( {m - 1} \right)\,} }
  - \left( {b + 1} \right)^{\,\underline {\, - \left( {m - 1} \right)\,} } } \right) =   \cr
  &  = {1 \over {m - 1}}\left( {{1 \over {\left( {a + 1} \right)^{\,\overline {\,m - 1\,} } }}
 - {1 \over {\left( {b + 2} \right)^{\,\overline {\,m - 1\,} } }}} \right) \cr}
$$
and taking the limit $b \to \infty$
$$
\sum\limits_{x = a}^\infty  {x^{\,\underline {\, - m\,} } } \quad \left| {\,\;1 < m \in R} \right.
 = {1 \over {\left( {m - 1} \right)m!}}
$$

Finally, in your particular case, the summand, the indefinite, definite and infinite sums are
$$
\eqalign{
  & {1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} = {1 \over {n^{\,\overline {\,3\,} } }}
 = \left( {n - 1} \right)^{\,\underline {\, - 3\,} }   \cr
  & \sum\nolimits_n {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_n {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr
  &  =  - {1 \over 2}\left( {n - 1} \right)^{\,\underline {\, - 2\,} }  + c
 =  - {1 \over {2n^{\,\overline {\,2\,} } }} + c =  - {1 \over {2n\left( {n + 1} \right)}} + c  \cr
  & \sum\limits_{n = 1}^m {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = \sum\nolimits_{n = 1}^{m + 1} {\left( {n - 1} \right)^{\,\underline {\, - 3\,} } }  =   \cr
  &  =  - {1 \over 2}\left( {{1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}
 - {1 \over {1 \cdot 2}}} \right) = {1 \over 2}\left( {{1 \over 2}
 - {1 \over {\left( {m + 1} \right)\left( {m + 2} \right)}}} \right)  \cr
  & \sum\limits_{n = 1}^\infty  {{1 \over {n\left( {n + 1} \right)\left( {n + 2} \right)}}}
  = {1 \over 4} \cr}
$$


问:
How to prove this logarithmic function?
\begin{equation}
-\log\left(\frac{(\sqrt{1+x^2}-1)}{x}\right) = \frac{1}{2}\log\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)
\end{equation}

You are not given the R.H.S . How to prove this?

Title: How to prove this logarithmic function?

Body: By the rules for logarithms we have $-\log(\frac{\sqrt{1+x^2}-1}{x})=\log(\frac{x}{\sqrt{1+x^2}-1})=\log((\frac{x}{\sqrt{1+x^2}-1}^2)^{\frac{1}{2}})=\frac{1}{2}\log(\frac{x^2}{(\sqrt{1+x^2}-1)^2})$. Now we simplify and use a little trick: $\frac{1}{2}\log(\frac{x^2}{(\sqrt{1+x^2}-1)^2})=\frac{1}{2}\log(\frac{1+x^2-1}{(\sqrt{1+x^2}-1)^2})=\frac{1}{2}\log(\frac{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)}{(\sqrt{1+x^2}-1)^2})=\frac{1}{2}\log(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1})$.


答:
In case 1, note that $1\lt|z|\lt2$ means that $1/|z|\lt1$ and $|z|/2\lt1$, hence the well known expansion
$1/(1-u)=\sum\limits_{n\geqslant0}u^n$, valid for every $|u|\lt1$, applied twice, yields
$$
\frac1{z^2+1}=\frac1{z^2(1+1/z^2)}=\sum\limits_{n\geqslant0}\frac{(-1)^n}{z^{2n+2}},$$
and
$$
\frac1{z-2}=-\frac12\frac1{1-z/2}=-\sum\limits_{n\geqslant0}\frac{z^n}{2^{n+1}}.
$$
The expansion of $f_1(z)$ as a Laurent series follows from the decomposition
$$
f_1(z)=\frac1{z-2}-\frac2{z^2+1}.$$</p>

Title: Laurent Series Expansion Problems

Body:
Expand in a laurent Series :
1- $f_{1} (z) =  \frac{z^{2} - 2z +5 }{(z^{2}+1) (z-2)}$

in the ring : $1 < |z| < 2 $

2- $ f_{2} (z) =  \frac{1 }{(z-3) (z+2)}$

In :
$i. 2 < |z| < 3
\\ ii. 0 < |z+2| < 5$

I managed to solve the second one but not sure if it is correct
For i. $2 < |z| < 3$ :

$ \frac{-1}{5} *  \frac{1}{z(1+ \frac{2}{z}) } +  \frac{1}{5} *  \frac{1}{-3(1- \frac{z}{3}) }
 = \frac{-1}{5} \sum_{n=0}^ \infty  (-1)^{n}  (\frac{2}{z})^{n} -  \frac{1}{15}\sum_{n=0}^ \infty    (\frac{z}{3})^{n}$

For ii. $0 < |z+2| < 5$ :
$ \frac{-1}{5} *  \frac{1}{z+2} +  \frac{1}{5} *  \frac{1}{(z+2 -5) }
 = \frac{-1}{5} *  \frac{1}{z+2} +  \frac{1}{25} *  \frac{1}{-5 (1- \frac{Z+2}{5} ) } \\ = \frac{-1}{5} *  \frac{1}{z+2}-  \frac{1}{25} \sum_{n=0}^ \infty (\frac{z+2}{5})^{n}      $


Title: How to prove this logarithmic function?

Body:
\begin{equation}
-\log\left(\frac{(\sqrt{1+x^2}-1)}{x}\right) = \frac{1}{2}\log\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)
\end{equation}

You are not given the R.H.S . How to prove this?

Body:
$\color{blue}{\dfrac{\sqrt{1+x^2}-1}{x}} = \dfrac{\sqrt{1+x^2}-1}{x}\cdot\dfrac{\sqrt{1+x^2} + 1}{\sqrt{1+x^2} + 1} = \color{blue}{\dfrac{x}{\sqrt{1+x^2} + 1}}$

Now if we square $\frac{\sqrt{1+x^2}-1}{x}$, we can get rid of $x$

$\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)^2 =\dfrac{\sqrt{1+x^2}-1}{x}\dfrac{x}{\sqrt{1+x^2} + 1} =\dfrac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}$

$\Rightarrow2\ln\left(\dfrac{\sqrt{1+x^2}-1}{x}\right) = \ln\left(\dfrac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)$

or

$-\ln\left(\dfrac{\sqrt{1+x^2}-1}{x}\right) = -\frac12\ln\left(\dfrac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right) = \frac12\ln\left(\dfrac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)$


问:
Is there any way to avoid Gödel incompleteness theorem?
I was wondering whether there have been successful attempts to avoid somehow the incompleteness theorem.
Many point out that every language that is powerful enough to model natural numbers can produce a godel-type sentence. But my point is that it's actually a METAmathematical strategy that Godel uses (as he says in the first pages of his paper). Arithmetic doesn't contain itself the predicate 'provable'. So basically I'm asking: can we avoid such inconsistencies just forbidding to represent any predicate like 'provable' unless moving to a higher-predicate metalanguage?</p>

答:
As Don Thousand and Mauro ALLEGRANZA pointed out, there are ways to circumvent the need for a provability predicate. A systematic method to go about constructing true but unprovable statements would use that every theory has an ordinal strength $\alpha$, then construct a set $S$ which is well ordered, with order type $\beta\ge\alpha$. Then, encode the statement "$S$ is well ordered" into that system. This gives a true but unprovable statement. An example of this can be found at https://arxiv.org/abs/2003.14342. I know of two ways to circumvent Gödel's incompleteness theorem:

A weak theory. Gödel's incompleteness theorem only applies to sufficiently strong systems. It is not applicable to systems like Presburger arithmetic and first-order logic.

An inconsistent theory. This pretty much goes without saying, see principle of explosion


问:
Uniform convergence of a sum of integral functions
I have to study punctual and uniform convergence of
$$g(x):=\sum_{n=1}^{\infty}\int_n^{n+x}\dfrac{dt}{e^{t^3}}.$$
If I define $g_n(x)=\int_n^{n+x}\frac{dt}{e^{t^3}}$, in order to study the punctual convergence of $g(x)$, I have to compute $\lim_{n\to \infty}g_n$ with $x\in \mathbb R$ fixed. Since for $n\to \infty$, $n+x\sim n$ $\forall x\in \mathbb R$ I wrote
$$\int_n^{n+x}\dfrac{dt}{e^{t^3}}=\int_0^{n+x}\dfrac{dt}{e^{t^3}}-\int_0^{n}\dfrac{dt}{e^{t^3}}$$
which (I think) tends to zero and I'd say that the limit function of $g_n$ is the zero function, but I'm not sure about this. If I want to deduce the convergence of the series I should find a function asymptotic to $g_n$. The sum has form $\int_1^{1+x}\frac{dt}{e^{t^3}}+\int_2^{2+x}\frac{dt}{e^{t^3}}+\dots$, so how could I get informations about the convergence of the series? I don't understand how to do with the $n+x$ term because for $n<\infty$ it can assume any real value.

I thank you in advance for your help.

答:
It converges uniformly on any set $[a; +\infty)$, but doesn't converges uniformly on $\mathbb R$.

It's enough to check case $a < 0$. If $n > -a + 1$, $|g_n(x)| = \int\limits_{n + a}^n \frac{dt}{\exp(t^3)} < \int\limits_{n + a}^n e^{-t}\,dt < e^{-n - a}$. Taking $M_n = e^{-n - a}$ note that $\sum\limits_{n=0}^\infty M_n$ converges, thus $\sum g_n(x)$ converges uniformly by Weierstrass M-test.

It doesn't converge uniformly on $\mathbb R$, because $g_n(-n - 1)$ doesn't even converges to $0$.


问:
By choosing a suitable binomial expansion of the form $(1±kx)^n$ and a suitable value of x. Find $(1.01)^{20}$ correct to 3 s.f.
I have tried replacing kx by 0.1, square rooting 0.1 and squaring 0.1, but none of these methods worked.
I don’t know where to start from. I would appreciate some assistance on this question. Thanks</p>

答:
The accepted answer has a major flaw in thinking that, if a real number $s$ and "true" value $S$ differs by less than 0.01 then they both round to the same 2dp numbers.  But this is easily shown to be false, e.g., $1.005-10^{-12}$ and $1.005$ differs by $10^{-12}\ll 0.01$ but rounds to different numbers to 2dp (or 3sf).  In other words,

Calculate $x$ to within error smaller than $10^{-n}$ (or half that).

and

Calculate $x$ to n decimal places

are different problems.

So you need to keep track of both the current partial sum and potential error.  Only if you can guarantee everything in range round to the same decimal number can you stop.

So:

First term 1, error $0<\varepsilon<\frac14$, potential range $(1,1.25)$ is too broad.

Two term sum $1+\binom{20}{1}\cdot 0.01=1.20$, error $0<\varepsilon<\frac14\cdot 0.2=0.05$, potential range $(1.20,1.25)$ again too broad

Three term sum $1+\binom{20}{1}\cdot 0.01+\binom{20}{2}\cdot 0.01^2=1.219$, error $0<\varepsilon<\frac14\cdot 0.019=0.00475$, potential range $(1.219, 1.22375)$ falls under what is represented by 1.22 to 3sf.


问:
Fundamental group of the union of the unit circle, $X$-axis, and $Y$-axis
I have been solving some past exam questions and I came across the following question.

Let  $X:= S_1 \cup (\mathbb R \times \{0\})\cup (\{0\}\times \mathbb R)$ be the union of the unit circle, the $X$-axis and the $Y$-axis. Compute the fundamental group of $X$ in terms of generators and relations.

I am not really sure how to go about this. My guess is that if I could find a retract of this space whose fundamental group is known then the question is solved. I am also wondering if the Seifert-van Kampen theorem would work but I have no clue. Any help would greatly be appreciated.

答:
Your space is $X=S^1\cup(\mathbb R\times \{0\})\cup(\{0\}\times\mathbb R)$, which is the union of the circle of radius $1$ centered in $(0,0)$ with the $x$ and $y$ axis.

The space $X$ is homotopic to $S^1\cup (\{0\}\times I)\cup(I\times \{0\})$, so a circle with two diameters.

Our space has now the form $\oplus$ and if we identify the two diameters with a point we obtain the wedge sum of four circles, whose fundamental group is the free product on four generators with no relations: $\langle a,b,c,d|\emptyset \rangle$.

Therefore $\pi_1(X)\cong\pi^1(S^1\vee S^1 \vee S^1 \vee S^1)\cong\mathbb Z*\mathbb Z*\mathbb Z*\mathbb Z$.


问:
Construct a set of local rules in plane such that all permitted tilings are non periodic.(there exists at least one tiling)
This is the problem given to me on a practice.

I imagine the problem is understandable but if there is anything unexplained, please tell me.

I have tried various things to solve this problem but all of them lead to a dead-end, and so far I don't think I have made any good progress to share.

Here are some definitions I think would be helpful to write.

"local rules: a set of $n.n$ squares which are forbidden in the tiling ($n$ is not a constant and can be any positive integer greater than $1$)",

"non periodic : means there is not a non_zero vector such that after adding the tiling to that vector we reach the same tiling",and the plane we are tiling the plane using $1.1$ black or white squares

答:
You can use a set of aperiodic Wang tiles as a starting point.

Wang tiles are normally depicted with colours, but you can choose to represent each tile by almost any n by n pattern of black and white cells, as long as they are unique enough.

Then enumerate all ways that you are allowed to make a patch of four Wang tiles which form a 2n by 2n square of black and white cells. List all n by n sub-squares of these allowed patches, giving you all the posssible n by n patterns that you will allow in the tiling as a whole. Then disallow all other n by n patterns.

This shows it can be done, though the list of local rules will be very long.


问:
Maximum distance from a point in hyperbola $ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes
The canonical hyperbola is
$$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

I have already derive the equation for the distance from any point $(x_0,y_0)$ in hyperbola to the asymptote $ y=\frac{b}{a}x$ as
$$ d =\frac{|bx_0-ay_0|}{c}$$
where
$a^2+b^2=c^2$ and $x_0 \geq a$.

I know that the maximum distance is when $x_0=a$ but I don't know how to proof it. Can you help me? Thanks.

答:
By definition distance to an asymptote  goes to zero for a point P on a hyperbola in first quadrant.

The asymptotes are a pair of straight lines that include $  2\alpha $ between them where $\tan \alpha = \frac {b}{a},$ seen in standard Wiki reference picture below.

Reckon distance $s$ from center ( intersection of asymptotes) in second quadrant. By dropping a perpendicular on the second asymptote from P we have  distance  $ p=s \tan 2\alpha $ which tends infinity as $s \to \infty.$

These two distances  green and blue can be taken as minimum and maximum distances respectively as P tends to the asymptotic limits.

( btw their product is finite.. which matter is outside this query).


问:
Basic question about cyclic groups and Galois theory
Let $F/K$ be a finite extension and suppose $F$ is the splitting field of a separable polynomial over $K$. Show that if Gal(F/K) is cyclic then for each divisor $d$ of $[F : K]$ there exists exactly one subfield $E$ of $K \subseteq E \subseteq F$ such that $[E : K]=d$.

Is this actually true? what I get is that one can find a subfield $E$ of $F$ such that $[F : E] = d$.

My work:

First, since $F/K$ is Galois then $[F : K] = |Gal(F/K)|$ so that if $d$ divides $[F : K]$ then d divides $|Gal(F/K|)$. By assumption the latter group is cyclic so for each divisor we can find a unique subgroup $H$ of $Gal(F/K)$ such that $|H| = d$.

But now the fundamental theorem implies $[F : F^{H}] = |H| = d$ so that $E=F^{H}$. I get then that $[F : E] = d$ and  not $[E : K]=d$. What am I doing wrong?

答: Posted as an answer by request: Let $n=[F:K]$, and do your argument for $n/d$. </p>


问:
Uniform convergance for $f_n(x)=x^n-x^{2n}$
the function $f_n(x)=x^n-x^{2n}$ converge to $f(x)=0$ in $(-1,1]$. Intuativly the function does not converge uniformally in (-1,1]. How can I prove it?
I tried using the definition $\lim \limits_{n\to\infty}\sup \limits_{ x\in (-1,1]}|f_n(x)-f(x)|$ function is continial fractional on $[-1,1]$ and $x=0,(\frac 1 2 )^{\frac 1 n}$ are the roots of the derivative. I found that the second derivative is negative in the second point. then $\sup=1/4$ and the function does not converge uniformally?</p>

答:
$f$ doesn't converge uniformly in $(-1,1]$ since
$$\lim_{x\to-1}\lim_{n\to\infty}f_n(x)=0\neq\lim_{n\to\infty}\lim_{x\to-1}f_n(x)=\lim_{n\to\infty}(-1)^n-1$$</p>


问:
Question about integral on the complex plane
That is the infernal integral:

$\int_{|z|=2}{\frac{e^{z+\frac{1}{z}}}{1-z^2}}dz$

I tried to solve it with residuals, but I can't find the expression of the function in its Laurent form.
Sorry if the question is already posted, i don't speak the language well yet.

答:
$\def\res{\operatorname{res}}$
A simple analysis shows that the function
$$f(z)=\frac{e^{z+\frac{1}{z}}}{1-z^2}
$$
has four singularities: $-1,1,0,\infty$.

Since the function $f(z)$ is everywhere else holomorphic the sum of the residues at these points is zero:
$$
\res(f(z),-1)+\res(f(z),1)+\res(f(z),0)+\res(f(z),\infty)=0.\tag1
$$

The important observation is:
$$
\res(f(z),\infty)=-\res\left(z^{-2}f(z^{-1}),0\right)
=-\res\left(\frac{e^{\frac{1}{z}+z}}{z^2-1}\right)=\res(f(z),0).\tag2
$$

Combining (1) and (2) one easily obtains the final result:

$$\frac1{2\pi i}\oint_{|z|=2}f(z)\,dz=-\res(f(z),\infty)=\frac{\res(f(z),-1)+\res(f(z),1)}2=-\frac{\sinh(2)}2.$$


问:
Integer solution to $x^4 + x^3 = y^4 +7$

Find an integer solution or show there are none of $$x^4+x^3=y^4+7$$

I have found using python that there are no solutions for x, y less than 100

but also that the equation has at least one solution modulo n for every n less than 100 so I can't attack in the usual way by showing it has no solutions mod n for some n.

Could someone suggest an alternate method/ approach I could use?

(not looking for a straight solution just some hints or tips)

Body:
Hint: $$x^4< x^4+x^3-7\leq (x+1)^4\iff x\geq 2$$

 If $x<0$ then $x=-t$ for $t>0$ and then $$x^4+x^3-7 = t^4-t^3-7$$

Now for $t> 2$ we have $$(t-1)^4< t^4-t^3-7 < t^4$$

 So you have to check for $x\in\{-2,-1,0,1,2\}$


问:
Equivalence relation on graphs identifying degrees
I consider a connected simple finite graph $G$ and I define the equivalence relation $v\sim w$ on its vertex set such that $v\sim w$ if and only if $\mathrm{deg}(v)=\mathrm{deg}(w)$. I found that it is possible that the quotient graph of $G$ under this equivalence relation may be isomorph to the path graph of length $\delta$ where $\delta$ is the number of distinct degrees in $G$. Ordering the degrees by $d_1<...<d_{\delta}$ the existence of such an isomorphism implies that any vertex of degree $d_i$ in $G$ is only connected to vertices with degree $d_{i\pm 1}$.

But can I say more about $G$? Is there a general class of graphs which satisfies this property?

I am thankful for any input I can get. I am going to use it for Markov chains because I managed to prove that a certain Markov chain is reversible if and only if said isomorphism exists for the hereinbefore defined quotient graph.

答:
Well, one thing we can do is ask for more structure. Write each $d_i$ as $a_i + b_i$ and ask that each vertex of degree $d_i$ is adjacent to $a_i$ vertices of degree $d_{i-1}$ and $b_i$ vertices of degree $d_{i+1}$. For the first degree $d_1$ we want $a_1 = 0$ and for the last degree $d_k$ we want $b_k = 0$; for the rest, we can choose $a_i$ and $b_i$ arbitrarily, but we want $a_i, b_i > 0$.

Then if there are $n_1, n_2, \dots, n_k$ vertices of degree $d_1 < d_2 < \dots < d_k$ respectively, we must have $n_i b_i = n_{i+1} a_{i+1}$ for the edge counts to work out, and we can do this just by having the proportion $n_1 : n_2 : \dots : n_k$ be right. We should scale up the proportion so that $n_i \ge \max\{b_{i-1}, a_{i+1}\}$ for all $i$, and then constructing the graph is possible...

...we just need a method for constructing a bipartite graph with $n_i$ vertices of degree $b_i$ on one side, and $n_{i+1}$ vertices of degree $a_{i-1}$ on the other side.

Here's one way. Just order the $n_i$ vertices on the first side cyclically. Give the first vertex on the other side edges to the first $a_{i-1}$ of them, then the next vertex edges to the next $a_{i-1}$, and so on, wrapping around when necessary. This will work out in the end. Actually, there are usually many ways to do this.

There are probably other graphs we can construct that don't have the regular $d_i = a_i + b_i$ split, but those are harder to find.


问:
Prove inequality involving integral
The exercise consists of proving the following:
$$\frac{4}{21}\cdot 2^{3/4}  \leq \int_{0}^{\sqrt{\frac{\pi}{2}}}t^{2}\cos(t^{2})dt$$

With $1-\frac{x^{2}}{2} \leq \cos(x) , \forall  x\in [0,\pi ]  $ (already proven, got it from Taylor polynomials) my idea was to use something like this $x^{2} (1-\frac{x^{4}}{2})  \leq \int_{0}^{\sqrt{\pi /2}}t^{2}\cos(t^{2})dt$. The problem is I don't see any of either $4/21$ or $2^{3/4}$ showing up.

Any hint or tip on this?

Edit:

I have the solution for this but it seems odd to me.
Basically it goes from this

$$cos(x^{2}) \geq 1-\frac{x^{4}}{2}, \ for \ 0 < x < 2^{1/4} \\ cos(x^{2}) \geq 0, \ for \ 2^{1/4} \leq x \leq \sqrt{\pi / 2}$$

and then it proceeds with integration (but from there it's easy). These steps look random to me, plus $cos(x^{2})$ is always greater than $1-\frac{x^{4}}{2}$ so why separate into that two cases?

答:
For simplicity, set $u=t^2$: $$\int_{0}^{\sqrt{\frac{\pi}{2}}}t^2\cos(t^2)~dt = \dfrac{1}{2}\int_0^{\frac{\pi}{2}}\sqrt{u}cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\cos(u)~du \geq \dfrac{1}{2}\int_{0}^{\sqrt{2}} \sqrt{u}\left(1 - \frac{u^2}{2}\right)~du$$

And I'm sure you can take it from here :)

Title: Prove inequality involving integral


问:
Equivalence of two seminorms
Let us consider the space $C^0([0,1])$ of real continuous functions over $[0,1]$.

Let $A = \{a_n : n \in \mathbb{N}\}$ be a countable set of $[0,1]$ and $\alpha_n$ be strictly positive real numbers such that $\sum_{n=0}^{+\infty} \alpha_n < +\infty.$ Now we define a seminorm by $$||f||_{A,\alpha} = \sum_{n=0}^{+\infty} \alpha_n |f(a_n)|.$$

If we have two such seminorms $||.||_{A,\alpha}$ and $||.||_{A',\alpha'}$ which are equivalent, I would like to show that necessarily $A=A'.$ I managed to obtain $\overline{A} = \overline{A'}$ but I have difficulties to obtain the desired equality.

答:
We argue by contradiction. Assume that $a_n \notin A'$ for some $n$. Let $f_{n, \delta} = \max \{ 0, 1- \delta^{-1}|x-a_n|\}$ (It's $1$ at $a_n$ and is zero outside $[a_n - \delta, a_n+\delta]$). Note that

$$\| f_{n, \delta}\|_{A, \alpha}\ge \alpha_n.$$

For each $\epsilon >0$, let $M_\epsilon$ be large so that
$$\sum_{M_{\epsilon}}^\infty \alpha_n' <\epsilon.$$

Then find $\delta$ small enough so that $|a_m' - a_n| > \delta$ for all $m=1, \cdots, M_\epsilon$ (here we used that $a_n \neq A'$). Then
$$ \|f\|_{A', \alpha'} = \sum_{M_\epsilon}^\infty |f(a_n')| \le \epsilon.$$

Thus for all $\epsilon>0$ we have found continuous functions so that $\| f_{n, \delta}\|_{A, \alpha}\ge \alpha_n$ and $\| f_{n, \delta}\|_{A', \alpha'}\le \epsilon$. Thus the two seminorms cannot be equivalent.

Title: Equivalence of two seminorms


问:
How many integers in the form $q^k$ mod by $p^d q$ for distinct prime $p$ and $q$?
Let $p$ and $q$ be distinct primes, and $d$ be a positive integer.
There are only finite many (at most $p^d q$) distinct integers in the form $q^k$ with non-negative integer $k$ mod by $p^d q$.
But how many of them?

When $p=2$ and $q=3$, I have calculated for $d=1, 2, \dots, 16$, and the answer is $2, 3, 3, 5, 9, 17, 33, 65, \dots, 16385$.
That is, the answer seems to be $2^{d-2} +1$ for all $d$ except $1$ and $2$.

Is this true for the case $p=2$ and $q=3$? And how about for general $p$ and $q$?

Thank you very much for your attention!

答:
Your answer for $p=2,q=3$ is correct.

There is the solution $q^0=1$ and then there are the solutions $3^{k+1}$ modulo $2^d3$ i.e. $3^k$ modulo $2^d$.

Now the least power $3^k\equiv 1 \pmod 8$ is $k=2$.
For $d>3$, by Hensel lifting,

https://en.wikipedia.org/wiki/Hensel%27s_lemma

the least power  $3^k\equiv 1 \pmod {2^d}$ is $2^{d-2}$.

Hence we get $2^{d-2}+1$ solutions:
$$1,3,3^2, ...,3^{2^{d-2}}.$$

Extension

Other values of $p$ and $q$ can be handled in the same way - give it a go!

In fact, for $p\ne 2$ the Hensel lifting is marginally easier since you can start with the solution for  $q^k\equiv 1 \pmod p$. (For $p=2$ we needed to start at modulo $p^3$.)


问:
Tournament bracket for a 4-player game and 13 players in total
We have a tournament in which 13 players participate, and matches are played in groups of 4 players (i.e. it's a 4-player game).

There are 13 rounds in total, each player skips one round, and the other 12 are divided into three matches of 4 competitors.

So, for example, if we denote players by 1-13:
in round 1, player 1 skips that round, 2-3-4-5 is one match, 6-7-8-9 the other one, and 10-11-12-13 the last one.

In the end, every player would play in 12 (out of 13) rounds.

Is it possible to make a bracket so that every player plays every opponent exactly three times in total across the duration of 13 rounds?

(In the example above, in Round 1, player 2 played once against players 3, 4 and 5, and zero times against all of the others.)

答:
How about this:

Round $1$: $(2, 3, 4, 5), (6, 7, 8, 9), (10, 11, 12, 13)$

Round $2$: $(1, 4, 5, 7), (3, 9, 12, 13), (6, 8, 10, 11)$

Round $3$: $(1, 5, 6, 11), (2, 9, 10, 13), (4, 7, 8, 12)$

Round $4$: $(1, 3, 8, 12), (2, 7, 11, 13), (5, 6, 9, 10)$

Round $5$: $(1, 2, 4, 9), (3, 7, 10, 11), (6, 8, 12, 13)$

Round $6$: $(1, 3, 9, 11), (2, 7, 10, 12), (4, 5, 8, 13)$

Round $7$: $(1, 2, 6, 12), (3, 5, 8, 10), (4, 9, 11, 13)$

Round $8$: $(1, 7, 9, 10), (2, 5, 6, 13), (3, 4, 11, 12)$

Round $9$: $(1, 7, 8, 13), (2, 5, 11, 12), (3, 4, 6, 10)$

Round $10$: $(1, 2, 8, 11), (3, 5, 7, 13), (4, 6, 9, 12)$

Round $11$: $(1, 3, 6, 13), (2, 4, 8, 10), (5, 7, 9, 12)$

Round $12$: $(1, 4, 10, 13), (2, 3, 6, 7), (5, 8, 9, 11)$

Round $13$: $(1, 5, 10, 12), (2, 3, 8, 9), (4, 6, 7, 11)$

Edit #1:

To see how to come up with this solution, consider the image below:

This image depicts the first round: The white node labelled "1" skips the first round while the three teams are made up of the blue nodes $2,3,4,5$, the green nodes $6,7,8,9$ and the red ones $10,11,12,13$.

In the second round, all quadrilaterals are rotated clockwise by $1/13$ turn (or, equivalently, the node labels are rotated counter-clockwise), so that node "2" gets to skip the round. The new blue team will then consist of nodes $11$, $6$, $8$, and $10$ and so on.

The reason why this construction works is that the $3\cdot {4\choose2}=18$ distances of players of the same team are evenly distributed among the numbers $1$ through $6$:

The 3 node pairs $(6,7)$, $(11,12)$ as well as $(12,13)$ each have distance $1$, the 3 node pairs $(4,5)$, $(7,8)$ and $(11,13)$ have distance $2$ etc - each distance is obtained exactly three times, so each pair of nodes will be included in a team in exactly three rotations.

Edit #2:

I just noticed that an equivalent scenario (even distribution of the distances of pairs of team members) can be achieved in a more symmetric way:


问:
Help with a particular solution of a differential equation
I want to find the solution of the following differential equation:
\begin{equation*}
x''(t)+\omega^2x(t)=a[K_1\cos(\omega t)+K_2\sin(\omega t)]+b[-K_1\omega\sin(\omega t)+K_2\omega\cos(wt)]
\end{equation*}
where $\omega>0$, $a,b \in \mathbb{R}$ and $K_1,K_2>0$. I have calculated the solution of the homogenous equation, but I don't know what particular solution of the ODE I have to take. Any help?

答:
Using Rem's idea, I have arrived to this system:
\begin{equation*}
A''(t)+2B'(t)\omega=aK_1+bK_2\omega
\end{equation*}
\begin{equation*}
B''(t)-2A'(t)\omega=aK_2-bK_1\omega
\end{equation*}
How can I find $A(t)$ and $B(t)$?


问:
What does it mean to divide a quaternion by its real part?
In one course a teacher said we were going to normalize a quaternion and did this:

$$\require{cancel} q = w + xi + yj + zk $$

$$ q` = \cancelto{1}{\frac{w}{w}} + \frac{x i}{w} + \frac{y j}{w} + \frac{z k}{w} $$

But on investigating on my own I found that a quaternion is normalized by dividing it by its norm. Is this another way to normalize a quaternion? And if not, does it have any other meaning?

答:
"Normalizing" means rescaling in a way that is useful or meaningful in a given context. If you have an application or problem that requires you to rescale a quaternion to have real part $1$, then the way to do that is to divide by the real part!

(Unless it's a pure imaginary quaternion, then you're out of luck.)

The natural thing to do, however, is to rescale a quaternion so that it has norm $1$. Recall

$$ |w+x\mathbf{i}+y\mathbf{j}+z\mathbf{k}|=\sqrt{w^2+x^2+y^2+z^2}. $$

If you divide a quaternion $p$ by $|p|$ to get $u:=p/|p|$ has unit norm, i.e. $|u|=1$.

Can you see why $u$ must have norm $1$? In general, $|rp|=r|p|$ for nonnegative reals $r$, and even more generally still the norm is multiplicative, $|pq|=|p||q|$ for any two quaternions $p,q$.


问:
Showing that Pell's equation has at least one integer solution
I'm studying Pell's equations of the form $x^2-dy^2=1$ for $d$ a square-free natural number and $x,y$ integers. In particular, I want to show that such an equation always has a solution.

I see that this is equivalent to finding units with positive norm in $\mathbb{Z}[\sqrt{d}]$. I know that there is always a fundamental unit $\epsilon$ which generates the multiplicative group of units, and $\epsilon$ is unique if we take it as the smallest unit with $\epsilon>1$. Clearly any even power of $\epsilon$ will have positive norm since $\mathbb{Q}(\sqrt{d})$ is real. Hence any solution of Pell's equation would be some power of $\epsilon$.

But why does there always exist an $n\in\mathbb{N}$ such that $\epsilon^n\in\mathbb{Z}[\sqrt{d}]$? Clearly this is the case when $\epsilon\in\mathbb{Z}[\sqrt{d}]$, but of course if $d\equiv 1$ (mod $4$) then it may be the case that $\epsilon\in\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$.

Computation of specific examples seems to indicate that the cube of any fundamental unit will be in $\mathbb{Z}[\sqrt{d}]$. I tried proving this directly, but the amount of calculation makes me think that there is a more elegant method. I see an argument in the final question of this document, but by the heuristic law of "conservation of difficulty", I think that once one uses Dirichlet's Diophantine approximation theorem to classify the units of real quadratic number rings, there must be a method that follows easily.

Looking forward to hearing your thoughts!

答:
If $d-1$ is divisible by $4$, an integral element not in $\mathbb{Z}[\sqrt{d}]$ is some $\omega=\frac{a+b\sqrt{d}}{2}$ with $a,b$ odd. Its norm is the integer $N=\frac{a^2-db^2}{4}$.

Then $\omega^2=-N+a\omega$, and $\omega^3=-N\omega+a\omega^2=-aN+(a^2-N)\omega$.

If $N$ is odd, then indeed $\omega^3 \in \mathbb{Z}[\sqrt{d}]$.

Note that for instance $\frac{1+\sqrt{5}}{2}$ is a unit in $\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$ of norm $-1$, and its square isn’t in $\mathbb{Z}[\sqrt{5}]$.


问:
Prove that $(x-1)^2 + (y-1)^2 \leq 1$ is convex with constraints
Let  $D = \{(x, y, z) \in \mathbb{R}^3: (x-1)^2 + (y-1)^2 \leq 1, \space x - z \geq 1,\space y + z = 0\}$ be a set. How do I show that it's convex or concave?
It should be easily solved by definition, but instead can we re-formulate it, setting $x\geq1-y$ as condition, so that the elements of $D$ are defined as $(x, y) \in \mathbb{R}^2$?

答:
$D$ is the intersection of the interior of a cylinder with a half-space and a plane. All 3 sets being convex, their intersection is convex.

Edit: If you want to $\mathbf {manually}$ check this, then let $A=(x_1, y_1, z_1)$ and $B=(x_2, y_2, _z2)$ be two points in $D$. Let $C=\lambda A + (1-\lambda) B=(x_3, y_3, z_3)$, with $0\leq \lambda \leq 1$, be a point on the $[A, B]$ line segment. We need to prove that $C\in D$:

Since $y_1+z_1=0$ and $y_2+z_2=0$, it's easy to show that $y_3+z_3=\lambda y_1+(1-\lambda)y_2+\lambda z_1+(1-\lambda)z_2=\lambda(y_1+z_1)+(1-\lambda)(y_2+z_2)=0$.

Likewise, it's easily verified that $x_3-z_3\geq 1$.

Finally, using the fact that $1=\lambda + (1-\lambda)$
$$\begin{split}
(x_3-1)^2 &= \lambda^2(x_1-1)^2 +2\lambda(1-\lambda)(x_1-1)(x_2-1)+(1-\lambda)^2(x_2-1)^2\\
(y_3-1)^2 &= \lambda^2(y_1-1)^2 +2\lambda(1-\lambda)(y_1-1)(y_2-1)+(1-\lambda)^2(y_2-1)^2
\end{split}$$
Summing the two and using the fact that $(x_1-1)^2+(y_1-1)^2\leq 1$
$$
(x_3-1)^2+(y_3-1)^2\leq\lambda^2+2\lambda(1-\lambda)\left [(x_1-1)(x_2-1) + (y_1-1)(y_2-1)\right]+(1-\lambda)^2$$

Now, notice that for any real numbers $a, b$, you have $2ab\leq a^2+b^2$, so $$\begin{split}
2(x_1-1)(x_2-1)&\leq (x_1-1)^2+(x_2-1)^2 &\leq 1\\
2(y_1-1)(y_2-1)&\leq (y_1-1)^2+(y_2-1)^2 &\leq 1\\
\end{split}$$
Plugging this above,
$$(x_3-1)^2+(y_3-1)^2\leq \lambda^2+2\lambda(1-\lambda)+(1-\lambda)^2 = 1$$


问:
Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$
Parametric equation of curve is $ x=2\sin\theta +\cos(2\theta) $ , $y=1+\cos(2\theta)$ , for $0\leq\theta\leq\pi/2$

Show that $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2\sin\theta}{2\sin\theta-1}$$
I drived $x$ and $y$ seperatly for $x$,

$$ x=2\cos\theta+\sin\theta\pm\sin\theta$$ and for the

derivative of $y$ i got $y=-2\sin(2\theta)$

and then I did

$$dy/dx= y=-2\sin(2\theta), \qquad   x=2\cos\theta+\sin\theta\pm \sin\theta$$

but I dont know if I did it wronge or I don't know how to continue. Any help would be much appreciated. Thank you!

答:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{dy/d \theta}{dx/d \theta}=\frac{-2\sin 2\theta}{2 \cos \theta-2\sin 2\theta}$$

Let $ \sin \theta =s, \cos \theta= c\;$
We have

$$ \dfrac{2\cdot-2 sc }{2c-2\cdot 2\cdot sc }=\dfrac{s}{2s-1}$$


问:
Convergence of a Maclaurin series.
Consider a function $$f(x)=\frac{4x^4}{x^2+6}.$$
Find its Maclaurin series and for what values of $x$ it converges?

My effort: I start with
$$\frac{4x^4}{x^2+6}=\frac{2x^4}{3}\bigg(1+\frac{x^2}{6}\bigg)^{-1}$$ $$=\frac{2x^4}{3}\sum_{n=0}^{\infty}(-1)^n\bigg(\frac{x^2}{6}\bigg)^n.$$
Above series converges only if $x^2<6$ (by Leibniz rule). Am I correct?

答:
You are correct. You identified part of the expression as the sum of a geometric series, obtained the development, and the interval on which the development is valid. Although this is debatable, since the mclaurin series is defined as $\sum_{n \ge 0} \frac{f^{(n)}(0)}{n!} x^n$, some might expect that you present you result in that form, i.e., something like
$$
f(x)= \sum_{n\ge 0} \dfrac{2 (-1)^n}{3 \cdot 6^n} x^{2n+4}.
$$

Title: Convergence of a Maclaurin series.


问:
Show that if $T$ is injective, then there exists a linear map $S:W\rightarrow V$ so that $ST=I_V$.
Assume $W$ is finite dimensional and $T:V\rightarrow W$ is linear.

Show that if $T$ is injective, then there exists a linear map $S:W\rightarrow V$ so that $ST=I_V$.

Can I have a hint on how to define $S$? If w = T(v), then I will define S(w) = v. But I'm not sure how to define S(w') for w' that are not hit by $T$. I feel like we need cases: (1) $w$ that are hit and (2) $w$ that are not hit.

EDIT: I've seen solutions in which people find the basis of W. What's the point in finding a basis? Can't we just say that $S(w)=v$ if $Tv=w$? That doesn't require us finding a basis, right? Or do we find a basis for the case in which there doesn't exist a $v$ such that $T(v)=w$?

I've noted my confusion in the comments below. Any help would be much appreciated.

答:
The image of $T$ is a subspace $U=T(V)$ of $W$.

On this subspace, define $S(w)=v$ if $T(v)=w$.

Since $T$ is injective, $S$ is a mapping from $U$ into $V$

Then one can extend this mapping to all of $W$ by setting $S(w)=0$ if $w\not\in U$.


问:
Show that $u = 0$ in $B_1 (0)$ if we only assume that $u \in C^1 (B_1 (0))$.
Let $n \geq  2$ and $B_1 (0)$ be the open unit ball in $\Bbb R^n$. Suppose $u \in C^1 (B_1(0))\cap C(\overline{B_1 (0)})$
is a solution of linear PDE
$$x \cdot Du=-u; x \in B_1 (0)$$

(a) Show that $u = 0$ in $B_1 (0)$.

(b) Show that $u = 0$ in $B_1 (0)$ if we only assume that $u \in C^1 (B_1 (0))$.

Here I can clearly see that $u=0$ is a solution as it satisfies the PDE. I was also trying to solve it using characteristic line method e.g. taking
$z=u(x(s))$ and $p=Du$ then we have $F(p,z,x)=x \cdot p+z$ and hence $\dot x(s)=x$ and $\dot z(s)=-z$. I don't think this method would be useful as there is no boundary condition.

How to attempt part a and part b then? Please help.

答:
For each $x\in B_1(0)$, consider

$$f(t) = u(tx).$$

Then

$$f'(t) = \frac{d}{dt} u(tx) = x\cdot Du(tx) = \frac{1}{t} (tx)\cdot Du(tx)  = -\frac{1}{t} f(t).$$

So $(tf)' = tf' + f= 0$ and so $t u(tx)$ is constant in $t$. Or, that

$$ u(x) = u \left( \| 2x\| \frac{x}{\|2x\|}\right) =\frac{1}{\|2x\|} \|2x\|u\left( \| 2x\| \frac{x}{\|2x\|}\right) = \frac{1}{2\|x\|}c\left(\frac{x}{2\|x\|}\right)$$

The only way that this $u$ is continuous at $x=0$ is that $c = u = 0$.


问:
how to solve this inequality,it's strange.
Prove that:
$$a^2+b^2+c^2+d^2+e^2+2ac+2ad+2bd+2be+2ce\ge 3(ab+bc+cd+de+ea),$$
where $a,b,c,d,e$ are nonnegative real numbers.

Body:
Counter example: if $a=b=0$ and $c=d=e=1$, then

$$\begin{align*}
LHS &= 0^2+0^2+1^2+1^2+1^2+2(0+0+0+0+1) = 5\\
RHS &= 3(0+0+1+1+0) = 6\\
LHS &\not\ge RHS
\end{align*}$$


问:
Determining the Lipschitz constants for this system of ODE
We consider the following system of ODE:

$$  u_1' = 4u_1 + 3u_2\\
    u_2' = 5u_1 - 6u_2 $$

I need to determine the Lipschitz constants for this system in the max norm $||.||_{\infty}$ and the $1$-norm $||.||_1.$

I understand that we're considering the vector valued function $F: \mathbb{R}^2 \to \mathbb{R}^2, F(u_1,u_2):= (4u_1 + 3u_2, 5u_1 - 6u_2),$ and we're supposed to find the Lipschitz constant for this $F$ in the max norm $||.||_{\infty}$ and the $1$-norm $||.||_1.$ But I'm a bit confused how to do that, or what theorems to use. Could you please help me out here? Perhaps at least point me to the right theorem? I think I need to consider the operator norm when we equip $\mathbb{R}^2$ with the max norm $||.||_{\infty}$ and the $1$-norm $||.||_1.$ Is that correct? But then, I'm having trouble finding the operator norms w.r.t. these two norms. Help appreciated!

答:
You have a linear system. The Lipschitz constant is the (operator) norm of the Jacobian is the norm of the system matrix. In the case of the max norm on the vector space you get the row-sum norm for the matrix, and for the one-norm of the vector space you get the column-sum norm for the matrix. The first gives $L=11$, the second $L=9$.

On the operator norms associated to the cited vector norms:

for the max norm
$$
\|Ax\|_\infty=\max_i\Bigl|\sum_j a_{ij}x_j\Bigr|\le\Bigl(\max_i\sum_j|a_{ij}|\Bigr)\|x\|_\infty=\|A\|_{op,\infty}\|x\|_\infty.
$$

and for the one norm
$$
\|Ax\|_1=\sum_i\Bigl|\sum_ja_{ij}x_j\Bigr|\le \sum_j\Bigl(\sum_i|a_{ij}|\Bigr)|x_j|\le \Bigl(\max_j\sum_i|a_{ij}|\Bigr)\sum_j|x_j|=\|A\|_{op,1}\|x\|_1
$$
In both cases vectors consisting of $0,\pm1$ can be constructed that realize the equality sign in every inequality.


问:
Local Preservation of Orientation Implies Global Preservation
I am trying to work through some problems in differential topology, and I came across one that I can't figure out, mostly because I'm having trouble understanding some of the definitions.

Let $f \colon M \to N$ be a diffeomorphism of connected oriented smooth ($C^\infty$) manifolds. If $df_x \colon T_xM \to T_{f(x)}N$ preserves orientation at a point $x \in M$, prove that $f$ preserves orientation globally.

I have a limited knowledge of what orientation means. I understand what it means for vector spaces, and the definition that I learned for a smooth manifold is that every transition function has positive determinant. An orientation-preserving map between two smooth manifolds is one with positive determinant (in local coordinates, of course). I assume then that this question is really asking to prove that if $f$ preserves orientation at one point, then it does on all of $M$. Even so, I don't know how to prove that. I have seen this question a few different times on this website, yet never found an answer that was simple enough for me to understand at this point; everyone seemed to be using differential forms. Could somebody help me out? Thanks!

答:
First you should look for the Euclidian case, supose that $U$ and $V$ are open connected sets on $\mathbb{R}^n$ and $f:U \to V$ is diffeo. So by definition $f$ preserve orientation in $x$, if the linear transformation $df_x: \mathbb{R}^n \to \mathbb{R}^n$ preserve orientation, or in a equialent way $\mathrm{det}(df_x)>0$. Now lets proof that if exists $x_0 \in U$ s.t. $f$ preserve orientation on $x_0$ then $f$ preserve orientation in every point in $U$, but as you should know the map $d: U \ni x \mapsto \mathrm{det}(df_x) \in \mathbb{R}$ is smooth, now note that because $f$ is a diffeo. the map $d$ is never zero, so if exists $y \in U$ s.t $d(y)<0$ the facts that $d(x_0)>0$ and $U$ is connected implies that should exists $y' \in U$ s.t. $d(y')=0$, absurd so $f$ preserve orientaion on every point.

Now for the general case $f:M \to N$. Define the set:
$$A=\{x \in M : \mbox{$f$ preserve orientation on x}\},$$
we will prove that $A$ is closed, open and non-empty, so $A=M$. First $A$ is non-empty by hypothesis. Now lets proof that is open, let $x_0 \in A$, so by definition exists charts $\phi:U\subset M \to U' \subset \mathbb{R^n}$ and $\psi:V\subset N \to V' \subset \mathbb{R}^n$ s.t the map $\bar{f}: U' \to V'$, given by $ \bar{f}=\psi^{-1} \circ f \circ \phi$, satisfies $\mathrm{det}(d\bar{f}_{\phi(x_0)})>0$. So by the Euclidian case exists a connected nbhd $W$ of $\phi(x_0)$ s.t $\bar{f}$ preserve orientation on $W$, so by definition $f$ preserve orientation for every point in $\phi^{-1}(W)$. By definition of $A$ we have that $\phi^{-1}(W) \subset A$, that is, $A$ is open as claimed. The proof that $A$ is closed is pretty similar.


问:
Exercise 2.17(b) in Eisendbud's Commutative Algebra
All rings are unital and commutative. Here are parts a. and b. of the exercise from Eisenbud. I don't think that part a. is needed for b., but I've included it just in case.

Suppose $R$ is a $\mathbb{Z}$-graded ring and $0 \neq f \in R_1$. Show that $R[f^{-1}]$ is again a $\mathbb{Z}$-graded ring. Let $S = R[f^{-1}]_0$.

a. Show that $R[f^{-1}] \cong S[x,x^{-1}]$, where $x$ is a new variable

b. Show that $S = R[f^{-1}]_0 \cong R/(f-1)$.

I am able to prove the first statement, where I defined the degree $n$ component of $R[f^{-1}]$ to be
$$
R[f^{-1}]_n = \left\{ \frac{r}{f^d} : r \ \text{is homogeneous}, \deg(r) - d = n  \right\},
$$
and I found part a. to be straightforward.

I am able to prove part b. when $f$ is not a zerodivisor as follows: define $\phi: R \to S$, by $r \mapsto \frac{r}{f^{\deg(r)}}$ for homogeneous $r$, and then extending to all of $R$ by linearity. It is easy to see the $\phi$ is onto, so it suffices to show that $\ker \phi = (f-1)$. The inclusion $(f-1) \subseteq \ker \phi$ is obvious. On the other hand, suppose that $r = r_1 + \cdots r_n \in \ker \phi$, where each $r_i$ is homogeneous of degree $d_i$. Moreover, assume that $d_1 < \cdots < d_n$. Then
$$
\phi(r) = \frac{r_1}{f^{d_1}} + \cdots + \frac{r_n}{f^{d_n}} = \frac{f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1} + r_n}{f^{d_n}} = 0,
$$
which implies $f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1} + r_n = 0$, since I am assuming that $k$ is not a zerodivisor (otherwise I would have to multiply this equation by some $f^k$, which would screw up the next step). But then
$$
r = r_1 + \cdots + r_n = (r_1 + \cdots + r_{n-1}) - (f^{d_n-d_1}r_1 + \cdots + f^{d_n-d_{n-1}}r_{n-1}) = (1 - f^{d_n - d_1}) r_1 + \cdots + (1 - f^{d_n - d_{n-1}}) r_{n-1},
$$
which is in $(f-1)$ as desired.

The claim is also true in the case where $f$ is nilpotent: then the localization at $f$ is $0$, and $1-f$ is a unit, implying that $R/(f-1) = 0$.

So really I'm after the case where $f$ is a zerodivisor, but not nilpotent.

答:
In general, when $\phi(r)=0$, it means that there is $k \geq 0$ such that $f^k(f^{d_n-d_1}r_1+\ldots+r_n)=0$. Then $r=(1-f^k)r+ f^kr=(1-f^k)r+f^k(r_1+\ldots+r_{n-1})-f^k(f^{d_n-d_1}r_1+\ldots+f^{d_n-d_{n-1}}r_{n-1})=(1-f^k)r+f^k\sum_{i=1}^{n-1}{(1-f^{d_n-d_i})r_i}$ and each term is in $(f-1)$, so $r \in (f-1)$.

Edit: here’s a somewhat simpler solution, probably easier to apply in general. Let $r \in R$ be in the kernel of $\phi$. It’s easy, writing $r$ in coordinates and “completing” the terms of lower degree with powers of $f$, to show that $r=(1-f)r_1+r’$ for some $r_1,r’ \in R$ and $r’$ homogeneous of degree $d$. Thus $\phi(r’)=0$. But $\Phi(r’)=\frac{r’}{f^d}$, so this means that there is a $k \geq 1$ such that $f^kr’=0$. But then $r’=(1-f^k)r’ \in (1-f)$.


问:
Two equal circles inscribed in a triangle
Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem  and algebraic manipulations are allowed to be used]

答:
Let the centres of the circles be $C_{1}$ and $C_{2}$. The segments $BC_{1}$ and $CC_{2}$,when extended, meet at $I$ , the incentre of $\triangle ABC$. Circle $C_{1}$ and $C_{2}$ touch the side $BC$ at points $E$ and $F$ respectively. The incircle of $\triangle  ABC$ touches the side $BC$ at point $D$. Draw $C_{1}E$, $C_{2}F$, $C_{1}C_{2}$ and $ID$. Let $ID$ intersect $C_{1}C_{2}$ at point $L$. Let the radii of the similar circles and the inradius of $\triangle ABC$ be $r'$ and $r$ respectively.

$\triangle IC_{1}C_{2}\sim \triangle IBC$ because $C_{1}C_{2}\parallel BC$.

$\Rightarrow  \frac{C_{1}C_{2}}{BC}=\frac{IL}{ID}$

$\Rightarrow   \frac{2r'}{21}=\frac{r-r'}{r}$

$\Rightarrow   r'=\frac{ar}{2r+a}=\frac{21r}{2r+21}$

Use $\Delta=rs$ to get $r$ and then plug it into the equation to get $r'$.