问:
Finding formula for parabolic cone with constant ratio of volume to surface area of each circular cross section
looking for some assistance with a problem I've been exploring with a friend for kicks.  I haven't taken uni level calculus or algebra in a couple of decades so I'm rusty!

I'm trying to express an equation for a parabolic cone (paraboloid?) where the ratio of volume to surface area of the circle through  covering every cross section throughout the height of the 'cone' is constant.  Presumably there is some parabola equation that creates a line that if spun around to create a parabolic cone shape would give this outcome.

Any takers?

答:
The ratio is a constant $c/2 =k ,\; c= 2k $, say

$$ \dfrac{\int\pi r^2 dy}{\pi r^2}= k  \tag {1a} $$

$$ \dfrac{\pi r^2 }{ \dfrac{d (\pi r^2)}{dy}}= k  \tag{1b} $$

Cancel $\pi$ and differentiate with respect to $y$ using Chain Rule as above  $ (u/v)'=0 \to \left( u/v \to u'/v' \right) $

$$ \dfrac{r^2}{2r \dfrac{dr}{dy}} = k $$

$$ \dfrac{dy}{2k}=\dfrac{dr}{r} \to \dfrac{dr}{dy}= \dfrac{r}{2k} \tag2 $$

Integrate with BC $y=0, r=a\; $

$$ \dfrac{y}{2k}=\log \dfrac{r}{a}  \tag 3$$

Or

$$ r= a e^\frac{y}{2k} \tag 4 $$

So it is an exponential curve rotated around $Y$ axis.

To verify the obtained result plug (2) into (1) and simplify

$$\dfrac{\text{Volume}}{\text{Cross-Section Area}}=\dfrac{\pi a^2 \int e^{ y/k} dy} {\pi(a^2 e^{y/(2k)})^2} =\dfrac{ e^{y/k} \cdot k} {(e^{y/(2k)})^2}= k \tag5 $$

which is a constant.


问:
Drake, Seven Axioms of the Algebra of Events
On page 3 of Drake Fundamentals of Applied Probability he lists The Seven Axioms of the Algebra of Events.

$1. A \cup B = B \cup A \\
2. A \cup (B \cup C) = (A \cup B) \cup C \\
3. A\cap( B \cup C) = A \cap B \cup A \cap C \\
4. (A')' = A \\
5. (A\cap B)' = A' \cup B'\\
6. A\cap A' = \phi \\
7.A \cap U = A  $

I have two questions.

First while Drake states that his choice is not unique is there a reason that he selected these seven and is seven the minimum possible number of axioms with which the Algebra of events can be expressed.

Second he further states that any other relation in the algebra of events can be proved directly from these seven axioms with no additional information and in fact in the chapter problem section asks the student to do so. Does the author mean that all the other set relations in the Algebra of Events can be proved by direct manipulation of the seven axioms with out resorting to the method of $x \in A$

For example, proofs of $A \cup A' = U$ are given in this posting
Proof of union of a set and its complement is equivalent to a universe

but these proofs use the method of $x \in A$ and not direct manipulation of Axiomatic statements as required by Drake. How would I prove $A \cup A' = U$ or $A \cup A = A$ using only direct manipulation of Drake's Seven Axioms of the Algebra of Events?

答:
remark

From 1-7 we deduce that $U$ is unique.  Suppose
$$A \cap U = A  \text{ for all }A\tag7$$
$$A \cap V = A  \text{ for all }A\tag{7'}$$
We claim $U = V$.

Proof. From $(7)$ we get $V \cap U = V$.

From $(7')$ we get $U \cap V = U$.

Then apply $(1)$ to get $V \cap U = U \cap V$, and conclude $U = V$.

ContentLicense: CC BY-SA 4.0


问:
Link between 'minimal' spanning set in Linear Algebra & 'minimal' in Set Theory
Theorem: Let $ V $ be a vector space and $ X = (v_1,...,v_s) \subseteq V $ be a subsequence of vectors. Then $ Span(X) $ is the minimal subspace that contains $X$.

Proof (what is written in my notes ):

[ we show $ Span(X) $ is indeed a subspace of $ V $ ]

We will prove minimality. Let $ U \subseteq V $ be a linear subspace s.t. $ X \subseteq U $ . Then also any linear combination of the vectors in $ X $ is in $ U $ ( since $ U  $ is a subspace ), hence $ Span(X) \subseteq U $. $ \Box $

In "How to prove it" by Velleman, there's the following definition of a minimal element of a set:

And later he adds the following:

I wanted to know if there is any connection/link between the Theorem above in linear algebra and the Definition I've provided that relates to Set Theory ( or are they totally different beings? ). If so, can you please explain? Relating to the defintion above, Is $ Span(X)$ a R-minimal element of some relation R? and what is the set $ B $ here s.t. $ Span(X) \in B $?

答:
There is indeed a connection, although there's a slight terminological mis-match: "minimal" in the original context corresponds to Vellman's "smallest," not Vellman's "minimal."

Specifically, the linear algebra result says that $Span(X)$ is smallest with respect to the subset relation amongst the collection of subspaces of $V$ containing $X$. That is:

$B$ is the set of all subspaces $S$ of $V$ such that $X\subseteq S$, and

$R$ is just $\subseteq$.

Note that as far as the choice of $R$ goes, this is exactly an instance of the situation described by Vellman's subsequent remark ("When comparing ..."). And elaborating on Vellman's remark, a phrase of the form "the minimal/smallest/least [structure]" generally means "the $\subseteq$-smallest element of the collection of all [structure]s." The only discrepancy is around the use of "minimal" rather than "smallest" (or "least," to mention another synonym); annoyingly, mathematicians aren't always careful about this, and when a term like "minimal" is used you should make sure the precise meaning is clear from context.

$\color{red}{\mathsf{WARNING}}$: keep in mind that minimal/smallest objects need not always exist! For example, if $X$ is finite and $V$ is infinite-dimensional, the set of infinite-dimensional subspaces of $V$ containing $X$ will not have a $\subseteq$-minimal element, let alone a $\subseteq$-smallest element.

(As an aside, it's worth noting that this characterization of $Span(X)$ does not depend on $X$ being finite: for every $X\subseteq V$ we can define $Span(X)$ to be the set of finite linear combinations of elements of $X$, and then $Span(X)$ is the smallest subspace of $V$ containing $X$ in exactly the same sense.)


问:
Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$
This is a math Olympiad problem.

my attempt :

By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:

$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$

By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.

Body:
hint

$$x^2+y^2-8x=9\iff$$
$$(x-4)^2+y^2=25$$
$$=(\pm 3)^2+(\pm 4)^2$$
$$=0^2+(\pm 5)^2$$


问:
More understandable explanation for Cat-Rabbit-Dog quiz
This is a quiz from Brilliant.org Algebra Fundamentals courses.

Quiz:

At Step 40, how many cats will there be in the 7th row?

Answer: 14

Explanation given by Brilliant.org is not clear to me so if someone can give a better explanation it will be helpful.

Explanation by Brilliant.org

In rows where dividing by 3 has a remainder of 1, the pattern is
cat-rabbit-dog.

7 when divided by 3 has a remainder of 1, so cats occur at 1, 4, 7,
etc., i.e. at points where dividing by 3 has a remainder of 1.

In step 40, the grid is 40 columns wide. The full cat-rabbit-dog
pattern happens for $39\div 3=13$ times, but since the pattern
at row 7 starts with a cat, column 40 has one extra cat. So there are
$13 + 1 = 14$ cats.

Note: A more compact way of saying "dividing $x$ by 3 has a remainder
of 1" is $x \mod 3 = 1$.

答:
By reading all the answers I able to understand more and here is my simple explanation.

Its a sequence.

1st row and 4th row were same in Step 4,
so the pattern is same in every 3rd row,
pattern in 1st and 4th row is Cat-Rabbit-Dog-Cat-Rabbit-Dog-...

In Step n we have n2 animals and n animals in each row.

Question is, in Step 40 how many cats are in row 7th?

In Step 40 each row has 40 animals.


Title: Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$

Body:
This is a math Olympiad problem.

my attempt :

By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:

$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$

By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.

Body:
From the equation, we have
$$(x-4)^2+y^2 = 25$$
Hence, we can deduce that  $(|x-4|,y) = (0,5),(3,4),(4,3)$ or $(5,0)$.
The solution is then
$$(x,y) = (4,5),(7,4),(1,4),(0,3),(8,3),(9,0)$$


问:
Show $\{x\in E: \|x\|\le 1\}=\bigcap\limits_{f\in E^{\star}\\ \|f\|\le 1}\{x\in E: |f(x)|\le 1\}$.
Let $(E,\|\cdot\|)$ be a normed vector space and $E^{\star}$ denote the dual space. I want to show that $\{x\in E: \|x\|\le 1\}=\bigcap\limits_{f\in E^{\star}\\ \|f\|\le 1}\{x\in E: |f(x)|\le 1\}$. I can show the inclusion of the first set in the second set but not the second set in the first set. Any hints?

答:
For the second inclusion:
$$A:=\bigcap_{f \in E^* , \ ||f|| \leq 1} \{ x \in E : |f(x)| \leq 1\} \subset \{x \in E : ||x|| \leq 1 \}$$
If $x\in A$ then for all $f \in E^*$ with $||f|| \leq 1 $ we have that $|f(x)| \leq 1$. Now, by a Corollary of the Hahn-Banach thm, there exists a functional $ f \in E^*$ with $||f||=1$ and $f(x) = ||x||$. Thus, $||x|| \leq 1$.


问:
Why is surface area not simply $2 \pi \int_{a}^{b} (y) dx$ instead of $2 \pi \int_a^b (y \cdot \sqrt{1 + y'^2}) dx$?
Geometrically speaking, it seems to me that if you have for example $y^2=8x$ revolved around the x-axis, taking the limit of the sum of $n$ surfaces of cylinders as $n$ approaches infinity should give you the surface area of that surface of revolution.

This is how the author initially derives the formula for finding the volume of solids of revolution. Take a rectangle under the curve over $\Delta x$ and revolve it around the axis to get an approximation of the volume of the solid over that interval. Add up those rectangles over $n$ changes in $x$ and take the limit as $n$ approaches infinity, which is the integral of the function that gives you the $y$ value (radius of that approximating cylinder) for each $x$ value.

Following the same principle, why wouldn't we be able to take those same cylinders, but instead of taking their volume, taking their surface area and take the limit as the number of those cylinders approaches zero?

In other words, in this case each $y$ value is given by $y = \sqrt{8x}$, which is the radius of that cylinder of height $\Delta x$ and an approximation of the surface area over that interval.

Why doesn't that work? Why do we need to deal with arc length? I don't understand why it doesn't work in this case, it seems to me that you're still getting a better and better approximation of surface area as those cylinders get smaller and smaller, eventually getting the exact surface area with the limit as their number goes to infinity.

PS: I saw this Surface area of a solid of revolution: Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ work? but it's still not making sense visually/geometrically.

答:
The answers above are sufficient I think. However, I just wanted to add this as another viewpoint, based on a traveling particle, that some may like, and some informal derivations using infinitesimals.

Imagine a particle traveling over time $t$ along the curve: $$(x(t),y(t))$$
Where we have:
$$y=f(x)$$
or
$$y(t)=f(x(t))$$

The velocity vector of the particle is:
$$(\frac{dx}{dt},\frac{dy}{dt})$$
and the speed of the particle:
$$v(t)=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$$
so that the infinitesimal quantity
$$dl=v(t)dt=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} \ dt$$
is speed multiplied by infinitesimal time and thus an infinitesimal length.

We can write:
$$\int_a^b f(x) \sqrt{((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)(dt)^2}=$$
$$\int_a^b f(x) \sqrt{dx^2+dy^2}=$$
$$\int_a^b f(x) dl$$

Then, just as in the answer by Christopher Carl Heckman, we have (not rigorously):
$$\int_a^b f(x) \sqrt{dx^2+dy^2}=$$
$$\int_a^b f(x) \sqrt{dx^2+dy^2}\frac{dx}{dx}=$$
$$\int_a^b f(x) \sqrt{\frac{dx^2+dy^2}{dx^2}}dx=$$
$$\int_a^b f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx=$$
The "average" radius of the frustum is:
$$r=y=f(x)$$
and the circumference
$$c=2 \pi r=2 \pi y = 2 \pi f(x)$$
so that the surface area becomes:
$$S=\int_a^b 2 \pi f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx=$$
$$\int_a^b 2 \pi y \sqrt{1+y'^2}dx$$
This makes it pretty clear that just having $$y dx=y\sqrt{1+0} dx$$ instead of $$y \sqrt{1+y'^2}dx$$ ignores the y-component of the velocity of the particle. In other words, if we are not careful then we are just measuring the distance traveled in the $x$-direction instead of the "Euclidian" distance.

The volume is a second order affect as can be seen by starting with the infinitesimal volume of a conical frustum between $y(x)$ and $y(x+dx)$:
$$dV=\frac{1}{3}\pi h(R_1^2+R_1R_2+R_2^2)=\frac{1}{3}\pi dx((y(x))^2+(y(x)y(x+dx))+(y(x+dx))^2)=$$
$$\frac{1}{3}\pi dx((y(x)^2+(y(x)(y(x)+y'(x)dx+...)+(y(x)+y'(x)dx+...)^2\approx$$
$$\frac{1}{3}\pi dx(y^2+y^2+y y' dx + y^2 + 2y y' dx + y'^2)=$$
$$\frac{1}{3}\pi dx(3y^2+3y y'dx+y'^2dx^2)=$$
$$\frac{1}{3}\pi dx(3y^2+ higher order terms )\approx$$
$$\pi y^2 dx$$

Finally, the volume is
$$V=\int_a^b \pi y^2 dx$$

ContentLicense: CC BY-SA 3.0


问:
Divisibility by 7 Proof by Induction
Prove by Induction that

$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$

Base case:

$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.

Now, having $n=k$, we assume that:

$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$

We have to prove that for $n=k+1$ that,

$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$

We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.

Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$

I get stuck here, please help me.

Title: Divisibility by 7 Proof by Induction

Body:
$b-a=4^{2\cdot2^k}-4^{2^k}+2^{2\cdot2^k}-2^{2^k}$

$=4^{2^k}(4^{2^k}-1)+2^{2^k}(2^{2^k}-1)$

$=4^{2^k}(2^{2^k}-1)(2^{2^k}+1)+2^{2^k}(2^{2^k}-1)$

$=(2^{2^k}-1)(8^{2^k}+4^{2^k}+2^{2^k})$

$=(2^{2^k}-1)2^{2^k}\color{blue}{(4^{2^k}+2^{2^k}+1)}$


问:
If $2^x=0$, find $x$.

If $2^x=0$, find $x$.

Solution: I know range of $2^x$ function is $(0,\infty)$.

So $2^x=0$ is not possible for any real value of $x$

Hence, equation is wrong. We can't find value of $x$. Am I right?

Please help me.

Can $x$ be in $[-\infty,\infty]$?

i.e is $2^x=0$ possible for $x=-\infty$?

答:
Yes. You are right, there is no $x \in \mathbb{R}$ such that $2^x = 0$. However, note that
$$\lim_{x \to -\infty} 2^x = 0$$
However, on the extended real line, i.e., $\mathbb{R} \cup \{-\infty\} \cup \{\infty\}$, there exists a solution, since one of the ways of defining "$2^{-\infty}$" on the extended real line is to define "$2^{-\infty}$" as the limit of $\lim_{x \to -\infty} 2^x$, which is $0$.</p>

Title: If $2^x=0$, find $x$.


Title: Divisibility by 7 Proof by Induction

Body:
Prove by Induction that

$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$

Base case:

$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.

Now, having $n=k$, we assume that:

$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$

We have to prove that for $n=k+1$ that,

$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$

We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.

Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot  2^{k}}+2^{2\cdot  2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$

I get stuck here, please help me.

Body:
$ b - a = 4^{2^{k+1}} + 2^{2^{k+1}} - 4^{2^k}-2^{2^k}$

An intermediate step: notice $4=2^2$, so we have that $b-a = 2^{2^{k+2}} + 2^{2^{k+1}} - 2^{2^{k+1}}-2^{2^k} = 2^{2^{k+2}} -2^{2^k} $

Notice that each term is divisible by $2^{2^k}$, and $7$ does not divide $2^{2^k}$.

Now $\frac{b-a}{2^{2^k}} = 2^{2^{k+2} - 2^k} -2^{2^k - 2^k} = 2^{3 \times 2^k} - 2^0 = 8^{2^k} -1$

Now it is more of a bonus fact that $8^m -1$ is always divisible by $7$.


问:
Lower bound for the upper limit of the sequence
Prove that for any sequence $ a_n > 0 $ the limit superior of $\left(\frac{1+a_{n+1}}{a_n}\right)^n$ is greater or equal to  $ e $.

It looks similar to the definition of $ e $, but I can't figure out where to start from. I think maybe it can be proved by contradiction, but I don't know how exactly.

答:
Suppose $$\frac{1+a_{n+1}}{a_n}\ge1+\frac1n\tag{1}$$ were true only for finitely many values of $n$: that means $$\frac{1+a_{n+1}}{a_n}<1+\frac1n\tag{2}$$ for $n\ge n_0$. But (2) is equivalent to $$ \frac{a_{n+1}}{n+1}<\frac{a_n}n-\frac1{n+1},$$ so by summing that up, we'd have
$$\frac{a_{n+1}}{n+1}<\frac{a_{n_0}}{n_0}-\sum^n_{k=n_0}\frac1{k+1}$$ for $n\ge n_0$ Since the harmonic series diverges, we'd have $a_{n+1}<0$ for sufficiently large $n$, contrary to our assumptions. So (1) must be true for infinitely many $n$, and that means
$$\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\ge\limsup_{n\to\infty}\left(1+\frac1n\right)^n=e.$$


问:
Limit of $n^c/b^n$ when $n$ tends to infinity
How do I prove that $\lim_{n\to\infty}\frac{n^c}{b^n}=0$, where $c\in\mathbb{R}$ and $b>1$?
I've managed to prove it using L'Hospital rule working with $\frac{x^c}{b^x}$, but is there a simpler way using only sequences?

答:
Let $m$ be any integer greater than $c$ and $0$. Since $b-1>0$ we have for $n\geq m$: $$b^n=(b-1+1)^n=\sum_{k=0}^n\binom{n}{k}(b-1)^k\geq\binom{n}{m}(b-1)^m=:g(n)$$
Now the right side $g(n)$ is a polynomial in $n$ of degree $m>c$ in $n$. It follows that $n^c/g(n)\to0$ as $n\to\infty$ and therefore by the squeeze theorem: $$\lim_{n\to\infty}\frac{n^c}{b^n}=0$$


问:
Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$
Here is a Math Tripos problem I cannot solve.

Eliminate $x$, $y$, $z$ from the equations
$$\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$ and
$$ax+by+cz=0$$

I dont know if there is a general (practical) method for such problem. I think you just manipulate the equations to eliminate $x,y,z$ but have been unable to find the path in this question.

The answer is $a^3+b^3+c^3=a^2(b+c)+b^2(a+c)+c^2(a+b)$ if that yields any insight.

答:
This looks like a computational problem. It can be routinely solved in SageMath using elimination in polynomial rings.
Here is a sample code to run.

Notice that since we are given just 3 equations (i.e. generators of the original ideal), we cannot request elimination of $x,y,z$ altogether (as this would yield zero ideal), but luckily elimination of $x,y$ results in the ideal generated by $z^3$ times a polynomial in $a,b,c$ -- this polynomial is what we need.

Alternatively, elimination can be done using resultants. Here is a sample code to run.


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:
The locus you are asking about is a subset of a cubic curve called an Apollonian strophoid (or strophoid of Apollonius).

The curve is the set of points $S$ for which either the internal or bisector of the angle $\angle ASB$ passes through $C$.  (The external bisector is the line through $S$ perpendicular to the internal bisector.)  As the strophoid $\Sigma$ comes in from infinity, $C$ is on the internal bisector but, as $\Sigma$ goes past $A$, $C$ is on the external bisector until $\Sigma$ passes $B$ after which $C$ is back on the internal bisector all the way to infinity.  So your locus is actually the strophoid minus the section between $A$ and $B$.

A GeoGebra applet that demonstrates the strophoid can be found here.  The strophoid is blue, the external bisector is green. You can move $S$ and $C$.  (Unfortunately the $S$ will jump around if you modify any of the points.  This is annoying but harmless.)  The strophoid is built using mathlove's equation.

There are two special cases where the cubic degenerates to a line $\ell$ and a circle $c$.   Both have been touched upon in other answers and comments for this question.

when $C$ is on the perpendicular bisector of $A$ and $B$, line $\ell$ is the perpendicular bisector and $c$ is the circumcircle of $A,B,C$.  For points $S$ on $c$ it is the external bisector that goes through $C$.  (Because, if $M$ is the midpoint of $A,B$ then the triangle $\triangle ECS$ is a right angle.

when $C$ is on the line $AB$, line $\ell$ is the line $AB$ and $c$ is an Apollonian circle of the segment $AB$.

For more about strophoids see the following references.

Gibert, Apollonian strophoids (part of a large catalog of so-called Triangle cubics)

mathcurve.com (Strophoid page)

Stachel, Strophoids, A Family Of Cubic Curves With Remarkable Properties (see Theorem 4).


答:
Another way to see it:
$$
\sum_{n=1}^\infty \frac{n^c}{b^n} < \infty
$$
by the ratio test, hence the sequence must converge to $0$.

edit:
$$
\left| \frac{\frac{(n+1)^c}{b^{n+1}}}{\frac{n^c}{b^n}} \right| = \frac{(n+1)^c}{b^{n+1}} \cdot \frac{b^n}{n^c} = \left(\frac{n+1}{n}\right)^c \cdot \frac 1 b \to \frac{1}{b} < 1
$$
as $n\to\infty$.

Title: Limit of $n^c/b^n$ when $n$ tends to infinity

Body:
How do I prove that $\lim_{n\to\infty}\frac{n^c}{b^n}=0$, where $c\in\mathbb{R}$ and $b>1$?
I've managed to prove it using L'Hospital rule working with $\frac{x^c}{b^x}$, but is there a simpler way using only sequences?


问:
Solving $x''(t) + \frac{2}{t}x'(t) + x(t) = 0$
I was asked the following question on my homework assignment:

We have the ODE $x''(t) + \frac{2}{t}x'(t) + x(t) = 0, t>0$.
We know that $x(t) = \frac{\sin (t)}{t}$ is a solution of this ODE. Rewrite this second order ODE to a system of first order ODE's and find the fundamental matrix. Finally, determine the solution corresponding to the initial value's $x(\pi) = 1, x'(\pi) = 0$

My attempt

So first I rewrote the second order ODE to a system of first order ODE's:
$$x''(t) + \frac{2}{t}x'(t) + x(t) = 0 \leftrightarrow x''(t) = -\frac{2}{t}x'(t) - x(t)$$
$$u(t) := \begin{pmatrix}x(t)\\x'(t)\end{pmatrix}, u'(t) := \begin{pmatrix}x'(t)\\x''(t)\end{pmatrix}$$
We get then the following system:
$$u'(t) = \begin{pmatrix}0&1\\-1&-\frac{2}{t}\end{pmatrix}u(t)$$
I know have to find the fundamental matrix $U(t) = \begin{pmatrix}\frac{\sin (t)}{t}& y(t)\\\frac{\cos (t)}{t} - \frac{\sin (t)}{t^2}& y'(t)\end{pmatrix}$

Or in other words, I need to find $y$. Using the Wronskian of $U$ (which is $w(t) = \frac{w(t_0)t_0^2}{t}$ with $t_0$ being the initial value) I can determine that $$y'(t) = (\cot(t) - \frac{1}{t}) y(t) +\frac{w(t_0)t_0^2}{\sin (t)} $$ Since this is a linear ODE $y$ is explicitly calculated as
$$y(t) = y(t_0)\exp\left(\int_{t_0}^t \cot (s) - \frac{1}{s} ds \right) + w(t_0)t_0^2\int_{t_0}^t \exp \left( \int_s^t \cot (r) - \frac{1}{r} dr\right)\frac{1}{\sin (s)} ds $$

But as far as I know, since $t_0 = \pi, \int_{\pi}^t \cot (s) ds$ is not defined.

The question is what is going wrong and/or what is supposed to be done to get $y(t)$?

I thank you in advance

答:
Letting $x(t)=\frac{\sin t}{t}y(t)$ be a solution gives
$$ \sin t y''+2\cos t y'=0. $$
This equation has solution
$$ y=c_1+c_2\cot t. $$
From this, it is easy to get the fundamental matrix.


问:
How to find $\operatorname{P.V.}\int_0^1 \frac{1}{x (1-x)}\arctan \left(\frac{8 x^2-4 x^3+14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16}\right) \textrm{d}x$?
The following problem is proposed by Cornel Valean:$$\operatorname{P.V.} \int_0^1 \frac{1}{x (1-x)}\arctan \left(\frac{8 x^2-4 x^3+14 x-8}{2 x^4-3 x^3-11 x^2+16 x+16}\right) \textrm{d}x=\log \left(\frac{5}{4}\right) \arctan\left(\frac{1}{2}\right).$$

The question is how can we prove this equality without using the main result in this paper and without the use of complex numbers?

答:
Too long for a comment.

Since
$$T(x)=\dfrac{-4x^3 +8x^2+14x-8}{2x^4-3x^3-11x^2+16x+16}
=\dfrac{R(x)-S(x)}{1+R(x)S(x)},$$
where
$$R(x)=\dfrac2{x^2-x-4},\quad S(x)=\dfrac{4x}{2x^2-x-4},\quad x\in[0,1],\tag1$$
(pointed by Sophie), then
$$\arctan T(x) = \arctan R(x) - \arctan S(x),$$
(see also WA plot)

$$I=\int\limits_0^1\arctan T(x)\; \dfrac{\text dx}{x(1-x)}
=\int\limits_0^1\arctan T(x)\;\left(\dfrac1x+\dfrac1{1-x}\right)\,\text dx,
$$$$
I=\int\limits_0^1\big(2\arctan R(x)-\arctan S(x)-\arctan S(1-x)\big)\; \dfrac{\text dx}{x},\tag2$$
wherein
$$S(1-x)=\dfrac{4-4x}{2x^2-3x-3}.\tag{1a}$$
This transformation makes the integral convergent, and the numerical integration gives
$$I\approx 0.10345,99740\,30782\,04062\,03377\,83301\,52783\,80081\,8.$$
Obtained result exactly corresponds to the OP value.

However, analytical result does not obtained.

We can additionally decompose arctangents of $\;R(x),S(x),S(1-x)\;$ to the elementary terms.

Assuming
$$R(x) = \dfrac{(a+2cx)+(b-2cx)}{1-(a+2cx)(b-2cx)},$$
and taking in account coefficients proportionality, one can get
$$
\begin{cases}
a+b=8c^2\\
2c(b-a)=4c^2\\
ab-1=16c^2\\
c>0
\end{cases}\Rightarrow
\begin{cases}
a=4c^2-c\\
b=4c^2+c\\
16c^4-17c^2-1=0\\
c>0
\end{cases}
$$$$
c=\sqrt{\dfrac{\sqrt{353}+17}{32}}\approx1.05753\,68526,\tag3
$$$$
a=4c^2-c,\quad b=4c^2+c,$$
$$\arctan R(x)=\arctan(a+2cx)+\arctan(b-2cx),$$
$$\arctan R(x)=\arctan(4c^2+c(2x-1))+\arctan(4c^2-c(2x-1)).\tag4$$

Similarly, assuming
$$S(x) = \dfrac{-4x}{-2x^2+x+4} = \dfrac{(gx+d)+(hx-d)}{1-(gx+d)(hx-d)},$$
one can get
$$
\begin{cases}
g+h=-(1+d^2)\\
4d(g-h)=1+d^2\\
2gh=1+d^2\\
d>0
\end{cases}\Rightarrow
\begin{cases}
(2g+1)+(2h+1)=-2d^2\\
(2g+1)(2h+1)=1\\
2d((2g+1)-(2h+1))=1+d^2\\
d>0,
\end{cases}
$$$$
\begin{cases}
(2g+1),(2h+1)=-d^2\pm\sqrt{d^4-1}\\
4d\sqrt{d^4-1}=1+d^2\\
d>0,
\end{cases}
\begin{cases}
16d^4-17d^2-1=0\\
g,h=\dfrac{-(1+d^2)\pm\sqrt{d^4-1}}2\\
d>0,
\end{cases}
$$$$
g,h=\dfrac{1+d^2}{-(1+d^2)\mp\sqrt{d^4-1}} =\dfrac{4d}{-4d\mp1},
$$$$
d=\sqrt{\dfrac{\sqrt{353}+17}{32}}=c,\quad g = -\dfrac{4c}{4c+1},\quad h = -\dfrac{4c}{4c-1},
$$$$
\arctan S(x) = \arctan(gx+d)+\arctan(hx-d)
$$$$
\arctan S(x) = \arctan\dfrac{4c^2+c-4cx}{4c+1}-\arctan\dfrac{4c^2-c+4cx}{4c-1} \tag5
$$$$
\arctan S(1-x) = \arctan\dfrac{4c^2-3c+4cx}{4c+1}-\arctan\dfrac{4c^2+3c-4cx}{4c-1} .\tag6
$$
Obtained integrals allow analytical calculation via integral logarithmic function of complex argument.

Taking the real parts, one can get
$$J(p,q,x) = \int \arctan(px+q)\,\dfrac{\text dx}{x}=J_1(p,q,x)+J_2(p,q,x)+J_3(p,q,x),\tag7$$
where
$$J_1(p,q,x) = (\ln p+\ln x)\arctan(px+q),$$
$$J_2(p,q,x)=\dfrac12\Im\left(\text{Li}_2\left(1+\dfrac{px}{q-i}\right)-\text{Li}_2\left(1+\dfrac{px}{q+i}\right)\right),
$$$$
J_3(p,q,x)=\dfrac12\Im\big(\ln(q+i)\ln(1-i(px+q))-\ln(q-i)\ln(1+i(px+q))\big).$$
From $(4),(5),(6)$ should that the term with $\;\ln x\;$ does not influent to the goal definite integral.
Therefore, we can assume
$$J_1(p,q,1)-J_1(p,q,0) = \ln p\arctan\,\dfrac p{1+(p+q)q}.\tag8$$
Also,
$$J_2(p,q,1)-J_2(p,q,0) = \dfrac12\Im\left(\text{Li}_2\left(1+\dfrac{p}{q-i}\right)-\text{Li}_2\left(1+\dfrac{p}{q+i}\right)\right),$$
$$J_3(p,q,x) = \dfrac12\Im\bigg(\ln(q+i)\ln(1-i(px+q))+\ln(q+i)\ln(1+i(px+q))
$$$$
-\ln(q+i)\ln(1+i(px+q))-\ln(q-i)\ln(1+i(px+q))\bigg)$$
$$= \dfrac12\Im\bigg(\ln(q+i)\ln(1+(px+q)^2)-\ln(q^2+1)\ln(1+i(px+q))\bigg),$$
$$J_3(p,q,1)-J(p,q,0)
= \dfrac12\Im\bigg(\ln(q+i)\ln\,\dfrac{1+(p+q)^2}{1+q^2}-\ln(q^2+1)\ln\dfrac{1+i(p+q)}{1+iq}\bigg)
$$$$
= \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln\dfrac{(1+i(p+q))(q+i)}{1+iq}\bigg)
$$$$
= \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln\dfrac{(1+i(p+q))(q+i)(1-iq)}{1+q^2}\bigg)
$$$$
= \dfrac12\Im\bigg(\ln(q+i)\ln\,(1+(p+q)^2)-\ln(q^2+1)\ln((q+i)^2(p+q-i)) \bigg)
$$

But this approach contradicts to the  OP requirements.


问:
Can we use Lebesgue dominated convergence theorem here?
Let $f_n(x)=\dfrac{1}{1+x^{\tfrac{n}{\ln(n+3)}}}$.

I'm trying to show that $\lim_{n \rightarrow \infty} \int \dfrac{1}{1+x^{\tfrac{n}{\ln(n+3)}}} d\mu = \int \lim_{n \rightarrow \infty}\dfrac{1}{1+x^{\tfrac{n}{\ln(n+3)}}} d\mu $.
I'm not sure if I can use the dominated convergence theorem here, because of the $\ln$ .

But here are some steps I've done:

Notice that

$f_n(x) \leq \frac{1}{1+x^2}$ for $n \geq 4$.

Take $g(x) =\frac{1}{1+x^2}$. Then we have $g \in L^1 $ (not sure about this step ).

Even if we can use the DCT, I'm not sure what $f$ would be.

Thanks in advance.

答:
$g(x)=\frac{1}{1+x^2}$ for $|x|\ge 1$ and $g(x)=1$ for $|x|\lt 1$, so $g(x) \in L^1$ can be used.  The limit is $f(x)=1$ for $|x|\lt 1$ and $f(x)=0$ for $|x|\gt 1$.  $f_n(1)=\frac{1}{2}$ not important.

Presume: $d\mu =dx$.


问:
Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.
Let $F$ be a field and let $f(x) \in F[x]$. Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the
quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.

$\textbf{My Attempt:}$

prove "$\implies$":

Assume $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$.

Then, $(f(x) + \langle x^2 − x \rangle) \cdot (f(x) + \langle x^2 − x \rangle) = f(x)f(x) + \langle x^2 − x \rangle = 1 + \langle x^2 − x \rangle$.

Then, $f(x)f(x) = 1$ $\implies$ $f(x) \neq 0$.

Then, neither $f(0)$ nor $f(1)$ can be equal to $0$. So does $f(0)f(1)$. So, $f(0)f(1) \neq 0$.

prove "$\Longleftarrow$":

Assume $f(0)f(1) \neq 0$. Then, $f(0) \neq 0$ and $f(1) \neq 0$.

Which I am getting stucked on how to related $f(0) \neq 0$ and $f(1) \neq 0$ to proving that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring.

Also, I don't know if the prove of "$\implies$" is correct as well.

答:
Hint: as for integers, $f$ is coprime to $\,x(x\!-\!1)\color{#c00}\iff f$ is coprime to $x$ & $x\!-\!1,\,$ cf. here, so

$$\begin{align}
&(1)=(f)\ \,{\rm in}\,\ F[x]/(x(x\!-\!1))\\[.2em]
\iff\ & (1)=(f,\,x(x\!-\!1))\,\ {\rm in}\,\ F[x]\\[.2em]
\color{#c00}\iff\ & (1)=(f,x)=(f(0),x),\ (1) = (f,x\!-\!1)=(f(1),x\!-\!1)\\[.2em]
\iff\ & f(0)\neq 0\neq f(1),\ {\rm by\ eval\ at}\ x=0,1,\ \&\,\ K\ \rm field\\[.2em]
\iff\ & f(0)f(1)\neq 0,\ {\rm by}\ K\ \rm domain
\end{align}\qquad$$

Remark $ $ employing CRT this multiplicativity of coprimality can be expressed more structurally, viz. the CRT factorization $\,F[x]/(x(x\!-\!1))\cong F[x]/x\times F[x]/(x\!-\!1)\,$  descends to unit groups, hence $f$ is a unit mod $\,x(x\!-\!1)\iff f$ is a unit mod $x$ & $x\!-\!1\iff f(0)\neq 0\neq f(1)$


问:
Is map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$ surjective on a hemisphere?
I have a map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$. It seems intuitve believe that $f(\mathbb{D}^2)=\mathbb{S}^2_+$ or $f(\mathbb{D}^2)=\mathbb{S}^2_-$ (hemispheres north and south), but I could not to prove it.

Could you give me a clue?

Thank you

答: Think of $\Bbb{S}^2$ as the surface of a tennis ball and think of $\Bbb{D}^2$ as a sock of about the same diameter as the tennis ball. Put the tennis ball in the bottom of the sock with the south pole at the toes and tie a piece of string around the sock so that it neatly traps the whole tennis ball with the knotted string at the north pole. Now fold the rest of the sock into tucks so that the edge fits snugly around the equator of the tennis ball (if you choose a short enough sock you will only need one fold). This gives you a picture of a surjection of $\Bbb{D}^2$ onto $\Bbb{S}^2$ that identifies $\partial\Bbb{D}^2$ with $\Bbb{S}^1$ in the standard way.


答: You are correct. The equation has no solutions. Not even for complex $x$. </p>

Title: If $2^x=0$, find $x$.

Body:
If $2^x=0$, find $x$.

Solution: I know range of $2^x$ function is $(0,\infty)$.

So $2^x=0$ is not possible for any real value of $x$

Hence, equation is wrong. We can't find value of $x$. Am I right?

Please help me.

Can $x$ be in $[-\infty,\infty]$?

i.e is $2^x=0$ possible for $x=-\infty$?


答:
Divisibility by $7$ is congruence to zero modulo $7.$
So we might get some insights by looking at the numbers' congruences mod $7$.

Note that $x^{2^{k+1}} = x^{2^k\cdot 2} = \left(x^{2^k}\right)^2.$
That is, every time we add $1$ to the exponent $k$ in
$2^{2^k}$ or $4^{2^k}$, we square the number.

So let's try applying these two ideas: do the arithmetic modulo $7$; and try the first few values of $n$ and see what happens.

So what happens is
\begin{align}
2^2 \equiv 4 \pmod7,\\
4^2 \equiv 2 \pmod7.\\
\end{align}

That is,
\begin{align}
\text{for } n &= 1,
& 4^{2^1} + 2^{2^1} + 1 = 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\text{for } n &= 2,
& 4^{2^2} + 2^{2^2} + 1 \equiv 2^2 + 4^2 + 1 &\equiv 4 + 2 + 1 \pmod7,\\
\text{for } n &= 3,
& 4^{2^3} + 2^{2^3} + 1 \equiv 4^2 + 2^2 + 1 &\equiv 2 + 4 + 1 \pmod7,\\
\end{align}
and so forth.

You should now know the congruence classes of $4^{2^n}$ and $2^{2^n}$ for every $n$ and be able to prove the obvious pattern of congruences for alternating odd and even $n$ by induction.
This is just a little less elegant that some of the other proofs in that you may find it necessary to have two cases in the inductive step, one for odd $k$ and one for even $k.$

Title: Divisibility by 7 Proof by Induction


答:
Other answers have shown how you could continue with what you had. But just for interest, here's another proof, which doesn't use the difference, but the quotient:

Let's call the formula

$$ a_n = 4^{2^n} + 2^{2^n} + 1 $$

You already showed that $a_1 = 21$ is a multiple of $7$. (So is $a_0 = 4^1+2^1+1 = 7$, so we can actually prove it for every non-negative integer $n$.)

With so many $2$'s involved, a substitution $4=2^2$ seems worth trying:

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

The multiple exponent operations are a bit confusing, but we can abstract them out by noticing $2^{2 \cdot 2^n} = (2^{2^n})^2$, so if we define $x_n = 2^{2^n}$, then

$$ a_n = x_n^2 + x_n + 1 $$

Take $7 | a_n$ as the inductive hypothesis, and look at the next iteration:

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

$$ a_{n+1} = x_{n+1}^2 + x_{n+1} + 1 = x_n^4 + x_n^2 + 1 $$

But this factors as

$$ a_{n+1} = (x_n^2 - x_n + 1) (x_n^2 + x_n + 1) = (x_n^2 - x_n + 1) a_n $$

Since $x_n$ is a positive integer, $(x_n^2 - x_n + 1)$ is also a positive integer. Since $7 | a_n$, we also have $7 | a_{n+1}$. Therefore by induction, $7 | a_n$ for every natural $n$.

Title: Divisibility by 7 Proof by Induction


问:
Placing coloured pieces on graph edges such that every cycle has at least one piece of each colour
Given a connected graph with N nodes, E edges, and Blue and Red pieces.

Any number (0, 1, 2, ...) of Blue and Red pieces can be placed on any edges.

What’s the minimum number of pieces needed to ensure that all cycles include at least one Blue piece and one Red piece?

Can anything be concluded about the pieces placing strategy to get minimum number of pieces?

Can anything be concluded about Blue-to-Red pieces ratio in the minimum pieces configuration? E.g. equal number of red and blue pieces.

答:
If we deleted every edge with a blue piece on it, we would be left with an acyclic graph; therefore we can be left with at most $N-1$ edges. Therefore at least $E-N+1$ edges need to have a blue piece on them, and similarly at least $E-N+1$ edges need to have a red piece on them.

This lower bound can be achieved, and in general the method to do it is to:

Pick a spanning tree $T$, and put a blue piece on every edge not in $T$.

Pick a spanning tree $T'$ (possibly the same as $T)$, and put a red piece on every edge not in $T'$.


问:
Probability of drawing balls from an urn when the ball color is not known
There are two balls in an urn. Their colors are unknown - they may be black or white. On the first turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.

On the second turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.

We now know that at least one of the balls in the urn is white. We then draw a third time. What is the probability that this ball is white.

To solve this problem - my first instinct is to say that it must be pretty close to fifty percent. It is definitely possible to draw the same ball twice in a row - 25% chance of this. And, there is a 100% chance one ball is white, with a 50% chance the other ball is white as well. How can I set up and solve a problem using this information?

答:
It's necessary to make some assumption about the a priori probability regarding the balls in the urn.  In the solution below, I assume that before beginning the draws, each of the two balls could be black or white with equal probability.

Consider $16$ equally likely trials.  In $4$ trials both of the urn's balls are black.  In $4$ trials both of the urn's balls are white.  In $8$ trials the urn has a black ball and a white ball.

In the $8$ trials where the urn has a black ball and a white ball, you will have $2$ trials where you have drawn $2$ white balls.  In the $4$ trials where the urn has two white balls, you will have $4$ trials where you have drawn $2$ white balls.

Therefore, there is a $\frac 13$ probability that you are in a trial where the urn has a white ball and a black ball, which means there's a $\frac 16$ probability that the next ball you draw will be black, so the probability that your third draw will be a white ball is $\frac 56$.


问:
Outside Temperature- First-Order, Linear, Initial Value Differential Equation
I am trying to solve a first-order initial value differential equation. Here is the equation:

Is this separable? I feel like you could solve it using an integrating factor where P(x)=k and Q(x) is equal to everything on the right side of the equation.  But is P(x) allowed to be a constant?
I can't seem to get the right answer.

答:
Write the equation $\dot{u}(t)+k u(t) = C(t)$, then
$e^{kt}(\dot{u}(t)+k u(t)) = e^{kt}C(t)$ and letting
$w(t)=e^{kt} u(t)$ we see that
$\dot{w}(t) = e^{kt}C(t)$, which is just an integration.


答:
There are three possibilities: both balls black, one white one black, or both balls white. We will name these possibilities BB, BW, WW.

For each possibility, we find the probability of the observed phenomenon happening. We call this observed phenomenon $\theta$.

For BB, the probability is clearly $0$. So, $p(\theta\mid BB)=0$.

For BW, the probability is $0.5\cdot0.5=0.25$. So, $p(\theta\mid BW)=0.25$.

For WW, the probability is $1\cdot1=1$. So, $p(\theta\mid WW)=1$.

Now, we have to start with some prior distribution, aka, what we think the likelihoods of each of the three possibilities occurring, if we weren't given the information about the first two draws. Let's say that my prior distribution is binomial, i.e, there's a 25% chance of BB, WW ($p(BB),p(WW)=0.25$), and 50% chance of BW ($p(BW)=0.5$). (Why am I choosing this? The question hasn't specified a prior distribution, making it ill-defined. But, I think this is a reasonable choice).

Given our prior, we can compute the posterior distribution by computing $p(\theta\mid X)p(X)$ for each possibility $X$, and then computing $$p(X)=\frac{p(\theta\mid X)p(X)}{\sum_Yp(\theta\mid Y)p(Y)}$$for each $X$. We have $p^*(BB)=0$, $p^*(BW)=\frac13$, $p^*(WW)=\frac23$.

To finish, observe that the probability of a white ball from $BW$ is $\frac12$, and $1$ from $WW$. So, our answer is $p^*(BW)\cdot\frac12+p^*(WW)\cdot1=\frac13\cdot\frac12+\frac23\cdot1=\frac56$.

Title: Probability of drawing balls from an urn when the ball color is not known

Body:
There are two balls in an urn. Their colors are unknown - they may be black or white. On the first turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.

On the second turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.

We now know that at least one of the balls in the urn is white. We then draw a third time. What is the probability that this ball is white.

To solve this problem - my first instinct is to say that it must be pretty close to fifty percent. It is definitely possible to draw the same ball twice in a row - 25% chance of this. And, there is a 100% chance one ball is white, with a 50% chance the other ball is white as well. How can I set up and solve a problem using this information?


问:
Express function as a difference of two non-negative functions
Given a function $f : X \to\mathbb{R}$, where $X$ is any set, one way to write $f$ as a difference of two non negative functions is defining $f_1, f_2 : X \to \mathbb{R}$ as

\begin{align*}
f_1 (x) := \max(f(x),0), \quad f_2(x) := \max(-f(x),0)
\end{align*}

and expressing $f(x) = f_1(x) - f_2(x)$ for all $x \in X$.

Is there any other way to do the same? And does the answer change if we let $(X,M,\mu)$ be a measure space, and assume $f$ to be measurable with respect to the measure space?

答: At least $f_1+g,f_2+g$ where $g$ is a non negative map will also work.


问:
Is there a formal definition of "continuous at a point for all practical purposes" or "continuous at a point within a given tolerance"?
Consider a function $f\colon D \to \mathbf{R}$ such that at some inner point $a$ of its domain $D\subseteq \mathbf{R}$
$$\lim_{x \to a^-}f(x) = L, \qquad \lim_{x \to a^+}f(x) =R, \qquad L \ne R \ ,$$
and the value $f(a)$ may be equal to either limit or have a yet different value.

Obviously the function isn't continuous at $a$.

But suppose that we are studying it with some numerical method, and the difference between $L$, $R$, and $f(a)$ is less than the precision used in the method. Say, for example, that $R=2+ \mathrm{e}^{-1000}$ and $L=2$. From the point of view of the numerical method the function can, so to speak, be considered continuous at $a$. A similar situation can arise with functions meant to fit empirical data.

Is there some notion of "approximate continuity at $a$" that applies in this case? An intuitive definition would be something like $\lvert \lim_{x \to a^-}f(x) - \lim_{x \to a^+}f(x)  \rvert <\delta$ and $\lvert \lim_{x \to a^-}f(x) - f(a) \rvert <\delta$ and so on, for some tolerance $\delta>0$. But I have no idea whether some definition of this kind is used in computational mathematics and how it is formally called.

I'd be thankful for any references you know of about some notion similar to this.

Note: The expression "approximate continuity" I use here has nothing to do with the definition of approximate continuity connected with measure theory, of course; but I couldn't find a better expression. Also, I'm aware that in computable analysis all computable functions are continuous.

答:
You can define the oscillation of a function at a point as follows:

Let $f: I \to \Bbb R$ be a bounded function where $I \subset \Bbb R$. Let $a \in I$. Then, for $\delta > 0$, define

$$M(f, a, \delta) = \sup_{|x - a| < \delta}f(x) \quad\text{and}\quad m(f, a, \delta) = \inf_{|x - a| < \delta}f(x).$$

Then, the oscillation is defined as $$o(f, a) := \lim_{\delta \to 0^+}[M(f, a, \delta) - m(f, a, \delta)].$$

(This limit must exist. Why?)

Note that $f$ is continuous at $a$ iff $o(f, a) = 0$. Now, for your purpose, you can define "continuous within tolerance $\epsilon$" by saying that "$o(f, a) < \epsilon$ for all $a \in I$."

Edit: If you don't particularly care about the value of the function at $a$ itself, you can modify the definitions of $M$ and $m$ to take the $\sup$/$\inf$ over those $x$ such that $0 < |x - a| < \delta$.


问:
Proving a property of relatively pseudocomplemented lattices
Goldblatt's "Topoi" suggests proving $(a \Rightarrow b) \sqcap (b \Rightarrow c) \leq (a \sqcup b) \Rightarrow c$, where $a \Rightarrow b$ is the pseudocomplement of $a$ relative to $b$ (exercise 8.3.16). I see why this holds in a distributive lattice, but how to prove this for a general lattice?

答:
Let $L$ be a lattice in which the expression
$$x \to y = \max \{ z \in L : x \wedge z \leq y \}$$
defines an operation, that is, for every $a,b \in L$, the set $\{ c \in L : a \wedge c \leq b\}$ has a maximum element.

Suppose that $L$ has a copy of either $N_5$ or $M_3$ as sub-lattices.

You can see versions of these lattices here.

With the same notation, let us try to compute $x \to z$, in $N_5$.
We have
$$x \to z = \max\{ u \in N_5 : x \wedge u \leq z \}.$$
Now,
$$\{ u \in N_5 : x \wedge u \leq z \} = \{ 0, y, z \},$$
and this set has no maximum element.

Also with the same notation, in $M_3$, we see that $x \to z$ is the maximum of the set $\{ 0, y, z \}$ which also has no maximum (with the order of $M_3$).
This means that these lattices are not pseudo-complemented.

Now, in the same section linked above, a characterization of distributive lattices is given has those lattices $L$ such that neither $M_3$ nor $N_5$ are isomorphic to sub-lattices of $L$.

So if $L$ is not distributive, it has either $M_3$ or $N_5$ as a sub-lattice (or an isomorphic copy).

Then, as we saw above, there is no pseudo-complement operation definable in the interval given by $[u,v]$, where $u$ is the minimum of that sub-lattice and $v$ the maximum.

But then $L$ is not relatively pseudo-complemented.

Hence, if $L$ is relatively pseudo-complemented, then (by definition) every interval in $L$ is pseudo-complemented, and so neither $N_5$ nor $M_3$ are sub-lattices of that interval, and therefore, not sub-lattices of $L$, and so $L$ is distributive.

Now, as you claim that you can prove the desired equality for distributive lattices, this is enough.


问:
Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too

答:
$f(x)$ has period $2$.

What happens in $[0,2]$ is repeated $50$ times in the same way, thus the ratio we have in this interval is the ratio of all the race.

Run points are $1/4$ and swim point are $3/4$, therefore the ratio $swim/run$ is $3$.

How did I find the points in the image below.
Solve
$$(\sin (2 \pi  x)-3) \sin \left(\frac{\pi }{4}-\pi  x\right)-\sin \left(3 \pi  x+\frac{\pi }{4}\right)=2 \sqrt{2}$$
solutions are $x=\frac12+2k;\;x=1+2k;\;k\in\mathbb{Z}$.

In the interval $[0,2]$ the are $\frac12,1$.

Solve
$$(\sin (2 \pi  x)-3) \sin \left(\frac{\pi }{4}-\pi  x\right)-\sin \left(3 \pi  x+\frac{\pi }{4}\right)=0$$
solutions are $x=\frac14+2k;\;\frac54+2k;\;k\in\mathbb{Z}$

In the interval $[0,2]$ the are $\frac14,\frac54$.

The set where $0\le f(x)\le 2\sqrt 2$ is $\left[\frac14,\frac12\right]\cup\left[1,\frac54\right]$

The length is $\frac12$ and represents $\frac14$ of the interval $[0,2]$ while the other part is $\frac34$. Thus the second part is three times the first part.

$$...$$


问:
Uniform continuous function is bounded by a linear function.
I was solving the next exercise:

Let $f:\mathbb{R}\to\mathbb{R}$ be an uniformly continuous function. Prove that there exist $a,b\in\mathbb{R}^{+}$ such that, for all $x\in\mathbb{R}$, $|f(x)|\leq a|x|+b$.

I found the proof here and is beautiful but I can't understand the first step, i.e., the part where the proof says that if $|y|< k\delta$, with $k\in\mathbb{N}$, by the triangle inequality, then $|f(0)-f(y)|<k$. I tried to prove using the next argument: $$|f(0)-f(y)|=\left|f(0)-f\left(\frac{y}{k+1}\right)+f\left(\frac{y}{k+1}\right)-f\left(\frac{y}{k+2}\right)+f\left(\frac{y}{k+2}\right)+\cdots-f\left(\frac{y}{2k}\right)+f\left(\frac{y}{2k}\right)-f(y)\right|$$Clearly $\left|\frac{y}{k+1}\right|<\left|\frac{y}{k}\right|<\delta$. Moreover, if $n\in\{1,\dots,k-1 \}$  $$\left|\frac{y}{k+n}-\frac{y}{k+n+1} \right|=\left|\frac{y}{(k+n)(k+n+1)} \right|<\left|\frac{y}{k}\right|<\delta$$If we apply triangle inequality a lot of times we obtain $$|f(0)-f(y)|\leq \left|f(0)-f\left(\frac{y}{k+1}\right) \right|+\left| f\left(\frac{y}{k+1}\right)-f\left(\frac{y}{k+2}\right)\right|+\cdots+\left|f\left(\frac{y}{2k}\right)-f(y) \right|$$All terms are less thant $\delta$ except the last one. In fact, I can't bound that difference. Not in a direct way. What I am missing? I really aprecciate any help. Thanks.

答:
Your idea is in the right direction, you just need the correct representation.
Try writing:

$$|f(0)-f(y)| = \\
\left|
\left(f(0)-f\left(\frac y{k}\right)\right)
+\left(f\left(\frac y{k}\right) - f\left(\frac {2y}{k}\right)\right)
+\left(f\left(\frac {2y}{k}\right) - f\left(\frac {3y}{k}\right)\right)
+\\ \dots
+\left(f\left(\frac {(k-1)y}{k}\right) - f\left(y\right)\right)
\right|$$

In other words, a telescoping sum:

$$|f(0)-f(y)| = \left|\sum_{n=0}^{k-1} f\left(\frac n k y\right) - f\left(\frac{n+1}k y\right)\right|$$


问:
Reduce formular to triangular number
This formula is given:

$$ n + (n-1) + (n-2) + (n-3) + (n-4) + \ldots + 1  \geq 100. $$

In the textbook it is written, the above is reduced using the triangular formula to

$$ n (n+1) / 2  \geq 100. $$

But why can we reduce it? The general triangular formula is:

$$ T_n = \sum^n_{k=1} k = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}. $$

In my case the formula would look like this:

$$ T_n = \sum^1_{k=n} k = n + (n-1) + (n-2) + \ldots + 1 = \frac{n(n+1)}{2}. $$

Can I just say that it is equivalent?

答:
Addition is commutative, so it doesn't matter in what order a finite set of numbers is added. What matters is the multiset of terms. In both cases in your question the multiset is

$$
\{1, 2, \dots, n-1, n\}.
$$

Title: Reduce formular to triangular number


答: Addition is commutative, so $1+2+\ldots+(n-1)+n = n+(n-1)+\ldots + 2+1$.

Title: Reduce formular to triangular number

Body:
This formula is given:

$$ n + (n-1) + (n-2) + (n-3) + (n-4) + \ldots + 1  \geq 100. $$

In the textbook it is written, the above is reduced using the triangular formula to

$$ n (n+1) / 2  \geq 100. $$

But why can we reduce it? The general triangular formula is:

$$ T_n = \sum^n_{k=1} k = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}. $$

In my case the formula would look like this:

$$ T_n = \sum^1_{k=n} k = n + (n-1) + (n-2) + \ldots + 1 = \frac{n(n+1)}{2}. $$

Can I just say that it is equivalent?


问:
Lebesgue integration and Riemann integration of a characteristic function
Let $E$ be a closed nowhere dense set on $[0,1]$.

a) Is it true that $\chi_E$ is Lebesgue integrable?

b) is it true that $\chi_E$ is Riemann integrable if and only if $m(E)=0$?

I know that $f$ is Riemann integrable if and only if $f$ is continuous except on a set of measure 0.

By definition of nowhere dense - the interior of the closure  is empty - I think that b) is true. But not sure about a).

Also, please let me know if you think my reasoning for b) is incorrect.

Thanks.

答: a. $E$ is closed and therefore measurable. As it is also included in $[0,1]$ which has a finite measure, $\chi_E$ is Lebesgue integrable.


问:
Understanding Quadratic Variation
I think part of the trouble a lot of people (or at least me personally) have with making the jump from calculus to stochastic calculus is the notion of quadratic variation. It doesn't have as much visual, or intuitive, appeal as "the area under a graph" or the "instantaneous rate of change".

Anyone come across a really good way to think about what quadratic variation is and what it measures?

I've heard that quadratic variation of a process over $[0,T]$ quantifies the accumulated randomness of the process, but even that is a bit difficult to unpack.

It's pretty easy to prove that a process which is continuously differentiable over an interval has zero quadratic variation over that interval. In the case of a stochastic process, if the process is cont. differentiable with respect to time over an interval $[0,T]$, doesn't this sort of imply a sense of "non-randomness" (probably bad choice of words)?

In other words if we have a stochastic process $M(t)$, and if we know that $lim_{s\rightarrow t^+} \frac{M(s) - M(t)}{s-t} = c<\infty$ (limit from the left), then if we're standing at $s<t$, we know how much $M(s)$ will move for an infinitesimally small $\Delta t$. In this sense, we can "predict" how $M(s)$ will move in the extremely short future based on the info we have at time $s$. Hence the "nonrandomness" interpretation. So a random process which has continuously differentiable sample paths over $[0,T]$ can be "predicted" for tiny, tiny future increments at each point $t\in [0,T]$ and accumulates no "randomness".

I'm sure this is wrong on many levels, but any insight or comments would be appreciated.

答:
Quadratic variation is the reason why you can't approximate a "jumpy" function as a 1st order taylor expansion, no matter how small you make the interval. It won't happen. Because the function "jumpiness" is always of higher "order" than your small interval. That is, it looks jumpy no matter at which micro/macroscopic scale you try to look at it. QV measures this jumpiness. In contrast, you can refine your scale for a well behaved function so as to make it small enough that the function looks absolutely smooth in that small enough interval, thus allowing you to approximate the increment with a first order expansion.

The practical consequence is that it is inadequate to describe the increment in this jumpy (say convex for arguement's sake) function in any arbitrarily small interval by just the first order expansion. The jumps (say for instance, 1 up and 1 down in the chosen interval) and convexity mix together, as the convexity places a higher value on the up jump rather than the down jump. These jump-convexity effects accumulate systematically from micro to macroscopic level. Note that the same essentially happens even for well behaved functions; but in those cases the small interval chosen squeezes this effect to a negligible amount. But in the jumpy case, the small interval cannot squeeze it - remember, the jumpiness is of higher order than the interval length.

This $jumpiness*convexity$ is exactly the extra drift term you get in Ito's lemma. Remember how we said that QV measures jumpiness. The extra term is simply $1/2*convexity*QV$.

In short, quadratic variation is the measure of "noisiness/jumpiness" of the function with respect to any arbitrarily chosen $x$ interval.

Title: Understanding Quadratic Variation


问:
Describe image under mapping $w=\frac{z-i}{z+i}$ of set $\{z=x+iy : x>0, y>0 \}$
I'm new to complex analysis. In polar form, when I plug in a general $z=re^{i\theta}$ to $w$, I end up with $$w(z)=\dfrac{r^2-2ir\cos\theta-1}{r^2+2r\sin\theta+1}$$

I know since $x,y>0$, I have $0 < \theta < \frac{\pi}{2}$, but that's all I can tell.

Where do I go from here?

答:
This is a bilinear or Mobius transformation. f(z) = $\frac{az+b}{cz+d}$. I'm a fan of Linear Geometry by Rafael Artzy, which covers this in detail.  For example, f(z) = az+b has w = b/(1-a) as an invariant (fixed) point.  It is a rotation through angle a about w.  An inversion is N(z) = 1/z. A translation T(z) = z+b is labeled $T_b$. If $\delta = ad-bc$, and f = $\frac{-\delta}{c^2}$, then a dilatation is $D_f$ and the full Mobius transformation B = $T_{d/c} N D_f T_{a/c} $. For your example, $\delta = 2i$.
Mobius transformations map the set of all straight lines and circles into the set of all straight lines and circles.


答:
Why even bother with polar coordinates?

$$w = \frac{x +(y-1)i}{x + (y+1)i} \times
\frac{x -(y+1)i}{x - (y+1)i} ~=~
\frac{(x^2 + 1 - y^2) - i(2x)}{x^2 + (y+1)^2}.\tag1$$

Note that by multiplying through by the complex conjugate of the denominator, I created a fraction whose denominator is a real number.

Addendum
Per OP's request

Under the assumption that $(x + iy) = z = re^{i\theta}$ convert equation (1) into polar coordinates, involving $r$ and $\theta$.

This is extremely ugly.  Frankly, the analysis in RobertTheTutor's answer is way over my head, so my approach would be to do things from scratch.

Since the original constraint is $0 < x,y$, you can assume that $0 < \theta < (\pi/2).$

Further, you have that $r^2 = |z|^2 = (x^2 + y^2)$,
while
$|w|^2 = \frac{(x^2 + 1 - y^2)^2 + 4x^2}{[x^2 + (y+1)^2]^2}.$

Assuming that you wish to express $w = se^{i\alpha}$,
you will have that $s^2 = |w|^2$ and that $\alpha$ is the unique angle, within a modulus of $(2\pi)$ such that

$\cos(\alpha) = \frac{x^2 + 1 - y^2}{|t|}$ and
$\sin(\alpha) = \frac{-2x}{|t|},~$ where
$t^2 = (x^2 + 1 - y^2)^2 + (-2x)^2.$

Here, since $r^2 = (x^2 + y^2)$, and since $\cos(\theta) = \frac{x}{r}, \sin(\theta) = \frac{y}{r}$, the challenge would be to somehow:

first express $s$ in terms of $(r)$ and trig functions involving $(\theta)$ and then express $\cos(\alpha), \sin(\alpha)$ in terms of $s,r$ and trig functions involving $(\theta).$

My response is based on nothing more than my surviving Chapter 1 of "An Introduction To Complex Function Theory" (Bruce Palka).  Palka does discuss mobius transformations, but the pertinent problems (involving mobius transformations) at the back of chapter 1 are structured very narrowly.

Palka does mention that he will discuss mobius transformations in later chapters, but it is a safe bet that this is no walk in the park.  I question the intent of your instructor.

Do you really think that if $z = re^{i\theta}$, that (for example) either $(z - i)$ or $\frac{z-i}{z+i}$ would be so easily expressible in polar coordinates involving $r$ and $\theta$?

If there is an elegant way to do this, I'd certainly like it explained to me, in terms compatible with someone who has only survived chapter 1 of the book.

This addendum is nothing more than an elaboration that if an expression for $w$ is required in polar coordinates involving $r$ and $(\theta)$ then I am way out of my depth.

About the only thing that I can say for sure, is that based on equation (1) above, $\sin(\alpha) < 0$, and the sign of $\cos(\alpha)$ will obviously be based on the sign of the expression $(x^2 + 1 - y^2).$

A cursory examination of equation (1) above does not yield any obvious insights to me, re constraints on $|w|^2.$

Title: Describe image under mapping $w=\frac{z-i}{z+i}$ of set $\{z=x+iy : x>0, y>0 \}$

Body:
I'm new to complex analysis. In polar form, when I plug in a general $z=re^{i\theta}$ to $w$, I end up with $$w(z)=\dfrac{r^2-2ir\cos\theta-1}{r^2+2r\sin\theta+1}$$

I know since $x,y>0$, I have $0 < \theta < \frac{\pi}{2}$, but that's all I can tell.

Where do I go from here?


问:
If $2^x=0$, find $x$.

If $2^x=0$, find $x$.

Solution: I know range of $2^x$ function is $(0,\infty)$.

So $2^x=0$ is not possible for any real value of $x$

Hence, equation is wrong. We can't find value of $x$. Am I right?

Please help me.

Can $x$ be in $[-\infty,\infty]$?

i.e is $2^x=0$ possible for $x=-\infty$?

答:
Yes. No solution for the equation $$ 2^x = 0 $$ is defined.

This can be proved using the definition of logarithms.

Title: If $2^x=0$, find $x$.


问:
Sudoku based problem. Counting with permutations?
I have been playing Sudoku lately, and I've seen a problem with this general form:

Given N items, each of which belongs alone (1-to-1) in one of N boxes, and given a list of statements of the form:
"item x may go in box y"
So that:

If no statement is in the list connecting item x to a box y, the item a certainly does not belong in box y.

Between each item and the box it belongs in, there is a statement in the list connecting them.

The statements described in 2 are a subset of all the statements in the list. (there are extra ones)

Is there an efficient process for reducing the number statements by getting rid of contradictory ones?

For instance, if a given item can only go in one box, all other statements about items that may go in that box can be erased. If a given box only has one item that can go in it, all statements connecting that item to other boxes can be erased. These generalize to subsets as well: If two items are the only items that may be in either of two boxes, all statements connecting those two items to boxes other than the two can be erased.

Are there more rules? I din't know.

It comes up in Sudoku when looking at a given block, row or column. The numbers not yet filled in are the items, the cells not filled in are the boxes, and the candidates (small numbers you write in cell that may be there) are the statements. Often, candidates can be reduced by seeing contradictions in them.

I'm just wondering if anyone knows what this problem is called, so I can read up on it. I'm not much of a math guy, and I haven't been able to find it through Internet, so any help is appreciated. I know the Sudoku version can be solved through enumeration, but I'm also curious about for N greater than 9, the general problem. Thanks!

答:
You are describing a Constraint-Satisfaction Problem (https://en.wikipedia.org/wiki/Constraint_satisfaction_problem). In the last few years, it was shown that every CSP is either $\textsf{NP}$-complete or in $\textsf{P}$. This result was proven independently by Zhuk and (IIRC) Bulatov.

The Generalized Sudoku Problem is also known to be $\textsf{NP}$-complete (http://www.cs.ox.ac.uk/people/paul.goldberg/FCS/sudoku.html), and determining the minimum number of clues needed also seems to be $\textsf{NP}$-hard (https://people.math.sc.edu/cooper/criticalsets.pdf).


问:
What is the lower bound of $k$ such that any $k$-regular graph $G$ on $n$ vertices contains a $t$-clique?
This is a Ramsey Theory problem I've been playing around with for a few days. It's possible there is a really obvious solution I haven't seen yet.

I am trying to find a lower bound $s$ such that $k > s$ implies any $k$-regular graph $G$ on $n$ vertices is guaranteed to have a $t$-clique. I denote the lower bound $s$ as $s(t)$.

It is not difficult to show that $s(3)=\frac{n}{2}$. So $k>s(3)=\frac{n}{2}$ implies all $k$-regular graphs on $n$ vertices contain a 3-clique. I conjecture that $\frac{n}{2}+1\leq s(4)\leq \frac{2n}{3}$. Clearly $s(t) \leq n-1$ for any $t$. Note that I am only focusing sufficiently large $n$ to avoid any errors, such as $n \geq t$.

答:
Without any restrictions on the structure of the graph, the graph with the most edges that avoids a $t$-clique is the Turán graph, which is a complete $(t-1)$-partite graph whose parts are as equal as possible.

In particular, if $n$ is divisible by $t-1$, we can have a complete $(t-1)$-partite graph with $t-1$ parts of size $\frac{n}{t-1}$, which is regular of degree $(1 - \frac1{t-1})n$. So in this case, we have $s(t) = (1 - \frac1{t-1})n$: any regular graph with higher degree will have more edges than the Turán graph, so it will contain a $t$-clique.

That's the easy answer. When $n$ is not divisible by $t-1$, this is still an upper bound on $s(t)$, but things can get messier...

Even $s(3)$ is interesting for odd $n$: it is not just $\frac{n-1}{2}$. I can prove that for odd $n$, $s(3) \le \frac25n$, which is achievable for multiples of $5$: just take a blow-up of the $5$-cycle. I don't know what happens when $n$ is odd and not divisible by $5$.

Pick a vertex $u$ in an $s$-regular triangle-free graph $G$. Its neighborhood $N(u)$ must be an independent set of size $s$. If $v \in N(u)$, then its neighborhood $N(v)$ is another, disjoint independent set of size $s$. Greedily extend $N(u)$ and $N(v)$ to a maximal pair of two disjoint independent sets $S, T$ with $|S|, |T| \ge s$: every vertex outside $S \cup T$ should have neighbors in both $S$ and $T$.

Because $n$ is odd and $G$ is regular, it cannot be bipartite, but $G[S \cup T]$ certainly is bipartite. So there must be some vertex $w \notin S \cup T$, with neighbors both in $S$ and in $T$.

Since $|N(w) \cap S| + |N(w) \cap T| \le |N(w)| = s$, without loss of generality $|N(w) \cap S| \le \frac s2$. Now to finish, let $x \in N(w) \cap S$.

The two sets $N(w)$ and $N(x)$ are disjoint (or else we'd get a triangle) and each has size $s$. Moreover, $N(w)$ misses at least $\frac s2$ vertices of $S$, and none of those can be in $N(x)$ either (since $S$ is an independent set and $x \in S$). So altogether $n \ge |N(w)| + |N(x)| + |S - N(w)| \ge \frac52s$, or $s \le \frac25n$.

Actually, this paper shows that for $t \ge 4$, $s(t) = (1 - \frac1{t-1})n - \Theta(1)$ no matter the value of $n$: we're off from the optimal result by only a constant error. Moreover, $s(3) = \frac 25n - \Theta(1)$ for all odd $n$, so nothing worse that usual happens when $n$ is not divisible by $5$.

In general, if we replace the $3$-clique by other graphs with chromatic number $3$, and we look at the corresponding problem, it is still hard for odd $n$.

An apparently well-known result, Andrásfai's theorem, generalizes my argument for $s(3)$ by showing that if $G$ is a triangle-free graph with minimum degree $\delta(G) > \frac{2n}{5}$, then $G$ is bipartite. So of course, if $n$ is odd and $G$ is $s$-regular, since $G$ cannot be bipartite, we must have $s \le \frac{2n}{5}$.


问:
Is the hemisphere bendable?
There is a theorem in classical differential geometry due to Cohn Vossen (1927) that compact, connected regular surfaces of everywhere positive gaussian curvature (ovaloids) are rigid - in the sense that if a regular surface in Euclidean space is isometric to a given ovaloid, then it is necessarily congruent to it. This is sometimes called the congruence theorem of ovaloids.

This has been generalised by Pogorelov and Alexandrov, and based on their work it is possible to conclude the torus is rigid as well.

I was wondering, is a hemisphere rigid in the above sense? Can anyone give an example of a regular surface in Euclidean space which is isometric to a hemisphere, but not congruent to it?

It is claimed in the book: Geometry and imagination (by Hilbert and Cohn-Vossen) on page 228, that there are pieces of the sphere which can be bent, and that it becomes bendable as soon as we remove a segment of a great circle on the sphere. However, I have not been able to find any example.

(I am not interested in infinitesimal bendings, but interested in all examples of surfaces which are isometric and not congruent)

答:
An important point is that by Minding's Theorem, if you have a (not necessarily compact!) surface of constant $K=1$, then it must be locally isometric to the sphere. But you can find two different families of these just among surfaces of revolution by explicit integration. Here's a sketch:

If you're interested in working it out, see Exercise 19 in section 2.3 of my differential geometry text.

Title: Is the hemisphere bendable?


问:
Boundary and closure of a set
I'm trying to show that a point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement.

Let $\overline{E}$ denote the closure of $E$ and $E^\circ$ be the interior of $E$. Then the boundary is defined to be $\overline{E} \setminus E^\circ$.

So what I need to show is $\overline{E} \setminus E^\circ = \overline{E} \cap \overline{(S \setminus E)}$.

After looking at this link: A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement. I am confused about how the OP concluded that he/she needed to show $S \setminus E^\circ = \overline{(S \setminus E)}$.

Could someone explain why?

答:
$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Suppose that you can show that $S\setminus\int E=\cl(S\setminus E)$. Then

$$(\cl E)\setminus\int E=\big((\cl E)\cap S\big)\setminus\int E=(\cl E)\cap(S\setminus\int E)=(\cl E)\cap\cl(S\setminus E)\;,$$

which is exactly what you want.

To see where the idea comes from, note that both $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$ are subsets of $\cl E$. If $A$ is any set disjoint from $\cl E$, therefore,

$$(\cl E)\setminus\int E=(\cl E)\cap\cl(S\setminus E)\quad\text{iff}\quad A\cup\Big((\cl E)\setminus\int E\Big)=A\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$

and a clever choice of $A$ might give us sets that are easier to think about than $(\cl E)\setminus\int E$ and $(\cl E)\cap\cl(S\setminus E)$. In particular, if we let $A=S\setminus\cl E$, we can compare the set,

$$(S\setminus\cl E)\cup\Big((\cl E)\setminus\int E\Big)\;,$$

which is simply all of $S$ except $\int E$, or $S\setminus\int E$, with the set

$$(S\setminus\cl E)\cup\Big((\cl E)\cap\cl(S\setminus E)\Big)\;,$$

which is simply $\cl(S\setminus E)$, and try to show that these are equal.

Title: Boundary and closure of a set


问:
Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.
Let $F$ be a field and let $f(x) \in F[x]$. Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the
quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.

$\textbf{My Attempt:}$

prove "$\implies$":

Assume $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$.

Then, $(f(x) + \langle x^2 − x \rangle) \cdot (f(x) + \langle x^2 − x \rangle) = f(x)f(x) + \langle x^2 − x \rangle = 1 + \langle x^2 − x \rangle$.

Then, $f(x)f(x) = 1$ $\implies$ $f(x) \neq 0$.

Then, neither $f(0)$ nor $f(1)$ can be equal to $0$. So does $f(0)f(1)$. So, $f(0)f(1) \neq 0$.

prove "$\Longleftarrow$":

Assume $f(0)f(1) \neq 0$. Then, $f(0) \neq 0$ and $f(1) \neq 0$.

Which I am getting stucked on how to related $f(0) \neq 0$ and $f(1) \neq 0$ to proving that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring.

Also, I don't know if the prove of "$\implies$" is correct as well.

答:
Let me analyze your attempt, rather than offer a solution.

It's easier if you write $I=\langle x-x^2\rangle$, so to reduce the clutter.

“$\Rightarrow$” Saying that $f(x)+I$ is a unit means there exists $g(x)+I$ such that $(f(x)+I)(g(x)+I)=1+I$. You're using $f$ again, but it's wrong.

The condition now reads $f(x)g(x)+I=1+I$ and therefore $f(x)g(x)-1\in I$, so
$$
f(x)g(x)-1=(x-x^2)h(x)
$$
for some polynomial $h$. Evaluate at $0$ and $1$ and finish up.

“$\Leftarrow$” The key here is to write $f(x)=(x-x^2)q(x)+ax+b$ using polynomial division. Evaluating at $0$ yields $b=f(0)$ and evaluating at $1$ yields $a+b=f(1)$.

Now this is much simpler: you just need to find the inverse of $ax+b+I$, using the fact that $b\ne0$ and $a+b\ne0$. Why? Because $f(x)+I=ax+b+I$.

Hint: the inverse has the form $cx+d+I$, for some $c,d\in F$. Moreover $x^2+I=x+I$.


问:
values of determinants?
Let $V,W$ be finite-dimensional $\mathbb{K}$-vector spaces of dimension $n \gt 0$ and $\varphi:V\to W$ a linear map.

a) Prove: for every ordered basis $B$ of $V$ and every ordered basis $C$ of $W$ - $\varphi$ is an isomorph map of vector spaces if $det(M_{C,B}(\varphi)) \neq 0$

b) Assume, $\varphi$ is an isomorph map. Prove that for every $\lambda\in \mathbb{K}$ \  {$0$} you can find ordered basis $B$ and $C$ so that $det(M_{C,B}(\varphi)) =\lambda$

c) prove that $det(\lambda A) = \lambda^{n}det(A)$ is true for every $\lambda \in \mathbb{K},A \in \mathbb{K}^{n\times n}$

(I translated from german so if anything doesn't make sense, let me know)

I would appreciate any hints on how to solve any of these.

Update

My attempt:

a) In our script it says that $\varphi:V\to W$ is bijective and an isomorph map of vector spaces only if the Matrix $M_{B_{W},B_{V}}(\varphi)$ is invertible. And if the determinant is not 0, the matrix is invertible.

b) I'm not sure about that yet

c) My idea is to use the Leibniz formula for determinants.

$det(\lambda A) = \lambda^{n} * det(A)$

proof

$\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} \lambda a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$

extracted the $\lambda$ from the multiplication

$\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\lambda^{n} \prod_{i=1}^{n}  a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$

factored out the $\lambda$

$\lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma) \prod_{i=1}^{n}  a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$

答:
$a)$ This is fine.

$b)$ Note that if $B = \{v_1,\dots,v_n\}$ is any basis for $V$, then $C = \{\varphi(v_1),\dots,\varphi(v_n)\}$ is a basis for $W$ (since $\varphi$ is an isomorphism) and $\textsf{M}_{C,B}(\varphi)$ is the identity matrix. Using the same idea, you can obtain the $n \times n$ matrix
$$
\begin{pmatrix}
\lambda & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{pmatrix}.
$$

$c)$ This is also fine, but I suggest not start with the equality $\det(\lambda A) = \lambda^n \det(A)$, and instead just explain the steps (as you did) to arrive at $\lambda^n \det(A)$ from the expression for $\det(\lambda A)$.

Title: values of determinants?


问:
Suppose a box contains six blue and four red balls. Five balls are selected at random w/o replacement, determine the probability mass function
I know this question is pretty basic, but I am just not getting the right answer no matter what I try. My thought was that the answer should be like:
5C4/(10C5) for 4 red balls selected, (5C3)/(10C5) for 3 red balls selected, down to (5C0)/(10C5) for 0 red balls selected. When I do this, my answers never add up to 1, as is necessary for a probability mass function. What am I missing?</p>

答:
This is a hypergeometric distribution.  Here is a simple example.  Suppose I wanted to know the probability of getting three red balls.  You wrote $$\frac{\binom{5}{3}}{\binom{10}{5}}$$ but this is wrong; the correct answer is $$\frac{\binom{4}{3} \binom{6}{2}}{\binom{10}{5}}.$$  Why?  Think of the $10$ balls as being labeled like this:
$$B_1, B_2, B_3, B_4, B_5, B_6, R_1, R_2, R_3, R_4.$$  Each ball has a distinct label.  Now say we want to pick five of them without replacement so that exactly three of the five are chosen from the balls labeled $R_1, R_2, R_3, R_4$ (so they are "red"), and the other two balls would be chosen from the $B$-labeled balls (so they are "blue").  There are obviously $\binom{4}{3}$ ways to pick the $R$-labeled balls, since there are $4$ such balls in the group and you want to pick exactly $3$ of them.  Similarly there are $\binom{6}{2}$ ways to pick the $B$-labeled balls.  And those choices are independent in the sense that they each lead to a distinct choice for the five balls chosen in total.  So their product $\binom{4}{3}\binom{6}{2}$ is the numerator of our probability and counts the total number of ways to choose $5$ balls out of the $10$ such that exactly $3$ are red.  Then the denominator is just $\binom{10}{5}$ as you wrote, for the number of ways to choose $5$ balls with no restriction.

You can easily generalize this counting argument for every possible outcome of $0$ to $4$ red balls chosen.  Even though you are picking $5$ balls, since there are only $4$ red balls in the group, it is impossible to get $5$ red balls.  Now check that the probabilities add up to $1$.


答:
There is no solution for this equation, note that's why log x is defined only to $x\gt 0$.

See the graph of $2^x$ and see that this function is never $0$:

Remark

The graph is not a proof this function is never $0$, it's just to illustrate what others have said in another answers graphically.

Title: If $2^x=0$, find $x$.


答:
I have re-written this answer because, with the help of WolframAlpha here, I was surprised to discover

$$0\le \big(3-\sin(2\pi x)\big)\sin(\pi x - \frac{\pi}{4})-  \sin(3\pi x +\frac{\pi}{4})\le 2\sqrt{2}\\
\implies (2 n - 1)<x\le \frac{2 \big(\pi n - \frac{3 π}{8}\big)}{\pi} \land n \in\mathbb{ Z}\\
\land \implies \frac{2 \big(\pi n + \frac{\pi}{8}\big)}{\pi}\le x\le\frac{(4 n + 1)}{2} \land n\in\mathbb{Z} $$

What the first "implies" means for $\space 0\le x\le 100\space$ is that there are discreet ranges of values that make this true as in these samples for $1\le n\le 6$

$$(1\le x \le \frac{5}{4}=1.25)\quad
(3\le x \le \frac{13}{4}=3.25)\quad
(5\le x \le \frac{21}{4}=5.25)\quad \\
(7\le x \le \frac{29}{4}=7.25)\quad
(9\le x \le \frac{37}{4}=9.25)\quad
(11\le x \le \frac{45}{4}=11.25)\quad $$

What the second "implies" means is that the following ranges apply for $\space x$ for $1\le n\le 6$.

$$\frac{9}{4}=2.25\le x \le \frac{5}{2}=2.5\qquad
\frac{17}{4}=4.25\le x \le \frac{9}{2}=4.5\\
\frac{25}{4}=6.25\le x \le \frac{13}{2}=6.5\qquad
\frac{33}{4}=8.25\le x \le \frac{17}{2}=8.5\\
\frac{41}{4}=10.25\le x \le \frac{21}{2}=10.5\qquad
\frac{49}{4}=12.25\le x \le \frac{25}{2}=12.5$$

We can see that each value of $n$ generates $\space x\space$ with a total "coverage of $\space 0.5\qquad $ Note that $\quad 1\le n \le 24\implies 1\le x \le 97\quad$ so there are $24\times0.5=12$ "full-numbers-worth" of $0\le x \le 100$ that satisfy $I.\quad$ However, the first "implies" above permits
$n=25\implies 99\le x \le 99.25\space $ and the second "implies" peermits  $n=0.0\implies 0.25\le x \le 0.5\quad $. There are a total of $100$ intervals of $0.25$ each so the members run $\frac14$ of the time.

This means that there members swim $\frac34$ of the time and the ratio of $\frac{swim}{run}$ is $\quad 3:1$

Title: Determining when $(\sin(\theta) - \cos(\theta))(2+\sin(\theta) \cos(\theta)) \leq 2$

Body:
Question:

$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2 \qquad \text{ LHS} \\ $

Answer key:

$\implies  \frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \qquad \text{ RHS} $

I verified RHS = LHS. However, to get to the RHS part, they have factored it somehow and I can't figure out an intuitive way to do so. If anybody has a more intuitive solution (does not have to be the same as what is given here), please provide it. From this step onwards we can easily solve the question because:

$\implies  (\sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0\\ $

$ \implies  (\sin(\theta)-\cos(\theta)-1) \leq0 \because  (3 + \sqrt{2}\sin\left(\frac{\pi}{4} - \theta\right)+\sin(2 \theta)) >0 \text{ }\forall \theta \in \mathbb{R} \\ $

Doubt:

I have no idea how the went from the LHS part to the RHS part. Is there an intuitive way to solve $(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) \leq 2$? In this solution they have simply said:
$$(\sin(\theta)-\cos(\theta))(2+\sin(\theta)\cos(\theta)) -2 \leq 0 \\ \implies
\frac{1}{2}(\cos(\theta)-\sin(\theta)+2\sin(\theta)\cos(\theta)+3)(\sin(\theta)-\cos(\theta)-1) \leq 0 \\$$
Please provide a logical way to solve it. Thanks in advance.

I'm looking for a no calculator solution where each step is motivated. And prove one factor is always positive or negative and use the other factor to find the region would be appreciated. Although, if there is another method, (as long as its intuitive) that works too


问:
How to write out the gradient of $f$ evaluated at $a$?
Suppose $a, b, c$ are all vectors of length $n$. Let $f(b, c) = ||b - c||^2_2$. I want to write down the gradient of $f$ wrt to $c$ evaluated at $a$. Is the following the correct notation?

$\nabla f(a) = \left(\frac{\partial f}{\partial c_1}(a_1), \ldots, \frac{\partial f}{\partial c_n}(a_n) \right) = -2(b - a)$

Is there a more concise way of writing it, or is the above the standard way to do it?

答:
If we consider $\underline{\mathrm{b}}$ to be a constant vector, then we can write our function $f$ as
$$f(\underline{\mathrm{x}};\underline{\mathrm{b}})=\sum_{i=1}^n(b_i-x_i)^2$$
Observe ($\partial_k$ is shorthand for $\partial/\partial x_k$)
$$\partial_k f(\underline{\mathrm{x}};\underline{\mathrm{b}})=2x_k-2b_k$$
So
$$\nabla f(\underline{\mathrm{x}};\underline{\mathrm{b}})=2\sum_{i=1}^n(x_i-b_i)\underline{\mathrm{e}_i}$$
$$\nabla f(\underline{\mathrm{x}};\underline{\mathrm{b}})=2(\underline{\mathrm{x}}-\underline{\mathrm{b}})$$
We can just replace
$$\nabla f(\underline{\mathrm{a}};\underline{\mathrm{b}})=2(\underline{\mathrm{a}}-\underline{\mathrm{b}})$$
Is anything unclear?


问:
Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$
This is a math Olympiad problem.

my attempt :

By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:

$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$

By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.

答: This is another perspective: If you didn't see how completing the square would give you a circle, you could also observe that $|x^2+y^2-8x|$ is at least $10$ as long as either $|x|, |y|$ is at least $10$. That is how many Diophantine equations are solved; observing that all solutions fall into a compact space that can be searched exhaustively, if there even is a solution.


问:
A linear function that returns 8096 when x is 8, and returns 1024 where x is 65536
I'm struggling to come up with a formula that returns 8096 when x is 8 and 1024 when x is 65536. And appropriately returns the correct y when x is between those values. I think this is a simple graph equation problem but I'm struggling to come up with something nice!

Thanks a lot

答:
You start with a general approach for linear functions
$$
f(x) = mx + b.
$$
You want
$$
f(8) = m\cdot 8 + b = 8096
$$
and
$$
f(65536) = m\cdot 65536 + b= 1024.
$$
This gives you a linear system of equations
$$
8m + b = 8096\\
65536 m+ b =1024
$$
which you want to solve for $m$ and $b$.
Now $\frac{65536}{8192}=8$ so this is equivalent to

$$
8m + b = 8096\\
8 m+ \frac{1}{8192}b =\frac 1 8
$$

now subtract one equation from the other. That leaves you with an equation only involving $b$ which you can therefore solve for $b$. Now plug this value in either of the two equations above and solve for $m$. Now since you know $m$ and $b$ you found your desired function.


问:
How many obtuse angle triangles are possible in a regular Heptagon by joining its vertices?
I am only able to make one possible case,

Where we take any 3-consecutive vertices, since one of the vertices contains angle of Heptagon, which is approximately 128.57°, we get 7 such triangles.

I am unable to any others cases where obtuse angle triangles are possible.

答:
Consider all the triangle with the center of the circle circumscribing heptagon lying inside it.

Take any vertex of the $△ABC$ (let's say we take vertex $A$)

If $∠A=α$ then $∠BOC=2α$

But $∠BOC < 180°$

⇒ $∠A<90°$ (as $A$ and  $O$ lie on the same side of  $BC$)

Similarly we can prove  $∠B<90°$  &  $∠C<90°$

Using the same logic we can prove  that the triangle is obtuse if $O$ lies outside the triangle.

Therefore for $△ABC$ to be obtuse the center O should lie outside the triangle.

There are only two non-congruent triangle satisfying the above property.

We can divide that into 2 cases  (as shown in figure).

$\underline{CASE-1} $

Let $C$ be the vertex containing the obtuse angle.

$C$ can take any of the seven vertices.

So there are total of $7$ possibilities in this case.

$\underline{CASE-2} $

Let $C$ be the vertex containing the obtuse angle.

There are two triangles with $C$ as vertex & $C$ can take any of the seven vertices.

So there are total of $7×2=14$ possibilities in this case.

Therefore, the total number obtuse triangles that can be drawn$=7+14=21$


问:
Is the local connectedness heritable over the connected subspaces?
Definition

A topological space $X$ is locally connected if each point $x\in X$ has a base of connected neighborhoods.

So let be $X$ a locally connected space and let be $Y$ a connected subspace. So if $\mathcal B(x)$ is a local connected base for any $x\in Y$ then the collection $\mathcal B'(x):=\big\{B_x\cap Y:B_x\in\mathcal B(x)\big\}$ is a local base with respect subspace topology on $Y$ and thus the statement follows proving that this base is locally connected but unfortunately I did not able to do this so that I think that the statement is false becouse after all the intersection between two connected set could be not connected however I did not find any counterexample showing that the local connectedness is not hereditable over the connected subspaces. So couldsome one help me, please?

答: In fact, every topological space is a subspace of a locally connected space.  To see this, let $X$ be a topological space and let $Y$ be $X$ together with one new point $*$, topologized such that a subset of $Y$ is open iff it is either empty or has the form $U\cup\{*\}$ where $U$ is an open subset of $X$.  Then any two nonempty open subsets of $Y$ intersect, so $Y$ is rather trivially locally connected, and the subspace topology on $X$ is the topology we started with.


问:
Estimate a sum of fractions
Are there any easy ways to prove that:
$$
\sum_{j=1}^{k-1} \left({j \over k - j}\right)^{k - 2j - 1} = O(\sqrt k)
$$
or give some other estimates to the sum.

答:
This is not a math proof - just some heuristic considerations.

We can observe first that terms of the sum decline very rapidly at $j\to1$ and $\to{k}$ (at the beginning and the end of the sum)- as $\sim\left(\frac{1}{k}\right)^{\sim{k}}$. In the "middle" of the sum (at $j\to\frac{k}{2}$) they have a "maximum" $\sim\left(\frac{k}{k}\right)^{\sim0}\sim1$, so we may conclude that main contribution comes from this "middle" part of the sum.

Having said that we change the order of summation and consider $j=\frac{k-1}{2}+l$, where l goes from some negative number through zero to some positive number.

We may suppose that $l<<k$ (at $l\sim{k}$ terms are small), but $l$ may be $>>1$ and $l^2$ may be $>>k$

At $k-j=\frac{k+1}{2}-l$ and $k-1-2j=-2l$

we get

$S(k)\sim\sum_l \left(\frac{\frac{k-1}{2}+l}{\frac{k+1}{2}-l}\right)^{-2l}=\sum_l \left(\frac{\frac{2l-1}{k}+1}{\frac{1-2l}{k}+1}\right)^{-2l}$$\sim\sum_l \left(1-\frac{2}{k}+\frac{4l}{k}\right)^{-2l}=\sum_l e^{-2l\ln(1-\frac{2}{k}+\frac{4l}{k})}\sim\sum_l e^{-2l(-\frac{2}{k}+\frac{4l}{k})}=\sum_l e^{-\frac{8l^2}{k}+\frac{4l}{k}}$.

Basically, with our calculation accuracy we don't need to keep $\frac{4l}{k}$ in the power of the exponent - we will see that it won't contribute into the main asymptotics term.

$S(k)\sim\sum_l e^{-\frac{8l^2}{k}+\frac{4l}{k}}\to\int_{-\infty}^{\infty}e^{-\frac{8l^2}{k}+\frac{4l}{k}}dl=\int_{-\infty}^{\infty}e^{-\frac{8}{k}(l-\frac{1}{4})^2}\exp({\frac{1}{2k}})dl=\sqrt{\frac{\pi{k}}{8}}\exp(\frac{1}{2k})\sim\sqrt{\frac{\pi{k}}{8}}$

Supplement

If we want to get a better approximation we have to dig a bit deeper.
We can easily show that the initial sum is equal to $S(k)=S_1(k)+S_2(k)=\sum_{l=0}^{\frac{k-3}{2}}(\frac{k-1-2l}{k+1+2l})^{2l}+\sum_{l=1}^{\frac{k-1}{2}}(\frac{k+1-2l}{k-1+2l})^{2l}$. Making change in the first sum ($l\to-l$) we get $$S(k)=\sum_{l=-\frac{k-3}{2}}^{\frac{k-1}{2}}(\frac{k+1-2l}{k-1+2l})^{2l}=\sum_{l=-\frac{k-3}{2}}^{\frac{k-1}{2}}e^{2l\log\frac{k+1-2l}{k-1+2l}}$$

We are looking for the asymptotics in powers of $k$, and because the terms of the sum decrease very rapidly at $l\to\frac{k-1}{2}$ and $l\to-\frac{k-3}{2}$ (as $\sim{k}^{-k}$) we can expand the summation to +-$\infty$. Next, we want to replace summation with integration by means of Euler-Maclaurin formula $\sum_a^b{f}(l)\sim\int_a^bf(l)dl+\frac{1}{2}(f(a)+f(b))+\frac{1}{12}(f^{(1)}(b)-f^{(1)}(a))+..  $.  But we easily see that all correction terms should be calculated on the boundaries of summation, where $f(\frac{k}{2})\sim{f^{(p)}}(\frac{k}{2})\sim{k}^{p}{k}^{-k}$ and, therefore, can be omitted.

So, to obtain the asymptotic correction terms we should only deal with logarithm:
$S(k)\sim\int_{-\infty}^{\infty}e^{2l\log\frac{k+1-2l}{k-1+2l}}dl=\frac{1}{2}\int_{-\infty}^{\infty}{e}^{l\left(\frac{1}{k}-\frac{l}{k}-\frac{(1-l)^2}{2k^2}+\frac{(1-l)^3}{3k^3}-...+\frac{1}{k}-\frac{l}{k}+\frac{(1-l)^2}{2k^2}+\frac{(1-l)^3}{3k^3}+...\right)}dl$

If we want to obtain the first asymptotic correction term ($\sim\frac{1}{\sqrt{k}}$) we don't need to keep in the power of exponent the terms with power of order $\frac{l^6}{k^5}$ or higher - they will only contribute into correction terms of order $\sim\frac{1}{k\sqrt{k}}$.

We get (keeping only the terms of order $\sim\sqrt{k}$ and $\sim\frac{1}{\sqrt{k}}$):

$S(k)\sim\frac{1}{2}\int_{-\infty}^{\infty}{e}^{l\left(\frac{2}{k}-\frac{2l}{k}+\frac{2}{3}\frac{(1-l)^3}{k^3}\right)}dl=\frac{1}{2}e^{\frac{1}{2k}}\int_{-\infty}^{\infty}{e}^{-\frac{2}{k}(l+\frac{1}{2})^2+\frac{2}{3}\frac{l(1-l)^3}{k^3}}dl$$\sim\frac{1}{2}e^{\frac{1}{2k}}\int_{-\infty}^{\infty}{e}^{-\frac{2}{k}t^2}\left(1+\frac{2}{3}\frac{(t-\frac{1}{2})(\frac{1}{2}-t)^3}{k^3}\right)dt\sim\frac{1}{2}e^{\frac{1}{2k}}\int_{-\infty}^{\infty}{e}^{-\frac{2}{k}t^2}\left(1-\frac{2}{3}\frac{t^4}{k^3}\right)dt$

Finally we have

$$S(k)\sim\sqrt{\frac{\pi{k}}{8}}\left(1+\frac{3}{8k}+O(\frac{1}{k^2})\right)$$

Asymptotics of higher order can be easily obtained in the same way.

Some figures

Title: Estimate a sum of fractions


问:
Inequality Induction Proof, how should I proceed?
I have been trying to prove inequalities using induction to no avail.

For example,

Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$

Base Case:

$\frac{1}{2^{1}} = 2$

$\frac{1}{2}<2\;$  Which is true.

We assume, for n=k, that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.

As induction step with n=k+1, we need to prove that

$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$

How should I proceed from here?

Body:
I finally managed to prove it with induction, which was the goal.

I followed the suggestion to prove that  $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}} = 2-\frac{k+2}{2^{k}}\;$

for k+1 we have that

$\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}+\frac{k+1}{2^{k+1}} =   2-\frac{k+3}{2^{k+1}}\;$

It follows from the inductive hypothesis that

$2-\frac{k+2}{2^{k}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$

$2-\frac{2k+4}{2^{k+1}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$

$2+\frac{-3-k}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$

$2-\frac{k+3}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$

Which verifies that $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}$ is 2 minus something else, which in turn means that it is less than 2 thus proving the original statement.


问:
The numbers in these patterns go together following the same rule. Find the missing number.
I don't see the pattern of the numbers, can someone tell me what the pattern is?

答:
Second number is the first prime greater than the first number, so $41$.

Or any of the other options by a suitable convoluted argument


答: Originally people did not have the concept of "negative numbers." Similarly there were no solutions for $x^2 = -1$ which nowadays we have $+i,-i$ as solutions. If you can come up with a number such that $a^x =0$ for some "new type of number x", maybe you can win the Fields Prize.</p>

Title: If $2^x=0$, find $x$.


问:
Properties of Unitary operators on Hilbert spaces
Is it true that for every unitary extension $U' \supset U$

$\psi \in \{ {\frak Dom}(U)\}^\perp \Longleftrightarrow U'\psi \in \{ U'({\frak Dom} (U))\}^\perp$

?

My question is: how can I be sure that $\{ {\frak Dom}(U)\}^\perp \subset \{{\frak Dom }(U')\}$? And why does $U' \psi$ belong to $\{ U'({\frak Dom} (U))\}^\perp$ and not simply to $\{{\frak Ran }(U')\}$?

Could you prove it?

答:
It is true that ${\frak Dom}(U)^\perp \subset {\frak Dom }(U')$ if we assume that orthogonal complement in ${\frak Dom}(U)^\perp$ is taken relative to the same space on which the extension $U'$ is defined. More precisely, suppose $B$ is a vector space and $U: A\to A$ is a unitary map defined on a subspace $A = {\frak Dom}(U)$ of $B$. Consider an extension $U':B\to B$ of $U$ defined on the whole space $B$. In this case, we have

$${\frak Dom}(U)^\perp = A^\perp \subset B = {\frak Dom}(U')$$

by construction.

Affirmative answer to the second question follows from the fact that the inverse of a unitary map is its Hermitian adjoint. Let $\psi \in {\frak Dom}(U)^\perp$. To prove that $U'\psi \in (U'{\frak Dom}(U))^\perp$ we need to show that $\langle U'\psi, \phi\rangle = 0$ for every $\phi \in U'{\frak Dom}(U)$. However,

$$
\langle U'\psi, \phi\rangle = \langle \psi, U'^{-1}\phi\rangle = 0
$$

where the last equality follows from the facts that $U'^{-1}\phi \in {\frak Dom}(U)$ and $\psi \in {\frak Dom}(U)^\perp$.


问:
Given matrices $A,B,C$ such that $BC=A$ find $\det(CB)$ and $\operatorname{Tr}(CB)$

Let $A = \begin{pmatrix} 2&-2 & 3\\ 0&0&3 \\ 0&0&3 \\ \end{pmatrix}$. $B$ is a $3\times 2$ matrix and $C$ is a $2 \times 3$ matrix such that $BC=A$. Find $\det(CB)$ and $\operatorname{Tr}(CB)$.

I read this thread but i have not learned about characteristic polynomials yet and the answer that does not use those is tedious.

I also read this but in my case it is the other way around, i know the determinant which equals $0$, but i do not know how to find the "other one".

Also, I am not that familiar with spaces and dimensions in linear algebra so i would prefer a solution that does not involve deep knowledge in this area.

Any tips ?

答:
A practical approach is to simply find suitable matrices $B$ and $C$ for which $BC = A$. If the question is well formulated, then it suffices to compute the trace and determinant of $CB$ for our particular choice of matrices $C$ and $B$.

To that end: we can take $B$ to be the matrix
$$
B = \pmatrix{1 & 1\\0&1\\0&1}
$$
because each column of $A$ is a multiple of one of these columns. With that in mind, we can take
$$
C = \pmatrix{2&-2&0\\0&0&3}.
$$
Verify that we indeed have $BC = A$. With that, we now compute
$$
CB = \pmatrix{2&-2&0\\0&0&3}\pmatrix{1&1\\0&1\\0&1} = \pmatrix{2&0\\0&3}.
$$
It is easy to compute the determinant $2 \cdot 3 = 6$ and the trace $2 + 3 = 5$.


问:
Lebesgue Integration and Riemann integration of a function
Let $f= \frac{\sin x }{\sqrt{x}}$ on $[\pi, \infty)$.

Is $f$ Lebesgue integrable?

Is $f$ (improper) Riemann integrable?

I think that $f$ is improper Riemann integrable since, $\int_0 ^\infty f(x) dx = \sqrt{\pi/2} $.

But not sure about the Lebesgue integration.

Also, I know that $\frac{\sin x}{x}$ is neither Riemann or Lebesgue integrable...

Thank you in advance.

答:
Both $\frac{\sin x}{x}$ and $\frac{\sin x}{\sqrt{x}}$ are Riemann integrable, that follows from the Dirichlet criterion. On the other hand, none of them is Lebesgue integrable. To see that consider part of the domain $[\pi, \infty]$ on which $|\sin x|>1/2$, call it $\mathcal{A}$. It is easy to see that
$$
\int_{\mathcal{A}}\Big|\frac{\sin x}{x}\Big|\,dx=\infty
$$
(write $\mathcal{A}$ explicitly and then estimate from below the integral by a suitable diverging series). It also shows that $\frac{\sin x}{\sqrt{x}}$ is not Lebesgue integrable on $[\pi, \infty]$, since
$$
\frac{|\sin x|}{\sqrt{x}}>\frac{|\sin x|}{x},\quad x\in[\pi,\infty).
$$