问:
Relationship between Turing bifurcation, saddle-node bifurcation, and Hopf bifurcation?
Quoting from

http://jxshix.people.wm.edu/2009-harbin-course/mississippi-bifurcation-2.pdf

a Turing bifurcation occurs when for an ODE and related PDE

$u' = f(u,v), v' = g(u,v)$

$u_t = d_1 \nabla u + f(u,v), v_t = d_2 \nabla v + g(u,v)$

a constant solution

$u(x,t) = u_0, v(x,t) = v_0$

remains a stable steady state for the ODE and becomes an unstable steady state for the PDE.

Stable steady states are generally destabilized either by one real eigenvalue (of the associated Jacobian matrix evaluated in the steady state) crossing the imaginary axis or by a pair of two complex eigenvalues crossing the imaginary axis (i.e. the real part changes from negative to positive values).

Generally, one real eigenvalue crossing the imaginary axis is called saddle-node bifurcation.
A pair of complex eigenvalues crossing the imaginary axis is referred to as Hopf bifuraction.

Are these two types of bifurcations exactly the ones that occur for the PDE (while the steady state remains stable for the ODE) in what is generally referred to as Turing bifurcation / Turing instability?

If so, what are the qualitative differences, if any, between the patterns formed past the Turing bifurcation?

答:
Generally, one real eigenvalue crossing the imaginary axis is called saddle-node bifurcation. A pair of complex eigenvalues crossing the imaginary axis is referred to as Hopf bifuraction.

This is true, however, some additional conditions have to be met to identify the bifurcation as saddle-node or Hopf. This is less important with Hopf bifurcation since the appearance of two purely imaginary eigenvalues in vast majority cases implies appearance (or disappearance) of a unique limit cycle, but in the case of simple real eigenvalue crossing the imaginary axis there are at least two more cases which are often met in applications: transcritical bifurcation and pitchfork bifurcation. At a qualitative level:

saddle-node bifurcation corresponds to the appearance "out of nowhere" two steady states, one is stable, another is unstable

transcritical bifurcation corresponds to the case when two steady states exchange, upon meeting each other, their stabilities

pitchfork bifurcation corresponds when from one, say stable, equilibrium, two more stable equilibria appear and the original one becomes unstable.

The exact conditions which type you encounter depend on the type of the normal form of the equation you consider. See, e.g, this book.

Are these two types of bifurcations exactly the ones that occur for the PDE (while the steady state remains stable for the ODE) in what is generally referred to as Turing bifurcation / Turing instability?

Yes, in the spatially explicit system all these four types can appear in a similar way how they appear in local systems described by ODEs.

If so, what are the qualitative differences, if any, between the patterns formed past the Turing bifurcation?

In the case of simple zero eigenvalue what is usually expected is an appearance of a spatially non-homogeneous steady state (or more than one). This steady state can be stable generating spatially non-homogeneous patterns, which are used in various problems of pattern formation and morphogenesis. A lot of details are given, e.g., in James Murray's book Mathematical Biology, together with some technical details. Hopf bifurcation yields not only spatially non-homogeneous solutions, but also temporally periodic oscillations.

P.S. And final remark. The name "Hopf bifurcation" is quite unfortunate. The fact that periodic solutions appear under parameter change when two eigenvalues cross imaginary axis

was known to the father of qualitative analysis of dynamical systems -- Poincaré. The exact statements of the theorem in the planar case, together with proofs, were given by Andronov. The main contribution of Hopf is generalization of this situation to $n$-dimensional case. So the correct name would be Poincaré-Andronov-Hopf bifurcation, or, in my opinion, a much better option is the bifurcation of the birth of limit cycle :)


问:
Factor $9x^4-37x^2+4$

Factor $9x^4-37x^2+4$ using difference of cubes, sum of cubes,
difference of squares, substitution, grouping or common factoring,
whichever is simplest method.

Usually factoring isn't to difficult for me, but for some reason I'm stuck on this one. I would usually use the factor theorem but the question specifically states to use one of the methods listed. And the only one this equation kind of works with is substitution, at least to my eyes, but I'm having trouble with doing it that way and I can't figure any other way to do it besides the factor theorem.

答: As Toby said, you can factor it down to $(9x^2 - 1)(x^2 - 4)$. However, note that both of these factors are themselves differences of squares. So, you may factor further to get $(3x-1)(3x+1)(x-2)(x+2)$ over the integers. If you allow rational numbers, you can pull out a factor of $3$ from the first $2$ factors to get $9(x-\frac13)(x+\frac13)(x-2)(x+2).$

ContentLicense: CC BY-SA 4.0

Title: Factor $9x^4-37x^2+4$


问:
Solve trigonometric equation $\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3=0$
I don't seem to have a credible way to solve the trigonometric equation below

$$\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3=0$$

I was able to guess quickly that $-\pi$ is a root, and, after staring at it long enough, was able to see that  $-\frac\pi2$ is another one. I graphed  the function and verified that these two are the only roots. However, I have not been able to solve it algebraically.

One thing I have tried is to convert it into a system of equations
$$3y^2=\pi^2 - x^2$$
$$\cos y +2\cos\frac{\pi-x}3=0$$
and hoped to explore the symmetry, as seen the plot

Edit:

Note that the plot reveals that one root is at an extreme, where the two curves are tangential to each other. This indicates that the root may be solved via calculus, albeit not algebraically. One easier example of such soluble equation is $\pi\cos x=\sqrt{\pi^2 -x^2}$ which has a single root at its extreme. The complication in the posted equation is that it has two roots.

答:
Let's set $$f:[-\pi,\pi]\to\Bbb R:f(x)=\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3$$

The two cosine arguments are roughly equal in size, so it actually turns out that the scalar $2$ is very useful. Indeed, for every $x\in(-\pi,\pi]$, we have that $$\bigg|2\cos \frac{\pi -x}3\bigg|>\bigg|\cos \sqrt{\frac{\pi^2-x^2}3}\bigg|\implies \operatorname{sgn}(f(x))=\operatorname{sgn}(\cos\bigg(\frac{\pi-x}{3}\bigg))$$
where $\operatorname{sgn}(x)$ is the sign function.

Our task of proving that $-\pi$ and $-\frac \pi2$ are the only solutions is then reduced to showing that $\cos(\frac{\pi-x}{3})$ is strictly non-zero across $(-\pi,-\frac\pi 2)\cup (-\frac\pi 2, \pi)$ -n.b. it is negative on the first half and positive on the second half.


问:
Solving linear Diophantine equation in 4 variables.
How can I solve (nontrivially) this equation in nonnegative integers:
$$ x - 2y + 3z - 4t = 0.$$

By inspection I found the set of solutions is: {(2,1,0,0),(4,0,0,1), (0,3,2,0), (1,0,1,1), (0,0,4,3), (0,1,2,1), (1,2,1,0),(6,1,0,1)}

1- Is my solution correct?

2- Also, I got a hint that this can be solved as a linear Diophantine equation but I only know how to do this for 2 variable and sometimes for 3 (where I express the third variable in terms of the other two)but not for 4 variables.

Any help will be appreciated.

答:
Clearly there are infinitely many solutions; scaling any solution yields another solution.

As for explicit solutions;for any $y,t\geq0$ and any $z\geq0$ such that $3z\leq2y+4t$ you have
$$x:=2y-3z+4t\geq0.$$


问:
Explanation about how to get a linear equation given a problem?
I have issues getting the linear equation of this kind of problem, can you explain to me what do I need first and why? I can't understand properly to my teacher, please

The owner of a bottled water franchise must pay $ 500 per month, plus 5% of the monthly income (I) for the use of the brand.

```<code>A) Determine the monthly expenses (E) in terms of (I)

B) How much should you pay if you had a monthly income of $ 15,450?
</code>```

NOTE: I need an explanation about how to get the equation not to solve it, thank you

答:
Well " must pay \$500 per month, plus...." so $E = 500 + .....$

"....plus 5% of the monthly income (I)".  So

$E = 500 + \text{5 % of } I$

and as "$x$% of K" means $\frac x{100}\cdot K$ we have

$E = 500 + \frac 5{100}I$.

That's it.  Ase $\frac 5{100} = \frac 1{20}$ we can rewrite it as $E = 500 +\frac 1{20}I$ but we don't have to.

That's A.

And B)

"How much should you pay (in expenses) if you had a monthly income of $ 15,450?"

Then Income $= I = 15450$.

So Expenses $=E = 500 + \frac {5}{100} I$ will be a matter of plugging in $I = 15450$.

$E = 500 + \frac 5{100}I =$

$500 + \frac 5{100}\cdot 15450$.

Which is what?


问:
What is an ideal in topology?
I am currently trying to prove a unique collection of subsets of I, where I is any nonempty set, is an ideal. I cannot find anything on the internet about an ideal in topology, I only find Algebra references. I can find the definition below but I cannot find any examples or anything. I was wondering if someone could help me understand an ideal in topology or point me to any books or articles that may help.

An ideal $\mathscr{L}$ is a nonempty collection of sets in a topological space X such that $\mathscr{L}$ is closed under heredity and finite additivity, i.e. $\mathscr{L}$ satisfies the following properties:

$T \in \mathscr{L}$ and $S \subseteq T$ implies $S \in \mathscr{L}$

$S \in \mathscr{L}$ and $T \in \mathscr{L}$ implies $S \cup T \in \mathscr{L}$

答:
That definition tells you exactly what an ideal is in this context: it’s a family of sets in some space that is closed under taking subsets and closed under unions. For example, let $X$ be any infinite space, and let $\mathscr{F}$ be the family of all finite subsets of $X$; then $\mathscr{F}$ is an ideal.

Subsets of finite sets are finite, so if $G\subseteq F\in\mathscr{F}$, then clearly $G$ is finite, and therefore $G\in\mathscr{F}$: $\mathscr{F}$ is closed under taking subsets.

If $F,G\in\mathscr{F}$, then $F\cup G$ is finite, so $F\cup G\in\mathscr{F}$.

In particular, if $\tau$ is the cofinite topology (sometimes called the finite closed topology) on an infinite set $X$, then $\mathscr{F}=\{X\setminus U:U\in\tau\setminus\{\varnothing\}\}$: the closed proper subsets of $X$ form an ideal.

An easy and useful exercise is to prove that if $\mathscr{L}$ is an ideal, and $\mathscr{A}$ is a finite subset of $\mathscr{L}$, then $\bigcup\mathscr{A}\in\mathscr{L}$: an ideal is not closed just under taking simple unions of two of its members, but also under taking the union of any finite number of its members.

If you’re acquainted with the notion of a filter on a set $X$, you should verify that a family $\mathscr{L}$ of subsets of $X$ is an ideal if and only if $\{X\setminus L:L\in\mathscr{L}\}$ is a filter on $X$, and $\mathscr{U}$ is a filter on $X$ if and only if $\{X\setminus U:U\in\mathscr{U}\}$ is an ideal on $X$: ideals and filters are dual notions.

It’s worth noting that neither of these notions actually involves the topology: you define ideals and filters in this way on any set. In a specifically topological setting we sometimes look at slightly more general notions of ideal and filter that do involve the topology. We might, for instance, speak of an ideal $\mathscr{C}$ of closed sets in a space $X$, meaning that $\mathscr{C}$ is a family of closed subsets of $X$ such that

if $H\subseteq C\in\mathscr{C}$, and $H$ is closed, then $H\in\mathscr{C}$, and

if $C,D\in\mathscr{C}$, then $C\cup D\in\mathscr{C}$. (Of course $C\cup D$ is automatically closed if $C$ and $D$ are.)

Added: Henno Brandsma reminds me that the nowhere dense sets are an excellent genuinely topological example of an ideal on a topological space. Let $\langle X,\tau\rangle$ be a topological space; a set $N\subseteq X$ is nowhere dense if $\operatorname{cl}N$ does not contain any non-empty open set. (For example, $\Bbb Z$ is nowhere dense in $\Bbb R$: it is already closed, and its interior is empty.) Let

$$\mathscr{N}=\{N\subseteq X:N\text{ is nowhere dense}\}\,;$$

then $\mathscr{N}$ is an ideal, though I’ll leave it to you to check that it satisfies the defining properties.

Title: What is an ideal in topology?


问:
Examples of two inequivalent definitions which were once thought to be equivalent
I once thought two definitions that I made in proof theory were equivalent, but then I discovered a counterexample. So I am now led to wonder, have there ever been cases in the history of mathematics when people made two inequivalent definitions and thought they were equivalent, but later a counterexample was discovered?</p>

答:
The answer depends on what exactly is meant by
"inequivalent definition" and "counterexample". For
example, if we have the system of equations
$$a = u+v,\; b = u-v $$ then is this equivalent to
the system $$ (a+b)/2 = u,\; (a-b)/2 = v? $$
This used to be accepted without question until fields with characteristic $2$ were introduced. In those fields, division by $2$ is not possible and
the second system does not even make sense.


问:
Planar and Spherical co-ordinates using just one number
I'd like to specify positions on a plane or in 3D space using just a single number

I can see one way to do it in a plane, using a spiral. I simply start at the centre and travel along the spiral a given distance.

I can clearly specify any point on the plane this way, to any required degree of precision, with a single length measurement.

Question

(a) Is there a more intuitive* function to use whilst still using only a single co-ordinate?

(b) What 3D function can I use to generalise this spiral (or similar function) to a 3D space?

*Note: A spiral is intuitive in that, the bigger the number, the further you are away radially from the origin. Of course, the actual distance from the origin and direction are not obvious without further manipulation of the measurement function.

答:
The spiral does not work in the strict mathematical sense, i.e.  there will always be gaps between successive loops of the spiral. In order to actually cover the entire plane (or even a small subset like a filled in square $[0,1]^2$ you need to use a relatively pathological object called a space-filling curve. But with a spiral, you do pass through say, every cube with integer coordinates if your spiral is tightly wound enough.

It would seem to me that this issue

Of course, the actual distance from the origin and direction are not obvious without further manipulation of the measurement function

disappears once you choose a certain fixed spiral, because the act of choosing the spiral also forces you to say what the angle and radius of each point is.

One choice, which does indeed pass through every box with integer coordinates (although some will be right on the border) is the spiral whose point at angle $\theta\ge 0$ is at distance $\theta/360º$. So after each round of 360º, the distance from the origin advances by $1$. (This sort of spiral that is linear in the angle is precisely what your picture is. There are other spirals that get more or less packed for larger angles..)

If you would like to have integer coordinates for your 'spiral', a reasonable choice is maybe something like the Ulam spiral.

As for a 3D generalisation of a densely packed spiral, I find it hard to visualise. Perhaps you can take some inspiration from yarn balls...?


问:
Triangulation that includes given subcomplex
I was reading Prasolov and Sossinsky's book Knots, Links, Braids and Three-Manifolds and came across the following statement in the proof of Theorem 9.2:

We can assume $L^3$ has a triangulation $K$ such that $T$ is a subcomplex of $K$.

The context is as follows: Let $M^3$ be a compact, oriented manifold with connected boundary $\partial M^3$, and $N^3$ be a handlebody with boundary homeomorphic to $\partial M^3$ and spine $T$ (i.e. a 1-dimensional simplicial subcomplex of $N^3$ such that a regular neighborhood of $T$ is homeomorphic to $N^3$). Attach $N^3$ to $M^3$ along an arbitrary homeomorphism of their boundary to yield $L^3$.

It's not clear to me why we can make the assumption above: is it to do with the way $\partial N^3$ meets $\partial M^3$? Can the same assumption be made for any three-manifold $M$? That is, can we choose a graph $T$ in $M$ and find a triangulation of $M$ that contains $T$ as a subcomplex?

Any advice is much appreciated!

答:
This claim is actually nontrivial and fails for general graphs embedded in 3-manifolds (even for topological circles in $S^3$; there are "wild knots" which cannot be realized as subcomplexes of any triangulation).

In order to prove the claim, note that $N^3$ can be triangulated so that $T$ is a subcomplex. Next, $M$ also admits a triangulation (this is quite nontrivial), so that $\partial M$ is a subcomplex. Moreover, if $M_1, M_2$ are homeomorphic triangulated 3-manifolds, then, after a suitable subdivision of triangulations, the underlying simplicial complexes are isomorphic.
Lastly, suppose that $S_1, S_2$ are triangulated surfaces and $f: S_1\to S_2$ is a homeomorphism. Then $f$ is isotopic to a homeomorphism which is piecewise-linear with respect to these triangulations. This is also somewhat nontrivial. Lastly, changing the map $f: \partial N\to \partial M$ by isotopy does not change the homeomorphism type of the manifold $N\cup_f M$.

Moise, Edwin E., Geometric topology in dimensions 2 and 3, Graduate Texts in Mathematics. 47. New York - Heidelberg - Berlin: Springer-Verlag. X, 262 p. DM 45.00; $ 19.80 (1977). ZBL0349.57001.


问:
Proof Multiplicative Group of Integers Modulo p is Cyclic
The problem:

Let p be a prime number.
Let$\;\omega$ be a pth root of unity.

Prove$\;Gal(\Bbb Q(\omega)/\Bbb Q)$ is cyclic.

The same basic question is here:
Galois Group of $x^p-1$ cyclic

but the answer just points out that$\;Gal(\Bbb Q(\omega)/\Bbb Q)$ is isomorphic to the multiplicative group of integers modulo p.

In another post, the issue of the proof that the multiplicative group of integers modulo p is cyclic is brought up:
Proof that the following multiplicative groups modulo m are cyclic

The poster said they'd already proven it for p being prime.  Another said the proof is quite long.

Is there another approach to the initial question of proving the Galois group is cyclic?  The book I'm working on (Pinter's Abstract Algebra) didn't have anything that complicated wrt this subject.

答:
$\Bbb{F}_p$ is the field with $p$-elements, $\Bbb{F}_p^\times$ is a group with $p-1$ elements,

All its elements satisfy $a^m=1$ for some $m$ dividing $p-1$.

Let $k$ be the smallest integers such that all the elements satisfy $a^k=1$. We have that $k \le p-1$.

Assume that $k<p-1$. All the elements are roots of $x^k-1\in \Bbb{F}_p[x]$, but this polynomial has at most $k$ distinct roots, contradicting that $\Bbb{F}_p^\times$ has $p-1$ elements.

Whence $k=p-1$ and we get some element of order $p-1$ ie. $\Bbb{F}_p^\times$ is cyclic.

$\omega$ is a root of $f=\frac{x^p-1}{x-1}$, thus any automorphism of $\Bbb{Q}(\omega)$ must send $\omega$ to another root of $f$ ie. to $\omega^a$ for some $a\in 1\ldots p-1$. As an automorphism is determined by where it sends $\omega$ this makes $Gal(\Bbb{Q}(\omega)/\Bbb{Q})$ a subgroup of the cyclic group $\Bbb{F}_p^\times$.

It remains to prove that $f$ is irreducible ($f(x+1)$ is Eisenstein at $p$) so that every $a$ gives an automorphism and $Gal(\Bbb{Q}(\omega)/\Bbb{Q})\cong \Bbb{F}_p^\times$.


问:
Other log solutions?
I am evaluating the expression:

$\ln(1)$

And I know the trivial solution is $0$.

Are there other solutions to this equation? I feel there should be, my logic is as follows:

if:

$\ln(1) = x \implies 1 = e^x \implies 1 = 1 + x + x^2/2! + x^3/3!... $
$\qquad\qquad\qquad\qquad\qquad\implies 0 = x + x^2/2! + x^3/3!...$

$\implies x = 0$ is one solution, the other solution is all $x$ such that: $$1 + x/2! + x^2/3! + x^3/4! ... = 0$$

There have to be other solutions, or limiting solutions...

What are they?

答: If $x$ is real, $1=e^x$ has the single solution $x=0$.  The easiest way to see this is that $e^x$ is monotonic, so there can only be one solution.  In the complex plane, because $e^{2\pi}=1$, we can solve $1=e^x$ with $x=2k\pi i$ for $k$ integral.</p>

ContentLicense: CC BY-SA 3.0

Title: Other log solutions?


问:
Question about the definition of faithful functors
In TOM LEINSTER's Basic Category Theory, Page 25, it says

I got very confused here. Doesn't the definition of injective says for distinct $f_1, f_2$ we have $F(f_1)\neq F(f_2)$? Why this is false? I think I must have some misunderstandings about the definition. Can anyone explain this for me and give some concrete examples?

答:
Let $f_1$ and $f_2$ be distinct morphisms. Then there must be objects $A_1,A_1',A_2,A_2'$ in $\mathcal A$ such that $f_1\in\mathcal A(A_1,A_1')$ and $f_2\in\mathcal A(A_2,A_2')$. Faithfulness of $F$ implies that if $A_1=A_2$ and $A_1'=A_2'$, we must have $F(f_1)\neq F(f_2)$.

But if $A_1\neq A_2$ or $A_1'\neq A_2'$ we can have that $F(f_1)=F(f_2)$ even though $F$ is faithful. For instance let $\mathcal A$ have two objects, $A_1,A_2$ and let the identity morphisms $1_{A_1}$ and $1_{A_2}$ be the only morphisms in $\mathcal A$. Fix an object $B\in\mathcal B$. We define $F:\mathcal A\to\mathcal B$ by $F(A_1)=F(A_2)=B$, and $F(1_{A_1})=1_{B}=F(1_{A_2})$. So we have that $F(1_{A_1})=F(1_{A_2})$ even though $1_{A_1}\neq 1_{A_2}$. Nevertheless, because the homsets $\mathcal A(A_i,A_j)$ for $i,j=1,2$ are either empty or a singleton set, we have that $\mathcal A(A_i,A_j)\to \mathcal B(F(A_i),F(A_j)), f\mapsto F(f)$ is injective, hence $F$ is faithful.


问:
F-distributions and chi-square distributions: Can someone help explain the answer to this problem?

Let $X_1, X_2, X_3, X_4$ be independent $\mathscr{N}(\mu,
 \sigma^2)$ random variables. Find the distribution of the ratio

$$R=\frac{(X_1-X_2)^2}{(X_3-X_4)^2}.$$

Note that $X_1-X_2$ and $X_3-X_4$ are independent $\mathscr{N}(0,
 \sigma^2)$ random variables. Thus $W_1=(X_1-X_2)^2/(2\sigma^2)$ and
$W_2=(X_3-X_4)^2/(2\sigma^2)$ are independent $\mathcal{x}^2(1)$
random variables. Since $R$ equals $W_1/W_2$, we conclude $R\sim
 F(1,1).$

I understand that an $F$ distribution is the ratio of two chi-squares with their respective degrees of freedom.  But for this one, why is the square of a non-standard normal distribution a chi-square?  Why is it only one degree of freedom?  Moreover, where does the $2\sigma^2$ come from on the denominators for $W_1$ and $W_2$?

答:
Note that $Y_1=X_1-X_2\sim N(0,2\sigma ^2)$ and $Y_2=X_3-X_4\sim N(0,2\sigma ^2)$. We can then normalize these into standard normal random variables: \begin{align*}
Z_1=\frac{X_1-X_2}{2\sigma^2}\sim N(0,1)\\
Z_2=\frac{X_3-X_4}{2\sigma^2}\sim N(0,1)
\end{align*}
Since $Z_1,Z_2$ are standard normal, we can say that $$\frac{Z_1^2}{Z_2^2}\sim F(1,1)$$
So \begin{align*}
\frac{\left(\frac{X_1-X_2}{2\sigma^2}\right)^2}{\left(\frac{X_3-X_4}{2\sigma^2}\right)^2}\sim F(1,1)\\
\frac{(X_1-X_2)^2}{(X_3-X_4)^2}\sim F(1,1)
\end{align*}


问:
Does two different extensions of a map between manifolds have the same derivative in the manifold-domain?
It is well know that the differential calculus is developed with respect open sets but a manifold not necessarily is an open set and actually many times it not. So in Differential Geometry it is usual give the following definition of $C^r$-function.

Definition

A function $f:S\rightarrow\Bbb R^n$ where $S\subseteq\Bbb R^k$ is of class $C^r$ if it can be extended to a $C^r$-function defined in an open set $U$ containing $S$.

Naturally if $\tilde {f_1}, \tilde{f_2}$ are two different extension of $f$ it is possible that even their derivatives are different in $S$ but fortunately this do not happens if $S$ is a manifold as Victor Gullemin and Peter Heine show in the text Differential forms part of which I summarise to follow but if you like you can read it here.

So let be $f:X\rightarrow Y$ a $C^r$-mapping from a k-manifold $X$ of $\Bbb R^n$ and an $l$-manifld $Y$ of $\Bbb R^m$ and thus let be $\tilde f$ any $C^r$-extension. So we define the derivative of $f$ in $X$ to be the restriction of the derivative of $\tilde f$ in $X$, that is
$$
Df(x):=D\tilde f(x)
$$
for any $x\in X$. So let's prove that this definition is consistent, that is if does not depend from the choice of a particular extension. Therefore let be $\phi:U\rightarrow V$ a local patch of $X$ and thus let be $h:=\tilde f\circ\phi$. Now it is possible to prove that
$$
D\tilde f(\phi(x))=Dh(x)
$$
for any $x\in U$ and since $h=\tilde f\circ\phi=f\circ\phi$ we conclude that in $X$ the derivatives of two different extension of $f$ are equal.

So I observe that if $\phi:U_1\rightarrow V_1$ and $\phi_2:U_2\rightarrow V_2$ are two different charts then what above showed says that
$$
D(f\circ\phi_2)(x_1)=D(f\circ\phi_2)(x_2)
$$
where $x_2=\phi_2^{-1}(\phi(x_1))$ but unfortunately to me it seems false this. Anyway since this was not clear to me I refer to the text Differential Topology by Victor W. Gullemin and Alan Pollack where I found this definition of derivative of a mapping between two manifold that I completely understand. However I do not understand if what showed in the link I posted implies that two different extesions of $f$ have the same derivative at $X$. So since well understand the last approach I primarily ask if with respect this it is true that two different extesions of $f$ have the same derivative at $X$ and then (only if you like) I ask to prove the equivalence of this approach with respect the first. So could anyone help me, please?

答:
Hints (pretty much the same as Deane's):

Step 1. First prove the claim in the case when $S\subset {\mathbb R}^k$ is actually an open subset of ${\mathbb R}^m\subset  {\mathbb R}^k$.

Step 2. Prove that if $S$ is an $m$-dimensional submanifold in ${\mathbb R}^k$ then for each $x\in S$ there exists a neighborhood $U$ of $x$ in  ${\mathbb R}^k$ and a diffeomorphism $h: U\to  {\mathbb R}^k$ such that $h(U\cap S)$ is an open subset of ${\mathbb R}^m\subset  {\mathbb R}^k$.

Step 3. Use the chain rule plus Step 1 in order to get the desired conclusion.


问:
Find all $n!$ that are products of two Fibonacci numbers
Find all positive integers $n$ such that $n!$ is the product of two Fibonacci numbers

I found this discussion, but it doesn't quite help my problem.
Check if the number is a result of multiplying two fibonacci numbers

答:
With the convention that $F_0=0$ and $F_1=1$, for $k>12$ the $k$-th Fibonacci number $F_k$ has a prime factor $p$ satisfying $p\geq k-1$, see here. It follows that if $n!=F_aF_b$ for some positive integers $a$ and $b$, then $a,b\leq n+1$. It is not hard to verify that $F_k\leq\varphi^k$ for all $k\geq0$, and so $a,b\leq n+1$ implies that
$$n!=F_aF_b\leq F_{n+1}^2\leq(\varphi^{n+1})^2=\varphi^{2n+2}.$$
Of course the left hand side increases more rapidly than the right hand side as soon as $n\geq\varphi^2$, i.e. $n\geq3$. Then a quick check shows that $7!>\varphi^{16}$, so any solution to $n!=F_aF_b$ must have $n\leq6$. From here it is a routine verification to find that
$$0!=F_1F_2,\quad 1!=F_1F_2,\quad 2!=F_2F_3,\quad 3!=F_3F_4,\quad 4!=F_4F_6,\quad 6!=F_5F_{12},$$
and that $5!=120$ is not a product of two Fibonacci numbers.


答:
There is no equation mentioned in the post. I assume you mean the equation $e^x=1$. This has no real solutions other than $x=0$. You can see this from the fact that the function $e^x$  is an increasing function, so can take on the value $1$ at most once.

The logarithm function can be extended to the complex numbers, where it is multiple-valued. In the complex numbers, the equation $e^z=1$ has the solutions $z=2n\pi i$, where $n$ ranges over the integers. So in the complex numbers, $\ln 1$ takes on infinitely many values.

Title: Other log solutions?

Body:
I am evaluating the expression:

$\ln(1)$

And I know the trivial solution is $0$.

Are there other solutions to this equation? I feel there should be, my logic is as follows:

if:

$\ln(1) = x \implies 1 = e^x \implies 1 = 1 + x + x^2/2! + x^3/3!... $
$\qquad\qquad\qquad\qquad\qquad\implies 0 = x + x^2/2! + x^3/3!...$

$\implies x = 0$ is one solution, the other solution is all $x$ such that: $$1 + x/2! + x^2/3! + x^3/4! ... = 0$$

There have to be other solutions, or limiting solutions...

What are they?


问:
Does $\sum_{n=1}^{\infty} \frac{1}{n^2x-1}$ converge uniformly on $(-\infty, -a), a>0$?
I'm currently using Rudins book and this problem is not one of the listed exercises, but I want to try to understand it.

My work:
Define $S_n(x)=\sum_{k=1}^n \frac{1}{k^2x-1}$. Then $\left\vert S_n-S_m\right\vert=\left\vert\sum_{k=n+1}^{m}\frac{1}{k^2x-1}\right\vert=\sum_{k=n+1}^{m}\left\vert\frac{1}{k^2x-1}\right\vert$. I want to find an upper bound for this difference but all I have is $|k^2x-1|=k^2x+(-1)|\leq |k^2x|+1$ which doesn't help.

Any tips?

答:
If $x\in (-\infty,-a)$ with $a>0$, then $\frac{1}{n^{2}x-1}$ is always negative and:

$$|\frac{1}{n^{2}x-1}|= \frac{1}{n^{2}|x|+1}$$

So $S_{k} = -\sigma_{k}$ where $\sigma_{k} = \sum^{k}_{n=1}\frac{1}{n^{2}|x|+1}$.

$$ \sigma_{k} = \sum^{k}_{n=1}\frac{1}{n^{2}|x|+1} <  \sum^{k}_{n=1}\frac{1}{n^{2}|x|} = \frac{1}{|x|}\sum^{k}_{n=1}\frac{1}{n^{2}} $$

Assuming the solution of the Basel problem (you will have a small follow-up exercise to prove boundedness of the comparison series if you can't assume this), you know that {$\sigma_{k}$} is a monotone bounded sequence and thus converges, therefore {$S_{k}$} wth $S_{k} = -\sigma_{k}$ also converges, either by the same reasoning (monotone bounded sequence) or by nature of the relationship.

To finish the uniformity given convergence of {$\sigma_{k}$} to \sigma' consider that:

$$ \sigma' - \sigma_{k} = \sum^{\infty}_{n=k+1}\frac{1}{n^{2}|x|+1} <  \sum^{\infty}_{n=k+1}\frac{1}{n^{2}|x|} = \frac{1}{|x|}\sum^{\infty}_{n=k+1}\frac{1}{n^{2}} <  \frac{1}{a}\sum^{\infty}_{n=k+1}\frac{1}{n^{2}}$$

Therefore given $\epsilon > 0$ There is a $M$ such that $|S_{j}-S'| < \epsilon$ for all $j>M$ for every $x\in (-\infty,-a)$.


问:
How to evaluate the integral $\int_{-\infty}^a e^{-x^2} dx$
How to evaluate the integral $\int_{-\infty}^a e^{-x^2} dx$?

I know that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$ because if I let $I = \int_{-\infty}^{\infty} e^{-x^2} dx$, then

\begin{align*}
I^2 &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-x^2 -y^2} dx dy \\
&= \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r} r dr d\theta\\
&= 2\pi \int_{0}^\infty e^{-r^2} r dr\\
&= \pi
\end{align*}
where the second equal sign follows from polar transformation. However, what if I'm interested in evaluating $\int_{-\infty}^a e^{-x^2} dx$? I'm having trouble translating $\int_{-\infty}^a$ into the correct bounds on $r$ and $\theta$.

答:
This integral is equal to
$$\int\limits_{-\infty}^a e^{-x^2}\,\mathrm{d}x = \dfrac{\sqrt{{\pi}}\left(\operatorname{erf}\left(a\right)+1\right)}{2},$$
where the error function is defined as
$$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int\limits_{0}^{x}e^{-\tau^2}\,\mathrm{d}\tau.$$


问:
$\int\sqrt{\sin(x)}(\cos(x))^3dx$
EDIT: @Andrew Chin's comment solved the issue

$\int(\sin(x))^{1/2}(\cos(x))^3dx$

=$\int(\sin(x))^{1/2}(\cos(x))^2(\cos(x))dx$

=$\int(\sin(x))^{1/2}(1-\sin^2(x))(\cos(x))dx$

Let $u=\sin(x)$, therefore $du=\cos(x)dx$, next:

=$\int(u)^{1/2}(1-u^2)du$

=$\frac{2}{3}(u)^{\frac{3}{2}}-\frac{2}{5}(u)^{\frac{5}{2}}+C$, substituting $u=\sin(x)$ we get:

=$\frac{2}{3}(\sin(x))^{\frac{3}{2}}-\frac{2}{5}(\sin(x))^{\frac{5}{2}}+C$

Except this is incorrect (the derivative of my answer clearly isn't equal to the integrand), but I don't know why. Can someone tell me where I went wrong?

答:
As I have pointed out in a comment, the mistake comes from the power rule of integration.
\begin{align}
\int\sqrt{\sin x}\cos^3x\,dx&=\int\sqrt{\sin x}\cos^2x\cos x\,dx\\
&=\int\sqrt{\sin x}(1-\sin^2x)\cos x\,dx\\
&=\int\sqrt u(1-u^2)\,du\\
&=\int\sqrt u\,du-\int\sqrt{u^5}\,du\\
&=\frac23u^{3/2}-\frac27u^{7/2}+C\\
&=\frac23(\sin x)^{3/2}-\frac27(\sin x)^{7/2}+C.
\end{align}

Differentiating the final line to check, we get
\begin{align}
\frac{d}{dx}\left(\frac23(\sin x)^{3/2}-\frac27(\sin x)^{7/2}+C\right)&=(\sin x)^{1/2}\cos x-(\sin x)^{5/2}\cos x\\
&=(\sin x)^{1/2}\cos x(1-\sin^2x)\\
&=\sqrt{\sin x}\cos x\cos^2x\\
&=\sqrt{\sin x}\cos^3x
\end{align}
as desired.

Title: $\int\sqrt{\sin(x)}(\cos(x))^3dx$


问:
For what $a$ would the graphs of $y=a^x$ and $y=\log_a x$ be tangent?
Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking).
Assuming there is a function

$$f(x) = a^x$$

Its inverse function would look like this:

$$f^{-1}(x) = \log_a x$$

Question:

For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?

答:
I found this to have a solution algebraically: $$a=e^{\frac{1}{e}}$$
For $f(x)$ to be tangent to $f^{-1}(x)$ and to $y=x$, it must have a tangent line with slope $1$ that has a $y$-intercept of $0$.
$$f(x)=\log_{a}(x)=\frac{\ln(x)}{\ln(a)}$$
$$f'(x)=\frac{1}{x\cdot\ln(a)}$$
To find $k$ such that $f'(k)=1$ (in terms of a):
$$f'(k)=\frac{1}{k\cdot\ln(a)}=1$$
$$k=\frac{1}{\ln(a)}$$
The equation of a tangent line at $(k,f(k))$ is $y-f(k)=f'(k)(x-k)$
$$y-\frac{\ln\left(\frac{1}{\ln(a)}\right)}{\ln(a)}=1\cdot\left(x-\frac{1}{\ln(a)}\right)$$
$$y=x-\frac{1}{\ln(a)}+\frac{\ln\left(\frac{1}{\ln(a)}\right)}{\ln(a)}=x-\frac{1}{\ln(a)}-\frac{\ln(\ln(a))}{\ln(a)}$$
Since the $y$-intercept must be $0$:
$$\frac{1}{\ln(a)}+\frac{\ln(\ln(a))}{\ln(a)}=0$$
$$1+\ln(\ln(a))=0$$
$$\ln(\ln(a))=-1$$
$$\ln(a)=\frac{1}{e}$$
$$a=e^{\frac{1}{e}}$$


问:
Question on the transition map of tangent bundle
This is a follow-up question on diffeomorphism of the bundle chart of a Tangent bundle

I don't understand how I can write $\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$. Shouldn't it be $\frac{\partial}{\partial x^i} = \sum\limits_{j}\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$, if I were to use chain rule?

How can I see that $\frac {\partial v} {\partial x}$ and $\frac {\partial y} {\partial u}$ is zero?

I would really appreciate any help.

答:
The sum over a repeated index is implied. This is known as the  Einstein convention.

Note that the $j$ in $$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}$$
is otherwise a "free variable."


问:
Floor function with Multiplier
I understand this: $f(x) = \lfloor x-4 \rfloor$
but how do you handle this $f(x) = 4\cdot\lfloor x-4 \rfloor$  . Do you distribute the $4$ then floor?

and the brackets denote floor.

Thank you

答:
Note that for every $x$ there is a unique $n$ so that $n \le x < n+1$.

And that means there is also a unique $r: 0 \le r < 1$ so that $x = n + r$.

$\lfloor x\rfloor = n$ and $\{x\} = x -\lfloor x\rfloor = r$.

And we always have $x = \lfloor x\rfloor + \{x\}$.

....

Okay get that under our belt.

.....

For any integer $n$ we have

$\lfloor x- n\rfloor= \lfloor x \rfloor -n$ because......

$x-n = \lfloor x \rfloor +\{x\} - n= (\lfloor x \rfloor  - n) + +\{x\}$ and $0\le \{x\} < 1$ and $(\lfloor x \rfloor  - n)$ is an integer so

$(\lfloor x \rfloor  - n)\le (\lfloor x \rfloor  - n) +\{x\} =x-n < (\lfloor x \rfloor  - n)+1$ so

$\lfloor x -n\rfloor =\lfloor x \rfloor  - n$.

.....

So $4\lfloor x -4\rfloor= 4(\lfloor x \rfloor  - 4)=4\lfloor x \rfloor -16$

Title: Floor function with Multiplier


问:
Sum of Legendre Symbols for evey number less than $p$
If $p$ is an odd prime, show that $\sum_{j=1}^{p-1}({\frac{j}{p}})=0$ with $(\frac{j}{p})$ the Legendre Symbol.

I'm not sure if it's enough to say that, because of the Euler's Criteria, there are exactly $\frac{p-1}{2}$ values for which $(\frac{j}{p})=1$ so the other values are $-1$, hence it's proved.

I somehow feel like it's not a good proof.

I've also seen that if $p\equiv 3 (4)$ it's true since $(\frac{-j}{p})=-(\frac{j}{p})$, so the terms cancel out, but following this for $p\equiv 1 (4)$ I don't achieve anything.

Thanks

答:
It's perfectly good. Another way: the nonzero squares mod $p$ are $1^2 \pmod p$,
$2^2 \pmod p$,..., $(p-1)^2 \pmod p$.  That would be $p-1$, but they're double-counted since $x^2 \equiv (p-x)^2$, so there are $(p-1)/2$ squares mod $p$.


问:
What is the probability the class lasts over $60$ minutes?

The Question

Classes are supposed to last a maximum of $60$ minutes, but usually
run over time. Let $X$ be the duration of the class, in minutes. The PDF
of $X$ is

$$f(x)=\left\{\begin{array}{ll} \frac{1}{300} & 0  \le x \lt 60 \\
 \frac{1}{60} & 60 \le x \lt 90\\ \frac{1}{100} & 90 \le x \lt 120 \\ 0
 & \text{otherwise.} \end{array}\right.$$

What is the probability the class lasts over $60$ minutes?\

What is the $0.9$ quantile of $X$?

My Understanding

My thought process was this, $1/60 +1/100 = 2.6\%$ it seems extremely low, but $1/300$ is a very low probability, making me think that $60-120$ minutes is much more possible.

Would this just be $115$ minutes? because its $90\%$ of $120$?

答:
$P(0 < X \le 60) = 60/300 = 0.2,\;$
$P(60 < X \le 90) = 30/60 = 0.5,\;$ and $\;$
$P(90 < X \le 120) = 30/100= 0.3.$

Maybe sketch the density function, essentially
consisting of three rectangles.

You will see that $P(X > 60) = 0.5 + 0.3 = 0.8$
and $P(X \le 60) = 0.2,$ so the 20th percentile is 60.

Also, the 70th percentile is 90.

With that start, can you find the 90th percentile,
which is $c$ such that $P(X \le c) = 0.9?$


问:
Show that $S(q^i,s^j,t)=\frac{1}{2t}||\mathbf{q}-\mathbf{s}||^2$ generates canonical at $t=0$
I'm currently working through Introduction to Mechanics and Symmetry, specifically the section on Generating functions of canonical transformations. I am having an issue with the following problem:

Show that $$S(q^i,s^j,t)=\frac{1}{2t}||\mathbf{q}-\mathbf{s}||^2$$ generates a canonical transformation that is the identity at $t=0$.

First off, I am somewhat confused as to the assumptions this problem is making. I think it would be right for me to assume that $$S:Q_1\times Q_2\times \mathbb{R}\cong \mathbb{R}^{n}\times \mathbb{R}^n\times\mathbb{R}\to \mathbb{R}$$ and if so, don't we have $||q-s||^2=||q||^2+||s||^2$ since $q$ and $s$ are orthogonal? Or am I supposed to take $||\mathbf{q}-\mathbf{s}||^2=\sum (q^i-s^i)^2$, like they share components.

I am also unsure how to deal with this time-dependent function, as the book only treated time-independent generating functions. I know that the canonical transformation generated by $S$ must satisfy $$i^*_\phi \Theta=dS$$ so if I go with $||\mathbf{q}-\mathbf{s}||^2=\sum (q^i-s^i)^2$ we get that $p_i=\frac{1}{t} q^i(q^i-s^i)$ and $r_i=\frac{1}{t}s^i(q^i-s^i)$, with $(q,p)$ and $(s,r)$ the coordinates on $T^* Q_1$ and $T^*Q_2$ respectively. I'm not sure what else I can do. Any suggestions would be greatly appreciated.

答:
You have to treat $S=\frac{1}{2t}\sum(q^i-s^i)^2$ as a function of fixed $t$ and then use it to generate the proper diffeomorphism as follows:

We have $p_i=\frac{1}{t}q^i(q^i-s^i)$, which gives $$\frac{t p_i}{q^i}=q^i-s^i$$ Then we come to $s^i=q^i-\frac{tp_i}{q^i}$. If we substitute this into $r_i=\frac{1}{t} s^i(q^i-s^i)$ we get \begin{align}r_i&=\frac{1}{t}(q^i-\frac{tp_i}{q^i})(q_i-q^i+\frac{tp_i}{q_i})\\
&=\frac{1}{t}(q^i-\frac{tp_i}{q^i})\frac{tp_i}{q^i}\\
&=p_i-\frac{tp_i^2}{(q^i)^2}\end{align}
So this generates a symplectomorphism $(q^i,p_i)\mapsto (q^i-\frac{tp_i}{q^i}, p_i-\frac{tp_i^2}{(q^i)^2})$ that is the identity at $t=0$, despite having a singularity there.


答:
The first step is to realise that the probability of the class lasting says exactly 61min, is zero, but there is a measurable probability of class taking anywhere between say 60 and 61 minutes. This probability is the area under the probability density function between 60 and 61.
To understand and solve these problem follow the steps

Draw the PDF.

Mark the points between which you whould like to calculate probability.

Find the area.

Body:
The Question

Classes are supposed to last a maximum of $60$ minutes, but usually
run over time. Let $X$ be the duration of the class, in minutes. The PDF
of $X$ is

$$f(x)=\left\{\begin{array}{ll} \frac{1}{300} & 0  \le x \lt 60 \\
 \frac{1}{60} & 60 \le x \lt 90\\ \frac{1}{100} & 90 \le x \lt 120 \\ 0
 & \text{otherwise.} \end{array}\right.$$

What is the probability the class lasts over $60$ minutes?\

What is the $0.9$ quantile of $X$?

My Understanding

My thought process was this, $1/60 +1/100 = 2.6\%$ it seems extremely low, but $1/300$ is a very low probability, making me think that $60-120$ minutes is much more possible.

Would this just be $115$ minutes? because its $90\%$ of $120$?


问:
$\aleph_{0} \times c=c$
I am wondering if my proof for the question: $\aleph_{0} \times c=c$ is correct.
$$\\
\exists \: f:\mathbb{N} \times (0,1) \:\to \:(0,1)\\
\owns \:(n,x) \:\to 0.nx_{1}x_{2}x_{3}....\\
card\mathbb{N} \times card(0,1) \le card(0,1) \\
\\
\exists g: (0,1) \to \mathbb{N} \times (0,1) \\
\owns x \:\to \:(n,x), \:(x=0.x_{1}x_{2}x_{3}...) \\
card (0,1)\: \le \:card\mathbb{N} x card(0,1) \\$$

$\therefore \: \aleph_{0}\times\:c=c$

答:
The idea of finding injections $f:\Bbb N\times(0,1)\to(0,1)$ and $g:(0,1)\to\Bbb N(0,1)$ is fine, but the execution has problems.

Assuming that by $0.nx_1x_2x_3\ldots$ you mean the decimal expansion whose first digits to the right of the decimal point are the digits of $n$ in its usual decimal representation, your function $f$ is not injective: it sends $\left\langle n,\frac19\right\rangle$ to $\frac19=0.1111\ldots$ for every $n\in\{1,11,111,1111,\ldots\}$. Your $g$ is not actually defined: for which value of $n\in\Bbb N$ is $g(0.1)$ equal to $\langle n,0.1\rangle$, for instance?

The latter problem is easy to fix: just let $g(x)=\langle 1,x\rangle$, for instance. Here’s a suggestion for fixing the former problem. You know that $|\Bbb R|=|(0,1)|=\mathfrak{c}$, so instead of mapping $\mathscr{N}\times(0,1)$ injectively into $(0,1)$, you could map it injectively into $\Bbb R$: how about letting $f$ send $\langle n,x\rangle\in\Bbb N\times(0,1)$ to $n+x\in\Bbb R$? You shouldn’t have any trouble showing that this map is an injection.

Title: $\aleph_{0} \times c=c$


问:
Solve Limit Tend To $\pi$, From Length Of Circle
Computing the length of a semicircle of radius 1. It's divided in $2^x$ equal arcs, I find the following formula for the total length of the chords created:

$$L=2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

When $x \to \infty$ we see, geometrically, that this expression tends to be  the length of the circle:

$$2^{x+1} \sin \left( \frac{45}{2^{x-1}} \right)$$

My goal is to define $\pi$ as the value of that expression when $x \to \infty$ (the limit). But how can I rigorously prove that this limit exists and that the expression tends to be the length of the circle? In other words, how can I write rigorously the geometric ideas?

Note: The number  $\frac{45}{2^{x-1}}$ is in degrees.

答:
Yes, your formula does converge to $\pi$, as Joshua Wang's answer nicely shows. However, in the comments there's some discussion about the problem of determining the value of $\sin \left(\frac{45°}{2^{x-1}}\right)$ without using a formula based on radians, since they are based on $\pi$.

The traditional approach is to simply bisect an angle whose sine is known from geometry (eg 30°, 45°, or 60°), and calculate the sine of the resulting half-angle. Algebraically, that comes down to solving a quadratic.

It can be shown that for any angle $\theta$,
$$\cos 2\theta = \cos^2\theta - \sin^2\theta$$
Thus
$$\sin^2\theta = (1-\cos 2\theta)/2$$
$$\sin^2\theta = (1-\sqrt{1-\sin^2 2\theta})/2$$
Or
$$\sin^2(\theta/2)=(1-\sqrt{1-\sin^2\theta})/2$$

So now we have a method of determining $\sin^2(\theta/2)$ from $\sin^2\theta$. Unfortunately, it involves taking a square root at every step (and an extra one to get the sine from its square).

This method can be used to quickly get a few decimals of $\pi$, but when we run this algorithm on a computer we'll soon run into arithmetic errors, mostly due to those nested square roots. Eg, using standard double-precision floating point numbers, which have 53 bits of precision, we can get a little over 26 bits of $\pi$, which is around 8 decimal digits.

Here's a short Python program to illustrate:

```<code>from math import sqrt

y2 = 1 / 2
m = 4
for i in range(1, 20):
    y = sqrt(y2)
    print(i, m * y)
    y2 = (1 - sqrt(1 - y2)) / 2
    m *= 2
</code>```

Output

```<code>1 2.8284271247461903
2 3.061467458920718
3 3.121445152258053
4 3.1365484905459406
5 3.140331156954739
6 3.141277250932757
7 3.1415138011441455
8 3.1415729403678827
9 3.141587725279961
10 3.141591421504635
11 3.141592345611077
12 3.1415925765450043
13 3.1415926334632482
14 3.141592654807589
15 3.1415926453212153
16 3.1415926073757197
17 3.1415929109396727
18 3.141594125195191
19 3.1415965537048196
</code>```

(Click here for a live version of the script, running on the SageMathCell server).

It works well up to the 14th step, but then the approximations get worse. We could use various tricks to reduce those errors, or use numbers with more precision, but there are much faster ways to compute $\pi$.

FWIW, here's a link to a similar pi calculator I wrote a few years ago, which uses $\tan$ rather than $\sin$, and which avoids the square roots. There are a couple of other pi programs on that page, including a link to the very fast Salamin-Brent algorithm. https://chat.stackoverflow.com/transcript/message/33735078#33735078

One simple way to improve the precision is to eliminate the subtraction in the numerator.

$$\sin^2(\theta/2)=\frac{1-\sqrt{1-\sin^2\theta}}2$$
So
$$\sin^2(\theta/2)=\left(\frac{1-\sqrt{1-\sin^2\theta}}2\right) \left(\frac{1+\sqrt{1+\sin^2\theta}}{1+\sqrt{1-\sin^2\theta}}\right)$$
$$=\frac{\sin^2\theta}{2(1+\sqrt{1-\sin^2\theta})}$$

Our Python code becomes

```<code>from math import sqrt

y2 = 1 / 2
m = 4
for i in range(1, 30):
    y = sqrt(y2)
    print(i, m * y)
    y2 /= 2 * (1 + sqrt(1 - y2))
    m *= 2
</code>```

Output

```<code>1 2.8284271247461903
2 3.0614674589207183
3 3.1214451522580524
4 3.1365484905459393
5 3.140331156954753
6 3.141277250932773
7 3.1415138011443013
8 3.1415729403670913
9 3.1415877252771596
10 3.1415914215111997
11 3.1415923455701176
12 3.1415925765848725
13 3.1415926343385627
14 3.1415926487769856
15 3.141592652386591
16 3.141592653288993
17 3.141592653514593
18 3.141592653570993
19 3.1415926535850933
20 3.141592653588618
21 3.1415926535894996
22 3.1415926535897203
23 3.1415926535897754
24 3.141592653589789
25 3.1415926535897927
26 3.1415926535897936
27 3.1415926535897936
28 3.1415926535897936
29 3.1415926535897936
</code>```

And now the value of $\pi$ is correct upto the 2nd last digit, but notice that the convergence slows down after step 20 or so. We're winning the battle against the limited precision of floating-point numbers, but it's a hard struggle. ;)

Body:
Yes, your formula does converge to $\pi$, as Joshua Wang's answer nicely shows. However, in the comments there's some discussion about the problem of determining the value of $\sin \left(\frac{45°}{2^{x-1}}\right)$ without using a formula based on radians, since they are based on $\pi$.

The traditional approach is to simply bisect an angle whose sine is known from geometry (eg 30°, 45°, or 60°), and calculate the sine of the resulting half-angle. Algebraically, that comes down to solving a quadratic.

It can be shown that for any angle $\theta$,
$$\cos 2\theta = \cos^2\theta - \sin^2\theta$$
Thus
$$\sin^2\theta = (1-\cos 2\theta)/2$$
$$\sin^2\theta = (1-\sqrt{1-\sin^2 2\theta})/2$$
Or
$$\sin^2(\theta/2)=(1-\sqrt{1-\sin^2\theta})/2$$

So now we have a method of determining $\sin^2(\theta/2)$ from $\sin^2\theta$. Unfortunately, it involves taking a square root at every step (and an extra one to get the sine from its square).

This method can be used to quickly get a few decimals of $\pi$, but when we run this algorithm on a computer we'll soon run into arithmetic errors, mostly due to those nested square roots. Eg, using standard double-precision floating point numbers, which have 53 bits of precision, we can get a little over 26 bits of $\pi$, which is around 8 decimal digits.

Here's a short Python program to illustrate:

from math import sqrt

y2 = 1 / 2
m = 4
for i in range(1, 20):
    y = sqrt(y2)
    print(i, m * y)
    y2 = (1 - sqrt(1 - y2)) / 2
    m *= 2

Output

1 2.8284271247461903
2 3.061467458920718
3 3.121445152258053
4 3.1365484905459406
5 3.140331156954739
6 3.141277250932757
7 3.1415138011441455
8 3.1415729403678827
9 3.141587725279961
10 3.141591421504635
11 3.141592345611077
12 3.1415925765450043
13 3.1415926334632482
14 3.141592654807589
15 3.1415926453212153
16 3.1415926073757197
17 3.1415929109396727
18 3.141594125195191
19 3.1415965537048196

(Click here for a live version of the script, running on the SageMathCell server).

It works well up to the 14th step, but then the approximations get worse. We could use various tricks to reduce those errors, or use numbers with more precision, but there are much faster ways to compute $\pi$.

FWIW, here's a link to a similar pi calculator I wrote a few years ago, which uses $\tan$ rather than $\sin$, and which avoids the square roots. There are a couple of other pi programs on that page, including a link to the very fast Salamin-Brent algorithm. https://chat.stackoverflow.com/transcript/message/33735078#33735078

One simple way to improve the precision is to eliminate the subtraction in the numerator.

$$\sin^2(\theta/2)=\frac{1-\sqrt{1-\sin^2\theta}}2$$
So
$$\sin^2(\theta/2)=\left(\frac{1-\sqrt{1-\sin^2\theta}}2\right) \left(\frac{1+\sqrt{1+\sin^2\theta}}{1+\sqrt{1-\sin^2\theta}}\right)$$
$$=\frac{\sin^2\theta}{2(1+\sqrt{1-\sin^2\theta})}$$

Our Python code becomes

from math import sqrt

y2 = 1 / 2
m = 4
for i in range(1, 30):
    y = sqrt(y2)
    print(i, m * y)
    y2 /= 2 * (1 + sqrt(1 - y2))
    m *= 2

Output

1 2.8284271247461903
2 3.0614674589207183
3 3.1214451522580524
4 3.1365484905459393
5 3.140331156954753
6 3.141277250932773
7 3.1415138011443013
8 3.1415729403670913
9 3.1415877252771596
10 3.1415914215111997
11 3.1415923455701176
12 3.1415925765848725
13 3.1415926343385627
14 3.1415926487769856
15 3.141592652386591
16 3.141592653288993
17 3.141592653514593
18 3.141592653570993
19 3.1415926535850933
20 3.141592653588618
21 3.1415926535894996
22 3.1415926535897203
23 3.1415926535897754
24 3.141592653589789
25 3.1415926535897927
26 3.1415926535897936
27 3.1415926535897936
28 3.1415926535897936
29 3.1415926535897936

And now the value of $\pi$ is correct upto the 2nd last digit, but notice that the convergence slows down after step 20 or so. We're winning the battle against the limited precision of floating-point numbers, but it's a hard struggle. ;)


问:
Find the domain and the area where the vector field is conservative!
So I have the following question where $$F=(x,y)/\sqrt{1-x^2-y^2}$$ is a vector field. I have to define the domain for $F$, I have rewritten the function as $1/\sqrt{1-x^2-y^2} $ multiplied with $(x,y),$ where $r^2= x^2+y^2$ and thus $x^2+y^2$ is not equal to  $1$. But is that it? If not, how do I get the domain for such a function please.

I have not started with the second part of the question yet, where i have to determine the area where $$F$$ is conservative.

Thereafter, calculate the line integral for $$\int_{\mathrm{C1}}F\, dr\, $$ where $$C1$$ is a circle with the center in origin and with radius $1/2$.

i have calculated param $x= 1/2cost$ and $y=1/2sint$  is what I got, where $$0\le t\le 2\pi$$.

It became really complicated when I tried to solve the problem, which got me wondering if it could be related to me skipping the second part of the question?

答:
$\vec{F} = \big(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}\big)$.

The domain for the vector field is all real values of $x,y$ where $x^2 + y^2 \lt 1$ which is all points inside a circle of radius $1$ centered at the origin (area of $\pi$).

Now we take derivative and check whether $P_y = Q_x$ where $P$ is the $x-$component of the vector field and $Q$ is the $y-$ component.

$P_y = \displaystyle \frac{\partial}{\partial y} \big(\frac{x}{\sqrt{1-x^2-y^2}}\big) = \frac{xy}{(1-x^2-y^2)^{3/2}}$ (for $x^2+y^2 \lt 1)$

Similarly find $Q_x$ and we see $P_y = Q_x$. That tells us the vector field is conservative where $x^2+y^2 \lt 1$.

As the vector field is conservative, its line integral over any closed curve (whose all points lie in the domain defined above) will simply be zero and that is also true for the circle of radius $\frac{1}{2}$ centered at the origin.


问:
What is the formal definition of ordered tree?
Hi I understand that for rooted tree, the definition is as follow:

Is a directed graph $D=<V,E>$

$\exists v\in V\rightarrow \deg^-(v)=0$ which indicate the in-degree of $v$ is $0$

$\forall v'\in(V/\{v\})\rightarrow\deg^-(v')=1$ which indicate the in-degree of nodes other than $v$ is $1$

I understand that an ordered tree is based on a rooted tree, which all sub-nodes of any node have an order. But I cannot find a formal mathematical expression for this.

May I ask how should I define it with formal mathematical expression??

答:
Let $T$ be such a tree and let us write $xTy$ for the unique path in $T$ between vertices $x$ and $y$. Let us denote $r$ as the root vertex in $T$.

We can impose a partial ordering on $V(T)$ by letting $x \leq y$ if $x \in rTy,\ \forall x,y \in V(T)$. In such a structure every leaf node is a maximal element of $T$.

Here is an example, generated via sage:

However, we can endow our tree structure with different orderings, e.g. horizontal orderings. Have a look on this exemplary collection of different partial orderings for rooted trees on wikipedia.


问:
A theorem in Linear Algebra; linear dependence - Axler

I really am having trouble understanding the statement and the proof. Why does the theorem pick $v_1 \neq 0 $? Why not $v_2$? Also in proving (a), why do we consider the largest $j$?

I do not understand the statement Note all of $a_2, a_3, \dots, a_m$ can be $0$ (because $v_1 = 0 $) Can someone teach me by example?

答:
First, this is an important and basic result.  Mildly rephrased, it says: if you have a finite linearly dependent set of vectors, then by looking from left to right you can always find one vector which lies in the span of the vectors to its left in your list.  Compare this with the statement that a finite set of vectors is linearly dependent iff you can always find one vector in the list which is a linear combination of the other vectors.  The latter statement is a more immediate consequence of the definition: if you have a nontrivial linear combination $a_1 v_1 + \ldots + a_n v_n = 0$, then by definition we must have $a_i \neq 0$ for at least one $i$, and then

$$v_i = \frac{-a_1}{a_i} v_1 + \ldots + \frac{-a_{i-1}}{a_i}v_{i-1} + \frac{-a_{i+1}}{a_i} v_{i+1} + \ldots + \frac{-a_n}{a_i} v_n.$$

Axler's statement is stronger in that it involves the ordering of the vectors.  This is surprisingly useful, and you should look ahead to see applications if you haven't already.

About the $v_1$: it's the first vector on the list, so that's the distinguished role it plays.  Note also that I would state the result a bit differently: I would allow $v_1$ to be a linear combination of the previous, empty, list of vectors, which happens iff $v_1 = 0$.  But when thinking about linear dependence early on it is probably best to concentrate on the case when none of the vectors are zero, because putting the $0$ vector into the list automatically makes the list linear dependent and is a rather degenerate case.

For your other questions:

We pick the largest $j$ because this definition enables us to write $v_j$ as a linear combination of previous vectors in the list only, which is what we are trying to do.

Not all of the $a_2,\ldots,a_m$ can be zero: since we've assumed that $a_1,\ldots,a_m$ are not all zero, if all but $a_1$ are zero then we must have $a_1 \neq 0$.  But then
$$0 = a_1 v_1 + a_2 v_2 + \ldots + a_m v_m = a_1 v_1,$$
and since $a_1 \neq 0$, we get $v_1 = 0$, and we've assumed it isn't.  (Note that you write "because $v_1 = 0$" where you should write "because $v_1 \neq 0$".)


问:
Proving that $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ for positive real $a,b,c$, $abc\ge 1$
I have been able to simplify the problem. Since $a^{4}\ge a^{3}$ iff $a\ge 1$ (and analogously for $b,c$), $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ holds iff $a+b+c\ge 3$. It is thus sufficient to prove $a+b+c\ge 3$.

I am not sure how to proceeds from there, however.

All insights are really appreciated.

答:
I think maybe the question should be $a^4+b^3+c^2\ge a^3+b^2+c$.

By AM-GM,

$a^4+a^4+a^4+1\ge 4a^3\Rightarrow a^4\ge \frac{4a^3-1}{3}$

Similarly, $b^3\ge \frac{3b^2-1}{2}$ , $c^2\ge 2c-1$

Hence, $a^4+b^3+c^2\ge \frac{1}{6}(8a^3+9b^2+12c-11)$

So we only need to prove that

$8a^3+9b^2+12c-11\ge 6a^3+6b^2+6c$ , which is equivalent to

$2a^3+3b^2+6c\ge 11$

By AM-GM, $2a^3+3b^2+6c=a^3+a^3+b^2+b^2+b^2+c+c+c+c+c+c\ge 11\sqrt[11]{a^6b^6c^6}\ge 11$

Hence, we are done.


问:
Local path connected and connected implies path connected
I am very close to proving that if a metric space $X$ is locally path-connected and connected, then it is path connected. I am stuck on the last step.

Let $X$ be connected and locally path connected. Let $x \in X$, and define $A$ to be the set of points path connected to $x$. $x$ is connected to itself, so $A$ is nonempty. Because $X$ is connected, if $A$ is open and closed, it must equal $X$ (since it is not $\emptyset$). We show that $A$ is open. Let $a \in A$. By local path connectedness, there exists an open ball $U \ni a$ that is path connected. Let $u \in U$, so $u$ is path connected to $a$. But $a$ is path connected to $x$, so $u$ is path connected to $a$. So $u \in A$, so $U \subset A$, so we've found an open ball around $a$ that's contained in $A$, so $A$ is open.

My difficulty is in showing $A$ is closed. It's enough to show that $A = \overline{A}$.

To show $A$ is closed, we prove $A = \overline{A}$. Clearly $A \subset \overline{A}$. For the reverse inclusion, let $y \in \overline{A}$. As $\overline{A} \subset X$ and $X$ is path-connected, there exists a open ball $U \ni y$ that is path=connected. Then $U \cap A$ is nonempty. Let $t \in U \cap A$. Since $t \in A$, $t$ is path-connected to $x$. Since $t \in U$ and $y \in U$, $t$ is path-connected to $y$. Therefore, $y$ is path-connected to $x$, so $y \in A$. So $\overline{A} \subset A$, so $A = \overline{A}$.

I think every step of this proof is fine, but to talk about such a $t$, I have to argue that $U \cap A$ is nonempty, but I can't think of an argument for this.

答:
The fact that $U \cap A \neq \emptyset$ follows immediately from the fact that $U$ is a neighbourhood of $y \in \overline{A}$. One of the many equivalent conditions for membership in the closure of $A$ is that every neighbourhood of the given point intersects $A$.

If you prefer it done sequentially, we know that $y \in \overline{A}$ if and only if there exists a sequence $a_n \to y$, meaning that, for all $\varepsilon > 0$, there exists some $N$ such that
$$n \ge N \implies d(a_n, y) < \varepsilon.$$
Taking $\varepsilon$ to be the radius of the open ball $U$, we get
$$n \ge N \implies a_n \in U \cap A.$$
Thus, $U \cap A \neq \emptyset$.


问:
How to show that this $2n \times n^2$ matrix has rank $2n-1$?
The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is

\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1
\end{bmatrix}

So, basically, it splits into two parts:

the upper part (from row $1$ to row $n$) has $n$ identity matrices $I_n$.

the lower part is $O_n^1, \ldots, O_n^n$, where each $O_n^i$ is an $n \times n$ matrix whose $i$-th row is all one whereas the other entries are zero.

By some examples and intuition, I am aware that the rank of such a matrix is $2n-1$, but how should I rigorously prove it?

答: Let $A$ be the $2n \times n^2$ matrix in your question. It is easy to check that $$AA^T = \begin{bmatrix}nI_n & 1_{n \times n} \\ 1_{n \times n} & nI_n \end{bmatrix}$$ where $I_n$ is the $n \times n$ identity matrix and $1_{n \times n}$ is the $n \times n$ matrix of all ones. (To see why this is true, try computing the dot product of each pair of rows of $A$.) Hence, $$AA^T - nI_{2n} = \begin{bmatrix}0_{n \times n} & 1_{n \times n} \\ 1_{n \times n} & 0_{n \times n} \end{bmatrix},$$ which clearly has rank-$2$. By inspection $\begin{bmatrix} 1_{n \times 1} \\ 1_{n\times 1}\end{bmatrix}$ and $\begin{bmatrix} 1_{n \times 1} \\ -1_{n\times 1}\end{bmatrix}$ are eigenvectors of $AA^T-nI_{2n}$ with respective eigenvalues of $n$ and $-n$. Hence, the eigenvalues of $AA^T-nI_{2n}$ are $n$, $-n$, and $0$ (with multiplicity $2n-2$). Therefore, the eigenvalues of $AA^T$ can be found by adding $n$ to each of the eigenvalues of $AA^T-nI_{2n}$, i.e. $2n$, $0$, and $n$ (with multiplicity $2n-2$). Since $AA^T$ is symmetric and has exactly $2n-1$ non-zero eigenvalues, $\text{rank}(AA^T) = 2n-1$. Therefore, $\text{rank}(A) = \text{rank}(AA^T) = 2n-1$ as well.


问:
A Ferris wheel has a radius of 20 m. Passengers get on halfway up on the right side.
A Ferris wheel has a radius of 20 m. Passengers get on halfway up on the right side.
The direction of rotation is counter clockwise. The bottom of the Ferris wheel is 2 m
off the ground. It rotates every 36 seconds.

a)  Graph your height     above the ground as a function of time

b)  Determine a sine function that expresses your height as h as function of elapsed time t
h(t)=20sin(πt/18)+22

c) Determine your height above the ground after 15 seconds algebraically.

After 15 seconds:
h=20sin(15π/18)+22
=20sin(5π/6)+22
=20*(1/2)+22
=32 m

d) Determine the first time (to the nearest tenth) when your height is 38 m above the ground
            algebraically.

When h=38 m
38=20sin(πt/18)+22
38-22=20sin(πt/18)
20sin(πt/18)=16
sin(πt/18)=16/20=4/5=.8
arcsin(.8)=0.927
πt/18=0.927 (radians)
t=(.927*18)/π≈5.31

Height above the ground after 15 seconds≈38 m
seconds elapsed when height is 38 m above ground≈5.3 seconds

答:
Your answer to part b is correct.  That is what you want to graph.  So plug in a number of $t$'s, get $h(t)$, and plot them.   If I were posing the problem, I would have asked b before a.

Added:  c and d are fine, as well.  A graph from Alpha is here


问:
If  is an odd function and lim→0 () exists then prove that lim→0 ()=0.

a/ Give an example of a function such that ∗(0) exists but lim→0 () does not exist.

b/ If  is an odd function (i.e., ()=−(−) for all ∗(0)=0) and lim→0 () exists then prove that lim→0 ()=0.

Hello, I am struggling with this question. Where to begin? Please help. Thank you!

答:
Let $C = \lim\limits_{x \to 0} f(x)$.  Obviously for any function $h$, we have

$$\lim\limits_{x \to 0^+} h(x) = \lim\limits_{x \to 0^-} h(-x).$$

Then

$$C = \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^-} f(-x) = \lim\limits_{x \to 0^-} -f(x) = - \lim\limits_{x \to 0^-} f(x) = -C  $$

so $C = 0$.


问:
Why is the expression for $\sin(\alpha)$ equal to $\frac{\delta V}{\delta x} + O(\frac{\delta V}{\delta x} )^3$
$\sin(\alpha) = \frac{\frac{\delta V}{\delta x}}{\sqrt{1 + (\frac{\delta V}{\delta x}})^2}$, $\frac{\delta V}{\delta x}\ll 1$.

$\sin(\alpha)$ can be approximated by:

$\sin(\alpha) \approx \frac{\delta V}{\delta x} + O(\frac{\delta V}{\delta x})^3$.

In this case, why is the additional terms in the order of $O(\frac{\delta V}{\delta x})^3$? i.e. why is the exponent 3 in this case?

答:
since $sin^{'}(\alpha)=cos(\alpha)\alpha^{'}$, $\alpha^{'}$ can be expressed as $\frac{\delta V}{\delta x}$, and $cos(\alpha)=1/sec(\alpha)=1/\sqrt{1+tan^2(\alpha)}$.

since $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-......$, and $(\frac{\delta V}{\delta x})^{4}$'s order is much lower than $(\frac{\delta V}{\delta x})^{3}$. so we can tarnsform $1/\sqrt{1+(\frac{\delta V}{\delta x}})^2$ to $\sqrt{1-(\frac{\delta V}{\delta x})^2}$.

thus the term in your question can be rewrite to $\frac{\delta V}{\delta x}\cdot({\frac{\delta V}{\delta x}})^2$, since $(1-x)^{a}=1+ax+\frac{a(a-1)}{2!}x^2+......$. now your term $O(\frac{\delta V}{\delta x})^3$ holds as a bound $M$ of ratio $[f(x)/(\frac{\delta V}{\delta x})^3]$.


问:
Proving infinite intersection has only one point
Given that:

$$ \begin{cases}a_1 = 1\\ a_{n+1} = \sqrt{a_n + 5} \end{cases} ~~~ \text{and} ~~~\begin{cases}b_1 = 4\\ b_{n+1} = \sqrt{b_n + 5} \end{cases}$$

Then:

$$ \bigcap_{n=1}^{\infty}[a_n,b_n] $$

Has only one point.

The thing is that I am stuck, it looks like the intervals are $$[1,4] \\ [\sqrt{5}, 3] \\ [ \sqrt{\sqrt{5} + 5}, \sqrt{8} ~] \\ \vdots $$

So it does go down (by going down I mean the length of the interval), and finally stops.. but where? How should I get the "final" point, is it just the limits?

$$ \lim a_n = \frac{1+\sqrt{21}}{2} \\ \lim b_n = \lim a_n$$

So the final point is $$ [ \lim a_n, \lim a_n]$$

I am really sure it is wrong, any help would be appreciated, Thanks!

答: Yes, finding the limits of the two sequences is the way to go. Be sure to show that both limits do in fact exist, i.e. that both sequences converge. Also verify that this limit is contained in all intervals.</p>


问:
Is this function Lipschitz? $f(y) = \sum_{k=1}^{\infty}\frac{1}{k^2}\frac{|y- q_k|}{\sqrt{1 + (y-q_k)^2}}$
Consider the function

$$f(y) = \sum_{k=1}^{\infty}\dfrac{1}{k^2}\dfrac{|y- q_k|}{\sqrt{1 + (y-q_k)^2}},$$ where $q_k$ is some sequence of rational.

Is this function Lipschitz with respect to $y$?

I am unable to simplify the function to get in the proper form, any hints will be helpful.

Thank you.

答:
Here's the outline of what you should do (I want you to do some of the work!):

Consider the function $g(x) = \frac{x}{\sqrt{1 + x^2}}$. Show $g$ is Lipshitz (possibly by appealing to $g'$).

Show $|g(x - q_k)|$ is Lipschitz by appealing to the fact that compositions of Lipschitz functions are Lipschitz.

Show that $\sum_k \frac{1}{k^2} |g(x - q_k)|$ is an infinite sum of Lipschitz functions, whose Lipschitz constants are summable. Conclude that $\sum_k \frac{1}{k^2} |g(x - q_k)|$ is a Lipschitz function.

Let me know if you get stuck again!


Title: Proving infinite intersection has only one point

Body:
Given that:

$$ \begin{cases}a_1 = 1\\ a_{n+1} = \sqrt{a_n + 5} \end{cases} ~~~ \text{and} ~~~\begin{cases}b_1 = 4\\ b_{n+1} = \sqrt{b_n + 5} \end{cases}$$

Then:

$$ \bigcap_{n=1}^{\infty}[a_n,b_n] $$

Has only one point.

The thing is that I am stuck, it looks like the intervals are $$[1,4] \\ [\sqrt{5}, 3] \\ [ \sqrt{\sqrt{5} + 5}, \sqrt{8} ~] \\ \vdots $$

So it does go down (by going down I mean the length of the interval), and finally stops.. but where? How should I get the "final" point, is it just the limits?

$$ \lim a_n = \frac{1+\sqrt{21}}{2} \\ \lim b_n = \lim a_n$$

So the final point is $$ [ \lim a_n, \lim a_n]$$

I am really sure it is wrong, any help would be appreciated, Thanks!

Body: We have: $0 \le |b_n - a_n|= \dfrac{|b_{n-1} - a_{n-1}|}{\sqrt{b_{n-1}+5}+\sqrt{a_{n-1}+5}}< \dfrac{|b_{n-1} - a_{n-1}|}{2\sqrt{5}}<...< \dfrac{|b_1 - a_1|}{(2\sqrt{5})^{n-1}} = \dfrac{3}{(2\sqrt{5})^{n-1}}\implies \displaystyle \lim_{n \to \infty} |b_n - a_n| = 0$. Further, you can show: $[a_n, b_n] \subseteq [a_{n-1}, b_{n-1}]\implies \bigcap_{n=1}^{\infty}[a_n,b_n]= \{a\}= \{\frac{1+\sqrt{21}}{2}\}$ ( the common limit of both sequences ).


问:
Defining topology on projective space via topology of a field
Given a topology on a field $K$ (in my case a non-archimedean local field with valuation $\nu$ and ring of integers $R$ with maximal Ideal $\mathfrak{m}$), the goal is to define a topology on $\mathbb{P}^n(K)$ that comes from $K$. I know that one way to do this is to look at the map
\begin{align}
K^{n+1} \setminus \lbrace 0 \rbrace &\longrightarrow \mathbb{P}^n(K) \\
(x_0, \dots, x_n) &\longmapsto [x_0, \dots,x_n]
\end{align}
which is surjective, so we can define the topology on the target to be the quotient topology from $K^{n+1}$. (The topology there being understood as product topology).

My question is the following: Since we know that $\mathbb{P}^n(K)$ admits an open cover by $n+1$ copies of $K^n$, is there a way to define a topology via this covering? Would this give us the same topology as the quotient?

Another question is, when we want to check the continuity of the specialization map
\begin{align}
\text{Sp }\colon \mathbb{P}^n(K) &\longrightarrow \mathbb{P}^n(k) \\
[x_0, \dots x_n] &\longmapsto [\lambda x_0 \text{ mod } \mathfrak{m}, \dots, \lambda x_0 \text{ mod } \mathfrak{m}]
\end{align}
where $k=R/\mathfrak{m}$ and $\lambda$ scales the coordinates so that all are in $R$ (but not all in $\mathfrak{m}$), with $\mathbb{P}^n(k)$ having discrete topology and $\mathbb{P}^n(K)$ the avove, can we check this using an open cover of $\mathbb{P}^n(K)$ by copies of $R^n$? In that case, the continuity of reduction mod $\mathfrak{m}$ is obvious via the metric on each space. For clarification: $\mathbb{P}^n(K)$ does admit an open cover by $n+1$ copies of $\mathbb{A}^n(R)=R^n$, because coordinates in $\mathbb{P}^n(K)$ can be scaled so that all of them are in the ring of integers R. This works for any local field, but thinking specifically of $K=\mathbb{Q}_p$, $R=\mathbb{Z}_p$ or even $K=\mathbb{Q}$ and $R=\mathbb{Z}$ should help intuition.

答:
Sure: you can define a topology on $\mathbb{P}^n(K)$ by saying a subset is open iff its intersection with each of the copies of $K^n$ is open in the product topology of $K^n$.  Or equivalently, we can take as a basis for our topology all the open subsets of any of the copies of $K^n$ that cover $\mathbb{P}^n(K)$.

This topology is in fact the same as the quotient topology.  To prove this, let $U_i\subset K^{n+1}\setminus\{0\}$ be the set of points whose $i$th coordinate is nonzero.  Each $U_i$ is open and saturated with respect to the quotient map $p:K^{n+1}\setminus\{0\}\to\mathbb{P}^n(K)$, so each $p(U_i)$ is open with respect to the quotient topology, and $p$ restricts to a quotient map $U_i\to p(U_i)$.  So, all we need to show is that the quotient topology on each $p(U_i)$ the the same as the product topology on $K^n$ when we identify them in the usual way.

For convenience of notation, we show this in the case $i=0$.  To show the quotient topology on $p(U_0)$ is the same as product topology on $K^n$, we need to show the map $q:U_0\to K^n$ given by $q(x_0,\dots,x_n)=(x_1/x_0,\dots,x_n/x_0)$ is a quotient map.  Now note that the map $f:U_0\to U_0$ defined by $f(x_0,\dots,x_n)=(x_0,x_1x_0,\dots,x_nx_0)$ is a homeomorphism, and $qf$ is just the projection that drops the first coordinate.  This projection is a quotient map, and thus so is $q$.


问:
How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$
I am a bit confused in this one, how do i do the required modification in this case?</p>

答:
As stated in the comments, $\sum_{r = 1}^{n} \frac{1}{r + 2}$ does not telescope. As for an example of a sum that does telescope, consider $\sum_{r = 1}^{n} \frac{1}{r(r + 1)}$. We can rewrite this summation as

$$\sum_{r = 1}^{n} \frac{1}{r(r + 1)} = \sum_{r = 1}^{n} \left(\frac{1}{r} - \frac{1}{r + 1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} -  \frac{1}{n + 1}\right),$$

and we can pair off terms to see that the entire sum ends up being $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
You might be interested to learn that we didn't have a very good definition of what a proof is until fairly recently. Before the birth of modern logic and set theory, mathematicians had managed to use accepted proof methods to derive contradictions (a famous example is Russell's paradox). A great deal of work was put into defining proof systems that didn't allow for such inconsistencies. Among other things, this involved creating much stricter rules for how you can construct sets.

Most people are content with our current proof methods, even though we rarely (if ever) use completely rigorous proofs. Unfortunately, it has been proven that any sufficiently advanced proof system can't prove its own consistency. In fact, if someone could prove that our current proof methods are consistent (i.e. can't generate contradictions), then that would imply that our proof systems is inconsistent. So in that sense, yes, the "false backbone of rigour" does fall appart a bit when you start digging. If our proof system is consistent, we will never know...

There are lots of strange and interesting phenomena related to proofs and logic. One thing you might have heard of is that certain statements can't be proven or disproven. They are, in a sense, unknowable. Unless I'm mistaken, the continuum hypothesis and the axiom of choice are two such statements. You can choose to assume they're true or false and write perfectly valid proofs in either case.

The same can't be said for provable statements, though. If we take your example and add Fermat's last theorem to our axioms we could face a problem. What if Fermat's last theorem wasn't true? (it is, but for the sake of argument suppose it isn't). You could use the other axioms to prove it false. You now have a system of logic that is inconsistent; you can use it to prove paradoxes. And if you can do that, you can literally prove anything. Not good...

(I'm essentially parroting what my logic prof said during our first lecture. I've fact-checked this to the best of my abilities, but this is still mostly from memory. If something is wrong, please let me know.)

Title: What is a proof?


答:
Suppose the sequence $a_{n}\to L$ or $b_{n}\to L$. Then $L$ must be nonnegative and real since $f(x)=\sqrt{x+5}$ has the property that $f([0,\infty))\subset [0,\infty)$. Moreover, $L$ must be a fixed point of the recursive equation. That is to say, if a solution of a recursive equation converges it must converge to an equilibrium solution.

$L=\sqrt{ L + 5 }$ so $L^{2}=L+5$, thus $L=\frac{1+\sqrt{21}}{2}$ is the unique non-negative equilibrium solution. Thus it suffices to show that $a_{n}$ and $b_{n}$ both converge.

The simplest way to prove convergence is through a well known result for first order difference equations. If you don't know it (here I mean can't use it in the exercise), the next easiest way is to observe that the sequences are monotone and bounded.


问:
KKT stationarity condition in simple SDP
This question is part of my continuing efforts to answer my original question in A puzzling KKT for LMI vs. scalar constraint.

Consider an SDP
\begin{align}
&\min  - \text{Tr}(P)\\
&\ \ \text{s.t.}
\begin{pmatrix}
A^TP + PA +C^TPC + Q & PB + C^TPD \\
B^T P + D^TPC& R+D^TPD
\end{pmatrix}\succeq0, P\succeq0
\end{align}

In a paper, they define a dual variable as
\begin{align}
Z= \begin{pmatrix}
S&U^T&0\\
U&T&0\\
0&0&W
\end{pmatrix},
\end{align}
and claim that
\begin{align}
AS+SA^T + CSC^T + BU + DUC^T + U^TB^T + CU^TD^T + D^TTD + W + I=0.
\end{align}
By computing the KKT stationarity condition, we get that the Trace of this equation is zero rather than its argument. What am I missing?

I will be happy for help about this step and two more related questions:

How will the equation change if we change to objective to $-\text{Tr}(R+D^TPD)$.

Is it a standard SDP? I learned that the objective should be expressed as $c^Tx$.

答:
You can write the primal as:
\begin{align}
&\min  - \text{Tr}(P)\\
&\ \ \text{s.t.}
\begin{pmatrix}
A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\
B^T P + D^TPC& R+D^TPD & 0 \\
0 & 0 & P
\end{pmatrix}\succeq0
\end{align}
The Lagrangian is:
\begin{align}
L(P,Z) = - \text{Tr}(P) - \langle
\begin{pmatrix}
A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\
B^T P + D^TPC& R+D^TPD & 0 \\
0 & 0 & P
\end{pmatrix} , Z \rangle
\end{align}
If you take the derivative with respect to $P$ you obtain the stationarity condition:
\begin{align}
0&=\frac{\partial L(P,Z)}{\partial P}\\
&=  - \frac{\partial}{\partial P} \left\{\text{Tr}(P) + \text{Tr} \left(S^T(A^TP+PA + C^TPC+Q) + U(PB+C^TPD))\right)\right\}\\
&\ \ \ \ \  - \frac{\partial}{\partial P} \left\{ \text{Tr}\left(U^T(B^TP+D^TPC) + T^T(R+D^TPD) + P^TW)\right)\right\}.
\end{align}
By using the general relation $\frac{\partial}{\partial P} \text{Tr}(\Gamma P \Lambda) = \frac{\partial}{\partial P} \text{Tr}(\Lambda\Gamma P ) = \Lambda^T\Gamma^T$ for any $\Lambda,\Gamma$ and the cyclic trace property, we get:
\begin{align}
0&= I + AS + SA^T + CSC^T + U^TB^T + CU^TD^T + BU + DUC^T + DTD^T+ W.
\end{align}

As for your other questions:

the $I$ at the end will become $DD^T$

there is no 'standard' SDP, but an SDP in standard form is:
$$\min_{X \succeq O} \{ \text{Tr}(CX) : \text{Tr}(A_iX)=b_i, \; i=1,\ldots,m\}$$
so clearly the given SDP is not in standard form. You can always reformulate it into standard form though (using a different choice for $X$).


问:
What exactly is sheafification?
I have recently learned about the very BASICS of sheaves, but I was wondering is there an easier definition for sheafification? I could not find anywhere an easier definition for sheafification. I kind of compare it to local rings, where in local rings, you collect the local data at a point for a variety or an algebraic set where some function is defined on that regular set. Am I heading in the right direction? If not, what is the rigorous definition of sheafification and what is some intuition one could use to understand it?

I would much more appreciate the intuition and a clear definition, rather than rigour.

答:
Sheafification is the left adjoint of the inclusion $i: Sh(X) \rightarrow Psh(X)$.

That's the important part anyway, the nitty gritty details of the construction are not so important (constructing the Étale space and whatnot). You should think of the sheafification $\mathcal F_+$ of $\mathcal F$ as identifying sections which are locally equal and adding new "formal" sections which are glued together by old ones in $\mathcal F$ in order to satisfy the sheaf conditions (gluing of sections exists and is unique).

For example: If you have a morphism $\phi:\mathcal F \rightarrow \mathcal G$ where $\mathcal F$ is a presheaf and $\mathcal G$ is a sheaf  and two sections $s,s' \in \mathcal F (U)$ which are locally equal we have that $\phi(s),\phi(s')$ are locally equal (since $\phi$ is compatible with restriction maps) and hence equal because $\mathcal G$ is a sheaf.

Thus it doesn't matter if we identify sections $s$ and $s'$ which are locally equal in $\mathcal F$ if all we care about are maps of $\mathcal F$ into sheaves.

Similarly for formal glued sections, if we have a morphism $\phi:\mathcal F \rightarrow \mathcal G$ then $\phi$ extends uniquely to these formal sections $s$ of $\mathcal F$ over $U$ by simply considering $\{s\mid_{U_i}\}_i$ which are actual sections of $\mathcal F(U_i)$, applying $\phi$ to get $\{\phi(s\mid_{U_i}) \}_i$ which are sections in $ \mathcal G(U_i)$ and then gluing these $\phi(s\mid_{U_i})$ to get a section $s'$ in $\mathcal G(U)$ and then set $\phi(s) = s'$.

We thus see that every morphism $\mathcal F \rightarrow \mathcal G$ gives us a unique morphism $\mathcal F_+ \rightarrow G$. And every morphism $\psi:\mathcal F _+ \rightarrow \mathcal G$ comes from the morphism given by precomposing $\psi$ with the apparent map $\mathcal F \rightarrow \mathcal F_+$.

Anyway, this is enough to say that $Hom(\mathcal F,\mathcal G) = Hom(\mathcal F_+,\mathcal G)$ for sheaves $\mathcal G$ and thus that $(-)_+$ is left adjoint to $i:Sh(X) \rightarrow Psh(X)$.

If this was too handwavy I'm sorry! I'll try and elaborate if you feel like it's necessary.

Title: What exactly is sheafification?


问:
Need calculation help! ( gram-schmidt formula)
I tried to use gram-schmidt formula.

Let $V=\operatorname{span}(S)$ with $\left\langle f,g\right\rangle=\int_{0}^{\pi}f(t)g(t)dt$ and $S$={$\sin t,\cos t, t,1$}.

Rename the elements of $S$ as $w_1, w_2, w_3, w_4$.

$v_1=w_1$.

$v_2=w_2-\dfrac{\left\langle w_2,v_1\right\rangle}{\left \| v_1 \right\|^2}v_1=\cos t$. (Thanks to AWertheim's help)

Using the same formula, $v_3=1-4/\pi \sin t$?

Then this problem becomes too complicated. Can anyone get $v_4$?

答:
Let $v_{1} = \sin t$. Then $$v_{2} = w_{2} - \frac{\langle w_{2}, v_{1}\rangle}{\parallel v_{1} \parallel ^{2}}v_{1}$$

Let's calculate each term individually.

$w_{2} = \cos t$, so:

$$\langle w_{2}, v_{1} \rangle = \int_{0}^{\pi}\sin t\cos t dt$$

Using the substitution $u = \sin t, du = \cos t dt$, we get:

$$\langle w_{2}, v_{1} \rangle = \int_{0}^{0} udu = 0$$

Hence, we get $v_{2} = w_{2} - 0v_{1} = w_{2}$. (N.B: strictly speaking, we need to also check that $\parallel v_{1} \parallel^{2}$ is nonzero, but you've already done this!)

(The other way to look at this is that with respect to this inner product, $\cos t$ and $\sin t$ are already orthogonal!)


答: As many people have mentioned, this sum does not telescope. A summation series telescopes only if there is addition and subtraction involved. The addition and subtraction causes terms to cancel out. For example, the summation series $\sum_{n=0}^\infty \frac1n - \frac1{n+1}$ telescopes because the subtraction allows parts of the series to cancel with parts of the series that come after. For a product series, it should have multiplication and division so that the operations cancel each other out. For example, the product $\prod_{n=0}^\infty \frac1n*({n+1})$ uses multiplication in the previous term to cancel out with division in the current term.

Title: How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$

Body: I am a bit confused in this one, how do i do the required modification in this case?</p>


问:
Problem about problem solving
I am having some problems on how to solve problems. When I read a chapter on say group theory or real analysis, I feel that I have understood the concepts quite well, but when I start solving exercises, after 5 or 6 problems I get stuck. When I see a tough problem, I get messed and cannot find how to come out of it. When I see the solution I have no problem in understanding it. I think about many things but cannot come up with a solution. Is it a problem of my basics not being right? I have an entrance exam & being in this situation is frustrating for me. Please help.</p>

答:
suppose {T} is a collection of bounded operators on a Hilbert space H, with  IT<1   for all k. Suppose also that.

Let SN - Σ Ν T k =-Ν

Show that SNf)     converges as N00 , for every Є Н . If  Tf denotes the limit, prove that  |T||< 1 . [Hint: Consider first the case when only finitely many of the  TK are non-zero, and note that the ranges of the TK are mutually orthogonal.]

Title: Problem about problem solving


问:
Micheal O'Searcoid, Metric Spaces: Problem 8.5:
There exists a function $f: X\to Y$ between metric spaces $X$ and $Y$ that is not continuous but has the property that, for each closed ball $B$ of $Y$, $f^{-1}(B)$ is closed in $X$.

What is an example in support of the above statement?

Is this result true for all metric spaces $X$ and $Y$?

If I replace "Closed"  with "Open" in the above problem, then the function is becoming continuous.

答:
HINT: Try $X=[0,2]$, $Y=[0,1)\cup[2,3]$, and

$$f:X\to Y:x\mapsto\begin{cases}
x,&\text{if }0\le x<1\\
x+1,&\text{if }1\le x\le 2\,.
\end{cases}$$

Here $X$ and $Y$ have their usual metrics as subspaces of $\Bbb R$ with its usual metric.


问:
Binary String Combinations
For this question, imagine finite binary strings consisting of $0$ and $1$.
For instance, $00111$ is a string.  Order does not matter, so the string $1100$ is
considered equivalent to $0011$.  We will therefore always write $0$s before writing $1$s.

For ease of notation, we will write $0^{n}$ when
we mean exaclty $n$ zeros.  Therefore,
$0^{4}1^{2}=000011$.  These exponents do
not indicate multiplication!

Here are three rules for manipulating a string:

Rule 1:  You may always replace a $1$ with $11$. We will write this as $1 \stackrel{1}{\rightarrow} 11$.

Rule 2: You may replace $11$ with a $0$.  More succinctly,
$11 \stackrel{2}{\rightarrow} 0$.

Rule 3:  You may erase $00$.  We can write this as
$00 \stackrel{3}{\rightarrow} \mbox{ null}$.

The following sequence is legal:
$0^41^{2}=\underline{\bf 00}0011 \stackrel{3}{\rightarrow} \underline{\bf 00}11 \stackrel{3}{\rightarrow}
\underline{\bf 11} \stackrel{2}{\rightarrow} 0. $

Therefore, the string $0^{4}1^{2}$ has been transformed into the string $0$, and
we can write $0^{4}1^{2} \rightarrow 0$.  For any bean counters out there, we used Rule 2 once
and Rule 3 twice.  We did not use Rule 1 at all.  Note that, for clarity, we have boldfaced
and underlined the characters that are being operated on.

Prove or disprove
$0^{3}1^{3} \rightarrow {\tt null}$.

Let $0^{n}1^{n}$ denote a string with $n$ zeros
followed by $n$ ones.  Prove or disprove $0^{n}1^{n}
\rightarrow {null}$.

What can you say about $0^{m}1^{n}$?

答:
What can you say about combining rule 1 and 2 ?

 It flips ones into zeroes, indeed $1\to 11\to 0$

So what does it do to $1^n$ ?

 prove by induction $1^n\to 0^n$

What does rule 3 applied repeatedly do to a string of only zeroes ?

 it removes any pair of zeroes, so $0^{2p}\to\tt{null}$ and $0^{2p+1}\to 0$

Can you modify the number of $1$ ?

 Yes notice that by rule 1 we can append as many additional $1$ as wanted, in particular if $n>0$ (if there is at least one $1$) then $1^n\to 1^{n+1}$

Conclude

 $0^m1^n\to 0^m0^n=0^{m+n}\to 0\text{ or }\tt{null }$ depending on the parity of $m+n$, but since we can turn $m+n$ even by appending some $1$ then we are always able to eventually land on $\tt{null}$.

Title: Binary String Combinations


问:
how to rotate a vector in 3d space around arbitrary axis
In my case I have two arbitrary vectors (suppose vector AB and CD ) and I am assuming that some rotation operation will happen to vector AB to get it the orientation of vector CD. so by using the knowledge the positions (cartesian coordinates) of both the vectors (already known) can I find the angle of rotation between them around some arbitrary axis of rotation in space?
Can I reproduce vector CD by using the knowledge of their positions and angle between them?

Basically I am not dealing with vectors but 3 dimensional objects with multiple cartesian coordinates in my system. so using the information of cartesian coordinates of both objects I want to do operations at first object to transform it to the second object. I hope the you are convinced with what I am trying to explain.

答:
The angle between two unit vectors $u$ and $v$ is given as

$$
\theta = \arccos\left(u^\top v\right)
$$

the shortest axis of rotation is the vector orthogonal to both vectors, and can be found using the cross product

$$
s = \frac{u \times v}{\lVert u \times v\rVert}
$$

In most cases, $s$ should be normalized, even if $u$ and $v$ are already both unit vectors.

(note that switching the order of the cross product will give you the negative of the axis.  The right-hand rule determines the order)

To create the rotation matrix that rotates $u$ into $v$, we can use the axis-angle formula for rotation matrices:  (taken from wikipedia)

$$
R = \begin{bmatrix}
\cos\theta + s_x^2 (1-\cos\theta) & s_x s_y (1-\cos\theta) -s_z \sin\theta & s_x s_z(1-\cos\theta + s_y \sin \theta \\
s_y s_x (1-\cos\theta)+s_z \sin\theta & \cos\theta + s_y^2 (1-\cos\theta) & s_y s_z (1-\cos\theta) - s_x\sin\theta \\
s_z s_x (1-\cos\theta) - s_y \sin\theta & s_zs_y(1-\cos\theta) + s_x \sin\theta & \cos\theta + s_z^2(1-\cos\theta)
\end{bmatrix}
$$

and now you can compute

$$
v = Ru
$$

and

$$
u = R^\top v
$$


问:
Generalization of inclusion-exclusion rule
This is taken from Pollard, A User's Guide to Measure Theoretic Probability problem 1.1.

Let $A_1, A_N$ be events in some probability space. Denote $\cap_{i \in J}A_i$ as $A_J$ for some set $J \subseteq [N]$. Also, denote the sum of the probabilities of the ways that $k$ events occur together as $S_k = \sum_{|J| = k}\mathbb P \{A_J\}$.

Show that the probability of exactly $m$ of the events $A_i$ occurring is given by:

$${m\choose m}S_m - {m+1\choose m} S_{m+1} ... +(-1)^{N-m} {N \choose m} S_N$$

What I've tried

I know that this is a generalization of the inclusion-exclusion rule which gives us the probability that none of the events occur. The inclusion-exclusion rule is easier to derive since it's clear what each term is adding and subtracting. Here I don't have strong intuition on what each of the terms represent.

The binomial coefficients remind me of the stars and bars result: the number of ways to distribute $m$ points into $k$ groups. So the first coefficient is the number of ways to distribute $m$ points into $1$ group, the second is the number of ways to distribute $m$ points into $2$ groups and so on until $N - m +1$ groups. This doesn't seem to be the right interpretation since it does not match up with the $S_k$ terms.

Any hints or intuition would be appreciated.

答:
Let $\mathbb1_B$ be the indicator function of an event $B$, so that $\mathbb1_B$ is a random variable that equals $1$ if the event $B$ occurs and $0$ otherwise; note that the expectation of $\mathbb1_B$ equals the probability that $B$ occurs. Let $E_m$ denote the event that exactly $m$ of the $A_j$ occur. Your desired formula would follow from
$$
\mathbb1_{E_m} = {m\choose m}\sum_{|J| = m}\mathbb 1_{A_J} - {m+1\choose m} \sum_{|J| = m+1} \mathbb1_{A_J} + \cdots + (-1)^{N-m} {N \choose m} \sum_{|J| = N}\mathbb 1_{A_J}
$$
by taking expectations of both sides.

The point is that this equality of functions can be checked separately on every possible combination of the events $A_j$. In particular, if you let $K$ denote the exact set of events that occur, then you can verify that the two sides of the above identity are equal one $K$ at a time. The argument will depend only on $\#K$, and will essentially be "if $\#K = m+k$, then how many terms in the first sum equal $1$, how many terms in the second sum, etc.".


问:
Using Schur Complement to prove a matrix is negative definite
Need to find range of values of $\epsilon$ for which $$\begin{bmatrix} A+\epsilon B & \epsilon C \\ \epsilon C^\top &  \epsilon D \end{bmatrix} \prec 0$$ given that $A,D \prec 0$ and $\epsilon > 0$.

The way I am going about it is using Schur Complement Lemma whereby this matrix is negative definite iff $ \epsilon D \prec 0$ and $A+\epsilon B-\epsilon CD^{-1}C^\top \prec 0$. Should the value of $\epsilon$ depend on the maximum eigenvalue of $A-\epsilon CD^{-1}C^\top$? What is the right way to go about it?

答:
Please correct me if I made any mistake in the following:

You can use the Rayleigh Quotient to establish that $\lambda_{\text{max}}(M+N) \leq \lambda_{\text{max}}(M) + \lambda_{\text{max}}(N)$ for any symmetric matrices $M,N$ with the same size.

From the comments above one can infer that each element of the Schur Complement must be symmetric. Since $\lambda_{\text{max}}(A) < 0$ (true with the given condition that $A \prec 0$), one can find a small enough $\epsilon > 0$ such that
$$
\begin{align*}
\lambda_{\text{max}}(A + \epsilon(B - CD^{-1}C^T)) &\leq \lambda_{\text{max}}(A) + \lambda_{\text{max}}(\epsilon(B - CD^{-1}C^T))\\
&= \lambda_{\text{max}}(A) + \epsilon\cdot\lambda_{\text{max}}(B - CD^{-1}C^T) \\
&< 0,
\end{align*}
$$
since $A,B,C,D$ are all known. So a feasible set of $\epsilon > 0$ is simply the ones that satisfy the inequality above, but note that the inequality is simply sufficient and the result may only be a subset of all feasible $\epsilon$'s.


答:
There is not complete agreement among mathematicians about which axioms or rules to use in every case. There are, however, no "loopholes" in a consistent set of axioms and rules. If they are inconsistent, then you can prove absolutely anything!

Quite often, however, mathematicians use axioms and rules that have been found to be very useful, but it is not known with certainty whether they are consistent or not, e.g. the ZFC axioms for set theory. After over a century of intensive study by the experts, no inconsistencies have been found in ZFC. As a foundation for mathematics, it just seems to work.

As for exam questions, the "usual axioms and rules" of logic and mathematics (those most widely used) can be assumed to be available to you unless otherwise stated on the exam paper or in the course materials. Yes, that is a bit vague, but you exploit such ambiguity at your peril. Most examiners have no sense of humour in this regard.

Title: What is a proof?

Body:
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.


问:
Properties of sequence
Sequence $S=(s_1,s_2,s_3,...)$ is a sequence of integers. What conditions must sequence $S$ satsify in order for the fraction $$\frac{s_ns_{n-1}...s_{n-k+1}}{s_ks_{k-1}...s_1}$$ to always be an integer for $1 \leq k \leq n$?

I tested some common sequences:
$cn$, $2^n$, $2^n-1$, $n^p$ all work, while $2n+1$, Fibonacci and Catalan numbers do not. One observation I made is that the sequences that work have finite differences that eventually become constant. However this is also true for some of the sequences that do not work.

答:
In what follows, I will denote the ratio you mentioned $\binom{n}{k}_s$, in analogy with the binomial coefficients. I do not know of a nice characterization of all sequences for which this property holds, but here are a few of infinite families of examples.

Let $a$ and $b$ be integers, and let $s_n$ satisfy the recurrence relation
$$
s_n=as_{n-1}+bs_{n-2},\\
s_1=s_2=1\qquad\;\;\;\;\;
$$
Then the ratio is always an integer. When $a=b=1$, then $s_n$ is the sequence of Fibonacci numbers, and the $\binom{n}{k}_s$ are called the Fibinomial coefficients.

Let $f$ be any polynomial with integer coefficients, and let $x_0$ be any fixed integer. Then another sequence which works is $$s_n=(\underbrace{f\circ f\circ\dots\circ f}_{\text{$n$ times}})(x_0)-x_0,$$ where $f$ is composed with itself $n$ times (Proof). The most famous example in this family is when $f(x)=qx$ and $x_0=1$, for some $q\neq 1$, in which case $s_n=q^n-1$, and the numbers $\binom{n}{k}_s$ are called the $q$-binomial coefficients. The same coefficients appear when $f(x)=qx+b$. For $f(x)=x+b$, the usual binomial coefficients appear. When $\deg f\ge 2$, $\binom{n}{k}_s$ grows quite quickly.

If $d$ is any fixed positive integer, then $s_n=n(n+1)\cdots (n+d-1)$ works. When $d=2$, these are the Narananya numbers. For general $d$, $\binom{n}{k}_s$ is the number of plane partitions which fit in a $k\times (n-k)\times d$ box. Equivalently, this is the number of tilings of an equiangular hexagon with side lengths $k,n-k,d,k,n-k,d$ by rhombi with side length $1$.

One final mention: Knuth and Wilf prove in The power of a prime that divides a generalized binomial coefficient that as long as the sequence $s$ satisfies the following property, then $\binom{n}{k}_s$ will always be an integer:
$$
\gcd(s_n,s_m)=s_{\gcd(n,m)}
$$

Title: Properties of sequence