问:
Is there a reason why this technique is invalid?

What is $ \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$? A simple way to evaluate this limit is to substitute $0$ for $x$ in the numerator to obtain

$ \displaystyle \lim_{x \rightarrow 0} \frac{1 - 1}{x} = \lim_{x \rightarrow 0} ( \frac{1}{x} - \frac{1}{x} )
= \lim_{x \rightarrow 0} (0) = 0 $

since $ \frac{1}{x} - \frac{1}{x} = 0$ since one quantity subtracted from the same quantity is 0. This technique circumvents the problem of division by zero while utilizing the fact that $\cos(0)$ is known.

答:
It's invalid.

You can not replace a variable with a constant in one part of an expression but leave it as a variable in another.

If you want to estimate a limit by replacing a variable with a constant you must replace it everywhere.  If you do that you ge $\frac {1 - \cos 0}{0} = \frac 00$ and that doesn't help us at all.

We must assume $x \ne 0$ and if we replace it we must replace it with $x = h\ne 0$ and we get $\lim_{x\to 0} \frac {1-\cos x}x \approx \frac {1-\cos h}{h}$ and we can't replace $h$ with $0$ in the top and not the bottom because $h$ ISN"T $0$.  And whatever the $x$ in the numerator is, the $x$ in the denominator must be the same thing.

.....

The reasoning of the error is that a little fudging in the top $x\approx 0$ means $\cos x \approx \cos 0$ won't affect much. But that is wrong.  The fudging in the bottom makes a huge difference.  $\frac 1x \not \approx \frac 10$.  That's a no-no.

Complete no-no.

And completely invalid.


问:
$\mathcal{F}(f)\star\mathcal{F}(g)(\xi)$
Can someone prove that $\mathcal{F}(f)*\mathcal{F}(g)(\xi)=
\mathcal{F}(f.g)(\xi)$?

where:
$\mathcal{F}f(x)=\int_{\mathbb{R^n}} f(x)e^{-2i\pi x\xi}dx$.
Or is there any link? And thank you so much.

答: The right-hand side is$$\begin{align}\mathcal{F}(f\cdot g)(\xi)&=\int f(x)g(x)\exp(-2\pi ix\xi)dx\\&=\iint f(x)\delta(x-y)g(y)\exp(-2\pi ix\xi)dxdy\\&=\iiint f(x)g(y)\exp(-2\pi ix(\xi-\eta)-2\pi iy\eta)dxdyd\eta\\&=\int\left(\int f(x)\exp(-2\pi ix(\xi-\eta))dx\right)\left(\int g(y)\exp(-2\pi iy\eta)dy\right)d\eta\\&=\int(\mathcal{F}(f))(\xi-\eta)(\mathcal{F}(g))(\eta)d\eta\\&=\mathcal{F}(f)\star\mathcal{F}(g).\end{align}$$

ContentLicense: CC BY-SA 4.0


问:
Evaluating $2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}$. Why can we say $2+\frac1x=x$?

The value of
$$2+\dfrac{1}{2+\dfrac{1}{2+\cdots}}$$ is ...

(a) $1-\sqrt{2}\quad$ (b) $1+\sqrt{2}\quad$ (c) $1\pm\sqrt{2}\quad$ (d) None of these

Now I have some question here from the way I saw it solved online.

He took
$$2+ \frac1x = x$$

Now, my question is that why do we say $2+\frac1x$ equals to $x$? Like, it should be equal to $y$, right?

If we do that m it means we already know that RHS is $x$. I think that is wrong to do.

When we are not given like here nothing in RHS. Just "is" is written there. So what should we assume RHS as in these cases?

Answer for x = 1 + $\sqrt{2}$

答:
Imagine that by some magic you have figured out the value of $x$. It's just a number. When you look at the first denominator in the expression you see exactly that expression again, so you can call it by the same name, $x$:
$$
x = 2 + \frac{1}{ x}
$$
and you can solve that equation for $x$.

The subtle problem with this argument is not that you used the same name $x$ twice, it's that you casually assumed that this expression that goes on forever actually makes sense. That assumption has to be justified - but not for this question. Since it's a multiple choice question the person who asked it is willing to assume the expression makes sense.

This kind of reasoning is common. For example, to express the number
$$
x = 0.131313\ldots
$$
as a fraction, you assume the infinite decimal makes sense. Then multiply by $100$ to see that
$$
100x = 13.1313\ldots = 13 + x.
$$
Solving for $x$ yields $13/99$.

In each of these examples what's hidden is justifying algebra in what looks like an infinite expression.


问:
Right triangles with integer sides
Most of you know these triples:

$3: 4 :5$

$5: 12 :13$

$8: 15 :17$

$7: 24 :25$

$9: 40 :41$

More generally we can construct such triangles such as
$$2x:x^2-1:x^2+1$$

My question is why one of the sides seems to be always prime? (When there is no common divisor)

答:
It's great that you are interested and looking for patterns, so don't get discouraged. As one of my teachers is fond of saying,

If all your conjectures are true, you aren't trying hard enough.

So, here are first ten counterexamples:

$$\begin{array}{c|c}
x& 2x:x^2-1:x^2+1\strut\\\hline
8 & 16:63: 65\strut\\
12 & 24:143: 145\strut\\
18& 36:323: 325\strut\\
22& 44: 483: 485\strut\\
28& 56: 783: 785\strut\\
30& 60: 899: 901\strut\\
32& 64: 1023: 1025\strut\\
34& 68: 1155: 1157\strut\\
38& 76: 1443: 1445\strut\\
42& 84: 1763: 1765
\end{array}$$

Mathematica code:

```
listofcounterexamples = {};
For[x = 1, x < 100, x++,
  If[Length[listofcounterexamples] == 10, Break[]];
  If[GCD[2x, x^2-1, x^2+1] == 1 && Not[PrimeQ[x^2+1]],
  AppendTo[listofcounterexamples, {x, 2x, x^2-1, x^2+1}]]];
listofcounterexamples
```

ContentLicense: CC BY-SA 3.0

Title: Right triangles with integer sides

Body:
It's great that you are interested and looking for patterns, so don't get discouraged. As one of my teachers is fond of saying,

If all your conjectures are true, you aren't trying hard enough.

So, here are first ten counterexamples:

$$\begin{array}{c|c}
x& 2x:x^2-1:x^2+1\strut\\\hline
8 & 16:63: 65\strut\\
12 & 24:143: 145\strut\\
18& 36:323: 325\strut\\
22& 44: 483: 485\strut\\
28& 56: 783: 785\strut\\
30& 60: 899: 901\strut\\
32& 64: 1023: 1025\strut\\
34& 68: 1155: 1157\strut\\
38& 76: 1443: 1445\strut\\
42& 84: 1763: 1765
\end{array}$$

Mathematica code:

listofcounterexamples = {};
For[x = 1, x < 100, x++,
  If[Length[listofcounterexamples] == 10, Break[]];
  If[GCD[2x, x^2-1, x^2+1] == 1 && Not[PrimeQ[x^2+1]],
  AppendTo[listofcounterexamples, {x, 2x, x^2-1, x^2+1}]]];
listofcounterexamples


答: @ChristinaDaniel OK, here is a counter example: Consider the expression $\frac{\sin 2x}{x}$ and let $x$ go to zero: The answer to this limit is $2$. Now consider the expression $\frac{\sin 2x-0}{x}$ for $x$ going to zero. The answer to this limit is still $2$. But $\sin0=0$ so we can now consider the expression $\frac{\sin 2x-x}{x}$, again with $x$ going to zero. But now this limit is $1$. So when you do a "partial" substitution, the answer changes. In other words, when you substitute for $x$, you need to do that for every $x$ in the expression.

Title: Is there a reason why this technique is invalid?

Body:
What is $ \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$? A simple way to evaluate this limit is to substitute $0$ for $x$ in the numerator to obtain

$ \displaystyle \lim_{x \rightarrow 0} \frac{1 - 1}{x} = \lim_{x \rightarrow 0} ( \frac{1}{x} - \frac{1}{x} )
= \lim_{x \rightarrow 0} (0) = 0 $

since $ \frac{1}{x} - \frac{1}{x} = 0$ since one quantity subtracted from the same quantity is 0. This technique circumvents the problem of division by zero while utilizing the fact that $\cos(0)$ is known.


问:
Re-arranging a derivative
If $T = \sqrt { \frac {2h} {g} }$ and $x(t) = \frac {1} {2} gt^2$.

Could someone kindly explain to me how $$ \frac {dt} {T} = \frac {dx}{gt} \sqrt{ \frac {g} {2h}} $$

I'm currently struggling with Griffiths Introduction to Quantum Mechanics. This worked example is on p. 10 of the 2nd edition.

答:
$\frac{dx}{dt}=gt$ so $\frac{1}{gt}\,dx=dt$. Dividing by $T$, we obtain
$$\frac{dt}{T}=\frac{dx}{gt}\cdot\frac{1}{T}=\frac{dx}{gt}\sqrt{\frac{g}{2h}}\,.$$

Title: Re-arranging a derivative


答: From the first equality $$\color{red}{\frac{1}{T} = \sqrt{\frac{g}{2h}}} $$ From the second equality $$ dx = gtdt \implies  \color{red}{dt = \frac{dx}{gt}}$$ Multiply the red equalities to get $$\frac {dt} {T} = \frac {dx}{gt} \sqrt{ \frac {g} {2h}}$$

Title: Re-arranging a derivative

Body:
If $T = \sqrt { \frac {2h} {g} }$ and $x(t) = \frac {1} {2} gt^2$.

Could someone kindly explain to me how $$ \frac {dt} {T} = \frac {dx}{gt} \sqrt{ \frac {g} {2h}} $$

I'm currently struggling with Griffiths Introduction to Quantum Mechanics. This worked example is on p. 10 of the 2nd edition.


问:
Interior of a set than sounds easy
Let $\Omega\subseteq\mathbb{R}^N$ an open set (it can be cosidered bounded if needed), and
$\phi:\Omega\to\mathbb{R}$ be a $C^1$ function with $\phi^{-1}(0)\neq\emptyset$ and $\nabla\phi(x)\neq 0,\ \forall\ x\in\phi^{-1}(0)$. Is it true that:

$$\mathrm{int}\overline{\phi^{-1}([0,\infty))}=\phi^{-1}((0,\infty))$$
?

Remark that one inclusion is obvious. I cannot find a counterexample.

答:
A counterexample is given by the function
$$\phi : \mathbb{R} \to \mathbb{R} : x \mapsto (x - 1)^2 - 1$$

Title: Interior of a set than sounds easy


答:
No, you cannot claim that $x=0$ in the numerator while $x\ne0$ at the denominator !

Using your method,  a simple way to evaluate this limit is to substitute $0$ for $x$ in the denominator to obtain
$$ \displaystyle \lim_{x \rightarrow 0} \frac{\cos x - 1}{0} =\lim_{x \rightarrow 0}\pm\infty$$ as the numerator is nonzero.


问:
Finite Presentations on Graded $A$-modules over Noetherian rings.
Let $R$ be a Noetherian ring and $A$ be a finitely generated graded $R$-algebra. Then we know $A$ is Noetherian by the Hilbert Basis theorem. Let $M$ be a finitely generated $A$-module. Then $M$ is a Noetherian $A$-module.

The text I am reading says that there exists free graded $A$-modules $L,L’$ and degree preserving maps $\varphi, \psi’$ such that $$L’\xrightarrow{\varphi} L\xrightarrow{\psi}M \rightarrow 0 $$
Is exact.

My question is: What exactly are the free-graded $A$-modules in this case? To say that a graded $A$-module is free is to say that it is of the form $A(n_1)\oplus A(n_2)\oplus \cdots A(n_l)$ for some $n_i \in \mathbb{Z}$ where $A(l)$ represents the twisting of the graded ring $A$ by $l$, i.e. $A(l)_m = A_{m+l}$ for all $m\in \mathbb{Z}$.

My attempt: We can construct $L$ easily: Since $M$ is finitely generated, choose homogeneous generators $x_1 \in M_{n_1} ,\cdots, x_r \in M_{n_r}$ and define the map $$\psi: A(-n_1) \oplus \cdots \oplus A(-n_r) \rightarrow M \text{ defined by} \psi(a_1,\cdots, a_r) = a_1 x_1 + \cdots + a_rx_r $$

Then clearly $\psi$ is a surjective, degree preserving map. So $L\xrightarrow{\psi} M \rightarrow 0$ is done.

It seems intuitive that $L’$ should be $\ker \psi$. It is evident that $L’$ is finitely generated but can we be sure that $L’$ is free as well?

答: Notice that $M'=\ker\psi $ is a finitely generated graded module. Hence you can repeat your argument replacing $M $ by $M'$. Then you get a surjective map from a free module $L'$ to $M'$. Composing this with the inclusion map $M'\subseteq L $ gives you the required exact sequence.


问:
Prove that the index of the set of left coset in a cyclic group is finite
If G is cyclic , show |G/H| < ∞ for any subgroup H except the identity.

I already know that any subgroup of a cyclic group is also cyclic but i have no idea how to prove a the quotient of G is finite especially when Lagrange theorem also only apply to finite set.

Thanks everyone for your helps, truly appreciate it!

答: HINT: Let $g$ be a generator of $G$, and pick $h\in H$ with $h\ne 1_G$. You know that $h=g^k$ for some $k\ne 0$. What’s the maximum possible number of left cosets that the subgroup $\langle h\rangle$ can have? Can $H$ have any more left cosets than $\langle h\rangle$?</p>


问:
Prove that for linear $T,S:U\mapsto V$, if rank($T$)=rank($S$) then there are bases $B,D$ of $U$ and $C,E$ of V for which: $[T]_C^B$=$[S]_E^D$
I'm out of ideas. I'd appreciate if someone could help with this proof.
I'm not sure if I'm on the right track, but I wanted to base my proof on the fact that since rank(T)=rank(S), this means that $rankC[T]_C^B$ = $rankC[S]_E^D$. Then there is the same number of columns that are linearly independent in both matrices, which means we can expand them to become bases... and then use a matrix to change bases. I feel like it doesn't make much sense, so I would like someone to write his proof.

答:
For $B$, start with a basis of the kernel of $T$, and complete it to a basis of $U$.

For $C$, start with the non-zero images of the elements of $B$, and complete to a basis of $V$.

Then
$$
[T]^{B}_{C}
=
\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0\\
0 & 1 & 0 & \dots & 0 & 0\\
0 & 0 & 1 & \dots  & 0 & 0\\
  &   &   & \ddots & \\
0 & 0 & 0 & \dots  & 0& 0\\
0 & 0 & 0 & \dots  & 0& 0\\
\end{bmatrix},
$$
where the number of $1$'s equals the rank of $T$.


问:
In Homotopy Type Theory, how do the continuous notions of spaces and paths, match the discrete notions of constructible terms and proofs?
Context: I'm starting with being interested in type theory as a framework within which to do mathematics (e.g. practically, using proof assistants based on type theory), and my understanding is that HoTT is motivated partially/primarily from that perspective. I and am not interested in homotopy theory per se, or  in using type theory to understand homotopy theory.

My understanding of homotopy type theory is limited, but I know that in HoTT we think of a type $T$ as a space, of instances $x,y:T$ of the type as points in that space, and of proofs $p:x=_Ty$ of equality as paths in the space.

How I can understand "proofs of equality as paths"

My interpretation of proofs of equality is that they are computational paths between terms. If we think of $x$ and $y$ as actual constructed terms in e.g. a proof assistant such as Coq, then the way to prove $x=_Ty$ is to show a sequence of computational steps (or at least, a program that would produce one, such as in an inductive proof) from $x$ to $y$. E.g. with the standard inductive definition of the natural numbers, it could be a transformation from $(1+1)+1$ (i.e. $x$) to $3$ (i.e. $y$), by repeatedly applying the addition rules to get $s (s (s (0)))$ and then applying the definition of $3$. i.e. proofs of $x=_Ty$ are computationally valid paths between terms, using the underlying computational system's (lambda calculus I'd presume) valid computation rules (e.g. beta-reduction, delta-reduction of lambda-terms).

So the way I think of it is: two terms may be different syntactically (e.g. $1+1$ and $2$), and are therefore different points in the "space belonging to their type" (e.g. $\mathbb N$). But they are semantically equal to each other, so that there is a computational path between them.

proofs of equality in homotopy type theory

In explanations of homotopy type theory that I've seen, the explanation usually asks you to think of types as spaces, where usually they are pictured as topological spaces (circles, donuts, etc). Then the point is made that two proofs $p,q:x=_Ty$ are fundamentally different, if there is no homotopy between them, and some image like this is shown to suggest that two proofs are not homotopic:

But afaik, a homotopy has to be continuous, meaning that the space has to have some kind of topological structure. However, the computational steps that we make in reasoning e.g. that $(1+1)+1 = 3$, are finite. My understanding is that it's fundamental to intuitionistic type theory that all the terms of a type are actually constructible terms, i.e. strings of symbols that we can write down, and hence that the "space" belonging to a type $T$ is discrete. But the only reasonable topology that I can think of on such a space would be the topology generated by the base set consisting of equivalence classes of terms based on $=_T$, which would I think imply that all paths $p,q:x=_Ty$ are homotopy equivalent (since any path consisting of points inside such a base set is continuous).

This makes me confused about homotopy type theory itself: It seems to me that in order to formalize the idea of proofs being fundamentally different, we'd have to think of a discrete object, probably a finite graph (where nodes are terms and links are basic computational steps), and concepts like homotopy and continuous functions and spaces are out of place.

Where is the mistake in my thinking?

答:
I would say there is no mistake in your thinking.  Rather, the mistake happened many decades ago when algebraic topologists gradually came to use the word "space" for a discrete object that should really be called an "$\infty$-groupoid".  The correct thing to say is then that HoTT interprets types as $\infty$-groupoids (though not, of course, necessarily finite ones).  This is even true semantically: all the formal mathematical interpretations of HoTT sending types to "spaces" actually use structures such as simplicial sets or cubical sets, which are discrete objects --- it's just that homotopy theorists have come to abuse the word "space" to refer to those things rather than to actual topological spaces.

The reason the word "space" took on this meaning is that topological spaces can be used as presentations of $\infty$-groupoids via a construction called the "fundamental $\infty$-groupoid of a space", and (at least, assuming classical logic) every $\infty$-groupoid is equivalent to one arising in this way.  Moreover, the fundamental $\infty$-groupoid of a space remembers (and is "determined by", in an appropriate sense) all of the algebraic invariants constructible from a topological space, and so algebraic topologists studying these invariants came gradually to identify a topological space with its fundamental $\infty$-groupoid --- and this was historically before anyone had given a precise definition of $\infty$-groupoid at all.

However, nowadays this conflation of spaces with $\infty$-groupoids has real dangers, one of which is the difficulty of intuition that you've run across.  Another is that we also sometimes want to study spatial $\infty$-groupoids, i.e. $\infty$-groupoids that also have an additional topology on their underlying sets that's unrelated to their $\infty$-groupoid structure, and if "space" means $\infty$-groupoid then that gets very confusing.

I have written some introductions to HoTT that emphasize this perspective, called A synthetic approach to higher equalities and The logic of space.


问:
Assertions about convergence/divergence of infinite series
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbot. I'd like someone to verify if my proofs/counterexamples about the below assertions are fine.

Consider each of the following propositions. Provide short proofs for those that are true and counterexamples for any that are not.

(a) If $\sum a_n$ converges absolutely, then $\sum a_n^2$ also converges absolutely.

(b) If $\sum a_n$ converges and $(b_n)$ converges, then $\sum a_n b_n$ converges.

(c) If $\sum a_n$ converges conditionally, then $\sum n^2 a_n$ diverges.

(d) If $\sum a_n$ with $a_n > 0$ is convergent, then is $\sum \sqrt{a_n}$ always convergent? Either prove it or give a counterexample.

(e) If $\sum a_n$ with $a_n > 0$ is convergent, then is $\sum \sqrt{a_n a_{n+1}}$ always convergent. Either prove it or give a counterexample.

Proof.

(a) We know that, for real numbers $a,b \in \mathbf{R}$
\begin{align*}
 \absval{a}^2 + \absval{b}^2 \le (\absval{a} + \absval{b})^2
\end{align*}

So, we can write,
\begin{align*}
 \absval{a_{m+1}}^2 + \ldots + \absval{a_{n}}^2 \le (\absval{a_{m+1}} + \ldots + \absval{a_{n}})^2
\end{align*}

As $\sum a_n$ is absolutely convergent, by the Cauchy criterion, given any $\epsilon > 0$, there exists $N \in \mathbf{N}$, such that
\begin{align*}
 \absval{\absval{a_{m+1}} + \ldots + \absval{a_{n}}} < \sqrt{\epsilon}
\end{align*}

for all $n > m \ge N$. Therefore,

\begin{align*}
 \absval{a_{m+1}}^2 + \ldots + \absval{a_{n}}^2 &\le  (\absval{a_{m+1}} + \ldots + \absval{a_{n}})^2\\
 &<(\sqrt{\epsilon})^2 = \epsilon
\end{align*}

Thus, $\sum a_n^2$ is absolutely convergent.

(b) This proposition is false. As a counterexample, consider $a_n := \frac{(-1)^{n+1}}{\sqrt{n}}$ and $b_n := \frac{(-1)^{n+1}}{\sqrt{n}}$. $\sum a_n$ and $(b_n)$ are convergent, but $\sum a_n b_n$ is the harmonic series, which is well-known to diverge.

(c) I was not able to find any counter-examples. For example, if we define $a_n := \frac{(-1)^n}{n}$ then the series $\sum n^2 a_n$ diverges.

(d) No. As a counterexample, consider $a_n = \frac{1}{n^2}$. $\sum a_n$ is convergent, but $\sum \sqrt{a_n}$ is divergent.

(e) The geometric mean of two numbers is always less than or equal to the arithmetic mean. \begin{align*}
 \sqrt{a_n a_{n+1}} \le \frac{a_n + a_{n+1}}{2}
\end{align*}

Both $(a_n)$ and $(a_{n+1})$ are convergent, and hence $\sum (a_n + a_{n+1})/2$ is also convergent by the algebraic limit theorem.

By the Comparison test, we must have $\sum \sqrt{a_n a_{n+1}}$ convergent.

答:
For a) I would have written :

Since $\sum |a_n|$ converges, there exists a rank $N$ such that $\forall n\geqslant N$, $|a_n|<1$. Then $\forall n\geqslant N, |a_n|^2 \leqslant |a_n|$. By theorem of comparison between positive terms series, we get that $\sum |a_n|^2$ converges.

For b), well done

For c), it's true

You can prove it by contraposition. If $\sum n^2 a_n$ converges then $n^2 a_n \rightarrow 0$ so there exists a rank $N$ such that $\forall n\geqslant N, n^2 |a_n| < 1$, so $\forall n\geqslant N, |a_n|<\frac{1}{n^2}$. By comparison, the serie converges absolutely.

For d), it's fine

For e), good too


答:
A counterexample:
$$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac12,\quad\enspace\text{not }0.$$
Indeed $\;1-\cos x=2\sin^2\tfrac x2$, so
$$\frac{1-\cos x} {x^2}= \frac{2\sin^2\frac x2}{4\bigl(\frac x2\bigr)^2}=\frac12\biggl(\underbrace{\frac{\sin\frac x2}{\frac x2}}_{\underset{\textstyle 1}{\downarrow}}\biggr)^2$$


问:
Multiplicity of Zero Eigenvalues of $A^*A$ and $AA^*$
We have a matrix $A_{m \times n}$. I know that the two matrices $A^*A$ and $AA^*$ share nonzero eigenvalues. This comes from multiplying $A^*Ax=\lambda x$ on both sides by $A$ ($\lambda$ is an eigenvalue of $A^*A$, and $x$ is an eigenvector of $A^*A$). Then, we have:

$$AA^*(Ax)=\lambda (Ax)$$

So now $\lambda$ is also an eigenvalue of $AA^*$, but now the eigenvector is $Ax$ (for $x \notin KerA$, so that $Ax\neq0$).

All other $\lambda$ not common to both matrices $AA^*$ and $A^*A$ is equal to zero.

My question is: when will the multiplicity of the zero eigenvalues be the same for both matrices $(AA^*)_{m\times m}$ and $(A^*A)_{n\times n}$?

Intuition tells me that this will only happen when both matrices $AA^*$ and $A^*A$ have the same dimensions, which implies $A_{m \times n}$ being square $(m=n)$. Otherwise, one matrix will always have more eigenvalues than the other (or more repetitions of eigenvalues, so higher multiplicity), the extra eigenvalues being equal to zero.

Is this correct? and how would you prove this more rigorously?

答:
Your intuition at the end of the post is correct.

One way to see this explicitly is to consider the SVD of $A$. Let $A=UDV^*$ where $U$ and $V$ are square orthogonal ($m\times m$ and $n\times n$ respectively), and $D$ is $m \times n$ rectangular diagonal with nonnegative diagonal entries in nondecreasing order. Noting that $A^*A=VD^* DV^*$ and $AA^* = UDD^*U^*$, you see that the eigenvalues of $A^*A$ and $AA^*$ are exactly the diagonal entries of $D^*D$ and $DD^*$ respectively. Walk through a few examples of rectangular diagonal $D$ to see how the diagonal entries of $D^*D$ and $DD^*$ differ. For example, if $D$ is a tall matrix ($m > n$), then $DD^*$ will have $m-n$ extra zeros on the diagonal compared to $D^*D$.


问:
Trying to figure out a fitting function for a curve where the fitting function always passes through the origin (0,0)
I have a curve that looks like this.

Acceleration v Velocity

I am trying to figure out a function I can use to fit to this curve. I have tried using a curve of the form $Ax^2 + Bx + C$ and while that does a good job of fitting the points it does not go through the origin. I am trying to construct a function of some sort that fits the points in the image and also goes through or is as close to the origin as possible.

I'd really appreciate some guidance!

Body:
Second degree best fit
$$f(x)=3.65823 x-0.235842 x^2$$
Third degree
$$g(x)=0.0197976 x^3-0.673543 x^2+5.83586 x$$


答:
I will outline a solution. I think you should be able to fill in the details.

Let $G$ be cyclic and generated by $g$. Then, as we know any subgroup is also cyclic, if $H$ is a subgroup, it is generated by $g^n$ for some $n\in \mathbb N$.

I claim that the index of $H$ in $G$ is $n$, with cosets $H, gH, \cdots, g^{n-1}H$. It suffices to show that any element lies in one of those cosets. Fix an element $g^a$ and find the remainder when $a$ is divided by $n$. Can you see how this tells you which coset it belongs to?

Title: Prove that the index of the set of left coset in a cyclic group is finite

Body:
If G is cyclic , show |G/H| < ∞ for any subgroup H except the identity.

I already know that any subgroup of a cyclic group is also cyclic but i have no idea how to prove a the quotient of G is finite especially when Lagrange theorem also only apply to finite set.

Thanks everyone for your helps, truly appreciate it!


问:
CDF - Cumulative distribution function
If the CDF is defined:
 $F_X\left(x\right)=\:1-e^{-λx}\::\:x\ge 0\:and\:0:\:x<0$

But when we substitute $x=0$ in the equation $1-e^{-λx}$ we got $0$.

So why we can't say that:
 $F_X\left(x\right)=\:1-e^{-λx}\::\:x>0\:and\:0:\:x\le 0$

答:
This is the CDF of an exponential distribution which is a continuous distribution. As such it’s CDF is expected to be continuous so the left- and righthand limits of $F_X$ at $0$ should coincide.

As a result both of your definitions are valid and equally correct.

Title: CDF - Cumulative distribution function


答:
Both statements are true. The reason we usually write the first is that although the CDF is $0$ at $x=0$, the density isn't. The PDF is $f(x)=\lambda e^{-\lambda x}$, which is $\lambda$ at $x=0$.

Writing the CDF in the first way makes clearer the distinction between $x\geq0$, where the denisty is positive, and $x<0$, where the density is $0$.

As Ian points out, the above is probably not technically rigorous (although it is a good intuitive reason as to why we should lump $x=0$ together with the points $x>0$, rather than with the points $x<0$).

A more rigorous answer, as Stefan hints at, is that the set $x\geq0$ is the support of the distribution, i.e. the smallest closed set $S\subset\mathbb R$ such that $\mathbb P(X\in S)=1$. So it makes sense to define the CDF separately for two types of points in $\mathbb R$: those in the support of $X$ (i.e. $x\geq0$), and those not in the support of $X$ (i.e. $x<0$).

Title: CDF - Cumulative distribution function

Body:
If the CDF is defined:
 $F_X\left(x\right)=\:1-e^{-λx}\::\:x\ge 0\:and\:0:\:x<0$

But when we substitute $x=0$ in the equation $1-e^{-λx}$ we got $0$.

So why we can't say that:
 $F_X\left(x\right)=\:1-e^{-λx}\::\:x>0\:and\:0:\:x\le 0$


问:
Showing that $B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$ is an open set

Show that $$B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$$ is an open set in the Euclidean space $(\Bbb{R^2},d_2),$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

Defining $A= \{(x,y) \in \Bbb{R^2} \mid x>0,y>0 \}$ and $E=\{  (x,y) \in \Bbb{R^2} \mid x<0,y<0 \}$ one sees that $B= A\cup E$ and thus $B$ will be open if $A$ and $E$ are open.

How can I show that $A$ and $E$ are both open? It seems that I would need to find open balls with arbitrary radius $B((x,y), r)$ for both cases that are contained in the set I'm working with?

答:
You can show $B^C$ is closed.

In fact $B^C = \{x,y\in \mathbb{R}^2,xy\leqslant 0\}$.

If you take $(x_n,y_n)$ a sequence of elements of $B^C$ that converges to $(x,y)$, then for all $n$, $x_ny_n \leqslant 0$.

When $n\rightarrow +\infty$, you get $xy \leqslant 0$.

So $(x,y)\in B^C$.

So $B^C$ is closed.

So $B$ is an open set.


答:
Note that $$B = (-\infty,0)\times(-\infty,0)\cup(0,\infty)\times(0,\infty)$$
Since each of these is an open set, so $B$ is an open set.

Title: Showing that $B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$ is an open set

Body:
Show that $$B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$$ is an open set in the Euclidean space $(\Bbb{R^2},d_2),$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

Defining $A= \{(x,y) \in \Bbb{R^2} \mid x>0,y>0 \}$ and $E=\{  (x,y) \in \Bbb{R^2} \mid x<0,y<0 \}$ one sees that $B= A\cup E$ and thus $B$ will be open if $A$ and $E$ are open.

How can I show that $A$ and $E$ are both open? It seems that I would need to find open balls with arbitrary radius $B((x,y), r)$ for both cases that are contained in the set I'm working with?


答:
Let $f(x) = \frac{1-\ln x}{e-x}$. We wish to find $\lim_{x\to e}f(x)$.

Using the proposed method would return the wrong answer.


答:
For $(x,y) \in A$ notice that
$$(x,y) \in B\left((x,y),\frac12\min\{|x|,|y|\}\right) \subseteq A$$
and the same thing for $E$.

For example, if $(x',y') \in B\left((x,y),\frac12\min\{|x|,|y|\}\right)$ we have
$$x-x' \le |x-x'| \le \sqrt{|x-x'|^2 + |y-y'|^2} \le \|(x,y)-(x',y')\| < \frac{x}2 $$
so $x'> \frac{x}2>0$. You can show the other inequalities similarly.


问:
Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Can I get a hint so that I can get started? I tried using the standard half angle formula, but they don’t work.

答: By the cosine rule,$$\cos^2\frac{R}{2}=\frac{1+\cos R}{2}=\frac{p^2+q^2-r^2+2pq}{4pq}$$so$$p\cos^2\frac{R}{2}=\frac{p^3r+pq^2r-pr^3+2p^2qr}{4pqr}.$$Add $p\leftrightarrow r$, then simplify.


问:
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$

答:
As $(a+ib)^{\frac13}$ is a multi-valued function, it is technically correct to state that one of the three values of $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}}$ is $7$

Method $1:$

Let  $\sqrt[3]{35 + 18i\sqrt{3}}=a+ib\iff \sqrt[3]{35 - 18i\sqrt{3}}=a-ib$

So,  $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 2a$

As this needs to be $7,$ let's try with $a=\frac72$

$$\text{So, }\left(\frac72+ib\right)^3=35+18i\sqrt3 $$

$$\implies (7+2i b)^3=8(35+18i\sqrt3)$$

$$\implies 7^3+3\cdot7^2\cdot2ib +3\cdot 7\cdot(2ib)^2+(2ib)^3=280+144i\sqrt3$$

Equating the real parts, $343-84b^2=280\implies 4b^2=3\implies b=\pm\frac{\sqrt3}2$

Equating the  imaginary parts, $294b-8b^3=144\sqrt3$

$\implies 2b(147-4b^2)=144\sqrt3\implies 2b=\sqrt3\implies b=\frac{\sqrt3}2$

As $\sqrt[3]{35 + 18i\sqrt{3}}=\frac{7+\sqrt3i}2,\sqrt[3]{35 - 18i\sqrt{3}}=\frac{7-\sqrt3i}2$

Again, $\sqrt[3]{35 + 18i\sqrt{3}}=\sqrt[3]{35 + 18i\sqrt{3}}\cdot \sqrt[3]1=\frac{7+\sqrt3i}2\cdot w$  where $w$ is a cube root of $1$

As the cube root of $1$ are $1,\frac{-1\pm\sqrt3i}2$

So, one of the other values of $\sqrt[3]{35 + 18i\sqrt{3}}$ is $\left(\frac{7+\sqrt3i}2\right)\cdot\left(\frac{-1+\sqrt3i}2\right)=\frac{-5+3\sqrt3i}2$

The corresponding value of $\sqrt[3]{35 - 18i\sqrt{3}}$ is $\frac{-5-3\sqrt3i}2$

Similarly, another value of  $\sqrt[3]{35 + 18i\sqrt{3}}$ is $\left(\frac{7+\sqrt3i}2\right)\cdot\left(\frac{-1-\sqrt3i}2\right)=-1+2\sqrt3i$

The corresponding value of $\sqrt[3]{35 - 18i\sqrt{3}}$ is $-1-2\sqrt3i$

Method $2:$

Let $\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = a$

$$\implies a^3=35 + 18i\sqrt{3}+35 - 18i\sqrt{3}-3\sqrt[3]{35 + 18i\sqrt{3}}\sqrt[3]{35 - 18i\sqrt{3}}a$$

$$\implies a^3-39a-70=0$$ as $(35-18i\sqrt3)(35+18i\sqrt3)=2197=13^3$

$$\implies a^3-7^3-39(a-7)=0$$
$$\implies (a-7)(a^2+7a+7^2-39)=0$$
$$\implies(a-7)(a+2)(a+5)=0 $$


问:
Linearize a second-order nonlinear ODE
How could the second-order equation $y′′=\sin(y)−y′y$ be linearized? I want to classify the equilibrium points, which I've been able to find, but I'm unsure how to linearize the equation. Would it look something like $y' = Ay + f(y)$?

答: No, if $y(x) = y_0 + h(x)$ (where $y_0$ is an equilibrium), then the linearization should be a second-order linear ODE for $h(x)$ (with constant coefficients that may depend on $y_0$). And to obtain it, just substitute $y(x)=y_0+h(x)$ into the equation and delete all terms which are nonlinear in $h$. (You can use Taylor expansion to handle the sine term.)

Title: Linearize a second-order nonlinear ODE


答:
Using the Half Angle Formula,

$$\cos^2 \frac{R}{2} = \dfrac{s(s -r)}{pq}$$

where $s = \dfrac{p+q+r}{2}$

$$\cos^2 \frac{P}{2} = \dfrac{s(s -p)}{rq}$$

$$p\cos^2 \frac{R}{2}+r\cos^2 \frac{P}{2} =\dfrac{s(s -r)}{q} +\dfrac{s(s -p)}{q}=\dfrac{s(2s-(r+p))}{q} = \dfrac{s(q)}{q}=s=\dfrac{p+q+r}2{}$$

Body:
Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Can I get a hint so that I can get started? I tried using the standard half angle formula, but they don’t work.


问:
Analize given data about a triangle PQR and solve

In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt 3, q=1$ and circumradius of $PQR$ is $1$.

I don’t want the complete answer for this. In the solution, there was point which mentioned area of $\Delta OQR =\frac 13 \Delta PQR$

Why is that the case?

答:
If you apply the sine rule on triangle $PQR$, which states that $$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$$ where $a,b,c$ are sides of the triangle opposite angles $A,B,C$ and $R$ is the circumradius, you get $$\dfrac{\sqrt{3}}{\sin P}=\dfrac{1}{\sin Q}=\dfrac{r}{\sin R}=2$$ so that $P=120^{\circ}$ and $Q=30^{\circ}$ which implies $R=30^{\circ}$ which implies $r=1$. Now, I hope you can continue from here.

Note that I have assumed that $p=QR$ and $q=PR$ which is not mentioned in the question.


问:
Given sum of infinite G.P, find the common ratio.
Sum of an infinite $G.P$ is $2020$. Each term of this $G.P$ is squared to make a new series whose sum is $20200$. If the common ratio of the original $G.P$ is ${a\over b}$ where $gcd(a,b)=1$, evaluate $b-a$.

My Work:-

Well sum of an infinte $GP$ = ${a\over 1-r}$ where $a$ is the first term and $r$ is the common ratio. Also we can simply observe that whenever we square the whole $G.P$ ; $a$ changes to $a^2$ and $r$ changes to $r^2$ as well. Thus basically we have $2$  equations :-

$a = 2020 + 2020r$

$a^2 = 20200 + 20200r^2$

Now I am stuck as to how to solve these equations.

$NOTE:-$ $G.P$ $STANDS$ $FOR$ $GEOMETRIC$ $PROGRESSION$

答:
Hint

$$2020=\dfrac a{1-r}\  \  \  \ (1)$$

$$20200=\dfrac{a^2}{1-r^2}\ \ \ \ (2)$$

Divide $(2)$ by $(1),$ to find $(3)$

Divide $(3)$ by $(1)$ and apply https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm to find $r$


问:
Computing the plane curves with given curvature
I want to compute the plane curves with curvature:

$\kappa (s) = 5/(s+1)$

I know I must apply the fundamental theorem of curves and solve some differential equation, but I am a bit confused about how to apply that theorem. Could you help with what equation must I solve and why?

答:
Consider the polar angle $\theta$ of a tangent vector. You can see that $\theta'(s) = \kappa(s)$. Thus you can solve ODE:

$$
\theta'(s) = \frac 5{s+1},\qquad\theta(0)=\theta_0,\\
\theta(s) = \theta_0 + 5\ln (s+1).
$$

Then you can use the fact that tangent vector $(dx,\ dy) = (ds\cos\theta,\ ds\sin\theta)$. In other words, you have a system of independent equations:

$$
x'(s) = \cos\Big(\theta_0+5\ln(s+1)\Big),\qquad x(0)=x_0,\\
y'(s) = \sin\Big(\theta_0+5\ln(s+1)\Big),\qquad y(0)=y_0,
$$

For brevity, I will use $\theta_0=0$, but it doesn't make the integral any harder:

$$
x(s) = \frac{s+1}{26}\Big(\cos \theta(s) + 5 \sin \theta(s)  \Big)-\frac{1}{26} + x_0,\\
y(s) = \frac{s+1}{26}\Big(\sin \theta(s) - 5 \cos \theta(s) \Big)+\frac{5}{26} + y_0
$$

You have 3 degrees of freedom here: 2 for the location of a starting point $x_0,y_0$ and 1 for the arbitrary rotation of the curve $\theta_0$.


问:
Can we deduce that set of $a,b,c$ is the whole set of integers?
We say that three positive integers $x,y,z$ are coprime if there exist three integers $a,b,c$ such that $$ax+by+cz=1$$

if we run $x,y,z$ over all possible coprime positive integers. Can we deduce that set of $a,b,c$ is the whole set of integers?

答:
Per OP's request, I am amplifying my comment into an answer.  However, I am not saying that the (constraint-1) shown below is the only constraint that $|a|, |b|, |c|$ may have to satisfy.  I am merely saying that (constraint-1) has to be satisfied.

We say that three positive integers $x,y,z$ are coprime if there exist three integers $a,b,c$ such that $$ax+by+cz=1$$

It is easy to establish that under the above definition, if for any given $a,b,c \in \mathbb{Z},$ there happens to exist positive integers $x,y,z$ with

$$ax+by+cz=1$$

then there must exist integers $x_1, y_1, z_1$ such that

$$(x_1 \times |a|) + (y_1 \times |b|) + (z_1 \times |c|) = 1. \tag{constraint-1}$$

Simply set

$x_1 = [x : ~\text{if}~ a > 0, ~\text{else}~ (-x)]$

$y_1 = [y : ~\text{if}~ b > 0, ~\text{else}~ (-y)]$

$z_1 = [z : ~\text{if}~ c > 0, ~\text{else}~ (-z)]$.

The only flaw in concluding that $|a|, |b|, |c|$ must therefore be coprime, is that one or two of $a,b,c$ could be equal to zero.  However, if you are given that each of $a,b,c$ is non-zero, then $|a|, |b|, |c|$ must be coprime, using the OP's definition.


问:
Definition of linear function
Suppose that each group of points below can be represented by a function, appropriately. Which of the functions below represents linear growth?

I believe there are 3, red orange and green. Are vertical lines linear functions?

答:
The orange dots and light red (pink?) dots represent linear growth as at any point $x$, $f(x)$ grows by the same amount per unit of time.

Using what I described above, it should be clear that the blue and maroon dots do not represent linear growth.

There are some ambiguities with respect to the green dots. I would not define those as representing a function since $f(1)$ does not map to a unique value.

Title: Definition of linear function


问:
Show that the ring of polynomials $k[x_{1},...,x_{n}]$ is an integral extension of the ring of symmetric polynomials

I want to prove that $A= k[x_{1},...,x_{n}]$, where $k$ is a field, is an integral extension of the subring $B$ consisting of the symmetric polynomials in $n$ variables.

It is sufficient by finding an expression with coefficients in B that nulls when evaluated the variables $x_{i}$ for every $i=1,...,n$. I tried it but I can’t find a suitable polynomial the satisfies this. Some help?

答: Consider the polynomial $p_n=(t-x_1)\dots(t-x_n)\in A[t]$. Clearly any permutation of the $x_i$ will leave $p_n$ unchanged, so every coefficient of $p_n$ must lie in $B$, and hence $p_n\in B[t]$. On the other hand, $p_n(x_i)=0$ for any $i\in\{1,\dots,n\}$, so $p_n$ is the desired monic polynomial satisfied by each $x_i$.


答:
\begin{align}
\dfrac{p+q+r}{2} &= p\cos^2\frac R2+r\cos^2\frac P2\\
&= \dfrac{p+p\cos R+r\cos P+r}{2}
\end{align}
which happens iff $$q = p\cos R+r\cos P$$
This is evident from the geometry in the following image:

Title: Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$


问:
How is 0 a limit point of $\{1/n\}_{n=1}^{\infty}$?
I'm working through Principles of Topology and can't wrap my head around how $0$ is a limit point of $\{1/n\}_{n=1}^{\infty}$. For reference, this is on page 44 of the 2016 edition.

The book provides the following definition/theorem & proof:

Definition: A point x in $\mathbb{R}$ is a limit point or accumulation point of a subset A of $\mathbb{R}$ provided that every open set containing x contains a point of A distinct from x.

Theorem 2.9: A real number x is a limit point of a subset A of $\mathbb{R}$ if and only if $d(x, A\setminus\{x\}) = 0$.

Proof: Suppose first that x is a limit point of A and let $\epsilon$ be a positive number. Then the open set $(x-\epsilon, x+\epsilon)$ contains a point y of A distinct from x. Since $y\in(x-\epsilon,x+\epsilon)$, then $d(x,y) < \epsilon$ so $d(x, A\setminus\{x\}) < \epsilon$. Since the latter inequality holds for all $\epsilon>0$, then $d(x, A\setminus\{x\}) = 0$.

Suppose now that $d(x, A\setminus\{x\}) = 0$ and consider an open set O containing x. Then O contains an open interval $(x-\delta, x+\delta)$ for some positive number $\delta$. Since $d(x, A\setminus\{x\}) < \delta$ and the interval $(x-\delta, x+\delta)$ consists precisely of all points at a distance less than $\delta$ from x, then $(x-\delta, x+\delta)$ must contain a point z in $A\setminus\{x\}$. Thus

$$z \in (x-\delta, x+\delta) \subset O$$

and $z \neq x$ since $z\in$ $A\setminus\{x\}$. Hence every open set containing x contains a point of A distinct from x, and x is a limit point of A.
$$\tag*{$\Box$}$$

Additionally, the book defines distance between a real number and a non-empty subset of $\mathbb{R}$ as:
$$d(a, B) = glb\{|a-b|:b \in B\}$$

and provides a few examples, e.g.:

$d(0, [1,2]) = d(0, (1,2)) = 1$

$d(1, [1,2]) = d(1, (1,2)) = 0$

So far, so good. The book then notes:

$0$ is the only limit point of $\{1/n\}_{n=1}^{\infty}$. Note in this case that the limit point is outside the set.

My question is... how? Say $x = 0$ and you choose $\epsilon = 3$, and you follow along the proof:
$$(x-\epsilon, x+\epsilon) = \{x \in \mathbb{R}: (x-\epsilon) < x < (x+\epsilon)\}$$

which certainly does contain a point $y = 1$ where $y \in \{1/n\}_{n=1}^{\infty}$, with $y$ being distinct from $x$.

What I don't understand: if $A = \{1/n\}_{n=1}^{\infty}$, how does $d(x, A\setminus\{x\}) = 0$?

As far as I can tell, since $0 \notin A$, then $A\setminus\{x\}$ is equivalent to $\{1/n\}_{n=1}^{\infty}$, and the distance between $0$ and $A$ would be $1$:

$$d(0, [1, \frac 12, \frac 13, \frac 14 ...)) = 1$$

What might I be missing here?

答:
You calculate this wrongly:

$$d(0, {1, \frac 12, \frac 13, \frac 14 ...}) = 1$$

It should be $0$. And as it has been stated in Theorem 2.9, $0$ is the limit point.


答:
Here's one way for square roots:

$$(\sqrt{x+y}+\sqrt{x-y})^2 = 2x + 2\sqrt{x^2-y^2}.$$

Perhaps this is easier to evaluate. For cube roots:

$$
(\sqrt[3]{x+y}+\sqrt[3]{x-y})^3 = 2x + 3\sqrt[3]{(x+y)(x^2-y^2)}+3\sqrt[3]{(x-y)(x^2-y^2)}.
$$

Of course, to get the coefficients in all of these I am using the binomial theorem. There really is no guarantee that these are easier to deal with in general.

Title: How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$?

Body:
The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate  the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$


问:
Intuition on ds v. dA in Stokes type problems
I'm slowly making my way through better understanding of Stokes Theorem-type problems. I struggle with two issues, the intuition about which I'm lacking.

First, what does it mean to integrate the curl over the surface? Is this the same thing as the flux, which I usually think of as the amount of "flow" over the surface? That's confusing to me.

Second, it is usually stated that Stokes relates the surface integral of curl dot dS (over S) to the double integral of the force dot dr (over C, the contour) as in the following:

When you see a model solution, as in the following, you will see that, to calculate the right side of this equation, dS is transformed into the cross product of the partial derivatives times dA.

Why is that? dS represents the infinitesmally small surface area. Why does that become the cross of the partials times dA? Doesn't make sense to me.

Here's an example, where you can see this in action.

答:
When you parametrize a surface by $\mathbf r(u,v)$, the parallelogram spanned by $\dfrac{\partial\mathbf r}{\partial u}$ and $\dfrac{\partial\mathbf r}{\partial v}$ has area $$\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|.$$
The cross product itself is a (non-normalized) normal vector and this area becomes the "fudge factor" (like the $r$ in $r\,dr\,d\theta$) that tells you how to relate area in the $uv$-plane to area on the surface. After that, it's just
\begin{align*}
\mathbf F\cdot\mathbf n\,dS &= \left(\mathbf F\cdot\frac{\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}}{\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|}\right)\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|du\,dv \\
&=\mathbf F\cdot \left(\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right) du\,dv.
\end{align*}


问:
How to solve matrix differential equation?
I have two coupled differential equations:
$$\tag{1}
\begin{array}{l}
\dot{z}_{1}=-H_{11} z_{1}-H_{12} z_{2} \\
\dot{z}_{2}=-H_{21} z_{1}
\end{array}$$
And I want to find the correlation functions:
$$\tag{2}
\left\langle z_{1}(t) z_{1}(0)\right\rangle, \quad\left\langle z_{2}(t) z_{2}(0)\right\rangle, \quad\left\langle z_{1}(t) z_{2}(0)\right\rangle$$

I rewrote the equations as a matrix equation to get:
$$\begin{array}{l}
z(t)=\left[\begin{array}{l}
z_{1}(t) \\
z_{2}(t)
\end{array}\right], \quad H=\left[\begin{array}{cc}
H_{11} & H_{12} \\
H_{21} & 0
\end{array}\right] \\
\frac{d}{d t}\left\langle z_{i}(t) z_{j}(0)\right\rangle=-H_{i k}\left\langle z_{k}(t) z_{j}(0)\right\rangle
\end{array}$$
Where $\left\langle z_{i}(t) z_{j}(0)\right\rangle$ is the correlation between $z_i(t)$ and $z_j(0)$.
I am unfamiliar with matrix differential equations, and I am not sure how to solve for $\left\langle z_{i}(t) z_{j}(0)\right\rangle$?

答:
The general solution to $\dot{X} = - H X$ is $X(t) = \exp(-tH) X(0)$, where $\exp(-tH)$ is a matrix exponential.  If $H_{11}^2 + 4 H_{21} H_{12} \ne 0$, your $H$ has distinct eigenvalues
$$ \lambda_{\pm} = \frac{H_{11}}{2} \pm \frac{\sqrt{H_{11}^2 + 4 H_{21} H_{12}}}{2}$$
and $$\exp(-tH) = e^{-t \lambda_+} \frac{H - \lambda_- I}{\lambda_+-\lambda_-} + e^{-t \lambda_-} \frac{H - \lambda_+ I}{\lambda_- - \lambda_+}$$


问:
Is it okay to raise to the power of $n$ (where $n\in\mathbb{R}$) on both sides in Euler's Identity?
So I was playing around with the identity and did this:
$$e^{i\pi}=-1$$
$$e^{(2\pi)i}=(-1)^{2}=1$$
$$e^{(2n\pi)i}=1^{n}=1\ \ \ \ \ (n\in \mathbb{R})$$
$$\Rightarrow\ e^{(2n\pi)i}=e^{(2\pi)i}\ \ (\forall n\in\mathbb{R})$$

Then, $$2n=2$$ So,$$n=1$$

Which is quite contradicting as any value of n should satisfy the condition. Which means I have made a mistake somewhere. Can anyone tell me where am I going wrong? Help would be appreciated! Thanks in advance.

答: No.  But the ever useful De Moivre's formula falls out by doing $e^{ni\theta}=\cos(n\theta)+i\sin(n\theta)$.


答:
The multiplication $f\colon
\begin{cases}\mathbb R^2&\to\mathbb R;\\
(x,y)&\mapsto x\cdot y
\end{cases}$ is known to be continuous.  Observe that $B$ is the preimage of the open set $(0,\infty)$ under $f$.


问:
Proof of this formula (multivariable function)
How to prove this:
$$ x^2 f_{xx} + 2xyf_{xy} + y^2 f_{yy} = n(n-1) f(x,y) $$

when f is a homogeneous function of the root n.

答: Since both sides are linear, you need only evaluate them with $f=x^ky^{n-k}$ for $0\le k\le n$. In other words, check $k(k-1)+2k(n-k)+(n-k)(n-k-1)=n(n-1)$.


问:
Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions

Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.

My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms  of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$

Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?

New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$

how can I solve this?

答:
Edit: @mathlove noticed problems with my initial solution. My effort to resolve those problems ended in a major revision of the solution. Please comment if you find mistakes in this new version.

Let's take off from where you calculated that $t^2-2t=a-\sqrt{a+t}$ . Our aim is to find a value (or values) for $a$ such that the above equation has exactly 2 real answers for $t$. Let's also keep in mind the implicit constraint that
$$a+t \ge 0 \Rightarrow t \ge -a \qquad(1)$$
This is because the given problem includes $\sqrt{a+t}$ . An important implication of (1) is that if $a$ is negative then $t$ should be positive, and if $t$ is negative then $a$ should be positive.

Another constraint that is derived from our starting equation is that
$$t^2-2t \le a \qquad (2)$$
where the equality holds when $a+t=0$. Besides, (2) means that if $a < 0$ then $t^2-2t=t(t-2) < 0$ , which implies that $0 < t < 2$.

And two other constraints, which we can derive from (2) are
$$t^2-2t+1 = (t-1)^2 \le a+1 \qquad (3)$$
$$0 \le (t-1)^2 \le a+1 \Longrightarrow a \ge -1 \qquad (4)$$

Alright, here we go:
$$t^2-2t=a-\sqrt{a+t}$$
$$t^2-t=a+t-\sqrt{a+t}$$
$$t^2-t+\frac{1}{4}=a+t-\sqrt{a+t}+\frac{1}{4}$$
$$(t-\frac{1}{2})^2 = (\sqrt{a+t}-\frac{1}{2})^2$$
$$t-\frac{1}{2} = \pm(\sqrt{a+t}-\frac{1}{2})$$
So we have two (quadratic) equations
$$t-\frac{1}{2} = +(\sqrt{a+t}-\frac{1}{2}) \Rightarrow t^2 = a+t \qquad (5)$$
and
$$t-\frac{1}{2} = -(\sqrt{a+t}-\frac{1}{2}) \Rightarrow (t-1)^2 = a+t \qquad (6)$$
Let's look more carefully at (5). The solutions to this quadratic equation are
$$t_1 = \frac{1 + \sqrt{1+4a}}{2} , t_2 = \frac{1 - \sqrt{1+4a}}{2}$$
For these solutions to exist, we should have $a \ge -\frac{1}{4}$.

Consider $t_1$. There seems to be no problem with $t_1$ for $-\frac{1}{4} \le a$ . In this range of $a$ , $t_1$ seems to satisfy constraints (1),(2),(3) and (4).

Now consider $t_2$. If $a > 0$ then by constraint (2) we should have:
$$t_2^2 - 2t_2 \le a$$
$$(\frac{1 - \sqrt{1+4a}}{2})^2 - 2(\frac{1 - \sqrt{1+4a}}{2})= \frac{-1+2a+\sqrt{1+4a}}{2} \le a$$
$$-1+2a+\sqrt{1+4a} \le 2a$$
$$\sqrt{1+4a} \le 1$$
$$1+4a \le 1$$
$$a \le 0$$
which is a contradiction with our assumption of $a > 0$. So, $t_2$ does not exist for $a > 0$. There seems to be no problem with $t_2$ for $-\frac{1}{4} \le a \le 0$ . In this range of $a$ , $t_2$ seems to satisfy constraints (1),(2),(3) and (4).

Note that if $a = -\frac{1}{4}$ then $t_1=t_2=\frac{1}{2}$.

Let's now look at (6). The solutions to this quadratic equation are
$$t_3 = \frac{3 + \sqrt{5+4a}}{2} , t_4 = \frac{3 - \sqrt{5+4a}}{2}$$
For these solutions to exist, we should have $a \ge -\frac{5}{4}$ , but we also have the stricter constraint (4), which requires that $a \ge -1$.

Consider $t_3$. If $a > 0$ then $t_3$ violates constraint (3). So, $t_3$ does not exist for $a > 0$. On the other hand, if $-1 < a \le 0$ then by constraint (1) t should be non-negative and by the implication of constraint (2) we should have $0 \le t \le 2$. But applying this range of $a$ in the calculation of $t_3$ results in $2 < t_3$, which is in contradiction with constraint (2). Finally, note that if $a=-1$ then $t_3=2$ , but this result violates constraint (3). Therefore, it seems that $t_3$ is not a solution to the equation for any value of $a$.

Now consider $t_4$. It seems that for $a \ge -1$ , $t_4$ satisfies constraints (1),(2),(3) and (4).

Note that although in the formula we get $t_3=t_4=\frac{3}{2}$ for $a=-\frac{5}{4}$ , these are not solutions of the main equation, because the value of $a=-\frac{5}{4}$ violates constraint (4).

Now let us see in what conditions $t_1$ or $t_2$ might be equal to $t_4$. From $t_1 = t_4$ it follows that $a=-\frac{1}{4}$. Also, from $t_2 = t_4$ it follows that $a=-\frac{1}{4}$. So at $a=-\frac{1}{4}$ , $t_1=t_2=t_4=\frac{1}{2}$.

Summing up our above analyses, we can see that

$t_1$ exists for  $-\frac{1}{4} \le a$

$t_2$ exists for  $-\frac{1}{4} \le a \le 0$

$t_3$ does not exist

$t_4$ exists for $-1 \le a$

Therefore, for $a > 0$ , the equation has exactly two distinct solutions ($t_1 , t_4$).


问:
Are $[0, 1)$ and $[0, 1]$ homeomorphic subspaces of the Sorgenfrey line?
My argument is that $\{1\}$ is a connected clopen subspace of $[0, 1]$ while the only connected subspaces of $[0, 1)$ are singular sets, which are not open in $[0, 1)$, so the spaces must not be homeomorphic. Is this reasoning correct?

Is there another argument that does not need require to test connectedness?

Thank you!

答: Your argument is correct, but you can just say that $[0,1]$ has a clopen singleton (which is $\{1\}$), whereas $[0,1)$ has none.


问:
Complex chain rule
Suppose $V_1$ and $V_2$ are open sets in $\mathbb{C}^m$ and $\mathbb{C}^n$, repsectively, $f:V_1\mapsto V_2$ and $g:V_2\mapsto \mathbb{C}^k$ are $\mathbb{R}-$ differentiable. We know that the real chain rule for the differential of $g\circ f$ at a point $z$ gives
$$D_z(g\circ f)=D_{f(z)}g\circ D_zf.$$ Can someone explain how can we prove that
$$\partial_z(g\circ f)=\partial_{f(z)}g\circ \partial _z f+\overline{\partial}_{f(z)}g\circ \overline{\partial}_zf,  $$and
$$\overline{\partial}_z(g\circ f)=\overline{\partial}_{f(z)}g\circ \partial _z f+\partial_{f(z)}g\circ \overline{\partial}_zf?  $$
We use the following notation:
$$\partial _z\varphi=\sum_p\frac{\partial \varphi}{\partial z_p}dz_p\quad \& \quad \overline{\partial}_z\varphi=\sum_p\frac{\partial \varphi}{\partial \overline{z}_p}d\overline{z}_p .  $$ Also that would be great if someone knows a reference for that kind of stuff. Thanks.

答:
I am going to avoid your notation and use more standard notation, but you can make the translation. I'm also going to do the case $\Bbb C\to\Bbb C$ and you can easily generalize it. So we start with
$$df = \partial f + \bar\partial f = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial\bar z}d\bar z.$$
Now, if $z=g(w)$, then
$$dz = \frac{\partial g}{\partial w}dw + \frac{\partial g}{\partial\bar w}d\bar w,$$
and we get $d\bar z$ by conjugating, so
$$d\bar z = \overline{\frac{\partial g}{\partial w}}d\bar w + \overline{\frac{\partial g}{\partial\bar w}}dw.$$
The chain rule tells us that to compute $d(f\circ g)$ we just substitute these latter expressions. We obtain
\begin{align*}
d(f\circ g) &= \frac{\partial f}{\partial z}\left(\frac{\partial g}{\partial w}dw + \frac{\partial g}{\partial\bar w}d\bar w\right) + \frac{\partial f}{\partial\bar z}\left(\overline{\frac{\partial g}{\partial\bar w}}dw+ \overline{\frac{\partial g}{\partial w}}d\bar w\right) \\
&= \left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial\bar w}}\right)dw + \left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial w}}\right)d\bar w\,.
\end{align*}
From this you read off that
\begin{align*}
\frac{\partial (f\circ g)}{\partial w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial\bar w}}\,, \quad\text{and} \\
\frac{\partial (f\circ g)}{\partial\bar w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial w}}\,.
\end{align*}

You will notice this is a lot like the Einstein summation convention in tensor analysis: We sum a product of a $w$ derivative and a $dw$ and a product of a $\bar w$ derivative and a $d\bar w$ in every case. It may be useful to keep in mind that $\overline{\dfrac{\partial g}{\partial w}} = \dfrac{\partial\bar g}{\partial\bar w}$ and $\overline{\dfrac{\partial g}{\partial \bar w}} = \dfrac{\partial\bar g}{\partial w}$.

Thus, we can finally rewrite this as:
\begin{align*}
\frac{\partial (f\circ g)}{\partial w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\frac{\partial\bar g}{\partial w}\,, \quad\text{and} \\
\frac{\partial (f\circ g)}{\partial\bar w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\frac{\partial\bar g}{\partial\bar w}\,.
\end{align*}

Title: Complex chain rule


答:
Consider $a^3 =35+ 18i\sqrt{3}$ and $b^3=35- 18 i\sqrt{3}$

You are supposed to find $a+b$
 $(a+b)^3=a^3+b^3+3ab(a+b) \implies(a+b)^3-3ab(a+b)=a^3+b^3$

Let $a+b=x \implies x^3-3ab(x)=70$ and $ab= [(35-18i\sqrt3)(35+18i\sqrt3)]^{1/3}=13$

You get $x^3-39x-70 =0$, product of the roots is $70$ and sum of the roots is $0$. You can do clever factorization or guess the roots(It has only 8 divisors). SO, number of possible should be just $8 \choose 3$. I kinda flicked your trick. ;)


问:
How many integers between $1$ and $1000$ are divisible by $2, 4,$ or $7?$
How many integers between 1 and 1000 are divisible by $2, 4,$ or $7?$

$A \cup B = A+B- A \cap B=500+142-71=571$

答: Hint: Find out how many numbers are divisible by $2$ and how many numbers are divisible by $7$. Note that $2 | 4$ and numbers are divisible by both $2$ and $7$ iff they're divisible by $14$ (why?). Find out which numbers are counted in both f the lists of multiples.


答:
Let $a=1$ be the circumradius and, per the sine rule
$$ \sin Q=\frac q{2a}=\frac12,\>\>\>\>\> \sin P=\frac p{2a}=\frac{\sqrt3}2$$
Given the non-right triangle, $Q=30^\circ$ and $P=120^\circ$. Thus, the triangle $PRQ$ is isosceles with $PE$ as a median as well. So,  $O$ is the centroid point, hence the area ratio $\frac13$.

Title: Analize given data about a triangle PQR and solve

Body:
In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt 3, q=1$ and circumradius of $PQR$ is $1$.

I don’t want the complete answer for this. In the solution, there was point which mentioned area of $\Delta OQR =\frac 13 \Delta PQR$

Why is that the case?


问:
Why does $\frac1\pi\int_0^\pi|\cos t|dt$ estimate the average size of $|u\cdot U_j$|, where $u$ and thirty $U_j$s are random unit vectors?
Why is the highlighted part true? and what is the hidden meaning behind average size of $\rvert u.U_j \rvert$?

Here is what I did to solve:

The solution that I got:

答:
Due to symmetry of the system, you can fix the direction of $\mathbf u$ e.g. to go along $x$. Then the average should be:

$$
\frac1{|\Omega|}\int_\Omega|U_x|d\Omega
$$

If we are talking about 2 dimensional vectors, then $\Omega$ is the unit circle, and with polar parameterization ($U_x=\cos\theta$):
$$
\frac{1}{2\pi}\int_0^{2\pi}|\cos\theta|d\theta=\frac{1}{\pi}\int_0^{\pi}|\cos\theta|d\theta = \frac{2}{\pi}=0.63662.
$$

However, we work in 3D, thus $\Omega$ is a unit sphere and with spherical parameterization ($U_x=\cos\theta$, $d\Omega=\sin\theta\, d\theta\, d\varphi$):
$$
\frac{1}{4\pi}\int_0^{2\pi}d\varphi\int_0^{\pi}|\cos\theta|\sin\theta\, d\theta=\frac{1}{2}\int_{-1}^1|z|dz = \frac12.
$$

So you should expect the answer to be closer to $0.5$. If you up the number of vectors from $30$ to $10^6$, you will see that.

In other words, the book should either ask you to have randn(2, 1)/randn(2, 30), or should give you an integral for 3D case. You might check that with 2D case the average is indeed $2/\pi$.


问:
Find: The expected number of urns that are empty
A total of $n$ balls, numbered $1$ through $n$, are put into $n$ urns, also numbered $1$ through $n$ in such a way that ball $i$ is equally likely to go into any of the
urns $1, 2, . . . , i$. Find the expected number of urns that are empty.

Can somebody help me? I don´t understand.

Thanks very much.

答: Let $X$ be the number of empty urns. Then $X=X_1+X_2+\dots+X_n$ where $X_i$ is the indicator variable whose value is $1$ if the $i$th urn is empty, $0$ otherwise. Then $E(X_i)$ is the probability that the $i$th urn is empty, i.e., $E(X_i)=\frac{i-1}{i}\frac{i}{i+1}\frac{i+1}{i+2}\dots\frac{n-1}{n}=\frac{i-1}{n}$ (telescoping product). By linearity of expectation, $$E(X)=E(X_1)+\dots+E(X_n)=\frac{0}{n}+\frac{1}{n}+\frac{2}{n}+\dots+\frac{n-1}{n}=\frac{{n\choose 2}}{n}=\frac{n-1}{2}.$$</p>


问:
Bounded non constant multivariable function
Show that there does not exist a non-constant continuous function $f:R^{2}  \to R$ such that $f(x,y)=5$ for all $(x,y)$ with $x^{2}+y^{2}<1$

My approach: I was able to reason out and prove this intuitively but not mathematically. I first drew a circle on the $XY$ axis of radius 1 centered at origin.  This bounded all the possible values of $x$ and $Y$ inside the circle. Now imagining that as a gradient plane where $z=5$ I took values from inside the circle. If the function is to give an output of 5 for all the values of $X\ and \ Y$ then somehow the X and Y need to be eliminated. I cannot use higher powers of X and Y as then the function would converge and never be equal to 5. If I was to take a function of the type $aX + bY = 5$ then it wouldn't be possible to select such real values of a and b to account for all the possible X and Y values. Therefore the function can only be a constant but since that isn't allowed, I can safely say that a non constant continuous function in the given domain $X,\ Y \ in\ [0,1) $ does not exist.

My question: How do I approach this mathematically so as to prove this ? I tried putting what I wrote above more "mathematically" but it didn't seem like a substantial proof or even resemble one.

答:
Of course there are lots of such continuous functions. You can even make them as smooth as you want. Here's one example. Define $f\colon\Bbb R^2\to\Bbb R$ by
$$f(x,y) = \begin{cases} 5, & x^2+y^2\le 1 \\ 6-(x^2+y^2), & x^2+y^2>1\end{cases}.$$
You could make something that looks conical by taking
$$f(x,y) = \begin{cases} 5, & x^2+y^2\le 1 \\ 6-\sqrt{x^2+y^2}, & x^2+y^2>1\end{cases}.$$
It might help to think about functions $f\colon\Bbb R\to\Bbb R$ and draw some pictures, and then move on to two variables.


问:
spectral norm alternate definition
The spectral norm of $A ∈ \Bbb R^{n\times n}$ is defined as $\|A\|_{*} = \max \{\|Ax\|_2 : \|x\|_2 = 1 \}$.

To show that $$\|A\|_{\ast} =\max \{y^{\top} Ax : \|x\|_2 = 1, \|y\|_2 = 1 \}$$

I have already proved that

$$\|A\|_{\ast} = \sigma_{1}$$

where $\sigma_{1}$ is the largest singular value of $A$. But I'm having difficulty proving above statement, can anyone give a hint to solve this?

Title: spectral norm alternate definition

Body:
For $x,y \in \Bbb{R}^n$ with $\|x\|_2=\|y\|_2 = 1$ by Cauchy-Schwarz we have
$$y^T Ax \le \|Ax\|_2\|y\|_2 \le \|A\|_*$$
so $\max\{y^T Ax : \|x\|_2=\|y\|_2 = 1\} \le \|A\|_*$. Conversely, for $x \in \Bbb{R}^n$ with $\|x\|_2=1$ we get
$$\|Ax\|_2 = \frac{(Ax)^TAx}{\|Ax\|_2} = y^TAx$$
where $y = \frac{Ax}{\|Ax\|_2}$ so $\|A\|_* \le \max\{y^T Ax : \|x\|_2=\|y\|_2 = 1\}$.


问:
mathematical operations with infinity
I suppose these are the equations with infinity that are universally considered correct:

∞ = ∞

∞ + n = ∞

∞ * n = ∞

n/∞ = 0

Where n can be any possible value.

These equations can be rearranged to give the following results:

∞ - ∞ = 0

∞ - ∞ = n

∞ / ∞ = n

∞ * 0 = n

Where n can be any possible value.

But can n also be infinite?

If so the following final derivations can be made(in no particular order):

∞ + ∞ = ∞

∞ - ∞ = n        (where -∞ <= n <= ∞)

∞ * ∞ = ∞

∞ / ∞ = n        (where -∞ <= n <= ∞)

Are these statements valid? Also the interesting thing here is that in the final equations, 1. and 3. both have a single value of infinity whereas 2. and 4. Can have any possible value, including -∞, 0 and ∞. Does that mean 1. and 3. are not undefined?

答:
Let us follow the convention that an expression with $\infty$ is "defined" (in the extended reals) if: when you replace each $\infty$ with any function/sequence whose limit is $\infty$, and each real number with any function/sequence with that limit, the limit of the entire expression is always the same real number or divergence to $\infty$ or $-\infty$.

Then you are right that $\infty+\infty$ and $\infty*\infty$ are defined and equal $\infty$. And you are right that $\infty-\infty$ and $\infty/\infty$ (and $0*\infty$) are undefined. (If you need proofs for these and can't find them elsewhere, let me know.)

Because of this, it is wrong to write something like "$\infty-\infty=n$ where $n$ is any number", because $\infty-\infty$ is undefined. It's not equal to anything. And we don't usually use the equals sign in a way that would allow multiple values at once.

But can n also be infinite?

I will interpret this as "are there functions where $\infty-\infty$, $\infty/\infty$ and $\infty*0$ correspond to limits that diverge to infinity?"
Yes. For examples: ${\displaystyle \lim_{n\to\infty}}(2n)-(n)=\infty$, ${\displaystyle \lim_{n\to\infty}}(n^2)/(n)=\infty$, and ${\displaystyle \lim_{n\to\infty}}(n^2)*(1/n)=\infty$.

Title: mathematical operations with infinity


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Combinatorics you have for example Szemerédi's regularity lemma and all sorts of "related" results, such as the Green-Tao theorem.</p>

Title: Landmarks of subjects of mathematics


问:
Find the volume of the region bounded by $z = x^2 + y^2$ and $z = 10 - x^2 - 2y^2$
So these are two paraboloids. My guess is I would want to find the intersection of these two which would be $2x^2 + 3y^2 = 10$ and construct a triple integral based on its projection. No idea how to do this but the answer comes out to be $\dfrac{50\pi}{\sqrt{6}}$.</p>

答:
$f(x, y) = x^2 + y^2$ and $g(x, y) = 10- x^2 -2y^2$. Now as you said you need to finds out where they intersect. That means we want $x^2 + y^2 = 10- x^2 -2y^2$ which means $2x^2+3y^2 = 10$. If you graph this in 2D you will see that it is an ellipse. Following the analogy with the 2D case, we have to compute

$\int \int g(x, y) − f(x, y)dxdy = \int \int_D (10 - 2x^2 -3 y^2) dxdy$, where $D$ is the interior  of the ellipse. Now all that you need to find out is the limits of integration. Can you do it?

Here is the intersection of $f$ and $g$


问:
The possible remainders when a multiple of 4 is divided by 6, and when a multiple of 2 is divided by 3
I'm currently studying for the GRE and a specific type of question has me stumped. I have a two part question, but first here is one problem and my work:

"If x is the remainder when a multiple of 4 is divided by 6, and y is the remainder when a multiple of 2 is divided by 3, what is the greatest possible value of x+y"

Since it says greatest I thought GCM and wrote(with arbitrary variables)

$$N = 24a + x$$

$$M = 6b + y$$
I thought that since 2 & 3 are factors of 4 & 6, respectively, maybe I could equate them:
$$24a+x=6b+y$$
Or maybe N is twice a large as M:
$$24a+x=2(6b+y)$$
1) How do I solve this? Am I even on the right track?

This kind of question has had me stumped for a while but the books explanation has me even more lost.

"The greatest possible remainder for a multiple of 4 being divided by 6, happens when 4 is divided by 6. When 4 is divided by 6 the result is 0 with a remainder of 6."

.....Don't remainders by nature have to be smaller than the divisor......

2)How is that possible? Shouldn't the result be 0 with a remainder of 4?

How can there be a remainder of 6 when the divisor is 6? I found a similar problem and solution here: Finding a number given its remainder when divided by other numbers and here: How to find the greatest remainder of a number that is a multiple of another number But the more I read the more lost I was.

At this point I feel like a child throwing spaghetti about trying to make art. I have no idea how to even start on this problem any help is appreciated.

答:
Multiples of 4 are: 4,8,12,16,20,24 etc.. and multiples of 2: 2,4,6,8,10,12,14 etc..
Divide the listed multiples of 4 by 6 to check the remainder by finding the remainder in each multiple. It is obvious that 6 goes into 16 with a remainder of 4 giving the value for X. Remember the question said the greatest possible value and in this case is remainder 4. Apply the same method to multiples of 2 divided by 3. 3 goes into 8, 2 times with a remainder of 2 giving the value for Y. Therefore X+Y=6</p>


问:
How many final placements are possible if the leader changed 0...5 times?
Six competitors A-F take part in cycling time trial. The race is about covering several dozen kilometers as quickly as possible. Following the rules of this sporting event, competitors start one by one, at regular intervals. Let's call the cyclist, who has ridden the entire route and at the moment he is in the lead (so he had the best time so far), the leader. The final placement is the list obtained by ranking the cyclists according to the final places they took in the race. How many final placements are there in which:

A) the leader didn't change from the start;

B) the leader changed 2
times;

C) the leader changed 3 time;s

D) the leader changed 4
times;

E)the leader changed 5 times;

I believe the key is understanding the mechanic of leadership. Also they don't wait for the previous one to finish so for example B can finish before A. Then A can no longer take the leadership. Subproblems A and E are trivial, A is just 5! and E is 1, if I am not mistaken.

EDIt: I got more info, answers for 3 changes and 4 changes are 85 and 15 respectively. Looking for a way that isn't bruteforcing it.

答:
Label the riders $1,2,\ldots,n$ according to their start order. Their final placings determine a permutation $\pi=r_1r_2\ldots r_n$ of $[n]$, where $r_1$ is the winner, $r_2$ has the second-best time, and so on. For $k\in[n]$ say that $r_k$ is a left-to-right minimum if $r_i>r_k$ for $i<k$. If $r_1r_2\ldots r_n$ has $m$ left-to-right minima, the lead changed hands $m-1$ times.

For example, the left-to-right minima of the permutation $463512$ of $[6]$ are $4$, $3$, and $1$. The lead changed hands first from $1$ to $3$ and then from $3$ to $4$.

If the left-to-right minima are $r_{k_1}=r_1,r_{k_2},\ldots,r_{k_m}$, insert $m$ pairs of parentheses to get

$$(r_{k_1}\ldots r_{k_2-1})(r_{k_2}\ldots r_{{k_3}-1})\ldots(r_{k_m}\ldots r_n)\tag{1}$$

and interpret $(1)$ as a permutation $\varphi(\pi)$ in cycle notation. It’s well-known and easy to see that the map $\varphi:S_n\to S_n$ is a bijection: given a permutation $\pi\in S_n$ in cycle notation, rotate each cycle to bring the smallest element to the front, order the cycles in descending order of smallest elements, and remove the parentheses to get $\varphi^{-1}(\pi)$. Thus, the number of outcomes in which the lead changed hands $k$ times is the number of permutations of $[n]$ with $k+1$ cycles.

This is given by $n\brack{k+1}$, a Stirling number of the first kind. These do not have a nice closed form, but they do satisfy a nice recurrence:

$${{n+1}\brack k}=n{n\brack k}+{n\brack{k-1}}$$

for $k>0$, with initial values ${0\brack 0}=1$ and ${n\brack 0}={0\brack n}=0$ for $n>0$.

For the present case with $n=6$ we have

$${6\brack 1}=120,{6\brack 2}=274,{6\brack 3}=225,{6\brack 4}=85,{6\brack 5}=15,{6\brack 6}=1\,.$$

To sum up, $n\brack k$ is the number of possible outcomes when there are $n$ riders, $k$ of whom are leaders at some point during the time trial; this is the number of possible outcomes for $n$ riders when the lead changes hands $k-1$ times.


问:
Uniform continuity regarding union of open sets
I am stuck at the following Real Analysis problem from my book:

Problem: Give an example of two open sets $A$ and $B$ and a continuous function $f: A \cup B \rightarrow \mathbb{R}$ such that $f\mid A$ and $f\mid B$ are uniformly continuous, but f is not uniformly continuous.

What have i tried?

I know that $f(x) = x^{2}$ from $\mathbb{R}$ to $\mathbb{R}$ is not uniformly continuous. With that in mind, i tried to solve the problem.

First, observe that $f\mid (0,1)$ is uniformly continuous (since f is Lipschitz in the bounded open interval $(0,1)$.

Now i am stuck. I am trying to choose the set $B$ as to obtain what was required in the problem, but i just cannot do it. Every time i pick an open interval $B$ such that $f\mid B$ is uniformly continuous, $f: A \cup B \rightarrow \mathbb{R}$ also turns out to be uniformly continuous, thus not passing the test.

Can someone help?

Thanks in advance, Lucas

答:
Consider $A = (0, 1),$ $B = (-1, 0),$ and $f:A \cup B \to \Bbb R$ given by

$$f(x) = \begin{cases} x + 1 & x > 0, \\ x - 1 & x < 0. \end{cases}$$

It is clear that $f \mid A$ is uniformly continuous. ($\delta = \epsilon$ works.) The same is true for $f \mid B$.

However, for $f$ on $A \cup B$, note that if we choose $\epsilon = 1$, then no matter which $\delta > 0$ is given, we can find $\eta > 0$ small enough so that $2\eta < \delta$ and $f(\eta) - f(-\eta) > 1$. Conclude from this.


答: Let$$A=\bigcup_{n\in\Bbb N,\ n\text{ odd}}\left(\frac1{n+1},\frac1n\right)\quad\text{and}\quad B=\bigcup_{n\in\Bbb N,\ n\text{ even}}\left(\frac1{n+1},\frac1n\right).$$Define $f\colon A\cup B\longrightarrow\Bbb R$ with$$f(x)=\begin{cases}1&\text{ if }x\in A\\0&\text{ if }x\in B.\end{cases}$$Then $f|_A$ and $f|_B$ are constant and therefore continuous. But $f$ is not uniformly continuous.

Title: Uniform continuity regarding union of open sets

Body:
I am stuck at the following Real Analysis problem from my book:

Problem: Give an example of two open sets $A$ and $B$ and a continuous function $f: A \cup B \rightarrow \mathbb{R}$ such that $f\mid A$ and $f\mid B$ are uniformly continuous, but f is not uniformly continuous.

What have i tried?

I know that $f(x) = x^{2}$ from $\mathbb{R}$ to $\mathbb{R}$ is not uniformly continuous. With that in mind, i tried to solve the problem.

First, observe that $f\mid (0,1)$ is uniformly continuous (since f is Lipschitz in the bounded open interval $(0,1)$.

Now i am stuck. I am trying to choose the set $B$ as to obtain what was required in the problem, but i just cannot do it. Every time i pick an open interval $B$ such that $f\mid B$ is uniformly continuous, $f: A \cup B \rightarrow \mathbb{R}$ also turns out to be uniformly continuous, thus not passing the test.

Can someone help?

Thanks in advance, Lucas


问:
Biparted graph uniqueness
There is a bipartite graph with parts $X$ and $Y$ such that:

$Y$ has 20 nodes

Each node in $X$ is connected to 8 nodes in $Y$ uniquely (i.e. the set of adjacents for each node in $X$ is unique).

I want to prove that we can connect each node in $X$ to another node in $Y$ such that the resulting graph will still satisfy (2).

答: Hint : Connect each node $x \in X$ to one of the nodes in $Y$ farthest from $x$. Consider distance between $x$ and $y$ to be infinity if there's no path connecting them.

Title: Biparted graph uniqueness


答:
What about $A=]0,1[$, $B=]1,2[$ and $f$ such that $f(x) =  x$ if $x\in ]0,1[$ and $f(x) = x+1$ if $x\in ]1,2[$?

$f$ is continuous on $A\cup B$, but clearly not uniformly since for $x_n = 1-1/n$ and $y_n = 1+1/n$, $x_n-y_n\rightarrow 0$, but $f(y_n)-f(x_n) \geqslant 1$ for all $n$.


答:
Alternative solution:

Denote $A_n$ as the set of integers between $1$ and $1000$ divisible by $n$. Then you can blindly follow the inclusion-exclusion principle

$$|A_2 \cup A_4 \cup A_7| \\= |A_2|+|A_4| + |A_7| - |A_2 \cap A_4| - |A_4 \cap A_7| - |A_7 \cap A_2| + |A_2 \cap A_4 \cap A_7| \tag 1$$

Note that $A_2 \cap A_4 = A_4, A_4 \cap A_7 = A_{28}, A_7 \cap A_2 = A_{14}, A_2 \cap A_4 \cap A_7 = A_4 \cap A_7 = A_{28}$ so some of the terms in $(1)$ will cancel out and you don't need to calculate all of them.

Can you take it from here?

Title: How many integers between $1$ and $1000$ are divisible by $2, 4,$ or $7?$

Body:
How many integers between 1 and 1000 are divisible by $2, 4,$ or $7?$

$A \cup B = A+B- A \cap B=500+142-71=571$


问:
Does improper integral exists or not?
The following question is from my real analysis quiz. I was unable to solve it in exam and unfortunately now failed to solve it again at my home.

$$\int_{0}^{\infty} e^{-x} \log(\cos^2x) dx $$

I am not able to find out how I would be able to do it. In my opinion, problem is both at $\infty$ and at 0. Finding the limit of integral at 0 and at $\infty$ doesn't seem to be an options as I think it would be really hard to compute the integral. Also, If I break the limit and apply the practical comparison test then in both at 0 and at $\infty$ I am getting the limit as 0.

So, please help me.

答:
From the convexity of $x\mapsto \lvert\cos x\rvert$ on $[0,\frac{\pi}{2}]$ and on $[\frac{\pi}{2}, \pi]$ respectively, we easily find that

$$ \left| \frac{2x}{\pi} - 1 \right| \leq \left| \cos x \right| \leq 1 \quad \text{for } x \in [0, \pi]. $$

The plot below demonstrates this comparison:

So by taking logarithm to both sides and multiplying by $-2$, we get

$$ 0 \leq -\log(\cos^2 x) \leq -2\log \left| \frac{2x}{\pi} - 1 \right|. $$

In particular, we know that $\log(\cos^2 x)$ is always non-positive. Then by substituting $u = \frac{2x}{\pi} - 1$,

\begin{align*}
\int_{0}^{\pi} \left| \log(\cos^2 x) \right| \, \mathrm{d}x
&= \int_{0}^{\pi} \left( - \log (\cos^2 x) \right) \, \mathrm{d}x \\
&\leq \int_{0}^{\pi} (-2)\log \left| \frac{2x}{\pi} - 1 \right|  \, \mathrm{d}x \\
&= -\pi \int_{-1}^{1} \log \left|u\right|  \, \mathrm{d}u \\
&= 2\pi.
\end{align*}

This shows that $\log(\cos^2 x)$ is integrable on $[0, \pi]$ by the comparison test. Finally, using the fact that $e^{-x} \leq e^{-n\pi}$ for each $n \geq 0$ and $x \in [n\pi, (n+1)\pi]$,

\begin{align*}
\int_{0}^{\infty} \left| e^{-x} \log(\cos^2 x) \right| \, \mathrm{d}x
&= \sum_{n=0}^{\infty} \int_{n\pi}^{(n+1)\pi} e^{-x} \left| \log(\cos^2 x) \right| \, \mathrm{d}x \\
&\leq \sum_{n=0}^{\infty} \int_{n\pi}^{(n+1)\pi} e^{-n\pi} \left| \log(\cos^2 x) \right| \, \mathrm{d}x \\
&= \frac{1}{1 - e^{-\pi}} \int_{0}^{\pi} \left| \log(\cos^2 x) \right| \, \mathrm{d}x \\
&< \infty,
\end{align*}

and therefore the improper integral converges.

In general logarithmic singularity does not pose any issue for local integrability. In OP's case, the singularities of the integrand $e^{-x}\log(\cos^2 x)$ at $x = n\pi + \frac{\pi}{2}$ for $ n \in \mathbb{Z}$ are "benign", and so, only the singularity at $x=\infty$ matters.

Title: Does improper integral exists or not?


问:
$\sum\limits_{\text{prime }p} 2^{-p}$ is an irrational number
I need help to prove the following result.

$\displaystyle\sum_{\text{prime }p} 2^{-p}$ is an irrational number.

答: Hint: Consider this number in base 2.  If it is rational, it must have a period.  What sort of consequence does this mean, that would lead to a contradiction?</p>


问:
Lines tangent to a parabola
Consider two points $A$ and $B$ on a parabola $p$. Call $I$ to the intersection point of the tangent lines to $p$ at $A$ and $B$ and let $P$ be the point on $p$ such that the line defined by $I$ and $P$ is parallel to the symmetry axis of $p$.

How can we prove geometrically that the tangent line to $p$ passing through $P$ is parallel to the line segment joining $A$ and $B$?

I was able to prove it using analysis (assuming that $p$ is the graph of a quadratic function and proving that the lines that I want to prove that are parallel have the same slopes), but I would like to have a more geometrical proof (or a reference to a textbook with such a proof).

答:
I just wanted to add to the collection a nice projective proof.

Let $\infty_{IP}$ be the point at infinity that lies on $IP$. Then the parabola $p$ is tangent to the line at infinity $\ell_{\infty}$ at $\infty_{IP}$.

As $I$, $P$, $\infty_{IP}$ are collinear, taking the polar dual implies that $AB$, the tangent at $P$, and $\ell_{\infty}$ are concurrent. So we're done.

Title: Lines tangent to a parabola


问:
Positive Semidefiniteness of a Hermitian Matrix
I came across this problem and I'm not sure if the proof I have found is OK. I may be overcomplicating things. Any suggestions or confirmation of correctness would be appreciated.

The Problem:

Let $A$ be an $n \times n$ Hermitian matrix such that $\det A_k > 0 $ for all $k = 1, 2, . . . , n − 1$ and $\det A \geq 0$. Prove that $A$ is positive semidefinite.

My shot:

Consider the matrix $A_{n-1}$. By Silvester's Criterion of Positivity, we know $A_{n-1}$ is positive definite $(A_{n-1}>0)$ and therefore it can be diagonalized by congruence using simple row operations (making sure to conserve the determinant) into a diagonal matrix $A_{n-1}'$ with positive entries.

We now consider the matrix $A'$, defined as the matrix $A$ but with the submatrix $A_{n-1}$ diagonalized as explained above:

$$A'=\begin{pmatrix} A_{n-1}' & *\\* & a_{n,\,n}\end{pmatrix}$$

$A'$ has the same determinant as $A$, so $\det A'\geq0$. We can now further diagonalize the matrix $A'$, keeping the determinant, to obtain:

$$D=\begin{pmatrix} A_{n-1}' & 0\\0 & a_{n,\,n}'\end{pmatrix},\:\det D\geq0$$

We compute the determinant of $D$:

$$\det D=\begin{vmatrix} A_{n-1}' & 0\\0 & a_{n,\,n}'\end{vmatrix}= a_{n,\,n}'\cdot\det A_{n-1}' \geq 0$$

As we have $\det A_{n-1}'>0$, we know that $a_{n,\,n}' \geq0$. Therefore, we have diagonalized the matrix A into a diagonal matrix with all its entries satisfying $a_{i,\,i}' \geq 0$, $\forall i=1,\;2,\;...,\;n$.

As stated by Sylvester's Law of Inertia, the number of positive, zero and negative entries of the diagonal matrix depends only on $A$ and not on the specific diagonalization that we use. As the matrix $A$ is Hermitian, one of its possible diagonal forms will have the eigenvalues of $A$ in its diagonal entries. Therefore, by Sylvester's Law of Inertia, the eigenvalues $\lambda_i$ of $A$ will also satisfy  $\lambda_i\geq0$, $\forall i=1,\;2,\;...,\;n$. This means that the matrix $A$ is positive semidefinite ($A\geq0$).

答: Just so this question does not go unanswered, I have confirmed that the above approach is correct. Any other suggestions are welcome.</p>


问:
Convex optimization books, except with a focus on problem solving and formalism later?
Let me start off by saying that I'm well aware of these books:

Convex Optimization, Boyd and Vandenberghe

Lectures on Convex Optimization, Nesterov

The books by Bertsekas

Convex Analysis by Rockafellar

Convex Analysis and Linear Optimization by Borwein and Lewis

I'd also like to mention that my math background is by no means deficient: I have a math bachelor's degree, a master's in statistics, and am doing a two-semester sequence in measure theory. The advice seems to consistently be that if you know linear algebra and multivariable calculus, you should be able to handle this stuff.

I personally can't stand these books. Here's why:

There is too much emphasis on formalism. Boyd and Vandenberghe, especially, seem insistent on defining term after term after term at you from the very beginning. Unless people who read this book are walking dictionaries, I don't see how one can reasonably learn from this book.

There is too little emphasis on problem solving. Why do I want to learn optimization? So that I understand how to minimize/maximize an objective function with constraints. I understand the need for formalism, but the fact that this is front and center in every one of these books makes them extremely difficult for self study. Furthermore, I do not want to just learn how to formulate a problem as a convex optimization problem; I would like to at least know how to solve such problems by hand (or enough theory so that I can program such solutions).

Now that all of that is out of the way, to my main question:

Does there exist a convex optimization book which

has a problems-first emphasis, and if it provides formal results, does provide them but primarily in the context of actually solving a convex optimization problem?

is suitable for self-study, without supplementary material necessary? (I am not going to watch videos on top of a textbook; the book should be able to stand on its own.)

is not just a definition-theorem-proof-type of book?

Please note that I am not looking for a linear optimization book, but a convex optimization one.

答: I would recommend to give Boyd and Vandenberghe another try, but this time start directly from the second part of the book (Applications, Chapter 6-), leaving the first part as reference. The covered applications are from machine learning and statistics, which is quite generic.</p>


问:
Evaluate:$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$
Evaluate:$$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$$

where $$\theta_{n}=\frac{\theta}{3^n}$$ and $0<\theta<\pi$

I did try to find relation between $\tan 3x$ and $\tan x$

$$\tan 3x-3\tan x=\frac{8\tan^3x}{1-3\tan^2x}$$

Also $$3\tan 3x-\tan x=\frac{8\tan x}{1-3\tan^2x}$$

so as to create some kind of telescopic action but couldn't split the expression given in the question

答:
Note that

$$ \frac{\tan x}{3-\tan^2 x} = \frac{\cot x - 3\cot(3x)}{8}. $$

So we have

$$ \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)}
= \frac{1}{8}\left( \frac{\cot \theta_n}{3^n} - \frac{\cot \theta_{n-1}}{3^{n-1}} \right). $$

Summing this from $n = 2$ to $\infty$ yields a telescoping series with the value

$$ \sum_{n=2}^{\infty} \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)}
= \frac{1}{8}\left( \lim_{N\to\infty} \frac{\cot \theta_N}{3^N} - \frac{\cot \theta_1}{3} \right)
= \frac{1}{8}\left( \frac{1}{\theta} - \frac{\cot (\theta/3)}{3} \right). $$


答:
If the $2$'s were replaced by $10$'s, the problem is clear. The decimal expansion of any rational number either terminates or repeats. The number in question is one that has $1$s at all prime places after the decimal point and zeros elsewhere. This does not terminate or repeat, so it must be irrational.

The problem you have is similar, except we consider the binary expansion. But the exact same argument works.

Title: $\sum\limits_{\text{prime }p} 2^{-p}$ is an irrational number

Body:
I need help to prove the following result.

$\displaystyle\sum_{\text{prime }p} 2^{-p}$ is an irrational number.


问:
Algebraic structure- associative property proof
We will define an algebraic structure in the following way:

$(\mathbb {N} ,*)$

$a*b=a+b-1$

And i have to prove associative property,i thought of doing it the following way:

$(a+b)-1=a+(b-1)$

$(a+b)=x,(b-1)=y$

$x-1=a+y$

$x-y=a+1$

$a+1=a+1$

But I'm not sure that's enough

Body:
Let $ (a,b,c)\in \Bbb N$.

$$(a*b)\color{red}{*}c=(a*b)+c-1$$
$$=(a+b-1)+c-1=a+b+c-2$$

on the other hand

$$a\color{red}{*}(b*c)=a+(b*c)-1$$
$$=a+(b+c-1)-1=a+b+c-2$$

Can you conclude ?


问:
Equivalence between two definition of bounded set in a metric and normed space.
If $X$ s a vector space over the field $\Bbb F$ then  the following notation well be used
$$
x+A=\{x+a:a\in A\}\\
x-A:=\{x-a:a\in A\}\\
A+B:=\{a+b:a\in A\wedge b\in B\}\\
\lambda A:=\{\lambda a:a\in A\}
$$
for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.

Definition

If $\tau$ is a topology on a vector space $X$ such that

every point of $X$ is a closed set

the vector space operation are continuous with respect to $\tau$

we say that $X$ is a topological vector space equipped with the vector topology $\tau$.

So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.

Definition

A subset $E$ of a topological vector space is said to be bounded if to every neighborhood $V$ of $0$ in $X$ corresponds a number $s>0$ such that $E\subseteq tV$ for every $t>s$.

Now a subset $S$ of a metric space $Y$ is bounded if there exist an open ball that contains it and this is equivalent to suppose that the supremum of the distances of the above set is finite. Now if $X$ is a normed space it is a vector topological space and a metric space so I ask to me if the definition two definition of bounded set agree in this particular case. So I observe that if $E$ is a bounded subset of normed space then for what above defined for any open ball $B(0,r)$ centered at $0$ there exist a $\sigma>0$ such that
$$
E\subseteq\rho B(0,r)
$$
for any $\rho\ge\sigma$ but $\rho B(0,r)$ is a bounded subset with the metric induced by the norm and so
$$
E\subseteq\rho B(0,r)\subseteq B(0,r')
$$
for some $r'>0$ so that $E$ is a bounded set with respect the metric induced by the norm. Now unfortunately I do not be able to prove the contrary, that is if $E$ is a bounded set with respect the metric induced by the norm  then it is a bounded set in the normed space too so that I ask to prove this. So could someone help me, please?

答:
If $E$ is a bounded subset (in the metric sense) then by the triangle inequality we may assume that $E$ is contained in some $B(0,r)$ for $r$ large enough. Now if $V$ is any neighborhood of $0$ it will contain some open ball $B(0,r')$ with $r'>0$. Now for any $\rho>\frac{r}{r'}$ we will have $E\subseteq B(0,r)\subseteq \rho B(0,r')$. So $E$ is bounded in the sense of topological vector spaces.

Edit:
If $\rho>\frac{r}{r'}$ and $x\in B(0,r)$, then $x=\rho x'$ with $x'=\rho^{-1}x$ and $\left\lVert x'\right\rVert=\rho^{-1}\left\lVert x\right\rVert<\rho^{-1}r<\frac{r'}{r}r=r'$ so $x'\in B(0,r')$ and therefore $x\in \rho B(0,r')$.


答:
In order to prove associative property, you need to show $$ (a*b)*c = a*(b*c)$$

Note that $$ (a*b)*c = (a*b)+c-1= (a+b-1)+c-1 = a+b+c-2$$

Also $$ a*(b*c)= a+(b*c)-1 = a+(b+c-1)-1 = a+b+c-2$$

Thus the operation is associative.

Title: Algebraic structure- associative property proof

Body:
We will define an algebraic structure in the following way:

$(\mathbb {N} ,*)$

$a*b=a+b-1$

And i have to prove associative property,i thought of doing it the following way:

$(a+b)-1=a+(b-1)$

$(a+b)=x,(b-1)=y$

$x-1=a+y$

$x-y=a+1$

$a+1=a+1$

But I'm not sure that's enough


答:
Let $\mathcal{F} $ be a familly on $9$ element subsets in $Y$ and let $x_1,x_2,...,x_n$ be a vertices in $X$. Then make new bipartite graph between  $\mathcal{F} $ and $X$ so that vertex $x_i$ is connected to $F\in  \mathcal{F} $ iff $N(x_i) \subseteq F$. Now try to prove there is a perfect matching.

Every node in $X$ has degree $12$ and every set $F\in\mathcal{F} $ has degree at most $9$.

Now this does work, use the result I most recently proved here: Hall's marriage theorem in bipartite graph

Title: Biparted graph uniqueness

Body:
There is a bipartite graph with parts $X$ and $Y$ such that:

$Y$ has 20 nodes

Each node in $X$ is connected to 8 nodes in $Y$ uniquely (i.e. the set of adjacents for each node in $X$ is unique).

I want to prove that we can connect each node in $X$ to another node in $Y$ such that the resulting graph will still satisfy (2).


问:
Curve through four points -- simple algebra??
The motivation for this is Bezier curves. But, if you don't know what these are, you can skip down to the last paragraph, where the problem is described in purely algebraic terms.

Suppose I want to construct a quadratic Bezier curve that passes through some given points.

First, the standard approach: I would use three points, say $\mathbf A$, $\mathbf B$, $\mathbf C$,  choose a "middle" parameter value, typically $h = \tfrac12$, and then construct a quadratic curve $\mathbf P(t)$,  such that $\mathbf P(0) = \mathbf A$, $\mathbf P(h) = \mathbf B$, and $\mathbf P(1) = \mathbf C$. Easy.

But, it's fairly well known that given four coplanar points, you can (usually) construct a parabola that passes through them. This problem goes back to Newton, even. But, every parabola is a quadratic Bezier curve. So, given the four coplanar points $\mathbf A$, $\mathbf B$, $\mathbf C$, $\mathbf D$, I'm hoping that there is a fairly simple calculation that gives us a quadratic Bezier curve passing through them.

More specifically, let's call the quadratic Bezier curve $\mathbf P(t)$, again, and let $\mathbf P_0$, $\mathbf P_a$, $\mathbf P_1$ be its control points, so that
$$\mathbf P(t) = (1-t)^2 \mathbf P_0 + 2t(1-t) \mathbf P_a + t^2 \mathbf P_1$$
We want to find control points $\mathbf P_0$, $\mathbf P_a$, $\mathbf P_1$ and parameter values $t=u$ and $t=v$ such that
$$\mathbf P(0) = \mathbf A \quad ; \quad
  \mathbf P(u) = \mathbf B \quad ; \quad
  \mathbf P(v) = \mathbf C \quad ; \quad
  \mathbf P(1) = \mathbf D \quad$$
It's obvious that $\mathbf P_0 = \mathbf P(0) = \mathbf A$, and $\mathbf P_1 = \mathbf P(1) = \mathbf D$, so the remaining problem is to find the middle control point $\mathbf P_a$ and the two parameter values $u$ and $v$. I'd like to know how to calculate these remaining unknowns. Ideally a nice simple way.

There is a paper on this subject, but the authors' approach seems oddly complex and esoteric, to me. I'm hoping for a simple series of elementary calculations that a smart high-school student could understand.

As far as I can tell, you don't need to know anything about Bezier curves to solve this problem. It can be considered purely algebraically: we are given $\mathbf A$, $\mathbf B$, $\mathbf C$, $\mathbf D$ in $\mathbb R^2$, and we want to find $\mathbf P \in \mathbb R^2$ and $u,v \in \mathbb R$ such that
$$(1-u)^2 \mathbf A + 2u(1-u) \mathbf P + u^2 \mathbf D = \mathbf B$$
$$(1-v)^2 \mathbf A + 2v(1-v) \mathbf P + v^2 \mathbf D = \mathbf C$$
So, it's really just an equation solving problem.

答:
Solved it myself, as below. I don't feel that this solution is "right" somehow, because it treats the interior points $\mathbf B$ and $\mathbf C$ very asymmetrically, so improvements are invited. But, anyway ...

Let's use the notation from the last few paragraphs of the question.

It's clear that $\mathbf P(0) = \mathbf A$, and $\mathbf P(1) = \mathbf D$, so these two points are interpolated, already, and we only have to worry about the other two points, $\mathbf C$ and $\mathbf D$.

First, we find numbers $h$ and $k$ such that
$$\mathbf C = \mathbf A + h(\mathbf B - \mathbf A) + k(\mathbf D - \mathbf A)
= (1-h-k)\mathbf A + h\mathbf B + k\mathbf D$$
This is possible provided that $\mathbf A$, $\mathbf B$, $\mathbf D$ are not collinear. Then, since $\mathbf P(v) = \mathbf C$, we have
$$(1-h-k)\mathbf A + h\mathbf B + k\mathbf D =
(1-v)^2 \mathbf A + 2v(1-v) \mathbf P + v^2 \mathbf D $$
But, since $\mathbf P(u) = \mathbf B$, we know that
$$ \mathbf B = (1-u)^2 \mathbf A + 2u(1-u) \mathbf P + u^2 \mathbf D$$
Substituting for $\mathbf B$ on the left-hand side, and equating coefficients of $\mathbf A$, $\mathbf P$, $\mathbf D$ gives
$$(1-v)^2 = 1 - h - k +h(1-u)^2 $$
$$2v(1-v) = 2hu(1-u)$$
$$v^2 = hu^2 + k $$
We can easily eliminate $v$ from these last three equations using the fact that $[2v(1-v)]^2 = 4[v^2][(1-v)^2]$. We get:
$$[2hu(1-u)]^2 = 4[hu^2 + k][1 - h - k +h(1-u)^2]$$
After a little algebra, this reduces to:
$$h(1-h)u^2 - 2hku + k(1-k) = 0$$

So, we solve this quadratic for $u$, and then get the unknown interior control point $\mathbf P$ from

$$ \mathbf P = \frac{\mathbf B - (1-u)^2 \mathbf A -u^2 \mathbf D}{2u(1-u)}$$

The number of solutions depends on the number of real solutions of the quadratic. Quite often, you can draw two different quadratic Bezier curves through the four given points, as explained in the paper cited in the question.


问:
Analize given data about a triangle PQR and solve

In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt 3, q=1$ and circumradius of $PQR$ is $1$.

I don’t want the complete answer for this. In the solution, there was point which mentioned area of $\Delta OQR =\frac 13 \Delta PQR$

Why is that the case?

答:
Please note that $PE$ is a median and hence $PE = 3OE$. So $\triangle PQR = 3 \times \triangle OQR$.

The question states that $\triangle PQR$ is not a right angled triangle otherwise $\angle R = 90^0$ is another obvious configuration for the given sides and circumradius.

As $p = \sqrt3 \gt q = 1, \angle Q \, $ is acute. That leads to only other possible configuration of $\angle P \,$ being obtuse.

If $C$ is the circumcenter, you can see why $\angle P = 120^0 \, PQ = PR$ and $E$ is the midpoint of $QR$.


问:
Why does $\frac1\pi\int_0^\pi|\cos t|dt$ estimate the average size of $|u\cdot U_j$|, where $u$ and thirty $U_j$s are random unit vectors?
Why is the highlighted part true? and what is the hidden meaning behind average size of $\rvert u.U_j \rvert$?

Here is what I did to solve:

The solution that I got:

答:
Let $U \sim U(\Bbb{S}^1)$ be a random vector with uniform distribution on the unit sphere. For a fixed unit vector $\mathbf{u} = (\cos\phi, \sin\phi)$ we have
\begin{align}
\Bbb{E}|\langle u,U\rangle| &= \int_{\Bbb{S}^1} |\langle \mathbf{u},\mathbf{x}\rangle| \,d\mathbb{P}_U(x)\\
&= \frac1{2\pi}\int_{\Bbb{S}^1} |\langle \mathbf{u},\mathbf{x}\rangle|\,d\lambda(\mathbf{x})\\
&= \frac1{2\pi} \int_0^{2\pi} |\langle (u_1,u_2),(\cos\theta, \sin\theta)\rangle|\,d\theta\\
&= \frac1{2\pi} \int_0^{2\pi} |u_1\cos\theta + u_2\sin\theta|\,d\theta\\
&= \frac1{2\pi} \int_0^{2\pi} |\cos\phi\cos\theta + \sin\phi\sin\theta|\,d\theta\\
&= \frac1{2\pi} \int_0^{2\pi} |\cos(\phi-\theta)|\,d\theta\\
&= \frac1{2\pi} \int_\phi^{\phi +2\pi} |\cos(\theta)|\,d\theta\\
&= \frac1{\pi}\int_0^{\pi} |\cos\theta|\,d\theta
\end{align}
since $\cos$ is $\pi$-periodic. More formally
$$ \int_\phi^{\phi +2\pi} |\cos(\theta)|\,d\theta = \int_\phi^{2\pi} |\cos\theta|\,d\theta + \int_{2\pi}^{\phi+2\pi} |\cos\theta|\,d\theta  = \int_\phi^{2\pi} |\cos\theta|\,d\theta + \int_{0}^{\phi} |\cos(\theta+2\pi)|\,d\theta \\
= \int_\phi^{2\pi} |\cos\theta|\,d\theta + \int_{0}^{\phi} |\cos(\theta)|\,d\theta = \int_0^{2\pi}|\cos\theta|\,d\theta = \int_0^\pi |\cos\theta|\,d\theta + \int_\pi^{2\pi} |\cos\theta|\,d\theta \\= \int_0^\pi |\cos\theta|\,d\theta + \int_0^{\pi} |\cos(\theta+\pi)|\,d\theta =  \int_0^\pi |\cos\theta|\,d\theta + \int_0^{\pi} |\cos(\theta)|\,d\theta = 2\int_0^{\pi} |\cos(\theta)|\,d\theta.
$$