问:
If $A,B$ are any $3 \times 3$ matrices, then $A+B=B+A$ and $AB=BA$
This question seems to be easy but not obvious, it consists of little bit of confusion.

What should be the $\text{negation of the following statement}$:

If $A,B$ are any $3 \times 3$ matrices, then $A+B=B+A$ and $AB=BA$.

The statement is not true. But how to negate it ?

If I say:

If $A,B$ are any $3 \times 3$ matrices, then $A+B=B+A$ and $AB \neq BA$.

This is a negation but not clear.

Any help

答:
The correct way to negate it is:

There exists $A,B$ which are $3 \times 3$ matrices, such that $A+B\neq B+A$ or $AB\neq BA$.

Draw a Venn diagram may help when meeting logics.


答:
More formally, we would say that original statement as

$$(\forall A,B \in M_{3 \times 3}(F))(A+B = B+A \; \land \; AB = BA)$$

For simplicity, define the predicate such that $P$ corresponds to $A+B=B+A$ and $Q$ so that $AB=BA$:

$$(\forall A,B \in M_{3 \times 3}(F))(P \land Q)$$

To negate a statement, you change all quantifiers around, and negate the predicate:

$$(\exists A,B \in M_{3 \times 3}(F))\Big( \neg (P \land Q) \Big)$$

Of course, the question remains, what is $\neg (P\land Q)$? If you'll remember, it is $\neg P \lor \neg Q$. Thus, the negation is

$$(\exists A,B \in M_{3 \times 3}(F))(\neg P \lor \neg Q)$$

i.e.

$$(\exists A,B \in M_{3 \times 3}(F))(A+B \ne B+A \; \lor \; AB \ne BA)$$

ContentLicense: CC BY-SA 4.0

Body:
This question seems to be easy but not obvious, it consists of little bit of confusion.

What should be the $\text{negation of the following statement}$:

If $A,B$ are any $3 \times 3$ matrices, then $A+B=B+A$ and $AB=BA$.

The statement is not true. But how to negate it ?

If I say:

If $A,B$ are any $3 \times 3$ matrices, then $A+B=B+A$ and $AB \neq BA$.

This is a negation but not clear.

Any help


问:
The convergence of the series $\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$
Is the series $$\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$$ convergent?

答:
$$\bullet \text{ Cauchy condensation criterion:}$$
$$ \text{Let $a_n$ be a decreasing sequence of non-negative numbers, then}$$
$$\sum_n a_n <+\infty \iff\sum_n2^na_{2^n}<+\infty $$

$$\implies \sum_{n=3}^{+\infty}\cfrac{1}{n^2\log\log(n)}<+\infty \iff \sum_{n=3}^{+\infty}\cfrac{2^n}{(2^n)^2\log\log2^n}<+\infty$$

$$\text{so,}$$
$$\sum_{n=3}^{+\infty}\cfrac{2^n}{(2^n)^2\log\log2^n}=\sum_{n=3}^{+\infty}\cfrac{1}{(2^n)\log\big(n\log(2)\big)}\le\sum_{n=3}^{\infty}\cfrac{1}{2^n}=\cfrac{1}{4}<+\infty$$

$$\text{Hence, }\quad \fbox{$\quad \sum_{n=3}^{+\infty}\cfrac{1}{n^2\log\log(n)}<+\infty$}$$


问:
Prove that $\left|\int_a^b f(x) dx - (b-a) f(a) \right| \leq \frac{(b-a)^2}{2}$ when $f$ satisfies $|f(u)-f(v)|\leq |u-v|$
this is a question from Apostol's Calculus, Volume I, p. 139 (Exercises 33) which I've gotten completely stuck on.

Let $f$ be a continuous function such that $|f(u) - f(v)| \leq |u -
 v|$ for all $u$ and $v$ in an interval $[a,b]$.

(b) Assume that $f$ is integrable on $[a,b]$.  Prove that $$ \left|
 \int_a^b f(x) dx - (b-a) f(a)\right| \leq \frac{(b-a)^2}{2}. $$

In part (a) we proved that $f$ was continuous at each point of $[a,b]$.  This proof is in the section on the basic limit theorems and definition of continuity, so essentially the only knowledge we have is of the definitions and basic theorems regarding limits.  We have also proved the continuity of indefinite integrals (though it is isn't clear to me how or if to apply that theorem).

Geometrically, I believe I understand in a non-rigorous way what is happening, and why it is true.  The quantity on the left is the area between $f(x)$ and the horizontal line at $f(a)$ from $a$ to $b$.  The quantity on the right is the area of the right triangle of base and height $b-a$ (or viewed as an integral it is $\int_a^b x dx$).  The condition on $f$ we are given in the hypothesis of the theorem roughly states that the change in $f$ over any interval must be less than the change in $x$ (or the width of the interval).  So, we have a function $f$ that is changing less than linearly over $[a,b]$; hence, the integral of the function will be $\leq$ than that of $f(x) = x$.

I cannot figure out at all how to turn this intuition/geometric argument into a rigorous analytical one.  Any help is appreciated.  Thanks.

答:
Another way to solve this problem:

Note that $$\lim\left| \dfrac{f(x)-f(y)}{x-y}\right|\le 1,\,\forall x,y\\ \implies \lim_{x \to y}\left| \dfrac{f(x)-f(y)}{x-y}\right|\le 1,\,\forall y \\ \implies |f'(y)|\le 1, \forall y $$
Now define $F(x)=\int_{a}^{x}f(t)\,dt$.

Now note that
\begin{align}
\int_{a}^{b}f(t)\,dt&=F(b) \\&=F(a)+(b-a)F'(a)+\frac{(b-a)^2}{2!}F''(c)\\&=0+(b-a)f(a)+\frac{(b-a)^2}{2!}f'(c)
\end{align}

So, $$\left|\int_a^b f(t) dt - (b-a) f(a)\right|\\=\frac{(b-a)^2}{2}|f'(c)|\le \frac{(b-a)^2}{2}$$

ContentLicense: CC BY-SA 3.0


问:
question about parameterization/Tangential vectors
the Semicubical parabola which has the function

$y^2=x^3$

can be parameterized as $(t^2, t^3)$ but also as $(t, t^{\frac{3}{2}})$. Now if I wanted to calculate the tangent vector

I would get $(2t, 3t^2)$ for the first and $(1, \frac{3}{2}t^{\frac{1}{2}})$ for the second.

I want to calculate the tangent vector at the origin (0,0), I would get

$(2\cdot0, 3\cdot0)=(0,0)$ for the first

and
$(1,0)$ in case of the second parameterization.

How can they differ? what mistake did I make?

答:
Because even though the trace of the two parametrizations (the curve they draw) is the same, their velocity is different.

We define the tangent (unit) vector to a curve as
$$ \mathbf{\tau}\ (t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} $$
when the velocity magnitude $|\mathbf{r}'(t)|$ is nonzero. So it is clear that you just need to pick the appropriate parametrization.


问:
An example of an infinite disjoint union $X = \bigsqcup_\alpha X_\alpha$ such that $H^n(X, G) \neq \bigoplus_\alpha H^n(X_\alpha, G).$
For an infinite disjoint union $X = \bigsqcup_\alpha X_\alpha$, it is at least intuitively clear why the homology of $X$ decomposes into a direct sum

$$H_n(X) \cong \bigoplus_\alpha H_n(X_\alpha),$$

since the singular chain groups $C_n(X)$ by definition contain only finite sums of singular simplices $\sigma: \Delta^n \to X_\alpha$, which corresponds to the finite sums of cycles on the right hand side.

However, Hatcher switches to using infinite products when he introduces Cohomology (for example, Example 3.14, and the section on axioms for Cohomology), and we instead have the relation

$$H^n(X) \cong \prod_\alpha H^n(X_\alpha). $$

Can someone provide an example where this isomorphism of Cohomology with the infinite product holds, but the infinite direct sum does not?
I would prefer a non-category theory explanation if possible. Thanks!

答:
If $X$ is a set equipped with the discrete topology then its cohomology is just $H^0(X, A) \cong A^X$ (the infinite product of $X$ copies of $A$). The natural map from the infinite sum $\bigoplus_X A$ to the infinite product $A^X$ is never an isomorphism once $X$ is infinite and $A$ has at least two elements. For example, if $X = \mathbb{N}$ and $A$ is at most countable then $\bigoplus_X A$ is countable but $A^X$ is uncountable.

This example suggests that we can think about cohomology as a generalization of taking functions on a space, which is true in various senses. In particular, we expect it to convert infinite coproducts into infinite products since taking functions has this property.

Note also that cohomology with coefficients in a commutative ring is again a ring, and for that to be true it needs to have a unit: if you replace the infinite product with the infinite sum then there won't be a unit.


答:
Since $C_n (X) = \bigoplus C_n(X_\alpha)$, then
$$
C^n(X) = \textrm{Hom} (C_n(X), \mathbb{Z}) = \textrm{Hom} (\bigoplus C_n(X_\alpha), \mathbb{Z}) = \prod \textrm{Hom} (C_n(X_\alpha), \mathbb{Z}).
$$
So for example if $X = \mathbb{Z}$ (with the discrete topology), then $H^0(X) = C^0(X) = \prod_{n \in \mathbb{Z}} \mathbb{Z}$.

$C^0$ (for example) gives all of the homomorphisms from $C_0$ to $\mathbb{Z}$, and this has to include maps which are nonzero on every summand in $C_0$; this is why $C^0$ has to be the direct product rather than the direct sum.

Title: An example of an infinite disjoint union $X = \bigsqcup_\alpha X_\alpha$ such that $H^n(X, G) \neq \bigoplus_\alpha H^n(X_\alpha, G).$

Body:
For an infinite disjoint union $X = \bigsqcup_\alpha X_\alpha$, it is at least intuitively clear why the homology of $X$ decomposes into a direct sum

$$H_n(X) \cong \bigoplus_\alpha H_n(X_\alpha),$$

since the singular chain groups $C_n(X)$ by definition contain only finite sums of singular simplices $\sigma: \Delta^n \to X_\alpha$, which corresponds to the finite sums of cycles on the right hand side.

However, Hatcher switches to using infinite products when he introduces Cohomology (for example, Example 3.14, and the section on axioms for Cohomology), and we instead have the relation

$$H^n(X) \cong \prod_\alpha H^n(X_\alpha). $$

Can someone provide an example where this isomorphism of Cohomology with the infinite product holds, but the infinite direct sum does not?
I would prefer a non-category theory explanation if possible. Thanks!


问:
Matrix Linearity and Orthogonal Vector Question Help
Question One: Let $v_1$ and $v_2$ $\in \mathbb{R^n}$. Let S be the set of all vectors in $\mathbb{R^n}$ that are orthogonal to both $v_1$ and $v_2$. That is, $S=\{x\in \mathbb{R^n}: x*v_1=0$ and $ x*v_2=0\}$. Is $S$ a subspace of $\mathbb{R^n}$?

My attempted solution: This may sound like a silly question but I'm not very sure why S would form a subspace for $\mathbb{R^n}$ in this case. If three vectors are to form a subspace for $\mathbb{R^3}$ for example, does it mean that the three vectors have to span $\mathbb{R^3}$ or does it simply mean that these three vectors exist in the $\mathbb{R^3}$ space while they might be dependent of each other? In other words, if three vectors are to span $\mathbb{R^2}$, can they still be a subspace for $\mathbb{R^3}$?

Question Two: Let $A$ and $B$ be two $n\times n$ matrices. Suppose that the first column of $AB$ is entirely zero, but the first column of B is not entirely zero, then which of the following are true?

There is not enough information on columns of A

$det(A)$ does not equal zero

The columns of A are linearly dependent

$det(A)$ cannot be computed with the given information

My attempted solution: For this question, I'm actually quite confused. The correct answer is #3 where the columns of A are linearly dependent. How exactly can this be proved?

Thanks in advance!

答:
What the first question is trying to differentiate is between a "subset" and a "subspace". The span of a finite number of vectors is always going to be a "subset" of the vector space due to closure, but it is not necessarily a "subspace" because then the subset would need to have the additional restrictions that would qualify it to be a subspace

Using your example, if three vectors in $\mathbb{R}^3$ span $S = \mathbb{R}^2$, it is indeed a subspace of $\mathbb{R}^3$, as it is a subset which is also in itself a vector space

For the second question, it requires looking at matrix multiplication as a linear combination of vectors

Let $A$ be composed of $n$ column vectors of dimension $n$, similarly for B, such that

$$A = [\textbf{a}_1, \textbf{a}_2, \textbf{a}_3,..., \textbf{a}_n]$$
$$B = [\textbf{b}_1, \textbf{b}_2,\textbf{b}_3,  ..., \textbf{b}_n]$$

Now, when we do the matrix multiplication $AB$, we can write it as

$$AB = [\textbf{x}_1, \textbf{x}_2,\textbf{x}_3,...,\textbf{x}_n]$$

where

$$\textbf{x}_i = \sum_{j=1}^n b_{ij}\textbf{a}_j$$

Here, $b_{ij}$ is the $j^{th}$ element of the column vector $\textbf{b}_i$

Can you see why the C option is right now?


问:
Normalization of a projective curve over a field doesn't add new constant functions
Suppose $k$ is a field (not necessarily algebraically closed, no characteristic assumptions) and $X\to\operatorname{Spec} k$ is a curve over $k$ (a projective integral separated scheme over $k$ of dimension one). Let $\nu:\widetilde{X}\to X$ denote the normalization. Is it true that $\mathcal{O}_X(X)\cong \mathcal{O}_{\widetilde{X}}(\widetilde{X})$?

My efforts:

I'm proceeding under the hope this is true, but I haven't gotten very far.

I already know that the global sections of any proper integral scheme over $k$ is a finite extension of $k$, so the goal is to explain why normalization doesn't add anything here. So restrict to an affine open $\operatorname{Spec} A\subset X$ with normalization $\operatorname{Spec} A'\subset \widetilde{X}$ - because restriction maps are injective in integral schemes, this exhibits the global sections of $X$ and $\widetilde{X}$ as subrings of $A$ and $A'$ respectively, each integral over the respective copy of $k$. Let $a,b$ be two elements of $A$ with $a/b$ integral over $\mathcal{O}_X(X)\subset A$. Clearly I need to argue that $a,b\in \mathcal{O}_X(X)$, but I don't see how to do this.

Alternatively, since normalization is birational, it must be an isomorphism on a dense open set. So if the constant functions can be detected from a dense open subset, then we'd be in business. But I'm worried about having some constant function defined on a dense open subset which doesn't extend (I know this seems silly, but I can't explain my way out of this). I guess that maybe we would want to say that the constant functions are the maximal subfield of the coordinate algebra of an affine open, but I don't know how to prove that or if it's correct. Or the constants are the maximal subfield contained in every residue field, but how do I know that this is birational? I suspect it isn't - take something like $\operatorname{Spec} \Bbb R[x,y]/(x^2+y^2)$ with just one real point.

Motivation:

I'm working on Hartshorne chapter 2 exercise 6.9a. The goal is to show that in the above situation, there's an exact sequence $$0\to \bigoplus \widetilde{\mathcal{O}_P}^\times/\mathcal{O}_P^\times \to \operatorname{Pic} X\to \operatorname{Pic} \widetilde{X}\to 0.$$
Via the hint, I can show that there's an exact sequence $$0\to \bigoplus \widetilde{\mathcal{O}_P}^\times/\mathcal{O}_P^\times \to \operatorname{CaDiv} X \to \operatorname{CaDiv} \widetilde{X}\to 0,$$ and the only thing I have left to do is to show that the image of the first term is disjoint from the principal Cartier divisors in order to get the requested exact sequence. If $k=\overline{k}$, this is easy and $\mathcal{O}_X(X)=\mathcal{O}_{\widetilde{X}}(\widetilde{X})=k$, but I'm wondering about the general case.

答:
Let $$X:U^2+V^2=0\subset \Bbb{P^2_Q}$$

$i=V/U=-U/V$ isn't regular at $[0:0:1]$ so it isn't in $O_X(X)$.

The normalization is $$Y: U^2+V^2=W^2+Z^2=UW-VZ=UZ+VW=0\subset \Bbb{P}^3_\Bbb{Q}$$

Looks a bit strange but it is an homogeneous prime ideal and with $v=V/U,w=W/U,z=Z/U$ the function field is just
$$Frac(\Bbb{Q}[v,w,z]/(1+v^2,w^2+z^2,w-vz,z+vw))=Frac(\Bbb{Q}[i,w,z]/(w-iz))$$

The morphism $Y\to X$ is

$$[U:V:W:Z]\to [U:V:W]$$

$i=V/U=W/Z$ is now regular at $[0:0:1:i]$ and $[0:0:1:-i]$ so it is in $O_Y(Y)$


问:
If $V$ is a $k$-dimensional linear subspace of $R^n$, then $V$ is the kernel of a linear transformation $T : R^n → R^{n−k}.$
If $V$ is a $k$-dimensional linear subspace of $R^n$, then $V$ is the kernel of a linear transformation $T : R^n → R^{n−k}.$

Basically just looking for whether this is True or False.

答: Consider the quotient map onto $\Bbb R^n/V\cong\Bbb R^{n-k}$.  (Halmos is a good reference on this.)


问:
Is there any bijective map $f: \mathbb R^{2} \to \mathbb R^{3}$?
The title pretty much says it all.
I'm trying to prove that if I have two maps $T:\mathbb R^{3} \to \mathbb R^{2}$ and $S: \mathbb R^{2} \to \mathbb R^{3}$, then $S\circ T$ is not invertible. I think I only need to show that there's no map $f:\mathbb R^{2} \to \mathbb R^{3}$ such that $f$ is bijective.

.
Is there any invertible map $f:\mathbb R^{2} \to R^{3}$

答:
There is a bijective function from $\mathbb{R}^2$ to $\mathbb{R}^3$.  There is no bijective linear function, though, and I assume from your choice of tags that this is relevant to the question you're trying to solve.  To see why, consider the dimensions of $\mathbb{R}^2$ and $\mathbb{R}^3$.

(I'm assuming here that you're thinking of $\mathbb{R}^2$ and $\mathbb{R}^3$ as $\mathbb{R}$-vector spaces, and looking for an $\mathbb{R}$-linear map.  If you only ask for $\mathbb{Q}$-linearity the situation is different).


问:
I am stuck with some prime number problems. Are there any hints you could give me?

Find all natural values of n which make the value of (n^3 - 1)/5 a prime number.

Find all values of p which make p, p + 2 and p + 4 prime numbers.

Find all prime numbers p which also make p + 10 and p + 20 prime too.

Thanks in advance!

答:
$\hspace0.5cm$   (1) It is rather easy to show that the only $n$ is $6$.

$\hspace0.5cm$ (2) A prime $p\gt3$ is of the form $3n+1$ or $3n+2$ so we have
$$3n+1,3n+3,3n+5 \text { or }3n+2,3n+5,3n+6$$ In both cases one of the three numbers is divisible by $3$. Thus the only example is $(p,p+2,p+4)=(3,5,7)$.

$\hspace0.5cm$ (3) Proceding as in (2) we have $$3n+11,3n+21\text { or }3n+12,3n+22$$ and again one of the numbers will be multiple of $3$.


问:
If $f:\omega_1 \to \omega_1$ is a bijection, then must the fixed points of $f$ contain a club?
I was reading some notes and we have the result that

If $f:\omega_1 \to \omega_1$ is a continuous bijection, then the fixed points of $f$ contain a club.

I followed that proof's constructive argument, but can it be proven that any bijection's fixed points contain a club? Here's my idea.

Assume there's no club $F$ such that $\forall \lambda \in F, \lambda \leq f(\lambda)$. Then the set $A = \{\alpha : \alpha > f(\alpha)\}$ is not a club. Then by Fodor's lemma, $f|_{A}:A \to \omega_1$ would give us some $\alpha < \omega_1$ such that $f^{-1}|_{A}(\alpha)$ is statioanry. This is a contradiction since $f$ is a bijection.

Now assume there's no club $F$ such that $\forall \lambda \in F, f(\lambda) \leq \lambda$. This means that $A = \{\alpha < \omega_1 : \alpha < f(\alpha)\}$ is stationary. In other words, $\{\alpha < \omega_1 : f^{-1}(\alpha) < \alpha\}$ is stationary. Then $f^{-1}|_{A}: A \to \omega_1$ is regressive. Apply Fodor's lemma again to get a contradiction.

Thus we get that $C_1 \cap C_2$ being the fixed points of $f$ which is a club. Is this proof true? It does not seem intuitively true at all that ANY bijection has this property, and I feel like there's some mistake I'm making.

答:
This means that $A = \{\alpha < \omega_1 : \alpha < f(\alpha)\}$ is stationary. In other words, $\{\alpha < \omega_1 : f^{-1}(\alpha) < \alpha\}$ is stationary.

This last deduction is wrong.  The latter set is $f(A)$, but the fact that $A$ is stationary does not imply that $f(A)$ is also stationary.

A bijection $\omega_1\to\omega_1$ can have no fixed points at all.  For instance, you can partition $\omega_1$ into pairs and have $f$ swap each pair.


问:
Solving $x^d \equiv a \pmod{29}$ using primitive roots
Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solutions, but I have no idea how I'd go about proving that no solutions exist, or that any solutions I find are comprehensive.

I know that $2$ is a primitive root $\bmod 29$, so $2^{28} \equiv 1 \pmod{29}$. For the first one I can manipulate to get $(2^{23})^5 \equiv 8 \pmod{29}$ to give $x \equiv 2^{23} \equiv 10 \pmod{29}$ as a solution, but this doesn't prove that no other solutions exist, does it? Also, I don't know how this approach would work with the third problem because $12$ cannot be written $2^n$.

Furthermore, I'm unsure of the general technique to show that the second is not soluble?

答:
Since $2$ is a primitive root modulo $29$, its multiplicative order is $28$, which means

$$2^y \equiv 2^z \pmod{29} \iff y \equiv z \pmod{28} \tag{1}\label{eq1A}$$

One way to solve your congruence equations is to first express the right hand sides as a congruent power of $2$. The first $2$ examples are quite easy. With the third one, i.e.,

$$x^7 \equiv 12 \pmod{29} \tag{2}\label{eq2A}$$

note $12 \equiv 3(4) \equiv (32)(4) \equiv 2^7 \pmod{29}$ (as J. W. Tanner's question comment states). Next, since $2$ is a primitive root and $x \not\equiv 0 \pmod{29}$, then there's an integer $1 \le a \le 28$ such that $x \equiv 2^a \pmod{29}$. Thus, \eqref{eq2A} becomes

$$\left(2^{a}\right)^7 \equiv 2^7 \pmod{29} \implies 2^{7a} \equiv 2^7 \pmod{29}  \tag{3}\label{eq3A}$$

Using \eqref{eq1A}, this gives

$$7a \equiv 7 \pmod{28} \implies a \equiv 1 \pmod{4} \tag{4}\label{eq4A}$$

This gives multiple answers of $x \equiv 2^{4b + 1} \pmod{29}$ for $0 \le b \le 6$, i.e., $x \equiv 2, 3, 19, 14, 21, 17, 11 \pmod{29}$.

Next, with your second example,

$$x^5 \equiv 8 \equiv 2^3 \pmod{29} \tag{5}\label{eq5A}$$

as done before, let $x \equiv 2^a \pmod{29}$ to get

$$2^{5a} \equiv 2^3 \pmod{29} \implies 5a \equiv 3 \pmod{28} \implies a \equiv 23 \pmod{28} \tag{6}\label{eq6A}$$

Thus, $x \equiv 2^{23} \pmod{29}$ is the answer, as J. W. Tanner's question comment also indicates.

With your final example of

$$x^4 \equiv 4 \equiv 2^2 \pmod{29} \implies x^2 \equiv \pm 2 \tag{7}\label{eq7A}$$

note $2$ is not a quadratic root modulo $29$ (which requires, as shown in this table that $p \equiv 1, 7 \pmod{8}$ but $29 \equiv 5 \pmod{8}$), and also $-2$ is not a quadratic residue (since that requires $p \equiv 1, 3 \pmod{8}$). In addition, using the method I show above, gives

$$2^{4a} \equiv 2^2 \pmod{29} \implies 4a \equiv 2 \pmod{28} \implies 2a \equiv 1 \pmod{14} \tag{8}\label{eq8A}$$

However, it's not possible to have an even value be equivalent to an odd value with an even modulo (e.g., since it would require $14 \mid 2a - 1$), so there are no solutions.


Title: Solving $x^d \equiv a \pmod{29}$ using primitive roots

Body:
Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solutions, but I have no idea how I'd go about proving that no solutions exist, or that any solutions I find are comprehensive.

I know that $2$ is a primitive root $\bmod 29$, so $2^{28} \equiv 1 \pmod{29}$. For the first one I can manipulate to get $(2^{23})^5 \equiv 8 \pmod{29}$ to give $x \equiv 2^{23} \equiv 10 \pmod{29}$ as a solution, but this doesn't prove that no other solutions exist, does it? Also, I don't know how this approach would work with the third problem because $12$ cannot be written $2^n$.

Furthermore, I'm unsure of the general technique to show that the second is not soluble?

Body:
$(x^5)^{17}\equiv x^{85}\equiv (x^{28})^3x\equiv x\bmod 29$,

so if $x^5\equiv8\bmod29$, then $x\equiv8^{17}=2^{51}=2^{28}2^{23}\equiv2^{23}\equiv10\bmod29$.

If $x^4\equiv4=2^2\bmod29$ had solutions, then $1\equiv x^{28}\equiv (x^4)^7\equiv(2^2)^7=2^{14}\bmod29$, so $2$ would not be a primitive root.

$2^7\equiv12\bmod29$, so if $x^7\equiv12\bmod29$, then $x^7\equiv2^7\bmod29$,

so $x\equiv2a\bmod29$, where $a^7\equiv1\bmod29$.

$a^7\equiv1\bmod29$ when $a\equiv2^0, 2^4, 2^8, 2^{12}, 2^{16}, 2^{20}, $ or $2^{24}$; i.e., $a\equiv1, 16, 24, 7, 25, 23, $ or $20$;

so $x\equiv2, 3, 19, 14, 21, 17, $ or $11\bmod29$.


问:
Integral limit $\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \; dx$
I need to compute the (Lebesgue) integral limit
$$\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \mathrm dx$$

I think I need to use the dominated conergence theorem or a similar theorem, but I'm not sure how to proceed.

Note on $[1, \infty)$ the application of DCT is obvious, but the singularity at $0$ has me confused.

答:
There actually is no singularity: $$\lim_{x \to 0}\frac{\sin\left(\frac{x}{n}\right)}{x^2+x} = \frac{1}{n}$$

And since $\sin(x) < x$ for $x > 0$, $$0<\int_0^1 \frac{\sin\left(\frac{x}{n}\right)}{x^2+x} dx \le \int_0^1 \frac{\frac{x}{n}}{x^2+x} dx = \frac{1}{n}\int_0^1 \frac{1}{x+1} dx = \frac{\ln(2)}{n}$$

Since both sides converge to $0$, the integral also converges to $0$. I ignored $x > 1$ partly because you said you got it already, but mainly because I'm not great with applying the convergence theorems.


问:
The numbers of Triangles in a graph equals $\frac{1}{6}tr(A^3)$
The numbers of Triangles in a graph equals $\frac{1}{6}tr(A^3)$ where A is a square n x n Adjacency Matrix. The proof that is presented in the textbook, Applications of linear algebra, from chapter 13, graph theory is as follows:

Proof. A walk of Length 3 from a vertex to itself is a triangle and the triangle actually yields two walks, one in each direction. It follows that if a vertex $i$ is contained in a triangle, then $(A^3)_{ii} = 2$. From there we see that $tr(A^3)$ equals twice the number of vertices contained in the triangles. However, since each triangle contains three vertices it follows that $tr(A^3)$ equals six times the number of triangles.

The author poses a question right after this proof asking the reader why the same approach above does not work for squares, pentagons, etc. What this means is why doesn't the number of squares in a graph equal  $Mtr(A^4)$ where M is some constant, and similarly for higher-order shapes.  My thought process is that a walk of length 4 from a vertex to itself need not be a square, nor does a walk of length 5, etc. I am not quite sure if I am on the right track, and was wondering if someone could give any insight as to why this is the case.

Note: the original proof comes from a textbook written by one of the (great) Math teachers at my university, I will leave a link to the book as it is free: https://www.math.umd.edu/~immortal/MATH401/

答:
Your reasoning is correct.

In general, the $(i, j)^{\text{th}}$ entry of the $n^{\text{th}}$ power of the adjacency matrix, $A^{n}$, represents the number of walks of length $n$ from $i$ to $j$. In other words, the $(i, j)^{\text{th}}$ entry in $A^{3}$ represents the number of walk of length $3$ in the underlying graph.

The argument works for triangles because a walk of length $3$ because this same symmetry guarantee is not there for higher-order shapes.

By the way, I read the same book when I took MATH 401 :)


问:
If $R$ and $S$ are fields, either prove or disprove that $R\times S$ is a field
That's the question from my homework. I am thinking $R\times S$ is not a field, but I'm not sure. I understand the definition of a field, but I am not sure how to proceed.</p>

答: Let $F_1$ and $F_2$ be two finite fields. Then $F_1 \times F_2$ has cardinality $|F_1| \cdot |F_2|$. But a finite field has cardinality a prime power, so it is sufficient to suppose that $F_1$ and $F_2$ have different characteristics to conclude that $F_1 \times F_2$ is not a field.</p>


问:
Number of ways to choose $p,q$ from $\{1,3,5,\cdots,399\}$ such that $q > p$?
List $M = \{1,3,5...397,399\}$

$M$ contains all positive odd integers from $1$ to $400$, this is not exact wording of the question but basically ask how many ways are there if we randomly choose two numbers $p$ and $q$ from $M$ such that $q > p$?

The answer is $19900$.

My approach was that we know there are $200$ odd integers within $400$, if we randomly picked two numbers, then it must either $q < p$ or $q > p$, so looks like we have $\frac{200\choose2}{2}$, dividing $2$ here means we rule out the case $q < p$.

But I precisely found that $200\choose2$ = $19900$ which is the answer.

I'm a bit confused here then, any suggestion?

答: You can pick any distinct pair, which you can do in $200 \choose 2$ ways.  You then call the larger one $q$ and the smaller one $p$.  You didn't already call them $p$ and $q$ when you picked them, so you do not divide by $2$.


答:
Alternative explanation

if we randomly picked two numbers,...

There are $(200 \times 199)$ ways of randomly picking two numbers.  This is what goes in the numerator of your analysis, re the denominator $= 2.$

With this formula (in the numerator) for picking two random numbers, any pair of distinct random numbers $(q,p)$ will be counted twice, once with $q$ picked first, and once with $p$ picked first.

Title: Number of ways to choose $p,q$ from $\{1,3,5,\cdots,399\}$ such that $q > p$?

Body:
List $M = \{1,3,5...397,399\}$

$M$ contains all positive odd integers from $1$ to $400$, this is not exact wording of the question but basically ask how many ways are there if we randomly choose two numbers $p$ and $q$ from $M$ such that $q > p$?

The answer is $19900$.

My approach was that we know there are $200$ odd integers within $400$, if we randomly picked two numbers, then it must either $q < p$ or $q > p$, so looks like we have $\frac{200\choose2}{2}$, dividing $2$ here means we rule out the case $q < p$.

But I precisely found that $200\choose2$ = $19900$ which is the answer.

I'm a bit confused here then, any suggestion?


问:
Locus of vectors of a given rank in the exterior algebra
The rank of $\alpha\in \bigwedge^2 \mathbb{C}^n$ is the minimal number of decomposable 2-vectors such that $\alpha$ can be written as a sum of them, see https://en.wikipedia.org/wiki/Exterior_algebra .
Consider the projectivization $\mathbb{P}(\bigwedge^2 \mathbb{C}^n)$: what is the dimension of the locus of classes of vectors of a given rank?

We clearly have the Grassmannian, whose dimension is $2(n-2)$, which is the locus of vectors of rank one.
Moreover, the maximal rank is $k$ if $n=2k$ or $2k+1$. I think the locus of vectors of maximal rank should be an open subset of $\bigwedge^2 \mathbb{C}^n$. How about the other cases? What is the link with the $PGL(n)$-orbits in $\bigwedge^2 \mathbb{C}^n$?

Thank you!

答:
Let $V = \mathbb{C}^n$. If $\alpha \in \mathbb{P}(\wedge^2V)$ has rank $k$ (though I would rather say it is $2k$ in this case) then there is a unique vector subspace $U \subset V$ of dimension $2k$ such that $\alpha$ is in the image of $\mathbb{P}(\wedge^2U) \to \mathbb{P}(\wedge^2V)$. This shows that the locus of rank-$k$ bivectors is isomorphic to an open subspace of the projective bundle
$$
\mathbb{P}_{\mathrm{Gr}(2k,V)}(\wedge^2\mathcal{U}),
$$
where $\mathcal{U}$ is the tautological bundle. Its dimension, therefore, is
$$
2k(n-2k) + k(2k-1) - 1.
$$
And indeed, these loci are precisely the orbits of $\mathrm{PGL}(V)$.


问:
Largest disc around which this complex function is one-to-one?
How would you determine explicitly the largest disc round the origin on which the function $f(z) = z^2 +z$ is one-to-one?

Is there a general method to do this for functions of this type?

答:
You need to find the largest disk containing the origin for which there is no pair $(z,\omega)$ with $z^2+z = \omega^2 + \omega$

Notice you can rearange
$$z^2 + z = \omega^2 + \omega \Longleftrightarrow (z+\omega)(z-\omega) = (\omega-z)$$

If $z$ and $\omega$ are distinct we must have $z+\omega = -1$, so what can you say about the disk?

I don't think there's a way of doing it in general that will save any time, you just have to look at a few and learn some tricks.


问:
Factors of $2n^2 \leq n$?
How many factors of $2n^2$ are less than or equal to $n$?
I know that the number of factors of $n^2$ less than $n$ is half the number of factors of $n^2$ (each factor $< n$ corresponds with one greater than $n$), but $2n^2$ is a whole different case, it seems. Is there any way to find an expression for this? And if not, is there an algorithm for it? I have looked into both combinatorics and prime factorization, but have arrived at dead-ends.

答:
This is a very interesting question. Assume $n=2^{a}(2k+1)$
for some integer $a$ and $k$. Let $f(x)=$numbers of positive divisors of the integer $x$ . Since factors of $2n^2\leq n$ we need number of factors of $2^{2a+1}(2k+1)^2$. Thus we have $f(2^a(2k+1))+c_{a}$.Where $c_{a}$ is the error factor has a small factor of bound, which needs to be deterministic. Although it's a rough idea, I am not finding the bound but for hints, you can try small cases. However
let $g(x)=$greatest integer lesser equal to x then, $$c_{a}\leq f(g(2^{\frac{2a+1}{2}}(2k+1)))-f(2^a(2k+1))$$.Where we know $f$ is famous divisor function or $\tau$ function and $g$ is the floor function.
Use this link for bound,
Lower bound for the sum of divisor function

Title: Factors of $2n^2 \leq n$?


答:
I see no general analytic solution, since it would seem to depend upon the prime factorization of $n$.

But the OP also asks for code.  That is very straightforward.  In Mathematica:

```<code>myfun[n_: Integer] := Length[
Select[Divisors[2 n^2], # <= n &]]
</code>```

So:

```<code>myfun[9098345]
</code>```

(* 27 *)

Here's a plot:

This isn't directly part of the problem, but seems to be the motivation of the problem.  If the above function is $f(n)$, calculate $F(N) = \sum\limits_{n=1}^N f(n)$, for $N = 10^{12}$.

I think the approach is the following:  Calculate the number of $2$s in that sum.  Then calculate the number of $3$s.  And so on, then add them up.

The number of $2$s is $10^{12}/2$.  The number of $3$s is $10^{12}/3$.  And so on.  But what is the maximum we add those up to in the total calculation?  I think it should be the largest factor allowed in the $10^{12}$ (last) term in the sum, i.e., $k_{max} = \sqrt{50} \cdot 10^5 = 707107$, obtained from the $2 n^2 = 10^{12}$ calculation.

If that's right, then:  $F(10^{12}) = 10^{12} \sum\limits_{k = 1}^{k_{max}} \frac{1}{k} = 10^{12}\ {\rm HarmonicNumber}(k_{max}) = 10^{12} \cdot 14.0461536491411$.

There are probably some rounding artifacts that must be included, but I think this is the right approach.  Someone should do this with greater care.

Title: Factors of $2n^2 \leq n$?

Body:
How many factors of $2n^2$ are less than or equal to $n$?
I know that the number of factors of $n^2$ less than $n$ is half the number of factors of $n^2$ (each factor $< n$ corresponds with one greater than $n$), but $2n^2$ is a whole different case, it seems. Is there any way to find an expression for this? And if not, is there an algorithm for it? I have looked into both combinatorics and prime factorization, but have arrived at dead-ends.

Body:
I see no general analytic solution, since it would seem to depend upon the prime factorization of $n$.

But the OP also asks for code.  That is very straightforward.  In Mathematica:

myfun[n_: Integer] := Length[
Select[Divisors[2 n^2], # <= n &]]

So:

myfun[9098345]

(* 27 *)

Here's a plot:

This isn't directly part of the problem, but seems to be the motivation of the problem.  If the above function is $f(n)$, calculate $F(N) = \sum\limits_{n=1}^N f(n)$, for $N = 10^{12}$.

I think the approach is the following:  Calculate the number of $2$s in that sum.  Then calculate the number of $3$s.  And so on, then add them up.

The number of $2$s is $10^{12}/2$.  The number of $3$s is $10^{12}/3$.  And so on.  But what is the maximum we add those up to in the total calculation?  I think it should be the largest factor allowed in the $10^{12}$ (last) term in the sum, i.e., $k_{max} = \sqrt{50} \cdot 10^5 = 707107$, obtained from the $2 n^2 = 10^{12}$ calculation.

If that's right, then:  $F(10^{12}) = 10^{12} \sum\limits_{k = 1}^{k_{max}} \frac{1}{k} = 10^{12}\ {\rm HarmonicNumber}(k_{max}) = 10^{12} \cdot 14.0461536491411$.

There are probably some rounding artifacts that must be included, but I think this is the right approach.  Someone should do this with greater care.


答:
You can compute the exact value of the integral
$$\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}=\frac{\sin \left(\frac{x}{n}\right)}{x}-\frac{\sin \left(\frac{x}{n}\right)}{x+1}$$
$$I(a)=\int\frac{\sin \left(\frac{x}{n}\right)}{x+a}\,dx=\cos \left(\frac{a}{n}\right) \text{Si}\left(\frac{x+a}{n}\right)-\sin
   \left(\frac{a}{n}\right) \text{Ci}\left(\frac{x+a}{n}\right)$$
$$J(a)=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x+a}\,dx=\text{Ci}\left(\frac{a}{n}\right) \sin \left(\frac{a}{n}\right)+\frac{1}{2}
   \left(\pi -2 \text{Si}\left(\frac{a}{n}\right)\right) \cos
   \left(\frac{a}{n}\right)$$
$$K=\int_0^\infty\frac{\sin \left(\frac{x}{n}\right)}{x^2+x}=J(0)-J(1)$$
$$K=\frac{1}{2} \left(\pi-2 \text{Ci}\left(\frac{1}{n}\right) \sin
   \left(\frac{1}{n}\right)-\left(\pi -2 \text{Si}\left(\frac{1}{n}\right)\right)
   \cos \left(\frac{1}{n}\right) \right)$$ Now, using asymptotics
$$K=\frac{\log (n)+1-\gamma }{n}+\frac{\pi }{4 n^2}+O\left(\frac{1}{n^3}\right)$$ which already a good approximation for small values of $n$.

For example, for $n=4$, the exact value is $0.494246$ while the above approximation gives $0.501357$.

Title: Integral limit $\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \; dx$

Body:
I need to compute the (Lebesgue) integral limit
$$\lim_{n \to \infty} \int_{\mathbb{R}^{+}} \frac{\sin(x/n)}{x^2+x} \mathrm dx$$

I think I need to use the dominated conergence theorem or a similar theorem, but I'm not sure how to proceed.

Note on $[1, \infty)$ the application of DCT is obvious, but the singularity at $0$ has me confused.


问:
Bijection between "circularly nonconsecutive" permutations and permutations with one fixed point
A permutation $\pi$ of $[n]:=\{1,2,\dots,n\}$ is called circularly nonconsecutive (CNC) if $\pi_{i+1}-\pi_i\neq 1$ for all $i=1,2,\dots,n-1$, and furthermore $\pi_1-\pi_n\ne 1$. In other words, $i+1$ does not occur immediately after $i$, for any $i\in \{1,2,\dots,n-1\}$, where we consider the first entry of the permutation as occurring immediately after the last in a circular fashion.

There are $3$ CNC permutations of $[3]$: $(1,3,2)$, and its two rotations, $(3,2,1)$ and $(2,1,3)$. In general, all $n$ rotations of a CNC permutation are also CNC.

It can be shown that the number of CNC permutations is equal to the number of permutations with exactly one fixed point. You can count both of these quantities with an inclusion exclusion argument and find the the resulting expressions are coincidentally the same, both equal to $$n!\sum_{k=0}^{n-1}\frac{(-1)^k}{k!}$$

Can you give a bijection between CNC permutations of $[n]$ and permutations of $[n]$ with one fixed point?

Equivalently, permutations with one fixed point can be partitioned into $n$ equal classes based on which point is fixed, just like CNC permutations can be partitioned into circular rotation classes. So it would be equivalent to find a bijection between CNC permutations for which $\pi_n=n$ and permutations where the only fixed point is $\pi_n=n$. The latter is obviously in bijection with derangements of $[n-1]$.

答:
OK, I found the answer in a wonderful blog post by Mike Spivey. It is implied by the following discussion, which is a better way of viewing the problem.

From now on, a succession in a permutation is an index $i\in \{1,\dots,n-1\}$ for which $\pi({i+1})=\pi(i)+1$. Furthermore, an extended succession is defined as follows; given a permutation $\pi:\{1,\dots,n\}\to \{1,\dots,n\}$, extend this to a function $\pi':\{0,\dots,n\}\to\{0,\dots,n\}$ by setting $\pi'(0)=0$ and $\pi'(i)=\pi(i)$ for $i\in \{1,\dots,n\}$. Then an extended succession of $\pi$ is defined to be a succession of $\pi'$. The only difference between the two concepts is that $\pi(1)=1$ counts as an extended succession, but not a succession.

The reason for this funny definition is that extended successions behave just like fixed points, in that for any $k\in \{0,\dots,n\}$, the number of permutations with $k$ fixed points equals the number of permutations with $k$ extended succesions. This is proven with the following bijection. In the rightmost column, $i$ represents a generic fixed point greater than $1$. Note how the fixed points, $3$ and $6$, become successions in the result. You also need to show that if $1$ is a fixed point, that it ends up at the beginning of the result (so it is an extended succession).

Bijection Step
Example
Effect on fixed points

Write $\pi$ in one line notation.
$[2,5,\def\r{\color{red}}\r3,1,4,\r 6]$
$\pi(i)=i$

Move each entry of $\pi$ one step to the left, and the first entry to the end, to get a new permutation in one-line notation.
$[5,\def\r{\color{red}}\r3,1,4,\r 6,2]$
$\pi(i-1)=i$

Write the result from the last step in cycle notation, so that each cycle has its largest element written last, and the cycles decrease in order of largest element.
$(2,3,1,5,6)(4)$
$i$ follows $i-1$ in a cycle.

Erase the parentheses, and interpret the result in one-line notation.
$[2,\r 3,1,5,\r 6,4]$
$i$ succeeds $i-1$.


问:
Showing two functions are a.e. equivalent in $L^3$ norm
Let $X$ be a measure space with measure $\mu$. Let $f,g \in L^3(X, \mu)$. Assume we have $$||f||_{L^3} = ||g||_{L^3} = \int_{X} ||f(x)||^2g(x) \ d\mu(x) = 1$$

How can we show $g=|f|$ a.e.?

答:
Let $p=3/2$, $q=3$, and define $h(x)\equiv|f(x)|^2$ for every $x\in X$. Note that $1/p+1/q=1$. Then, Hölder’s inequality actually holds as an equality:
\begin{align*}
1&=\int_X|f(x)|^2g(x)\,\mathrm d\mu(x)\leq\int_X|f(x)|^2|g(x)|\,\mathrm d\mu(x)=\|hg\|_1\leq\|h\|_p\|g\|_q\\&=\|f\|_3^2\|g\|_3 =1.
\end{align*} This is possible only if $|h|^p$ (that is, $|f|^3$) and $|g|^q$ (that is, $|g|^3$) are “linearly dependent” functions in the sense that there exist constants $\alpha$ and $\beta$, not both zero, such that $\alpha|f(x)|^3=\beta|g(x)|^3$ for $\mu$-almost every $x\in X$.

Without loss of generality, assume that $\alpha\neq0$ and let $\gamma\equiv\beta/\alpha$. Therefore, $|f(x)|^3=\gamma|g(x)|^3$ for $\mu$-almost every $x\in X$. But $\|f\|_3=\|g\|_3=1$ forces $\gamma$ to be $1$.

Finally, the fact (see the chain of inequalities above) that $$\int_X|f(x)|^2g(x)\,\mathrm d\mu(x)=\int_X|f(x)|^2|g(x)|\,\mathrm d\mu(x)$$ implies also that $|f(x)|^2g(x)\geq 0$ for $\mu$-almost every $x\in X$. This, together with $|f(x)|^3=|g(x)|^3$ for $\mu$-almost every $x\in X$, is sufficient to conclude that $|f(x)|=g(x)$ for $\mu$-almost every $x\in X$ (in particular, $g$ is non-negative $\mu$-almost everywhere).


问:
Show y is odd in the equation $y^3 =x^2 +2$
Found a question which asks you to find all integer solutions $y^3 =x^2 +2$ .    In order to do so, it first asks you to show that y is odd.  How would one go about showing this?

Many Thanks

答:
If $y=2k$ then $\color{brown}{x^2=}8k^3+2=\color{brown}{2(4k^3+1)}$.

Is this possible?


答:
If the first number picked was $1$, there are no winning moves.

If the first number picked was $3$, there is one winning move, namely $1$.

In general, if the first number picked was $2k+1$, there are $k$ winning moves.

$k$ ranges from $0$ to $199$, and $0+1+\dots+199=\binom{200}{2}$.


答:
if we randomly picked two numbers, then it must either q
 p

Um,.... what if $p =q$?  Isn't that possible?

dividing 2 here means we rule out the case q<p

Why do you think that there are ${200 \choose 2}$ pairs?  Why not $200^2$ pairs.  Or $200 P 2$ pairs.  Or any other number?

========

${200 \choose 2}$ are number of pairs regardless of position.  But you are chooses specific $p$s and $q$s with regard to order.  SO there are $200 P 2  = \frac {200!}{198!} = 200\cdot 199$ and there are not ${200\choose 2}=\frac {200!}{198!2!}$ ways to choose an ordered (but unequal) pair.

Now a nasty hard-nosed professor will ask you to justify that exactly half of the pairs have $p < q$ and exactly half of the pairs have $p > q$.  Couldn't it be possible that there are more pairs where $p < q$ then there are pairs where $q > q$?  Or fewer?

=======

Alternatively.  There are $200^2$ pairs of of $(p,q)$ ordered pairs.

Either $p < q$.  Or $p = q$.  Or $p > q$.

If $p = q$ then $p$ may be $1,3....,399$ so there are $200$ pairs where $p=q$.

So there are $20^2 - 200 = 200(200-1) =200\cdot 199$ pairs $(p,q)$ where $p \ne q$.

And for every pair $(p,q)$ where the first term is less than the second,  we can map it to a pair $(\rho, \tau) = (q, p)$ where the first term is greater than the second.  Likewise for ever pair $(p,q)$ where the first term is greater we can map it to a pair $(\rho,\tau) = (q,p)$ there the first term is smaller.

So pairs where the first term is less, are in one-to-one cooresponde to the pairs where the first term is more.  So the number of pairs $(p,q)$ where $p < q$ are half of all the pairs $(p,q)$ where $p \ne q$.

So there are $\frac {200\cdot 199}2$ such pairs.

======

Third alternative.  If the second term is $2k+1$ and the first term is $< 2k+1$ then the first term can be anything between $1=2\cdot 1-1$ to $2k-1$.  There are $k$ such pairs.

So the total numbers of such pairs is $\sum_{k=1}^{199} k =\frac {199\cdot 200}2$.

Body:
if we randomly picked two numbers, then it must either q < p or q > p

Um,.... what if $p =q$?  Isn't that possible?

dividing 2 here means we rule out the case q<p

Why do you think that there are ${200 \choose 2}$ pairs?  Why not $200^2$ pairs.  Or $200 P 2$ pairs.  Or any other number?

========

${200 \choose 2}$ are number of pairs regardless of position.  But you are chooses specific $p$s and $q$s with regard to order.  SO there are $200 P 2  = \frac {200!}{198!} = 200\cdot 199$ and there are not ${200\choose 2}=\frac {200!}{198!2!}$ ways to choose an ordered (but unequal) pair.

Now a nasty hard-nosed professor will ask you to justify that exactly half of the pairs have $p < q$ and exactly half of the pairs have $p > q$.  Couldn't it be possible that there are more pairs where $p < q$ then there are pairs where $q > q$?  Or fewer?

=======

Alternatively.  There are $200^2$ pairs of of $(p,q)$ ordered pairs.

Either $p < q$.  Or $p = q$.  Or $p > q$.

If $p = q$ then $p$ may be $1,3....,399$ so there are $200$ pairs where $p=q$.

So there are $20^2 - 200 = 200(200-1) =200\cdot 199$ pairs $(p,q)$ where $p \ne q$.

And for every pair $(p,q)$ where the first term is less than the second,  we can map it to a pair $(\rho, \tau) = (q, p)$ where the first term is greater than the second.  Likewise for ever pair $(p,q)$ where the first term is greater we can map it to a pair $(\rho,\tau) = (q,p)$ there the first term is smaller.

So pairs where the first term is less, are in one-to-one cooresponde to the pairs where the first term is more.  So the number of pairs $(p,q)$ where $p < q$ are half of all the pairs $(p,q)$ where $p \ne q$.

So there are $\frac {200\cdot 199}2$ such pairs.

======

Third alternative.  If the second term is $2k+1$ and the first term is $< 2k+1$ then the first term can be anything between $1=2\cdot 1-1$ to $2k-1$.  There are $k$ such pairs.

So the total numbers of such pairs is $\sum_{k=1}^{199} k =\frac {199\cdot 200}2$.


问:
Evaluate eigenvalues and algebraic multiplicity of an eigenvalue efficiently (with hand)
For
$$
A=\left[\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{array}\right]
$$

Using MATLAB, one can easily show that $\lambda_1 = 0\,\,\,\,\,\&\,\,\,\,\lambda_2 = 5$ are the eigenvalues of the $A$.

However, I would like to do it by hand.

Since $det(A) = 0$, we know that $\lambda_1 = 0$ is an eigen value.

However, I am having difficulty finding the other value. I know if we take $det(A-\lambda I) = 0$, then we get the characteristic equation and can find eigenvalues.

But, I think it is too much to do for such a special matrix.

On the other hand, if I can establish that algebraic multiplicity of $\lambda_1 = 0$ is $4$, then I can say that since $A$ must $5$ eigenvalues (not necessarily different), and $4$ of them being equal to $0$, then the other eigenvalue $\lambda_2 = \sum{a_{ii}} - 4 * \lambda_1 = 5 - 4*0 = 5$.

Is it possible to efficiently compute the algebraic multiplicity of $\lambda_1 = 0$ as $4$ by observing the matrix?

Body:
The columns are eigenvectors
$$
\left(
\begin{array}{rrrrr}
  1  &  -1  &  -1  &  -1  &  -1    \\
  1  &  1  &  -1  &  -1  &  -1    \\
  1  &  0  &  2  &  -1  &  -1    \\
  1  &  0  &  0  &  3  &  -1    \\
  1  &  0  &  0  &  0  &  4    \\
  \end{array}
\right).
$$


问:
Question about the proof of the Banach fixed point theorem
I was working on the following problem:

Problem:
Let f be a continuous function defined on all $\mathbb{R}$ s.t $$|f(x)-f(y)| \leq c|x-y|$$
for c $\in  (0,1)$
. Pick a point $y \in \mathbb{R}$ and construct the sequence $$(y,f(y),f(f(y)),...)$$
show that the sequence is cauchy.

My attempt: I showed that $$|y_n-y_{n+1}|< \frac{c^n}{c}|y_1-y_2|$$ therefore because
$c^n \to 0$ $$|y_n-y_{n+1}| \to 0$$

My question: Why is this not enough, I generalized this to the case where $m \neq n+1$ just to make sure I got it right but, I dont understand why this is not enough for a proof.
According to rudin:

if all we need is for $m > N$ then why is my argument not good enough, since what I showed above was that there exists some $N$ such that when $n>N$ and $m=n+1 > N$ $$|y_n-y_m|=|y_n-y_{n+1}|< \epsilon$$

Thank you in advance!

答:
The condition $|y_n - y_{n+1}|\to 0$ is not enough to deduce $y_n$ is Cauchy as can be seen by considering the following counterexample: $y_n = \log n\to\infty$, but $|y_n - y_{n+1}| = \log(1+\frac1n) \to 0$. So you need to use the explicit rate $|y_n - y_{n+1} | \le ac^n$ where $c<1$. This of course gives for $m>n$
$$ |y_n-y_m| \le |y_{n}-y_{n+1} | + \dots + |y_{m-1} -y_m| \le a \sum_{k=n+1}^m c^k  \le a\sum_{k=n+1}^\infty c^k \to 0   $$


答: To prove that $(y_n)$ is Cauchy you have to find $N$ such that $|y_n-y_m| <\epsilon$ for all $n$ and all $m$ greater than $N$.  You cannot choose $m$ to be $n+1$.

Title: Question about the proof of the Banach fixed point theorem

Body:
I was working on the following problem:

Problem:
Let f be a continuous function defined on all $\mathbb{R}$ s.t $$|f(x)-f(y)| \leq c|x-y|$$
for c $\in  (0,1)$
. Pick a point $y \in \mathbb{R}$ and construct the sequence $$(y,f(y),f(f(y)),...)$$
show that the sequence is cauchy.

My attempt: I showed that $$|y_n-y_{n+1}|< \frac{c^n}{c}|y_1-y_2|$$ therefore because
$c^n \to 0$ $$|y_n-y_{n+1}| \to 0$$

My question: Why is this not enough, I generalized this to the case where $m \neq n+1$ just to make sure I got it right but, I dont understand why this is not enough for a proof.
According to rudin:

if all we need is for $m > N$ then why is my argument not good enough, since what I showed above was that there exists some $N$ such that when $n>N$ and $m=n+1 > N$ $$|y_n-y_m|=|y_n-y_{n+1}|< \epsilon$$

Thank you in advance!


问:
If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$?

If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$?

I'm sure there is some logarithmic rule that can allow me to solve this in one or two steps, but as I am not very familiar with logarithms, I have decided to use the definition of logarithms to convert them to exponential equations.

So from $\log_ax=3$, we have $$x=a^3$$Similarly, from $\log_bx=4$, we have $$x=b^4$$Then I multiply those two equations together:$$\begin{align}x^2&=a^3b^4\\x^2&=(ab)^3\cdot b\\2\log_{ab}x&=3+\log_{ab}b\\\log_{ab}x&=\frac{3+\log_{ab}b}2\end{align}$$The only problem now is the extra $\log_{ab}b$ term. If I can get the value of that, then I will solve the problem. But right now I can't seem to find a way to complete this problem. Can anyone provide some insight on this problem? Thanks.

答: Actually from $x=a^3$ and $x=b^4$ we immediately have $a=b^{4/3}$. Thus $\log_{ab} b = \log_{b^{7/3}}b= \dfrac{3}{7}$. Thus we get $\log_{ab}x=\dfrac{12}{7}$.


Body:
If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$?

I'm sure there is some logarithmic rule that can allow me to solve this in one or two steps, but as I am not very familiar with logarithms, I have decided to use the definition of logarithms to convert them to exponential equations.

So from $\log_ax=3$, we have $$x=a^3$$Similarly, from $\log_bx=4$, we have $$x=b^4$$Then I multiply those two equations together:$$\begin{align}x^2&=a^3b^4\\x^2&=(ab)^3\cdot b\\2\log_{ab}x&=3+\log_{ab}b\\\log_{ab}x&=\frac{3+\log_{ab}b}2\end{align}$$The only problem now is the extra $\log_{ab}b$ term. If I can get the value of that, then I will solve the problem. But right now I can't seem to find a way to complete this problem. Can anyone provide some insight on this problem? Thanks.

Body:
$$a^3=x=b^4\implies b=a^{3/4}$$
Then
$$\log_{ab}b=\log_{a^{7/4}}a^{3/4}=\frac{3/4}{7/4}=\frac37$$
and hence $\log_{ab}x=\frac{12}7$.


问:
Throwing a die 4 times and calculating the number of outcomes
Okay, so this is my problem:
How many outcomes where exactly ONE number occurs exactly TWO times
ex (2, 1, 2, 5) or (3, 6, 6, 1) but NOT (1, 2, 1, 2) or (3, 1, 3, 3)

My solution:
So the first throw has 6 possibilities, then this must be followed by one throw with only ONE possibility, then followed by a throw with 5 possibilities, followed by a throw with 4 possibilities.
Then this gives me: 6 * 1 * 5 * 4 = 120 outcomes

This solution gave me 0 points, so my question is where do I go wrong?

答:
Although a correct solution has been given assuming it isn't a die, and the problem is simple enough to be solved in an elementary way, it would be good for you to know how to solve a bigger die problem in an almost mechanical way as the product of two multinomial coefficients.

Suppose the problem is to find in how many ways a die  rolled $11$ times yields two triples, one double and three singles, ie of the pattern $3-3-2-1-1-1$

Of the $6$ "slots", we need to assign two for the $3s$, one for the $2s$ and three for the $1s$ which is given by the multinomial coefficient $\dbinom{6}{2,1,3}$

And we need to permute the result to account for multiple hits in a slot which is given by the multinomial coefficient $\dbinom{11}{3,3,2,1,1,1}$

so answer = $\dbinom{6}{2,1,3}\dbinom{11}{3,3,2,1,1,1}$

You may not yet have learnt of the various interpretations of the multinomial coefficient, but arithmetically it is essentially permutations with repeated elements, which you must have learnt, so you can simply use $\;\;\dfrac{6}{2!1!3!}\cdot\dfrac{11!}{3!3!2!1!1!1!}$

I hope this is of help to you as you advance into more complex problems of this type.


答:
$x = a^3 \Rightarrow a = x^{1/3}$  and $x = b^4  \Rightarrow b = x^{1/4}$

So, $ab = x^{\frac13 +\frac14} = x^{\frac7{12}} \Rightarrow x = (ab)^{12/7} \Rightarrow \log_{ab}(x) = \frac{12}7$

Title: If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$?


问:
More Questions from Mathematical Analysis by Apostol
I was solving the exercise questions of the book "Mathematical Analysis - 2nd Edition" by Tom Apostol and I came across the questions mentioned below. While I was able to solve a few questions, the others I did not even get any hint of!

1.
(a) By equating imaginary parts in DeMoivre's Formula prove that
$$\sin {n\theta} = \sin^n\theta \left\lbrace \binom{n}{1} \cot^{n - 1}\theta - \binom{n}{3} \cot^{n - 3}\theta + \binom{n}{5} \cot^{n - 5}\theta - + \cdots \right\rbrace$$
(b) If $0 < \theta < \dfrac{\pi}{2}$, prove that
$$\sin{\left( 2m + 1 \right)\theta} = \sin^{2m+1}\theta . P_m\left( \cot^2 \theta \right)$$
where $P_m$ is the polynomial of degree $m$ given by
$$P_m(x) = \binom{2m + 1}{1} x^m - \binom{2m + 1}{3} x^{m - 1} + \binom{2m + 1}{5} x^{m - 2} - + \cdots$$
Use this to show that $P_m$ has zeros at $m$ distinct points $x_k = \cot^2 \left( \dfrac{k\pi}{2m + 1} \right)$ for $k = 1, 2, \dots, m$.
(c) Show that the sum of zeros of $P_m$ is given by
$$\sum\limits_{k = 1}^{m} \cot^2 \dfrac{k\pi}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$
and that the sum of there squares is given by
$$\sum\limits_{k = 1}^{m} \cot^4 \dfrac{k\pi}{2m + 1} = \dfrac{m\left( 2m - 1 \right) \left( 4m^2 + 10m - 9 \right)}{45}$$

Prove that $z^n - 1 = \prod\limits_{k = 1}^{n} \left( z - e^{\dfrac{2ki\pi}{n}} \right)$ for all complex $z$. Use this to derive the formula
$$\prod\limits_{k = 1}^{n - 1} \sin \dfrac{k\pi}{n} = \dfrac{n}{2^{n - 1}}$$

As far as the solutions are concerned, I am through with the 1st part of 1st question and even half of the second part. But, in the second question, proving the zeros and their sum (and the sum of their squares) is getting really difficult. I am not getting any sort of hint as to how to prove it further.

And for the second question, I could do the first half part since it was essentially finding the $n$ roots of unity. But for the second part, I have nearly proved everything but what was asked. Many times I came to the conclusion that $$\prod\limits_{k = 1}^{n} \sin \dfrac{k\pi}{n} = 0$$ which is obvious because at $k = n$, we have a term of $\sin \pi$ which is equal to $0$. I am not getting how to remove that last term from the product using the result we just proved above!

Help will be appreciated!

答:
Way late, but here are hints for the 1st question:

1(b) Having proved that
$$\sin{\left( 2m + 1 \right)\theta} = \sin^{2m+1}(\theta) P_m\left( \cot^2 \theta \right),$$
we observe that the LHS is zero for $\theta_k:=\frac{k\pi}{2m+1}$, $k=1\ldots m$. Check that these $\theta$'s take $m$  distinct values, and lie between $0$ and $\frac\pi2$, so $\sin(\theta_k)>0$ for all $k$, hence $x_k:=\cot^2(\theta_k)$, $k=1,\ldots m$, are points where the polynomial $P_m$ is zero. Since $\cot^2$ is one-to-one on the interval $(0,\frac\pi2)$, conclude the $x_k$ are distinct and therefore they are the $m$ zeros of $P_m$.

1(c) If you've determined that the roots of the $m$th degree polynomial
$$f(x):=a_m x^m + a_{m-1} x^{m-1} +\cdots+a_2 x^2 + a_1x + a_0$$
are $r_1, r_2,\ldots, r_m$, then you have the equality
$$
a_m(x-r_1)(x-r_2)\cdots (x-r_1)=f(x).
$$
Expand out the LHS and match coefficients on $x^{m-1}$ to deduce
$$
r_1 + r_2 +\cdots+r_m = -\frac{a_{m-1}}{a_m}.
$$
This is the first of Vieta's formulas. By matching coefficients on $x^{m-2}$, deduce the second Vieta formula, involving the sum of all products of the form $r_ir_j$ where $i\ne j$. How can you combine these two formulas to obtain $\sum r_i^2$? If you're not seeing it, try playing with these formulas for a few small values of $m$, say $m=3$ or $m=4$.


答:
As $\log_ab=\dfrac{\log b}{\log a}$ when all the logarithms remain defined,

$$3=\log_ax=\dfrac{\log x}{\log a}\implies\log a=?$$

Similarly, $\log b=\dfrac{\log x}4$

$$\log_{ab}x=\dfrac{\log x}{\log a+\log b}$$

Replace the values of $\log a,\log b$ in terms of $\log x$


问:
Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$
Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$

My attempt: multiply both sides by the denominators, and it is equivalent to prove:
$$
n\sum_{k=1}^n{\frac{1}{a_k}}-n\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n+1}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right) \\
\sum_{k=1}^n{\frac{1}{a_k}}-\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n\left( n+1 \right)}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right)\\ \sum_{k=1}^n{\left( \frac{1}{a_k}-\frac{1}{\frac{1}{k}+a_k} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\frac{1}{a_k}}\sum_{p=1}^n{\frac{1}{\frac{1}{p}+a_p}}
\\ \sum_{k=1}^n{\left( \frac{\frac{1}{k}}{a_k\left( \frac{1}{k}+a_k \right)} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\sum_{p=1}^n{\frac{1}{a_k\left( \frac{1}{p}+a_p \right)}}}
$$

But how to proceed further?

答:
I know the question has already been answered, and the answer is quite nice, but I was thinking along a different line:

$$\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)} \geq \frac{2}{n(n+1)}\left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$

Since $\displaystyle \sum\limits_{k=1}^{n}k = \dfrac{n(n+1)}{2}$, we have

$$\left(\sum\limits_{k=1}^{n} k \right)\left(\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)}\right) \geq \left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$

$$\left(1+2+3+ \cdot \cdot \cdot +n\right)\cdot \left(\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)} \right) \geq \\ \left(\frac{1}{a_1}+\frac{1}{a_2}+\cdot\cdot\cdot+\frac{1}{a_n}\right)\cdot\left(\frac{1}{1+a_1}+\frac{1}{\frac{1}{2}+a_2}+\cdot\cdot\cdot+\frac{1}{\frac{1}{n}+a_n}\right)$$

These multiplications can be expanded into sums which can be organized into rows and columns, like an $n \times n$ matrix, say '$B\geq C$'
$$
\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)}\\
\frac{2}{a_1(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{2}{n}+ a_n)}
\\
\vdots\\
\frac{n}{a_1(1+a_1)}+\frac{\frac{n}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}\\
\geq \\
\frac{1}{a_1(1+a_1)}+\frac{1}{a_1(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_1(\frac{1}{n}+ a_n)}\\
\frac{1}{a_2(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_2(\frac{1}{n}+ a_n)}\\
\vdots\\
\frac{1}{a_n(1+a_1)}+\frac{1}{a_n(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}
$$

Now note that the diagonals are equal, so we don't need to worry about them.
However, take for example the opposite entries of the diagonals,

First the entries in the first 'matrix' $B$ are of the form $b_{ij}=\dfrac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}$ while the entries in the second 'matrix' $C$ are of the form $c_{ij}=\dfrac{1}{a_i\left(\frac{1}{j}+a_j\right)}$

then $c_{ij}+c_{ji} \leq b_{ij} + b_{ji}$:

$$\frac{1}{a_i\left(\frac{1}{j}+a_j\right)}+\frac{1}{a_j\left(\frac{1}{i}+a_i\right)} \leq \frac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}+\frac{\frac{j}{i}}{a_i\left(\frac{1}{i}+a_i\right)}\\
a_j\left(\frac{1}{i}+a_i\right)+a_i\left(\frac{1}{j}+a_j\right) \leq \frac{i}{j}a_i\left(\frac{1}{i}+a_i\right)+\frac{j}{i}a_j\left(\frac{1}{j}+a_j\right)\\
\frac{1}{i}a_j+a_i a_j+\frac{1}{j}a_i + a_i a_j \leq \frac{1}{j}a_i+\frac{i}{j}a^{2}_i+\frac{1}{i}a_j+\frac{j}{i}a^{2}_j\\
2ija_i a_j \leq i^2 a^{2}_i+ j^2 a^{2}_j$$

and

$$0 \leq i^2 a^{2}_i - 2ija_i a_j + j^2 a^{2}_j = (ia_i-ja_j)^2$$

the inequalities hold because $a_k>0$ for $k=1,2,...,n$

Therefore the sum of the upper and lower triangular parts of $B$ is $\geq$ to the sum of the of the upper and lower triangular parts of $C$.

Sorry for the extra long answer.


问:
Circular motion trig
We have $x_P = -2 + 4 \cos (-\pi t)$ and $y_P = 1 + 4 \sin ( - \pi t)$ with $t$ in seconds.

We have to find the coordinates of the intersection with the y-axis. So I use trig and I eventually end up with:

$ t = -\dfrac{1}{3} + k \cdot 2$ and $ t = \dfrac{1}{3} + k \cdot 2$

The correction model does this too, and then they just say 'the corrosponding points are ...' and give the 2 points. But what I don't understand is how this can be correct. If we fill in for k, we get 6 points within $[0, 2\pi]$. This would imply 6 intersections, right? What am I missing here?

p.s. - $t$ would be 1/3, 7/3, 10/3, 5/3, 11/3, 17/3

答:
$$\text{If }t=2k+ \frac13,$$

$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right)=-\frac{\sqrt3}2$$

$$\text{If }t=2k- \frac13,$$

$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi-\frac\pi3\right)=-\sin\left(-\frac\pi3\right)=\sin\left(\frac\pi3\right)=+\frac{\sqrt3}2$$

So, the two coordinates of the intersection with the y-axis are $\left(0,1+4\cdot\pm \frac{\sqrt3}2 \right)=\left(0,1\pm2\sqrt3\right)$

Alternatively, as $-1\le \cos(-\pi t)\le 1\implies-3\le x_p\le 2$

If $x_p$ assumes any value in the above range $=c$(say),
$-2+ 4\cos(-\pi t)=c\implies \cos(-\pi t)=\frac{c+2}4$

$$\implies \sin (-\pi t)=\pm\frac{\sqrt{\left(4^2-(c+2)^2\right)}}4=\pm\frac{\sqrt{\left(12-c^2-4c\right)}}4$$

$$\implies y_p=1+4\left( \pm\frac{\sqrt{\left(12-c^2-4c\right)}}4\right)=1\pm \sqrt{12-c^2-4c}$$

So, the intersections will be $\left(c,1\pm \sqrt{12-c^2-4c}\right)$

Title: Circular motion trig


答:
Just remember

$$ \log_vu= \dfrac{\log_p u}{\log_p v}$$

where $p$ can be any arbitrary real number. Writing as power to base

$$ x=a^3 = b^4$$

The given quantity

$$ \dfrac{\log x}{\log a+ \log b}  = \dfrac{\log x}{\log x^\frac13+ \log x^\frac14}  $$

$$ = \dfrac{\log x}{\frac13 \log x+ \frac14\log x} = \dfrac{1}{\dfrac13+\dfrac14} = \dfrac{12}{7}$$


答:
Yes you can. Note that it is a symmetric matrix, there are four linearly dependent row/column vectors i.e. all the row(column) vectors can be linearly expressed by a single row(column) vector. So treating the matrix as a transformation matrix we see that as the row vectors spans only one dimension, the transformed vectors will also span only one dimension. So there is only one non zero real eigenvalue. Rest all eigenvalues has to be zero.

Also for a matrix like above it is easy to see the the eigen vector for the non zero eigenvalue is type $\begin{bmatrix} 1\\1\\.\\.\\1 \end{bmatrix}$ for ex. $\begin{bmatrix} 1 & 2 & 1\\1 & 2 & 1\\1 & 2 & 1\end{bmatrix}\begin{bmatrix} x\\y\\z\end{bmatrix} = \lambda\begin{bmatrix} x\\y\\z\end{bmatrix} $ then

$x+2y+z = \lambda x, x+2y+z = \lambda y, x+2y+z = \lambda z$,

so $x=y=z$. This also independently confirms that there is only one non zero eigenvalue.

If you want more technical view you can see this. What is the relation between rank of a matrix, its eigenvalues and eigenvectors

Title: Evaluate eigenvalues and algebraic multiplicity of an eigenvalue efficiently (with hand)

Body:
For
$$
A=\left[\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{array}\right]
$$

Using MATLAB, one can easily show that $\lambda_1 = 0\,\,\,\,\,\&\,\,\,\,\lambda_2 = 5$ are the eigenvalues of the $A$.

However, I would like to do it by hand.

Since $det(A) = 0$, we know that $\lambda_1 = 0$ is an eigen value.

However, I am having difficulty finding the other value. I know if we take $det(A-\lambda I) = 0$, then we get the characteristic equation and can find eigenvalues.

But, I think it is too much to do for such a special matrix.

On the other hand, if I can establish that algebraic multiplicity of $\lambda_1 = 0$ is $4$, then I can say that since $A$ must $5$ eigenvalues (not necessarily different), and $4$ of them being equal to $0$, then the other eigenvalue $\lambda_2 = \sum{a_{ii}} - 4 * \lambda_1 = 5 - 4*0 = 5$.

Is it possible to efficiently compute the algebraic multiplicity of $\lambda_1 = 0$ as $4$ by observing the matrix?


问:
Reference request: a formula for the prime-counting function
Let $\pi(n)$ denote the prime counting function, which returns the number of primes less than or equal to $n$. When one asks WolframAlpha for $\pi(10,000)$ (or any suitably small number), it displays the result, along with several formulas it uses to calculate this result. The final listed formula caught my eye - WolframAlpha claims that $$\pi(n) = -\sum_{k=1}^{\log_2(n)}\mu(k)\sum_{l=2}^{\lfloor \sqrt[k]{n} \rfloor} \left\lfloor \frac{\sqrt[k]{n}}{l}\right\rfloor \mu(l)\Omega(l)$$ where $\mu(k)$ is the Mobius function and $\Omega(l)$ is the function that gives the number of prime factors counting multiplicities in $l$.

Question: Does this formula have a name? Where is it's correctness proven? Alternatively, I'd be interested in a proof of it's correctness.

答:
(for $s$ large enough)

$$F(s)=\int_1^\infty \sum_l \lfloor y/l\rfloor \mu(l)\Omega(l) y^{-s-1}dy = \sum_l \mu(l) \Omega(l)l^{-s}\int_1^\infty \lfloor y\rfloor y^{-s-1}dy$$ $$=\frac{\zeta(s)}{s}\sum_l \mu(l) \Omega(l)l^{-s}=\frac{1}{s}\sum_n n^{-s}\sum_l \mu(l) l^{-s} \sum_{p | l}1 $$
$$ = \frac1s \sum_p \sum_{n,l, p |l} \mu(l) n^{-s} l^{-s} = \frac1s \sum_p \sum_d d^{-s}\sum_{ p L | d} \mu(pL)$$

If $v_p(d)=0$ then $\sum_{ p L | d} \mu(pL)=0$.

If $v_p(d)>0$ then $$\sum_{ p L | d} \mu(pL)=\sum_{L| dp^{-v_p(d)}}\mu(p)\mu(L)=-1_{d = p^{v_p(d)}}$$

Thus $$F(s)=-\frac1s \sum_p \sum_{r\ge 1} p^{-rs}$$

$$-\int_1^\infty \sum_k \mu(k)\sum_l \lfloor x^{1/k}/l\rfloor \mu(l)\Omega(l) x^{-s-1}dx=-\sum_k \mu(k) F(ks)k $$ $$=  \frac1s\sum_k \mu(k)\sum_p \sum_{r\ge 1}p^{-rks}= \frac1s \sum_p p^{-s}= \int_1^\infty \pi(x)x^{-s-1}dx$$
Which proves your formula $-\sum_k \mu(k)\sum_l \lfloor x^{1/k}/l\rfloor \mu(l)\Omega(l)=\pi(x)$.


问:
Exterior derivative of a coordinate function
I'm starting to learn about differential forms.  From what I understand the coordinate differential forms $dx^1, \dots, dx^n$ are actually the exterior derivatives of the coordinate functions $x^1, \dots, x^n$ (not necessarily Cartesian coordinates).  How is this shown?

The only way I know how to take the exterior derivative of a function ($0$-form) $f$ is to use the formula $df = \frac {\partial f}{\partial x^1}dx^1 + \cdots + \frac {\partial f}{\partial x^n}dx^n$, but applying this to a coordinate function $x^1$ would give $$d(x^1) = \frac{\partial x^1}{\partial x^1}dx^1 + \cdots + \frac{\partial x^1}{\partial x^n}dx^n = dx^1$$ which is wholly unconvincing to me.  If someone showed me this proof I'd think that they'd come up with the method just because they knew what the answer should be.

I think the problem is that I need a definition of the exterior derivative that doesn't already contain the coordinate functions and then I can confirm that for any arbitrary smooth $f$ the exterior derivative $df$ does in fact equal $\frac {\partial f}{\partial x^1}dx^1 + \cdots + \frac {\partial f}{\partial x^n}dx^n$ where $dx^i$ is the exterior derivative of the coordinate function $x^i$.

So what is the rigorous way to show that the things that we expand the $1$-form $df$ in ($dx^1, \dots, dx^n$ in the usual notation) are the same things that we get be taking the exterior derivative of each coordinate function ($x^1, \dots, x^n$)?

答:
If you have an open subset $O \subset \mathbb{R}^n$ and a smooth function $f: O \rightarrow \mathbb{R}$, then the exterior derivative of $f$ at $x \in O$ is defined to be the linear function $df(x): \mathbb{R}^n \rightarrow \mathbb{R}$ such that, given any $v \in \mathbb{R}^n$, $\langle df(x), v\rangle$ is the directional derivative of $f$ in the direction $v$. More precisely, if $c: (-\delta, \delta) \rightarrow M$ is any smooth curve such that $c(0) = x$ and $c'(0) = v$, then
$$ \langle df(x), v\rangle = \left.\frac{d}{dt}\right|_{t=0}f(c(t)) $$
If you apply this definition to the functions $x^i: O \rightarrow \mathbb{R}$, then it is easy to check that given any smooth function $f$, the formula
$$
df = \frac{\partial f}{\partial x^1}\,dx^1 + \cdots + \frac{\partial f}{\partial x^n}\,dx^n
$$
So this formula is not a definition but a consequence of the definitions of $x^1, \dots, x^n$ as functions and the definition of the exterior derivative of a function I wrote above.


答:
Just go back to definitions.

$\log_a x = 3\iff a^3 = x$.

And $\log_b x = 4 \iff b^4=x$.

So $a = x^{\frac 13}$ and $b=x^{\frac 14}$.

So $ab = x^{\frac 13}x^{\frac 14}= x^{\frac 13 + \frac 14}$.

So $(ab)^{\frac 1{\frac 13 + \frac 14} } = x$.

Which means $\log_{ab} x = \frac 1{\frac 13 + \frac 14}$.


问:
Solving the domain and range of a region satisfying two inequalities?
The question I was provided was:

"Find the domain and range of the region satisfied by the following inequalities:
i) $y \ge (x-1)^2$
ii)$y \le2x+1$

Any help would be greatly appreciated. Would you recommend graphing or solving algebraically?

答: I am giving you a very basic way to find out the regions graphically. For $$y\ge(x-1)^2$$ Note that $y=(x-1)^2$ is a polynomial which has $\mathbb R$ as domain and obviously $\mathbb R_{\ge 0}$ as its range. To find out what that inequality tells you, you can pick two points in and out of the region the parabola made in $\mathbb R^2$. As you see $P$ is in and $Q$ is out with respect the parabola. Now satisfy the coordinates of $P$ into the inequality. You see $$(1/2-1)^2=1/4$$ and it is smaller than $2$. The same for $Q$ tells us $$(3-1)^2=4$$ is greater than $1$. So, the desired region satisfying $$y\ge(x-1)^2$$ is the parabola an the region which is enclosed by it.


问:
Get lower bound using the chebyschev's inequaility.
I tried to solve a problem about chebyschev's inequality, but I don't know where i got wrong!

Here's the problem :

Find a lower bound for the probability that the average number shown on a die in 1000 tosses differs from 3.5 by 0.01.

And I attach the image that show how I solved.

the problem is that I got negative value for lower bound!
It makes sense since the probability is between [0,1], but the solution tells the correct answer is 0.29.

Please give me some advice!

答:
The original inequality tells that
$$
\mathbb{P}[|X-\mu|\ge k \sigma] \le \frac1{k^2}
\implies
\mathbb{P}[|X-\mu|< k \sigma] \ge 1-\frac1{k^2}
$$
Here, $X = \sum_{k=1}^{1000} X_k/1000$ where $X_k$ denotes the $k$th die roll. It's easy to see that $\mu = \mathbb{E}[X] = \mathbb{E}[X_k] = 3.5$ as needed for our problem. Last step is to determine the standard deviation.
$$
\sigma
 = \mathbb{Var}(X)
 = \frac{\mathbb{Var}(X_k)}{1000}
 = \frac{35/12}{1000}
 = \frac{7}{6000}.
$$
So your bound uses $k\sigma = 0.01 \iff k = 0.01/\sigma$ and we get
$$
\mathbb{P}[|X-3.5|\ge 0.01]
 \le \frac1{(0.01/\sigma)^2}
 = (100 \sigma)^2
 = \left(100 \cdot \frac{7}{6000} \right)^2
 = \left(\frac{7}{60} \right)^2
 = \frac{49}{3600}.
$$
Hence,
$$
\mathbb{P}[|X-3.5|< 0.1] \le 1-\frac{49}{3600} \approx 98.64 \%.
$$


问:
Find $\lim\limits_{n\rightarrow \infty} \left| \ln \frac{1}{\left| x_n-A \right|} \right|^{1/n}$
The sequence $\{x_n\}$ satisfies $x_{n+1}=x_n+\sin x_n,n=1,2,\cdots $.

Prove $\{x_n\}$ is convergent and find its limit $A$.

Let $x_1\neq A$. Find the limit $$\lim_{n\rightarrow \infty} \left| \ln \dfrac{1}{\left| x_n-A \right|} \right|^{1/n}$$

My Attempt: For the first question, let $k\in \mathbb Z$ and $f(x)=x+\sin x$. When $x=k\pi$, it is trivial that $A=k\pi$. When $x\in (k\pi,(k+1)\pi)$, since $f(x)\in (k\pi,(k+1)\pi)$ and $f(x)$ monotonously increases on $\mathbb R$, obviously $\{x_n\}$ monotonously increases and is bounded from above. Take the limits of both sides of the recurrence, and we get $A=A+\sin A$, so $A=(k+1)\pi$.

My Question is, did I correctly do the first question? And how to solve the second one? I don't have any ideas yet.

答: To solve the second question, use the fact that for small $\delta$, $\sin(\pi-\delta)=\sin(\delta)\approx \delta-\frac{\delta^3}{6}$. Thus, starting at $\pi-x_1=\delta$ for small positive $\delta$, we obtain $\pi-x_2\approx\frac{\delta^3}{6}$, $\pi-x_3\approx\frac{\left(\frac{\delta^3}{6}\right)^3}{6}=\frac{\delta^9}{6^4}$, $\pi-x_4\approx\frac{\left(\frac{\delta^9}{6^4}\right)^3}{6}=\frac{\delta^{3^3}}{6^{3^2+3+1}}$ and so on. So $\pi-x_n\approx\frac{\delta^{3^n}}{6^{\frac{3^n-1}{2}}}$. You can make the approximations rigorous using Taylor series expansions, and show that the errors don't accumulate. Then you can solve for $\log(\pi-x_n)$ and $|\log(\pi-x_n)|^{1/n}$. By taking $n \to \infty$, you'll notice that it doesn't matter where $x_1$ is in the range $(0,\pi)$.


问:
Show $se(3)$ is the tangent space of $SE(3)$ the matrix group of $3D$ rigid motions

$se(3)=\begin{pmatrix}A&c\\0&0\\\end{pmatrix}:A\in so(3), c\in \mathbb{R}^3\}$

Show that $se(3)$ is the tangent space of $SE(3)=\begin{pmatrix}A&c\\0&1\\\end{pmatrix}:A\in SE(3), c\in\mathbb{R}^3\}$

$so(3)$ is the set of skew symmetric matrices.

I have a definition for the tangent space as $T_p(M)=\{X\in M:\exists \gamma:(-\epsilon,\epsilon)\to M\text{ which is differentiable and } \gamma(0)=p, \gamma'(0)=X\}$

But I'm not sure how I would show $se(3)$ would satisfy this for $SE(3)$

答:
I'll assume you meant $A\in{\rm SO}(3)$ in your definition of ${\rm SE}(3)$.

Anyway, fix an element
$$M = \begin{bmatrix}A_0 & c_0 \\ 0 & 1\end{bmatrix}$$
in ${\rm SE}(3)$.

Let $\gamma(t)$ be a differentiable path in ${\rm SE}(3)$ defined near $t=0$ with $\gamma(0)=M$ and $\gamma'(0) = X$ for some matrix $X$. We see that $\gamma$ must have the form
$$\gamma(t) = \begin{bmatrix}A(t) & c(t) \\ 0 & 1\end{bmatrix}$$
where $A(0)=A_0$ and $c(0)=0$. Thus
$$X = \gamma'(0) = \begin{bmatrix}A'(0) & c'(0) \\ 0 & 0\end{bmatrix}.$$
Since $A(t)$ was a differentiable path in ${\rm SO}(3)$, $A'(0)\in T_{A}{\rm SO}(3) = {\rm so}(3)$, the set of skew-symmetric matrices. The vector $c'(0)$ is an element of $\mathbb{R}^3$, so we conclude $X\in {\rm se}(3)$. Thus $T_M{\rm SE}(3)\subset {\rm se}(3)$. On the other hand, the dimension of ${\rm SE}(3)$ and hence its tangent space is $\dim{\rm SO}(3)+3 = 3+3=6$ while the dimension of ${\rm se}(3)$ is also $3+3=6$, so we get equality $T_M{\rm SE}(3) = {\rm se}(3)$.


问:
Spectra of matrices $A$ and $A^\top A$
For a square matrix $A$, I see that $\lambda$ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^2$ since $A x = \lambda x \Rightarrow A^2 x = A(\lambda x) =\lambda^2 x$. But for a non-symmetric $A$, can we say anything about the spectrum of $A^\top A$ relating to the spectrum of $A$? For example, if $\lambda_{\max}$ is the maximum eigenvalue of $A$, will the spectrum of $A^\top A$ be bounded by $\lambda_{\max}^2$ somehow (and vice versa)? I ask this because of a remark (page 4, footnote 3) in this paper.

In the footnote, it claims that the eigenvalues of $(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)$ are less than 1 since the eigenvalues of $(W W^\top - \frac{1}{n} {\bf 1}{\bf 1}^\top)=(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)^\top$ are less than 1. Their claim seems to be correct, but I don't see the rationale of that.

答:
This is a bit complicated. You have misunderstood the authors' intent. Meanwhile, the authors have overlooked the fact that $W$ is symmetric. Although one can simply consider $W-\frac1n\mathbf1\mathbf1^T$, the authors had opted to consider $(W-\frac1n\mathbf1\mathbf1^T)(W-\frac1n\mathbf1\mathbf1^T)^\top$ unnecessarily.

First, you wrote

In the footnote, it claims that the eigenvalues of $(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)$ are less than 1 since the eigenvalues of $(W W^\top - \frac{1}{n} {\bf 1}{\bf 1}^\top)=(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)^\top$ are less than 1.

That wasn't what the authors meant. What they wanted to prove is not that the eigenvalues of $A=W-\frac{1}{n}\mathbf1\mathbf1^\top$ are less than $1$ in size, but that the spectral norm of $A$ is less than $1$. In other words, they wanted to prove that the eigenvalues of $AA^\top$ are strictly less than $1$.

However, by definition (see sec. II A on p.3), the doubly stochastic matrix $W\in\mathbb R^{n\times n}$ in the paper is a weight matrix of an undirected connected graph $E$ such that $w_{ii}>0$ for each $i$. Hence it is necessarily symmetric and irreducible. The eigenvalues of $AA^\top$ are just the squares of the eigenvalues of $A$. So, it does suffice to prove that the eigenvalues of $A$ are less than $1$ in sizes. Although you have misunderstood the authors' intent, your (mis)interpretation of their strategy also works.

At any rate, since $W$ is a symmetric irreducible doubly stochastic matrix with a positive diagonal, so is $WW^\top$. (In general, $WW^T$ is not necessarily irreducible, even if $W$ is primitive, but in our case, $W$ is an irreducible matrix with a positive diagonal. Therefore $WW^T$ is primitive and hence irreducible.) We want to show that all eigenvalues of $S-\frac1n\mathbf1\mathbf1^\top$ are less than $1$ in sizes when $S=WW^\top$ (the authors' intent) or when $S=W$ (your interpretation).

Since $S$ is irreducible and doubly stochastic, by Perron-Frobenius theorem, $1$ is a simple eigenvalue of $S$ and the moduli of all other eigenvalues of $S$ are strictly less than $1$. Thus we can arrange the eigenvalues of $S$ as $\lambda_1(\,=1)>\lambda_2\ge\cdots\ge\lambda_n\,(>-1)$. Since $S$ is also symmetric, we can pick an orthogonal eigenbasis $\{v_1=\mathbf1,v_2,\ldots,v_n\}$ with $Sv_i=\lambda_iv_i$. Then $\mathbf1^\top v_j=v_i^\top v_j=0$ when $j\ge2$ and
\begin{cases}
(S-\frac1n\mathbf1\mathbf1^\top)v_1=S\mathbf1-\mathbf1=0,\\
(S-\frac1n\mathbf1\mathbf1^\top)v_j=Sv_j=\lambda_jv_j\text{ when } j\ge2.
\end{cases}
Hence the eigenvalues of $S-\frac1n\mathbf1\mathbf1^\top$ are $0,\lambda_2,\ldots,\lambda_n$, which are strictly less than $1$ in moduli.

Title: Spectra of matrices $A$ and $A^\top A$


问:
How is the general formula for the square of a polynomial reached at in this paper?
In the following paper in equation 104, we have the following general formula for the square of a polynomial:

https://dspace.mit.edu/bitstream/handle/1721.1/106948/Roy_Aggresive%20flight.pdf?sequence=1&isAllowed=y

But on this Stack Exchange answer, the square of a polynomial has a different general formula.

General square form of Taylor expansion polynomial

Which ones correct?

Can you explain how the convolution in the paper is reached at?

答:
$\left(P^2\right)_n$ denotes the coefficient on $x^n$ of $P(x)^2$. Applying the linked answer, $$P^2=\left(\sum_{k=0}^Np_kx^k\right)^2 = \sum_{k=0}^N\sum_{j=0}^N p_kx^kp_jx^j$$

The $x^k \cdot x^j$ can be combined as $x^{k+j}$ and instead of summing over $k$ and $j$, you can sum over $n = k+j$ and $j$ to get $$\sum_{n=0}^{N}x^n \sum_{j=0}^Np_jp_{n-j}$$

The coefficient on $x^n$ is then $\sum_{j=0}^Np_jp_{n-j}$.


问:
Is $\frac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ a consequence of the geometrical definition of $\sin\theta \equiv \frac{opp}{hyp}$?

Is $\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ a consequence of the geometrical definition of $\sin\theta \equiv \dfrac{opp}{hyp}$?

I do not know whether $\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = 1$ is an additional constraint on the sine function or not.

答:
$$\dfrac{d \sin\theta}{d \theta} \big|_{\theta=0} = \lim _{h \to 0} \dfrac{\sin(0+h) - \sin(0)}h
\\ \;  \\ \;
\\= \lim _{h \to 0} \dfrac{\sin(h) }h
$$
The geometric definition of $\sin(h)$ allows us to set up the inequality
$$| \sin(h)| \le |h|\le |\tan(h)|  $$
($h$ is in radians)
now dividing through by $| \sin(h)|$ ...
$$1\le \bigg | \dfrac{ h}{ \sin(h)}  \bigg |  \le |\sec(h)|  $$
So by the squeeze theorem we have
$$\lim _{h \to 0} \bigg | \dfrac{ h}{ \sin(h)}  \bigg | =1$$
Hopefully you can take it from here


答:
It is obvious that $A$ has $\mbox{rank}(A)=1$. Therefore, by the rank-nullity theorem, the Nullspace of $A$ has dimension $4$.

This means that the eigenspace for the eigenvalue $\lambda=0$ is 4-dimensional.

It is a fundamental result of Linear algebra that the multiplicity of an eigenvalue is greater or equal to the dimension of the eigenspace. [This is typically stated as "algebraic multiplicity" $\geq$ "geometric multiplicity"]

This means that $\lambda=0$ has multiplicity $4$.

It follows that $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$.

Finally,
$$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5= \mbox{tr}(A)=5$$

This gives $\lambda_5=5$.


问:
Why is it okay to take out the constant when finding asymptotes or hyperbola?
$\require{cancel}$I have been tasked to find the equation(s) of the asymptotes for the graph of the following hyperbolic equation:$$16x^2-25y^2-400=0$$My teacher explains that you can just take out the constant term to find the asymptotes. For this specific case:$$16x^2-25y^2\cancel{-400}=0\implies16x^2-25y^2=0$$I have graphed both $16x^2-25y^2-400=0$ and $16x^2-25y^2=0$ and confirmed that $16x^2-25y^2=0$ indeed does graph the asymptotes of $16x^2-25y^2-400=0$.

Can someone provide a simple explanation as to why this is the case?

答:
Here's what's happening

The equation of the hyperbola in standard form is

$$b^2x^2 - a^2y^2 -a^2b^2 = 0$$

Divide by $x^2$

$$b^2 - a^2\frac{y^2}{x^2} - \frac{a^2b^2}{x^2} = 0$$

Now, as a point on the hyperbola approaches infinity, we have the following as the curve is continuous

$$\lim_{x \to \infty} \left(b^2 - a^2\frac{y^2}{x^2} - \frac{a^2b^2}{x^2}\right) = 0$$

The constant term will go to zero as x grows large, therefore we have

$$\lim_{x \to \infty} \left(\frac{y}{x}\right) = \pm\frac{b}{a}$$

Hence we have the pair of lines that are asymptotes to the hyperbola are described the pair of lines equation obtained by removing the constant from the equation. This only works in standard form however, and will have toappropriately have to be translated rotated for the general case


Body: $$\log_{ab} x = \frac{1}{\log_x ab} = \frac{1}{\log_x a + \log_x b} = \frac{1}{\frac{1}{\log_a x} + \frac{1}{\log_b x}} = \frac{1}{\frac{1}{3} + \frac{1}{4}} = \frac{12}{7}$$


问:
Are the proofs of Caratheodory's extension theorem and Caratheodory's lemma typically included in a graduate probability class?
By graduate probability class I mean the typical "Probability I" required by statistics PhD programs.

Both Caratheodory's extension theorem and Caratheodory's lemma are crucial for the construction of probability spaces. However, as stated in Williams' Probability With Martingales, page 20: "once we have our model, we have no use of the theorem", it seems to me that there are no significant use of both theorems in the later part of such class. In Amir Dembo's notes on probability theory, he also stated Caratheodory's extension theorem without a proof.

If the proofs are omitted, more time can be used on topics like weak convergence. Any thoughts?

Thanks in advance.

答: It depends on what one wants out of that education. For either pure or applied statistical work you don't need it. But if you are likely to need to do some heavy probability theory then it is useful to have a good understanding of the measure theoretic background. It likely isn't necessary to teach it in a statistics program but it is much more important in a mathematics program.</p>


问:
Finding matrices whose multiplication does not change the trace
Let $C$ be an $n\times n$ matrix. Then find matrices $A$ of size $n \times n$ such that $trace(AC)=trace(C)$. One obvious choice for $A$ is the Identity matrix $I$, but are there other choices?

A similar question is, if $A$ and $C$ are $m\times m$ matrices and $B$ and $D$ are $n\times n$ matrices such that $trace(C)=trace(D)$ and $trace(AC)=trace(BD)$, then what is the explicit relationship between $A$ and $B$? Here also we can choose $A$ and $B$ to be Identity matrices, but are there other choices of matrices?

答:
The answer to this question is "find it with this possible hint" Choose $A-I=B$ a non invertible matrix such that $BC=0$ then $det(A-I)=0$.
This is answer to the first question.
For 2nd question try this, might work.
$$
A=C=\left[\begin{array}{ll}
I_{n} & O\\
O & O \\
\end{array}\right]
$$
and $B=D=I_{n}$


问:
Integrate: $\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz$

Q. Show that : $$\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz = \frac{\sin pt}{p}$$

I considered the following contour

$$\int_\Gamma \frac{e^{tz}}{z^2 + p^2}dz + \int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz  = 2 \pi i (\text{Residue}[f(z), z=ip] + \text{Residue}[f(z), z=ip]) \; \; \; \;\; \;(1)$$
The first integral goes to zero.
$$\text{Residue}[f(z), z=ip]  + \text{Residue}[f(z), z=-ip]= \frac{\sin (pt)}{p} \hspace{1cm} (2)$$

From $(1)$ and $(2)$ we get
$$\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz = 2 \pi i \frac{\sin pt}{p}$$

Did I make any mistake or (answer according to) question is wrong?

答:
I do not use the contour shown in the figure.  I prefer going, from the top of the vertical line at $a+i R$, to the left horizontally to $z=i R$, and then going in a semicircular arc between $\theta \in [\pi/2,3 \pi/2]$.  In this case, the integral over the horizontal portion to the left goes as

$$\frac{e^{i R t}}{R^2-p^2} \frac{1-e^{a t}}{t} $$

as $R \to \infty$.  We get a similar expression for the horizontal segment at $z=-i R$.

It then becomes crystal clear that the integral along the circular arc vanishes, because the integral there looks like

$$i R \int_{\pi/2}^{3 \pi/2} d\theta e^{i \theta} \frac{e^{R t \cos{\theta}} e^{i R \sin{\theta}}}{R^2 e^{i 2 \theta}+p^2}$$

The magnitude of this integral vanishes as $R \to \infty$ because the cosine term is negative over the integration region. So, at least along this contour, the integral vanishes.


问:
Division of a polynomial in three variables
I would like to demonstrate the following

Let $f(x,y,z) \in K[x,y,z]$ a polynomial such that $f(a^n,b^n,ab)=0$; then f is divisible by $xy-z^n$.

Hint: use a polynomial division algorithm

EDIT
At this link Showing that $\mathbb{C}[x,y]^{\mu_n}$ and $\mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic as rings
I found the same problem. The answer suggest:
$$ $$
If $f(x,y,z)=\sum_{i,j,k} a_{i,j,k}x^iy^jz^k$ so
$$0=f(u^n,v^n,uv)=\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$$
The equation $\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$ give some dependence relation amongst the $a_{i,j,k}$ (since the exponent of $u$ and $v$ are the same for some $i,j,k$ and since the element $u^av^b$ are indipendent over $K[x,y]$.

Can anyone tell me what are these relations between coefficients
$a_{i,j,k}$ they are talking of? How do they obtain? How do these
allow me to prove that $f(x,y,z)=g(x,y,z)(xy-z^n)$? TO PLEASE SOMEONE
CAN HELP ME?

答:
I'm going to assume that $K$ is a field and $K[x, y, z]$ is the associated ring of polynomials. Unfortunately, I don't see a polynomial division solution to this...

The problem is equivalent to showing that $f(x, y, z)$ is an element of the $0$ coset of the quotient ring $K[x, y, z]/ (xy - z^n )$. Let's first consider $K[a^n, b^n, ab]$. Observe that there is a surjective homomorphism $\varphi : K[x, y, z] \rightarrow K[a^n, b^n, ab]$ where $(x, y, z) \mapsto (a^n, b^n, ab).$ Moreover, $\ker (\varphi) = (xy - z^n)$ (show why). Then by the First Isomorphism Theorem for rings, it follows that
$$K[x, y, z] / (xy - z^n) \cong K[a^n, b^n, ab].$$
(can you give the isomorphism explicitly?) The conclusion that $f(x, y, z) \in 0 + (xy - z^n)$ follows.

This form shows up in some fairly interesting results.


问:
Limit of a Logarithmic Function
Consider a function $f(x)=\lim_{x \to \infty} \mathrm{log}_T(x^a+x^b)$ for $x>1$ and $a,b$ are real numbers satisfying $a>b$. Find:

$\lim_{x \to 1+} \mathrm{log}_x(x^a+x^b)$

$\lim_{x \to \infty} \mathrm{log}_x(x^a+x^b)$

For the first part, I tried writing converting it to natural logarithm and then apply the limit. But it just diverges. Not sure if it is correct.

$\mathrm{log}_x(x^a+x^b)=\frac{\ln(x^a+x^b)}{\ln x}$.

The numerator tends to $\ln 2$ but the denominator tends to 0.
Any help is appreciated. Thanks!

答:
Hints: for the first part the answer is $+\infty$.

For the second part you have to consider various cases. For example, if $a>b>0$, then you can apply L'Hopital's Rule to get $\lim_{x \to \infty} \frac {ax^{a}+bx^{b}} {x^{a}+x^{b}}$. Divide numerator and denominator by $x^{a}$ to see that the limit is $a$.

Title: Limit of a Logarithmic Function


问:
How find MVUE for $g( \theta)$?
let $X=(x_1,...,x_n)$ ,$n>1$ random sample of $B(1, \theta)$ find MVUE of $g( \theta)=P_ \theta(x_1\geq x_2 )$

i did something like that:

$\prod_{i=1}^{n}f_x(x;\theta )=e^{\sum_{i=1}^{n}x_iln(\theta )+\sum_{i=1}^{n}(1-x_i)ln(1-\theta )}$

and that tells me that there exist a unique MVUE but i dont know how to find $T(X)$ that  such that $E(T(X))=g(\theta)$

i am a bit confused can someone explain  how to do it?

答:
One easy unbiased estimator is the following:

$$T=\mathbb{1}_{\{X_1\geq X_2\}}$$

You surely know that $S=\sum_{i=1}^n X_i$ is complete and sufficient, thus you can use Rao - Blackwell and Lehmann - Scheffé together finding that the UMVUE is

$$\mathbb{E}[T|S=s]=\mathbb{P}[T=1|S=s]=\frac{ \binom{n-2}{s} + \binom{n-2}{s-1}+ \binom{n-2}{s-2} }{\binom{n}{s}}=1-\frac{\binom{n-2}{s-1}}{\binom{n}{s}}=1-\frac{s}{n}\cdot\frac{n-s}{n-1}$$

The unbiased estimator I showed is not the only one you can use..I leave to you as an exercise to find the other obvious and more suitable unbiased estimator for $g(\theta)$

Title: How find MVUE for $g( \theta)$?


答:
Asymptotes are essentially the lines passiing through the center of  the hyperbola. So if $$ax^2+by^2+2hxy+2gx+2fy+c=0, h^2>ab~~~~(1)$$ represents a hyperola.
Then by changing $c$ to $c'$, we demand the same quadratic to represent pairs of straight lines such that $\Delta'$ is zero namely
$$\Delta'=2fgh+abc'+af^2+bg^2+c'h^2=0$$ then
$$ax^2+by^2+2hxy+2gx+2fy+c'=0~~~~(2)$$
will be the combines equation of the asymptotes which can be separated out. The question why to change constant only, can be answered by noting that
if hyperbola, (1) can be written as  $(lx+my+n)(px+qy+r)=\pm s^2 \implies XY=\pm s^2$ represents a hyperbola then putting the constant $s=0$, $(lx+my+n)(px+qy+r)=0$, will represent the asymptotes.

Note that $L_1 L_2=\pm s^2$ is the most general equation of a hyperbola, where $L_1$ and $L_2$ are equations of two non-parallel lines.

Title: Why is it okay to take out the constant when finding asymptotes or hyperbola?

Body:
$\require{cancel}$I have been tasked to find the equation(s) of the asymptotes for the graph of the following hyperbolic equation:$$16x^2-25y^2-400=0$$My teacher explains that you can just take out the constant term to find the asymptotes. For this specific case:$$16x^2-25y^2\cancel{-400}=0\implies16x^2-25y^2=0$$I have graphed both $16x^2-25y^2-400=0$ and $16x^2-25y^2=0$ and confirmed that $16x^2-25y^2=0$ indeed does graph the asymptotes of $16x^2-25y^2-400=0$.

Can someone provide a simple explanation as to why this is the case?


问:
Solve a simple equation with log in it
I'm stuck with solving this equation,

$$2 \log x = \log 9 $$

This is how far I made it:

\begin{align}
\log x &= \log 4,5 \\
x &= ?
\end{align}

I'm a beginner at logarithms so I appreciate ways to solve it and not just an answer.

答:
Note that $\frac{1}{2}\log{9}\ne\log\frac{9}{2}$. Instead $\frac{1}{2}\log{9}=\log{9}^{1/2}=\log{3}$ - constants become powers when you take them inside the $\log$.

Once you're in the situtation where you have the equation $\log{x}=\log{y}$, then taking exponentials of both sides gives you $x=y$.

Title: Solve a simple equation with log in it


答:
Yes, thanks for the reply. I see my problem now. I have mistaken the definition of spectral norm of a matrix as the maximum eigenvalue, but it is the maximum singular value instead. So everything works perfectly. Btw, I don't think $W$ needs to be symmetric though. But it needs to be doubly stochastic and irreducible.

Since a product of stochastic matrices is stochastic, we have $W^n$ to be stochastic for any $n$. So its eigenvalue is bounded by 1 (otherwise, we have $\lim_{n\rightarrow \infty} W^n x = \lambda^n x$, where elements on the LHS are bounded and that on the RHS can be unbounded and this leads to a contradiction). And the eigenvector with eigenvalue 1 exists, as it is just the all-one vector $\bf 1$.

Now, since $W W^\top$ is also stochastic, it also has maximum eigenvalue of 1. Moreover, since $W$ is doubly stochastic, we have $W W^\top {\bf 1} = W ({\bf 1}^ \top W)^\top= W {\bf 1}= 1 \cdot {\bf 1}$. So the eigenvector of $W W^\top$ with eigenvalue 1 is also $\bf 1$. Since we also assume $W$ to be irreducible,

(Update: the irreducibility of $WW^\top$ is not guaranteed even if $W$ is irreducible as many here have kindly pointed out. So I guess we have to assume $W W^\top$ to be irreducible here.)

by Perron-Frobenius Theorem, the eigenvector with the maximum eigenvalue is distinct. Therefore, we have $W W^\top - 1 \frac{{\bf 1}}{\sqrt{n}} \frac{{\bf 1}^\top}{\sqrt{n}}=W W^\top -\frac{{\bf 1 1}^\top}{n}$ (with the largest component removed) having the its eigenvalues less than 1.

And as $\underset{A}{\underbrace{(W-\frac{{\bf 1 1}^\top}{n})}}(W-\frac{{\bf 1 1}^\top}{n})^\top =W W^\top -\frac{{\bf 1 1}^\top}{n}$, the maximum singular value of $A$ has to be less than 1 since the set of singular value of $A$ is equal to the set the square rooted eigenvalues of $A A^\top$.

Title: Spectra of matrices $A$ and $A^\top A$

Body:
For a square matrix $A$, I see that $\lambda$ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^2$ since $A x = \lambda x \Rightarrow A^2 x = A(\lambda x) =\lambda^2 x$. But for a non-symmetric $A$, can we say anything about the spectrum of $A^\top A$ relating to the spectrum of $A$? For example, if $\lambda_{\max}$ is the maximum eigenvalue of $A$, will the spectrum of $A^\top A$ be bounded by $\lambda_{\max}^2$ somehow (and vice versa)? I ask this because of a remark (page 4, footnote 3) in this paper.

In the footnote, it claims that the eigenvalues of $(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)$ are less than 1 since the eigenvalues of $(W W^\top - \frac{1}{n} {\bf 1}{\bf 1}^\top)=(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)(W-\frac{1}{n} {\bf 1} {\bf 1}^\top)^\top$ are less than 1. Their claim seems to be correct, but I don't see the rationale of that.


问:
Why do we divide by standard deviation when standardizing a normal distribution?
We have this random variable $Y= \frac{x - μ}{\sigma}$ to convert a normal distribution $N(\mu, \sigma)$ to a $N(0, 1)$. It is quite intuitive to subtract $\mu$, since you move all the values in the $x$-axis, and thus move the mean $\mu$ to the origin of coordinates. But it does not seem intuitive to divide by the standard deviation.

This answer is basically this, however, I have not understood the answers, specifically this:

$$E[Y] = \frac{E[X] - \mu}{\sigma} = \frac{\mu-\mu}{\sigma} = 0.$$

$$\text{Var}(Y) = \frac{1}{\sigma^2}\text{Var}(X) = \frac{1}{\sigma^2}\sigma^2 = 1.$$

And I would also like to get an intuitive answer.

答:
Another method could be by understanding its reverse.

Say the times taken by students to get up in the morning can be modelled by a normal distribution with mean 20.0 minutes and standard deviation 3.0 minutes.

And label those who take more than 1.5 standard deviations from the mean as "lazy"
So, how late is "lazy" exactly?

```<code>                 (1.5 x 3) + 20 = 24.5 minutes
</code>```

You get it? You just reverse-coded(?) the coding. So in order to get back to the z format, you need to minus the mean and divide by the standard deviation

Hope this helps. :D

Body:
Another method could be by understanding its reverse.

Say the times taken by students to get up in the morning can be modelled by a normal distribution with mean 20.0 minutes and standard deviation 3.0 minutes.

And label those who take more than 1.5 standard deviations from the mean as "lazy"
So, how late is "lazy" exactly?

                 (1.5 x 3) + 20 = 24.5 minutes

You get it? You just reverse-coded(?) the coding. So in order to get back to the z format, you need to minus the mean and divide by the standard deviation

Hope this helps. :D