问:
Solving $\tau^4 +1=u^2 +v^2$ over the integers
I'm trying to show, that the solutions $t_i$ from Pell's equation $t^2-2s^2=\pm 1$ are not squares (only for the trivial cases). For the even solutions $t_{2i}$ (solutions for the + sign) this is easy, but for the odd solutions $t_{2i+1}$ (solutions for the - sign) this is a little bit harder.

This question leads to the diophantine equation $\tau^4-2s^2=-1$ and the new question, whether it has nontrivial solutions or not.

I do the following approach. First of all I take the equation
$\tau^4+1=u^2+v^2$ (*),
try to find solutions $(u, v, \tau)$ and set in the end $s=u=v$. With

$\begin{align}
u&=u_2 a^2 +u_1 a+u_0 \\
v&=v_2 a^2 +v_1 a+v_0 \\
\tau&=\tau_1 a+\tau_0
\end{align}$

and comparison of the coefficients one gets

$\begin{align}
u&=2a^2 b^2 (b^4 +1)+4ab^3 +1 \\
v&=a^2 (b^8-1)+2ab(b^4 -1)+b^2 \\
\tau&=a(b^4 +1)+b
\end{align}$

with natural numbers $a$ and $b$.

Now the question: are there any other solutions for (*) and if not, why?
Maybe there is also an easy way showing, that the $t_{2i+1}$ are not squares, then I would prefer this proof.

thank you

答:
I'll assume your parametrization attempts to cover all solutions up to sign, and up to switching $u$ and $v$, because in your parametrization clearly $u$ and $v$ are both nonnegative, and $u$ is odd. Even then the parametrization is incomplete; there are no natural numbers $a$ and $b$ corresponding to the solution $(\tau,u,v)=(8,31,56)$.

Here's the first approach that springs to my mind; it's not necessarily much easier, and the final step is a bit unsatisfying, but it's certainly too much for a comment:

If $\tau$ and $s$ are integers such that $\tau^4-2s^2=-1$ then clearly $\tau$ is odd, and
$$-\tau^4=1-2s^2=(1+s\sqrt{2})(1-s\sqrt{2}).\tag{1}$$
This suggests an argument in $\Bbb{Z}[\sqrt{2}]$, which is a UFD with fundamental unit $1+\sqrt{2}$.  Indeed the two factors on the right hand side are coprime, so both are fourth powers up to units. Then without loss of generality $1+s\sqrt{2}=(1+\sqrt{2})^kr^4$  with $k\in\{0,1,2,3\}$, and comparing norms in $(1)$ shows that $k=1,3$. This yields four quartic Thue equations, two of which have no solutions by reducing mod $4$. The other two have only the trivial solutions $s=\pm1$, found with the aid of a computer package.

Edited this section to provide excruciating detail and the absolutely minimal amount of theory:

If $\eta\in\Bbb{Z}[\sqrt{2}]$ divides both factors on the right hand side of $(1)$, then $\eta$ divides their sum, which is
$$(1+s\sqrt{2})+(1-s\sqrt{2})=2.$$
Because $\Bbb{Z}[\sqrt{2}]$ is a unique factorization domain, and $2$ factors as $2=\sqrt{2}^2$ where $\sqrt{2}\in\Bbb{Z}[\sqrt{2}]$ is prime, it follows that $\eta=u\sqrt{2}^m$ for some unit $u\in\Bbb{Z}[\sqrt{2}]^{\times}$ and some $m\in\{0,1,2\}$. If $m\neq0$ then $\sqrt{2}$ divides both factors, and hence $-\tau^4$ is divisible by $\sqrt{2}^2=2$. But we already noted that $\tau$ is odd; a contradiction. Hence $m=0$ and so $\eta=u$ is a unit in $\Bbb{Z}[\sqrt{2}]$. This shows that the two factors on the right hand side of $(1)$ have no common prime factors. Factoring the two factors as
$$1+s\sqrt{2}=v\cdot p_1^{e_1}\cdots p_k^{e_k}
\qquad\text{ and }\qquad
1-s\sqrt{2}=w\cdot q_1^{f_1}\cdots q_l^{f_l},$$
for units $v,w\in\Bbb{Z}[\sqrt{2}]^{\times}$, pairwise coprime primes $p_i,q_j\in\Bbb{Z}[\sqrt{2}]$, and positive integers $e_i,f_j\in\Bbb{Z}_{>0}$, we see that
$$\tau^4=-(1+st\sqrt{2})(1-s\sqrt{2})=-vw\cdot p_1^{e_1}\cdots p_k^{e_k}q_1^{f_1}\cdots q_l^{f_l}.$$
Because the left hand side is a fourth power, it follows that all $e_i$ and $f_j$ are multiples of $4$. That is to say, we have
$$1+s\sqrt{2}=vr^4\qquad\text{ and }\qquad1-s\sqrt{2}=ws^4,$$
for some units $v,w\in\Bbb{Z}[\sqrt{2}]^{\times}$ and $r,s\in\Bbb{Z}[\sqrt{2}]$. The unit group of $\Bbb{Z}[\sqrt{2}]$ is generated by $-1$ and the fundamental unit $1+\sqrt{2}$, so $v=\pm(1+\sqrt{2})^m$ for some integer $m$, and of course $r=x+y\sqrt{2}$ for some integers $x,y\in\Bbb{Z}$, so we find that
$$1+s\sqrt{2}=\pm(1+\sqrt{2})^m(x+y\sqrt{2})^4,$$
for some $m,x,y\in\Bbb{Z}$. Without loss of generality we may assume that $m\in\{0,1,2,3\}$, because
$$(1+\sqrt{2})^m(x+y\sqrt{2})^4=(1+\sqrt{2})^{m-4n}\big((1+\sqrt{2})^n(x+y\sqrt{2})\big)^4,$$
so we can reduce $m$ modulo $4$ and adjust $x$ and $y$ accordingly. Because the map
$$\Bbb{Z}[\sqrt{2}]\ \longrightarrow\ \Bbb{Z}[\sqrt{2}]:\ x+y\sqrt{2}\ \longmapsto\ x-y\sqrt{2},$$
is an isomorphism, it follows that
$$1-s\sqrt{2}=\pm(1-\sqrt{2})^m(x-y\sqrt{2})^4,$$
and so
\begin{eqnarray*}
\tau^4&=&-(1+s\sqrt{2})(1-s\sqrt{2})\\
&=&-\big(\pm(1+\sqrt{2})^m(x+y\sqrt{2})^4\big)\big(\pm(1-\sqrt{2})^m(x-y\sqrt{2})^4\big)\\
&=&-\big((1+\sqrt{2})(1-\sqrt{2})\big)^m\big((x+y\sqrt{2})(x-y\sqrt{2})\big)^4\\
&=&-(-1)^m(x^2-2y^2)^4,
\end{eqnarray*}
where of course  $\tau^4$ and $(x^2-2y^2)^4$ are positive, so $m$ is odd, i.e. $m=1$ or $m=3$. This shows that either
$$1+s\sqrt{2}=\pm(1+\sqrt{2})(x+y\sqrt{2})^4
\qquad\text{ or }\qquad
1+s\sqrt{2}=\pm(7+5\sqrt{2})(x+y\sqrt{2})^4,$$
for some integers $x,y\in\Bbb{Z}$. Expanding the fourth power and comparing coefficients shows that
\begn{eqnarray*}
\pm1&=&x^4+8x^3y+12x^2y^2+16xy^3+4y^4,\
\pm1&=&7x^4+40x^3y+84x^2y^2+80xy^3+28y^4,
\end{eqnarray*}
and then reducing mod $4$ shows that we must have the $=$-sign in the first case, and the $-$isgn in the second case.emphasized text These are two Thue equations, for which there exist effective methods to determine all integral solutions. With the help of a computer I found that $(x,y)=\pm(1,-1)$ and($(x,y)=(\pm1,0)$ are the only integral solutions, and these all correspond to $s,\tau=\pm1$.


问:
Finding the Flow of a Vector Field on a Plane
Consider the following autonomous vector field on the plane:

\begin{align}
\dot{x}=-\omega y\\
\dot{y}=wx
\end{align}

where $(x,y)\in\Bbb{R}^{2}$ and $\omega>0$. Show that the flow generated by this vector field is given by

$$
    \begin{pmatrix}
    x(t) \\
    y(t) \\
    \end{pmatrix}
=
    \begin{pmatrix}
    \text{cos}\hspace{1mm}\omega t & -\text{sin}\hspace{1mm}\omega t \\
    \text{sin}\hspace{1mm}\omega t & \text{cos}\hspace{1mm}\omega t \\
    \end{pmatrix}
    \begin{pmatrix}
    x_0 \\
    y_0 \\
    \end{pmatrix}.
$$

I have attempted to solve the above ODEs by implicit integration and obtained $x(t)=\dfrac{x_{0}}{1-\omega^{2}yt}$ and $y(t)=-\dfrac{x_{0}\omega^{2}yt}{1-\omega^{2}yt}+y_0$, and am not sure how to proceed from here - appreciate any pointers I can get, thanks.

答:
You have
\begin{align}
\dot x &= -\omega y \\ \dot y &= \omega x
\end{align}
If $x$ and $y$ are differentiable, then the equations above show they are twice differentiable. By differentiating again one has
\begin{align}
\ddot x &= - \omega^2 x \\ \ddot y &= - \omega^2 y
\end{align}
Thus, $x(t) = x_0\cos\omega t + \frac{x'(0)}{\omega}\sin\omega t$ and $y(t) = y_0\cos \omega t + \frac{y'(0)}{\omega}\sin\omega t$. The coupling of $x$ and $y$ show in fact that $x'(0) = -\omega y_0$ and $y'(0) = \omega x_0$, and thus
\begin{align}
\begin{pmatrix}x(t) \\y(t)\end{pmatrix} = \begin{pmatrix}\cos\omega t & -\sin\omega t \\\sin\omega t  & \cos \omega t \end{pmatrix}\begin{pmatrix} x_0 \\ y_0\end{pmatrix}
\end{align}

A more technical but direct way is to solve the differential equation for vectors:
\begin{align}
X' = MX
\end{align}
where $M = \begin{pmatrix}0 & -\omega \\ \omega & 0 \end{pmatrix}$. This is a constant coefficient differential equation so $X(t) = \exp(Mt)X(0)$. Just compute $\exp(Mt) = \sum_{n=0}^{\infty}\frac{(Mt)^n}{n!}$: one can easily show it is $\begin{pmatrix}\cos\omega t & -\sin\omega t \\\sin\omega t  & \cos \omega t \end{pmatrix}$.

ContentLicense: CC BY-SA 4.0


问:
The Maurer-Cartan invariant for the gauge theory in 4D
Suppose the 4D Euclidean gauge theory with group $G \simeq SU(n), \ n \geqslant 2$. The vacuum solution (i.e., the solution giving zero gauge connection $F = DA$) is given by
$$
\tag 1 A_{i}(x) = g\partial_{i}g^{-1},
$$
where $g$ is the gauge group element. It is characterized by the integer $n$ belonging to the homotopic class of the homotopic group $\pi_{3}(G)$, and hence by the Maurer-Cartan invariant
$$
n = \frac{1}{24\pi^{2}}\int\limits_{S^{3}}\text{tr}\big[l^{3}\big], \quad l = g\partial g^{-1}
$$
However, often people write
$$
\tag 2 n = \frac{1}{16\pi^{2}} \int \limits_{S^{4}} \text{tr}\big[F^{2}\big] = \frac{1}{24\pi^{2}} \int \limits_{H^{+}}dK = \frac{1}{24\pi^{2}}\int\limits_{S^{3}}\text{tr}\big[l^{3}\big],
$$
where $H^{+}$ is an upper patch of the sphere $S^{4}$ (and $H^{-}$ is the lower one), and the sphere $S^{3}$ is given as $S^{3} = H^{+} \cap H^{-}$.

But I don't understand why $F^{2}$ isn't identically zero for $(1)$, and therefore don't understand $(2)$. Can You help me?

答:
(1) only describes the behavior of the gauge field on the boundary of ${R}^4$, i.e., on $S^3$. (In quantum field theory, this equation arises when one is looking for Euclidean classical field configurations of finite action; the finiteness of the action requires the field to go to a pure gauge on the boundary of $R^4$.) In the bulk of $R^4$, the gauge field can have any value compatible with this boundary condition (only gauge fields that belong to the topological sector $n=0$ can have the form (1) also in the bulk).

Thus, there is no contradiction in (2), since $tr(F^2)\not=0$ in the bulk. The stated equalities follow from the fact  that $tr(F^2)$ is a total derivative $\partial_\mu K^\mu$, whose in integral can be recast as an integral of $K$ on $S^3$.

ContentLicense: CC BY-SA 3.0


问:
Showing that an ideal is equal to $\mathbb{Z}[i]$
Let $a$, $b$, and $c$ be three elements of $\mathbb{Z}[i]$, the ring of Gaussian integers. We define the following to be  subset of $\mathbb{Z}[i]$:

$$(a,b,c)=\{ax+by+cz:\ x,y,z \in\mathbb{Z}[i]\}.$$

We know that $(a,b,c)$ is an ideal of $\mathbb{Z}[i]$. Show that the ideal $(-1+3i, -1+5i,-4+7i$) is equal to $\mathbb{Z}[i]$ .

What I have done so far:

I have factorised the three elements into a product of irreducible elments in $\mathbb{Z}[i]$:

$(-1+3i)=(1+i)(1+2i)$, $(-1+5i)=(1+i)(2+3i)$,$(-4+7i)=(2+3i)(1+2i)$

However I don't know how to show that the ideal is equal to  $\mathbb{Z}[i]$. I think it might have something to do with the norms of the three elements in the ideal...

答:
There are very many ways of handling a specific question like this. Let me show you the technique I most often use, which has nothing to do with the fact that $\Bbb Z[i]$ has unique factorization.
\begin{align}
(−1+3,−1+5,−4+7)&=(-1+3i,2i,-4+7i)\\
&=(-1+3i,2,7i)\\
&=(-1+3i,2,7)=(-1+3i,1)=(1)\,.
\end{align}
If it’s not self-explanatory, I can elaborate.

EDIT: Elaboration.

Let’s call the three Gaussian numbers $U, V, W$, in the order you gave.
Step 1 (first line) replaces $(U,V,W)$ by $(U,V-U,W)$. Do you see why the two triples generate the same ideal? (If you don’t see this immediately, show that every number of form $aU+bV+cW$ is also of the form $a'U+b'(V-U)+c'W$, and also show the converse. This  understanding is the most important part of the deal. Make sure you absorb it.)

Second step combines two steps, replacing the generator $2i$ by a number that is $2i$ times the unit $-i$. Do you see why this ideal is the same as the preceding? Once you’ve done that, do the same thing as in step 1, by keeping the second generator but adding twice it to the third generator. Do you see why $(-1+3i,2,-4+7i)$ is the same ideal as $(-1+3i,2,7i)$?

Next step replaces $7i$ by $7$, same as the first part of step 2.

Final step uses the fact that if $2$ and $7$ are in the ideal, so is $1$, which is enough to make the ideal the whole ring.

I hope it’s clear now.


问:
If $\{v, w\}$ is a basis for $V$, is it necessarily true that $\{v+w, v-w\}$ is a basis for $V$? Hint: No
I'm really stuck: I tried to do a proof by contradiction to prove the claim false but I couldn't get a contradiction. So I suppose I need to come up with a counter example. In terms of general problem-solving heuristics, how does one conceive of a counter example?</p>

答:
A basis for an n dimensional vector space has three properties

It spans the space.

The vectors are independent.

There are n vectors.

And any two of those is enough to prove the third.   Since v and w form a basis for V, V has dimension 2. v+w and v-w are are 2 vectors so it is only necessary to show either that v+ w  and v- w span the space or that they are independent.

To show that they are independent look at a(v+ w)+ b(v-w)= 0 for some a and b.  a(v+ w)+ b(v- w)= av+ aw+ bv- bw= (a+ b)v+ (a- b)w= 0.  Since v and w are independent we must have a+ b= 0 and a- b= 0.  Adding those, 2a= 0 so a= 0 and then b= 0, proving that v+ w and v- w are independent and so a basis for V.


问:
$\lim_n nr^n$ if $|r| < 1$
In an exercise I'm asked to study the limit of the following sequence:

$$(nr^n)_{n\in \Bbb N}$$

I'm aware of the following:

If $|r|>1$ then the sequence diverges if $r$ is positive, or it has no limit if $r$ is negative.

If $r=1$ then it also diverges

If $r = -1$ then it has no limit

But I'm not being able to find out what the limit is when $|r| < 1$. I know that if that's the case then $\lim_n r^n = 0$, but I don't know how to do this for this sequence.

答:
Take

$$n_0>\frac r{1-r}$$

so that for all $n\ge n_0$
$$\frac{n+1}nr\le s<1.$$

Then by telescoping on $n_0+1,n_0+2,\cdots n_0+n$,

$$\frac{n_0+n}{n_0}r^n<s^n$$ and $$nr^n<n_0s^n.$$

Title: $\lim_n nr^n$ if $|r| < 1$


问:
Limit at non continuous point or arctangent
For an integral I need to calculate, I need to calculate the following limit:

$$\lim_{t \rightarrow \pi^-} \arctan \left(\frac{\tan(\frac{t}{2})}{3}\right).$$
I have used the formula:

$$\tan(\frac{t}{2})= \frac{\sin(t)}{\cos(t)-1}$$
and then applied L'Hopital's rule to show that the limit inside of the arctangent is $\infty$ but, although I know $\tan(\frac{\pi}{2})$ is also $\infty$, I need to justify that the limit is $\frac{\pi}{2}$ rigorously.

答:
As an alternative we can use that for $\theta>0$

$$\arctan \theta = \frac \pi 2 -\arctan\left(\frac 1 \theta\right)$$

that is

$$ \arctan \left(\frac{\tan(\frac{t}{2})}{3}\right) = \frac\pi 2 - \arctan \left(\frac{3}{\tan(\frac{t}{2})}\right) \to \frac \pi 2-\arctan 0=\frac \pi 2$$


问:
Difference quotient for higher derivatives
For a differentiable function $f:\mathbb R\to\mathbb R$, we have, by definition,

$$f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}h.$$

Is there a similar construction for higher derivatives, i.e. is it possible to write $$f^{(n)}=\lim_{h\to0} \Delta_f^n(x,h),$$

where $\Delta_f^n(x,h)$ is a rational function depending only on $$h, f(x), f(x+h), f(x+2h),\dots, f(x+n h)$$ ?

My attempt: See my answer below.

答:
See the Grünwald–Letnikov derivative.

There are further  differ integrals.


问:
Laplace Inversion with Residue Theorem doesn't satisfy IC of IVP
I have the following initial value problem

$$
\frac{d\theta}{dt} = A(p-\theta) + B(\omega-\theta)
$$
subject to the initial condition

$$
\theta(0)=\theta_0
$$

and the constitutive set of equations
$$
\omega(t)=\theta(t)-h\phi(t)
$$

$$
\phi(t)=-\sum_{j=1}^N C_j\int_0^t e^{-\lambda_j^2}(t-x)\omega'(x)dx
$$

$$\omega(0)=0$$
where the prime denotes a temporal derivative, and $\theta_0$, $A$, $B$, $p$, and $h$ above are all real constants, as are the Fourier-Bessel coefficients $C_j$ and the eigenvalues $\lambda_j$.

With the use of the Convolution Theorem, and skipping details, the system above can be Laplace-Transformed to

$$
\Theta(s)=\frac{s\theta_0+Ap}{sF(s)}
$$

where

$$
F(s)=\Big(1+hs\sum_{j=1}^N \frac{C_j}{s+\lambda_j^2}\Big)(s+A+B)-B
$$

The denominator has a pole at $s=0$ and the $N$ other poles are the zeroes of $F(s)$ all of which lie on the Negative Real Axis between $0$ and $-\lambda_N^2$, and can be easily determined from a Bisection algorithm, as they all occur between the asymptotes to $\pm \infty$ at $s=-\lambda_j^2$ and the crossing of the real axis is due to the property $C_jC_{j-1}<0$ $\forall j$. Accordingly the solution in physical space can be written using the Cauchy Residue Theorem as

$$
\theta(t)=p+\sum_{j=1}^N \Big(\frac{\bar z_j \theta_0 +Ap}{\bar z_j F'(\bar z_j)}\Bigr)e^{\bar z_j t}
$$

where the prime denotes a derivative w.r.t. $s$ so that

$$
F'(s)=\Big(1+hs\sum_{j=1}^N \frac{C_j}{s+\lambda_j^2}\Big)+h(s+A+B)-B\Big[\sum_{j=1}^N \frac{C_j}{s+\lambda_j^2}-s\sum_{j=1}^N \frac{C_j}{(s+\lambda_j^2)^2}  \Bigr]
$$

and $\bar z_j, j=1\cdots N$ are the zeroes of $F(s)$. So far, so good.

The Issue with the Solution:

The solution above does not satisfy the initial condition $\theta(0)=\theta_0$, despite this having been taken into account with the Laplace Transform of the derivative in the original ODE. The very short time solutions therefore are not accurate, but the solution does tend to the correct long term solution.

Interestingly, as the number of terms in the Fourier-Bessel series for the flux increases, the initial condition does tend to be satisfied. But shouldn't it be satisfied regardless? Am I missing some crucial detail?

If I use a function-sampling method like Gaver-Stehfest, the inverse transform yields the correct short term solutions even at very short elapsed times from $t=0$. See the graphic below

Any suggestions would be deeply appreciated. Thanks in advance.

答:
DAMN! 15 minutes after posting a $50$ point bounty, I discovered the answer to my problem. I had assumed that there were only $N$ zeroes of $F(s)$. It turns out that there is an isolated pole far from the asymptote to $+\infty$ at $s=-\lambda_N^2$ as indicated in the graphic below (note that the abscissa is $-s$). Once I incorporated this additional pole, the solution worked out perfectly.

I still have a question however:

At the initial condition, the solution is

$$
\theta(0)=p+\sum_{j=1}^{N+1} \Big(\frac{\bar z_j \theta_0 +Ap}{\bar z_j F'(\bar z_j)}\Bigr)
=p\Big(1+A\sum_{j=1}^{N+1}\frac{1}{\bar z_j F'(\bar z_j)}\Bigr)
+\theta_0\sum_{j=1}^{N+1}\frac{1 }{F'(\bar z_j)}
$$

which would imply that the following conditions have to hold:
$$
1+A\sum_{j=1}^{N+1}\frac{1}{\bar z_j F'(\bar z_j)}=0
$$

and
$$
\sum_{j=1}^{N+1}\frac{1 }{F'(\bar z_j)}=1
$$

and while I can verify numerically that they do, I'm unable to prove analytically that this is invariably the case. Any pointers?

Thanks.


问:
Non-orientable 6-manifold with certain Betti numbers
My question is about an aside that came about when answering the following question Closed oriented manifold with middle Betti is one with odd degree..

Is there any sequence $(1,a_1,a_2,a_3,a_4,a_5,0)$ $a_i \geq 0$ integers which cannot be the Betti number sequence of a compact non-orientable $6$-manifold?

I can think of examples (from products of real projective spaces, non-orientable surfaces etc.) with $(1,0,0,0,0,0)$, $(1,1,1,1,1,1,0)$, $(1,0,1,0,0,0,0)$, $(1,1,0,0,0,0)$ , $(1,0,0,0,1,0,0)$ then connect sum gives almost every possibility. But for some sequences I not able to think of an example, for example $(1,0,0,1,0,0,0)$.

Sorry if this is a naïve question I haven't thought about non-orientable manifolds for a long time. My expectation is that probably everything can be realised but I don't know enough examples (perhaps fibre bundles are enough to give everything)?

答:
I claim that in every even dimension, any sequence of rational Betti numbers can occur.
As you noted, in each dimension $k$, it's sufficient to find examples $N^k_m$ with $0 < m < k$ with $H^\ast(N^k_m;\mathbb{Q})\cong H^\ast(S^m;\mathbb{Q})$, for then connect summing gives everything.  EDIT:  This is wrong.  connect summing two non-orientable manifolds adds rational cohomology in degree $k-1$.  End Edit

Now, if $m$ is even, one can simply use $N^k_m = S^m\times \mathbb{R}P^{k-m}$.  (Since $m$ and $k$ are both even, so is $k-m$, and $\mathbb{R}P^{even}$ has the rational cohomology of a point.)

So, we need only handle the case where $m$ is odd.  Set $n:=k-m$, which is also odd since $k$ is even and $m$ is odd.  Consider the $\mathbb{Z}_2$ action on $S^m \times S^n$ with $-1\ast(x,y) = (-x, r(y))$ where $r:S^n\rightarrow S^n$ is any reflection in a hyperplane.

This action is free since it is free on the first factor.  Call the resulting quotient manifold $M$.  I claim that $N^k_m = M$ works.

Because $m$ is odd, the antipodal map is orientation preserving.  Since $r$ reverses orientation, it follows that the $\mathbb{Z}_2$ action reverses orientation.  In particular, $M$ is non-orientable.

I claim that $H^\ast(M;\mathbb{Q}) \cong H^\ast(S^m;\mathbb{Q})$.  To this this, first note that since $\mathbb{Z}_2$ is finite, we have a transfer homomorphism $f:H^\ast(M;\mathbb{Q})\rightarrow H^\ast(S^m\times S^n;\mathbb{Q})$ for which the composition $f\circ \pi^\ast$ is multiplication by $2$ (which is an isomorphism with $\mathbb{Q}$ coefficients).  In particular, $\pi^\ast$ is an injection, so the rational Betti numbers of $M$ are zero except possibly in degrees $0,m,n,m+n$.  Further, the Betti numbers are $a_0, a_m, a_n, a_{m+n}$ are all bounded by $1$ (except when $m = n$, in which case $a_{m}$ is bounded by $2$).

Now the antipodal action on $S^m$ is free with quotient $\mathbb{R}P^m$.  The associated bundle construction now shows that $M$ is the total space of an $S^n$ bundle over $\mathbb{R}P^m$.  Pulling back this bundle along the double cover $S^m\rightarrow \mathbb{R}P^m$, we obtain a commutative diagram $$\begin{array} AS^n & \longrightarrow & S^n \\
\downarrow & & \downarrow \\
S^m\times S^n & \longrightarrow & M\\
\downarrow &  & \downarrow \\
S^m & \longrightarrow & \mathbb{R}P^m
\end{array}$$

The induced map on cohomolgoy $H^m(\mathbb{R}P^m;\mathbb{Q})\rightarrow H^m(S^m;\mathbb{Q}))\rightarrow H^m(S^m\times S^n;\mathbb{Q})$ is easily seen to be non-trivial.  It follows from commutativity that $H^m(\mathbb{R}P^m;\mathbb{Q})\rightarrow H^m(M;\mathbb{Q})$ must be non-zero, so $H^m(M;\mathbb{Q})$ is non-trivial.

In addition, if $H^n(M;\mathbb{Q}) \neq 0$ (or $H^m(M;\mathbb{Q})$ has dimension $2$ when $m=n$), then we have a problem:  If $x\in H^m(M;\mathbb{Q})$ is non-zero and $y\in H^n(M;\mathbb{Q})$ is non-zero (and $x$ and $y$ are independent if $m = n$), then $\pi^\ast(xy) = \pi^\ast(x)\pi^\ast(y)\neq 0$, which implies $H^{m+n}(M;\mathbb{Q}) = 0$.  Since we already know $M$ is non-orientable, this is absurd.  Thus, we conclude that $H^\ast(M;\mathbb{Q})\cong H^\ast(S^m;\mathbb{Q})$.


问:
Prove that the order of the preimage of N is the order of the kernel times the order of N.
I cannot seem to make any progress at all with this proof. Suggestions and hints. No full proofs, please. Thanks!

Suppose $\phi:G\to G'$ is a homomorphism of finite groups, and define $K=\ker(\phi)$.  If $|K|=k$ and $N$ is a subgroup
of $G'$, show that
$$
\left|\phi^{-1}(N)\right|=k\cdot|N|
$$
That is, prove that for every $n\in N$, there are precisely $k$ elements in $G$ that map to $n$. One might say in this case that
$\phi$ is a $k$-to-one homomorphism.

答:
This was 2015 and the question shows up in search results, so it's time for a full proof.

The statement is not true (in general) if $\phi$ is not surjective. Consider the embedding. $\mathbb Z_2\to \mathbb Z_2 \times \mathbb Z_2$ of finite abelian groups. $\mathbb Z_2\times \mathbb Z_2$ has order $4$, its preimage has order $2$, and the kernel has order $1$.

In the case of a canonical quotient $\pi: G \to G/N$ the statement is clearly true. $\pi$ sends cosets of size $|N|$ to points, so for each subset $S$ of $G/N$ one gets $|\pi^{–1}(S)| = |S|\, |N|$. Let $\phi: G \to H$ be a surjective morphism of finite groups. The first isomorphism theorem states that $\phi$ is, up to composition with an isomorphism, a canonical quotient map as above, and the result follows immediately.


答:
When writing a formal proof, it's best to immediately write assumptions and what you want to prove formally. In your case the assumptions are:

$(\forall \alpha,\beta\in F)\quad\alpha v + \beta w = 0 \implies \alpha = \beta = 0,$

$V = \operatorname{span}_F \{v,w\}$.

What you would like to prove or disprove is:

$(\forall \alpha,\beta\in F) \quad \alpha(v+w) + \beta(v-w) = 0 \implies \alpha = \beta = 0,$

$V = \operatorname{span}_F \{v+w,v-w\}$.

First of all, since we know that $\{v,w\}$ is a basis, we know that $\dim V = 2$, and having that in mind, 3. and 4. are equivalent. Thus, it is enough to prove or disprove one of them.

I will concentrate on 3., so take $\alpha,\beta\in F$ such that $\alpha(v+w) + \beta(v-w) = 0$. We want to manipulate this expression:

\begin{align}
\alpha(v+w) + \beta(v-w) = 0 &\implies (\alpha + \beta)v+(\alpha - \beta)w = 0\\
&\stackrel{1.}{\implies} \alpha + \beta = 0\ \wedge\ \alpha -\beta = 0\\
&\implies \alpha = \beta\ \wedge \ 2\alpha = 0.
\end{align}

At this point, the field $F$ becomes relevant. If $\operatorname{char} F \neq 2$, then $2\in F$ is invertible, and $2\alpha = 0\implies \alpha = 0$, and therefore $\beta = \alpha = 0$. So, in this case, we started from $\alpha(v+w) + \beta(v-w) = 0$ and concluded $\alpha = \beta = 0$ and thus proved that 3. is true, i.e. $\{v+w,v-w\}$ is indeed a basis for $V$.

However, if $\operatorname{char} F = 2$, then $2\alpha = 0$ is true for all $\alpha\in F$, and we can't conclude that $\alpha = 0$. But that doesn't disprove 3. yet. For that, we need a counterexample. Choose $\alpha = \beta = 1 \in F$. Then $$\alpha(v+w) + \beta(v-w) = (v+w)+(v-w) = 2v = 0,$$ but $\alpha \neq 0$ and $\beta \neq 0$. Therefore, we provided a counterexample to 3., so $\{v+w,v-w\}$ is not a basis for $V$.

Title: If $\{v, w\}$ is a basis for $V$, is it necessarily true that $\{v+w, v-w\}$ is a basis for $V$? Hint: No

Body: I'm really stuck: I tried to do a proof by contradiction to prove the claim false but I couldn't get a contradiction. So I suppose I need to come up with a counter example. In terms of general problem-solving heuristics, how does one conceive of a counter example?</p>


问:
Is the matrix of an hermitian (real) endomorphism symmetric?

Definition

Let be $V$ and $U$ real vector spaces equipped with an inner product. So given a linear transformation $f:V\rightarrow U$ a function $f^*:U\rightarrow V$ is called the adjoint of $f$ if
$$
\langle\vec u,f(\vec v)\rangle=\langle f^*(\vec u),\vec v\rangle
$$
for all $\vec v\in V$ and for all $\vec u\in U$. In particular an endomorphism $f\in\mathscr L(V,V)$ is called Hermitian if $f=f^*$ and skew-Hermitian if $f=-f^*$.

Now let be $\mathscr B:=\{\vec e_1,...,\vec e_n\}$ an orthonormal basis for $V$. So we observe that
$$
\langle f(\vec e_i),\vec e_j\rangle=\langle f^*(\vec e_i), \vec e_j\rangle=\langle \vec e_i,f(\vec e_j)\rangle
$$
for each $i,j=1,...,n$ and we conclude that the matrix computed using the basis $\mathscr B$ is symmetric.
So using the previous definition I ask if the matrix $A$ of an hermitian endomorphism is necessarly symmetric also if we don't compute the matrix using an orthonormal basis. Indeed generally if $\mathscr B$ is not orthomormal then the $a_{i,j}$ element of $A$ is given by the equation
$$
a_{i,j}=\langle f(\vec e_j),\vec e^{\, i}\rangle
$$
where $\vec e^{\, i}$ is the $i$-th element of the reciprocal basis of $\mathscr B$ so that it seems to me that generaly
$$
a_{i,j}=\langle f(\vec e_j),\vec e^{\, i}\rangle\neq\langle f(\vec e_i),\vec e^{\, j}\rangle=a_{j,i}
$$
that implies that $A$ is not symmetric. So could someone help me, please?

答: No: the matrix of a Hermitian with respect to a basis that is not orthonormal is not necessarily symmetric.</p>


问:
Summation using binomial coefficient
I am trying to calculate the following using binomial coefficients and summation, but my memory is standing in the way:

$$
   \sum_{k=1}^n {k} * 2 ^ {k - 1}
$$

Thanks!

With great help I got to:

$$ (n - 1)* 2 ^{n} - 1 $$

Can you please confirm this?

答:
Clement C's solution is the best approach. There is another approach, however.

Notice that
$$
\sum_{k=1}^n k2^{k-1} = \sum_{k=1}^n \sum_{i=1}^k 2^{k-1}
$$
because $k=\sum_{i=1}^k 1$. By reversing the order of summation, you get
$$
\sum_{k=1}^n k2^{k-1} = \sum_{i=1}^n \sum_{k=i}^n 2^{k-1}
$$
A closed form solution can now be found easily for the inner sum, which will produce a form that can be used to find a closed form solution for the outer sum.

Title: Summation using binomial coefficient


问:
Can an ordered set of vectors guarantee that the first half are larger in real and imaginary components than the second half?
Recently, on one of my problem sets I wrote a solution to a question that I feel should be correct. Me and my supervisor spoke about it but could not come up with a way to show my method worked or did not.

Question: Red sweets and blue sweets are randomly distributed across the 99 bags. Can you always pick 50 bags in such a way that you have at least half of the red sweets, and at least half of the blue sweets?

My attempt: Let each bag be represented by $ z_k=r_k+b_k i$ with $\sum_{k=1}^nr_i=1$ and $\sum_{k=1}^nb_i=1$

ie, $r_i$ and $b_i$ represent the proportion of total red/ blue sweets.
Note: only considering case where $r_k<0.5, b_k<0.5$ $∀k$ if not true then problem is trivial.

Now, Take $$|0.5+0.5i-z_k|$$ and order $z_k$ from smallest to largest value of above. such that $$|0.5+0.5i-z_k|\le|0.5+0.5i-z_{k+1}|$$
My claim; If we let $$u=\sum_{k=1}^{50}z_k ,v=\sum_{k=51}^{99}z_k$$
Then by the way we ordered $z_k$ $$Re(v)\le0.5, Im(v)\le0.5$$
and thus $u$ has $Re(u)>0.5$ and $Im(u)>0.5$ as required.

My question; Is this correct/ valid. If so, how do I prove that $v$ lies within the $(0.5,0.5)$ square

Note: I am a 1A student. I know how to solve this problem using other means but am still interested in this method.

答:
Your approach (as is) doesnt' work.

Here is a counter example.

Let 50 bags contain $\frac{1}{50}$ of the blue sweets each.

Let 49 bags contain $ \frac{1}{49}$ of the red sweets each.

Then, your selection will pick up all 50 bags with blue sweets, giving you $ Re(u) = 1, Im (u) = 0$.


问:
Zeros of Prime Zeta Function
Let $P(s)=\sum_{p}\frac{1}{p^s}$ be the prime zeta function. It is clear that $P(s)$ converges absolutely for $\sigma>1$, where $s=\sigma+it$, and can be analytically continued to the strip $0<\sigma\leq 1$ (Fröberg 1968)https://link.springer.com/article/10.1007%2FBF01933420.

Now, I have two questions about the zeros of $P(s)$:

(1) Are there any zeros of $P(s)$ on the line $\Re(s)=1$?

(2) Is $s=1$ the pole or the essential sigularity of $P(s)$?

If $s=1$ is the pole of $P(s)$, what is the Laurent expansion of $P(s)$ at $s=1$?

In fact, if $P(s)$ has no zeros on the line $\Re s=1$ and
$$P(s)=\frac{\alpha}{s-1}+\alpha_0+(s-1)h(s)$$
for some complex number $\alpha,\alpha_0$ and a holomorphic function $h(s)$, then we can prove by the theorem 3.4.2 of Jameson's book: The Prime Number Theorem, that
$$\lim_{x\to \infty}\frac{\pi(x)}{x}=\alpha$$
which is absurd. So where it goes wrong?  Here, $\pi(x)$ is the prime counting function.

答:
You are asking here about zeros of $P(s)$ at $\Re(s)=1$, but showing that $P(s)$ has no zeros for $\Re(s)>1$ is already not trivial, since the prime indicator function has no Dirichlet inverse so constructing $\frac{1}{P(s)}$ as a dirichlet series which is absolutely convergent for $\Re(s)>1$ is not possible. One "trivial" zero free zone from this Math.SE question is by noting that $|P(s)|$ will be at least $\frac{1}{2^{\sigma}}-\sum_{p\geq3 }\frac{1}{p^{\sigma}}$ where $s=\sigma+it$.

This leads us to conclude (from numerical computations) that $P(s)$ has no zeros for $\Re(s)>1.76$. If we look at the relation function $P^*(s)=\sum_{p^k}\frac{1}{p^{ks}}$ which is the "prime power zeta function", the situation is even worse. Using the same bounding method as before, we get that $|P^*(s)|>\frac{1}{2^{\sigma}}-\sum_{p^k\geq3}\frac{1}{p^{k\sigma}}$ which gives the much worse zero free region of $\Re(s)>2.21$. In this answer I will outline a procedure to reduce the zero free region of $P^*(s)$ to $\Re(s)>1.11$, along with proving that for $\Re(s)>\frac{1}{2}$ the function $P^*(s)$ will only have zeros if $\zeta(2s)=\zeta(s)$.

To begin the proof, I will examine the divisor sum $\sum_{d|n}\frac{\lambda(d)}{\Omega(d)+1}$ where $\Omega(n)$ counts the number of prime factors of $n$ (with multiplicity) and $\lambda(n)=(-1)^{\Omega(n)}$. Since $\frac{\lambda(d)}{\Omega(d)+1}$ only depends on the number of prime factors of $d$ and there are ${\Omega(n)\choose k}$ divisors of $d$ with exactly $k$ prime factors, we can split our sum over all these classes to get that

\begin{align*}
\sum_{d|n}\frac{\lambda(d)}{\Omega(d)+1}&=\sum_{k=0}^{\Omega(n)}{\Omega(n)\choose k}\frac{(-1)^k}{k+1}\\
&=\int_{0}^{1}\sum_{k=0}^{\Omega(n)}{\Omega(n)\choose k}(-1)^kx^kdx\\
&=\int_{0}^{1}(1-x)^{\Omega(n)}dx\\
&=\frac{1}{\Omega(n)+1}
\end{align*}

Thus, we can apply this formula to get that

\begin{align*}
\left(\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}\right)\left(\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}\right)&=\sum_{n=1}^{\infty}\frac{\sum_{d|n}\frac{\lambda(d)\Omega(n/d)}{\Omega(d)+1}}{n^s}\\
&=\sum_{n=1}^{\infty}\frac{1-\sum_{d|n}\lambda(d)}{n^s}
\end{align*}

Since $\sum_{d|n}\lambda(d)$ is the indicator function for squares, this means that

\begin{align*}
\left(\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}\right)\left(\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}\right)&=\zeta(s)-\zeta(2s)\tag{1}
\end{align*}

We now work on simplifying our first sum, namely we see that

\begin{align*}
\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}&=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{\lambda(n)z^{\Omega(n)}}{n^s}dz\\
&=\int_{0}^{1}\prod_{p}\frac{1}{1+zp^{-s}}dz\\
&=\int_{0}^{1}\zeta^{-z}(s)\zeta^{z}(2s)\prod_{p}\frac{\left(1+p^{-s}\right)^z}{1+zp^{-s}}dz
\end{align*}

We now let

$$G(s;1-z):=\zeta^{z}(2s)\prod_{p}\frac{\left(1+p^{-s}\right)^z}{1+zp^{-s}}$$

so that

\begin{align*}
\sum_{n=1}^{\infty}\frac{\frac{\lambda(n)}{\Omega(n)+1}}{n^s}&=\int_{0}^{1}G(s;1-z)\zeta^{-z}(s)ds\tag{2}
\end{align*}

Substituting (2) into (1) and noting that $\sum_{n=1}^{\infty}\frac{\Omega(n)}{n^s}=\zeta(s)P^*(s)$ gets us that

\begin{equation}
\frac{1}{P^*(s)}=\frac{1}{\zeta(s)-\zeta(2s)}\int_{0}^{1}G(s;z)\zeta^z(s)ds
\end{equation}

The definition of $G(s;z)$ is absolutely convergent for $\Re(s)>\frac{1}{2}$, and thus since $\zeta^z(s)$ is defined in any zero free region of the zeta function we get (under RH) that $P^*(s)$ can only be zero of $\zeta(s)=\zeta(2s)$ for $\Re(s)>
\frac{1}{2}$. This result holds without RH or anything special for $\Re(s)>1$ which is what we will use in the rest of the proof.

What we will simply do is note that

\begin{align*}
P(s)+P(2s)&=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(ks))+\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2ks))\\
&=\frac{3}{2}\ln(\zeta(2s))+\sum_{k=3}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2s))+\sum_{k=2}^{\infty}\frac{\mu(k)}{k}\ln(\zeta(2ks))
\end{align*}

We can now lower bound $\ln(\zeta(2s))=-\sum_{p}\ln\left(1-\frac{1}{p^{2s}}\right)$ by the same methods as in the first paragraph, and we will get a greater than zero result for values of $s$ even with $\Re(s)<1$ because we are working with the doubled $\zeta(2s)$, due to $\zeta(s)=\zeta(s)$ holding. Including more terms and numerically computing the prime zeta function, we arrive at the desired bound of $P^*(s)$ having no zeros for $\Re(s)>1.11$. The acutal computations are tedious, but I will share them if anyone is interested.

This same attack won't work for $P(s)$, since the associated functions $\mu(n)$ and $\omega(n)$ are not completely multiplicative/additive which is what we used in the crucial proof of (1).

Title: Zeros of Prime Zeta Function


答:
With odd $n\gg 0$, take

three bags with $\frac13i$

$n-3$ bags with $\frac1{n-3}$

The first three are closer to $\frac12+\frac12i$, hence are picked into $u$. That means that $\frac{n-1}2$ of the latter make your $v$ and give it a total real part of $\frac{n-1}{2n-6}=\frac12+\frac{1}{n-3}>\frac12$.

Title: Can an ordered set of vectors guarantee that the first half are larger in real and imaginary components than the second half?

Body:
Recently, on one of my problem sets I wrote a solution to a question that I feel should be correct. Me and my supervisor spoke about it but could not come up with a way to show my method worked or did not.

Question: Red sweets and blue sweets are randomly distributed across the 99 bags. Can you always pick 50 bags in such a way that you have at least half of the red sweets, and at least half of the blue sweets?

My attempt: Let each bag be represented by $ z_k=r_k+b_k i$ with $\sum_{k=1}^nr_i=1$ and $\sum_{k=1}^nb_i=1$

ie, $r_i$ and $b_i$ represent the proportion of total red/ blue sweets.
Note: only considering case where $r_k<0.5, b_k<0.5$ $∀k$ if not true then problem is trivial.

Now, Take $$|0.5+0.5i-z_k|$$ and order $z_k$ from smallest to largest value of above. such that $$|0.5+0.5i-z_k|\le|0.5+0.5i-z_{k+1}|$$
My claim; If we let $$u=\sum_{k=1}^{50}z_k ,v=\sum_{k=51}^{99}z_k$$
Then by the way we ordered $z_k$ $$Re(v)\le0.5, Im(v)\le0.5$$
and thus $u$ has $Re(u)>0.5$ and $Im(u)>0.5$ as required.

My question; Is this correct/ valid. If so, how do I prove that $v$ lies within the $(0.5,0.5)$ square

Note: I am a 1A student. I know how to solve this problem using other means but am still interested in this method.


问:
Idempotency in Matrices and deducing Linear Map Compositions

Hello, could anyone help me out with these two questions.

With question d) I know that because D is an idempotent matrix, then In - D must also be idempotent through:

(In-D)(In-D) = In-2Dn+D = In - D

And also the relation between the linear maps:

S: Rn -> Rn has matrix D

S' : Rn -> Rn had matrix In - D

Then the dot product of S and S' is equal to the multiplication of D and In-D.

But that's as far as I can get with d)

And with e) the only thing I have proved so far is the E is idempotent due to when it's squared it returns E but I don't know what to do next.

答:
Hint for d: $S \circ S' = D \circ (I_n - D) = D \circ I_n - D \circ D$.

Hint for e: For every $n$, $T^{n} = T \circ T^{n-1}$. For example,
$$
T^3 = T \circ T^2 = T \circ T = T.
$$


问:
Prove that $x ^ {1 - x} (1 - x) ^ {x} \le \frac {1}{2}$
How do I prove $x ^ {1 - x}(1 - x) ^ {x} \le \frac{1}{2}$, for every $x \in (0, 1)$.

For $x = \frac {1}{2}$ the LHS is equal to one half. I tried studying what happens when $x \lt \frac {1}{2}$ and its correspondent, but to no result.

Body:
let $a=x,b=1-x$,

$a+b=1$,

By  AM-GM$$\frac{1}{2}=\frac{{(a+b)}^2}{2}\ge 2ab=\frac{ab+ba}{a+b}\ge \sqrt[a+b]{a^b b^a}=a^bb^a=x^{1-x}{(1-x)}^{x}$$


问:
Evaluate Limit two variables , how to tackle these questions?
I'm aware this is a beginner question :D but im new to the subject.
I thought about known limits like $\tan(x)/x$ as $x\to0$ but im stuck, I can't find a way to include both known limits because of the denominator... maybe using the fact that $|\sin x|<|x|<|\tan x|$ ? ...
a hint would be much appreciated :D
$$
\displaystyle
\lim\limits_{\substack{x\to 2 \\ y\to 1}}
\frac{\tan(y-1)\sin^2(2 y -  x)}{(x-2)^2 +(y-1)^2}=
$$

furtermore, any tips on how do i tackle these types of questions?

答:
Let consider for simplicity the change of variable $u=x-2 \to 0$ and $v=y-1 \to 0$ then

$$\lim_{(x,y)\to(2,1)}
\frac{\tan(y-1)\sin^2(2 y -  x)}{(x-2)^2 +(y-1)^2}=\lim_{(u,v)\to(0,0)}
\frac{\tan (v)\sin^2(2v-u)}{u^2 +v^2}$$

and then use

$$\frac{\tan (v)\sin^2(2v-u)}{u^2 +v^2}=\frac{\tan (v)}{v}\frac{\sin^2(2v-u)}{(2v-u)^2}\frac{v(2v-u)^2}{u^2 +v^2}$$

which can be solved using standard limits for the first two terms and polar coordinates for the last one.


问:
Are there any Combinatoric proofs of Bertrand's postulate?
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).

I was trying to prove one by the Sieve on this grid

```<code>    1     2     3     4 ... p
  p+1   p+2   p+3   p+4 ... 2p
 2p+1  2p+2  2p+3  2p+4 ... 3p
...........................
...........................
........................... p^2
p^2+1 p^2+2 p^2+3 p^2+4 ...
</code>```

but it did not work.

Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?

答:
Edited: Here is a very relevant link on Robin Chapman's page.  It is the combinatorial proof which Erdos originally used to prove Bertrands postulate.  (Mentioned in Pete L. Clark's comment)

The PDF is very similar to the proofs I mentioned here, (and here)  and is based entirely on the primes dividing the central binomial coefficient.

I will leave in the proof regarding a lower bound for $\pi(x)$, since it is still relevant.

What follows is a proof Chebyshev's Theorem based on the primes dividing the central binomial coefficient.  First, two well known facts:

Fact 1: Let $r(n,p)$ denote the highest power of a prime $p$ that divides $n!$. Then $$r(n,p)=\left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{N}{p^{2}}\right\rfloor+\left\lfloor\frac{N}{p^{3}}\right\rfloor+\cdots$$

Fact 2: For every integer $N\geq1$ we have that $$\frac{4^{N}}{2N}\leq\binom{2N}{N}.$$

Theorem: We have that $$\pi(x)\geq\ln(2)\frac{x}{\ln x}-2.$$

Proof. Consider $\binom{2N}{N}.$ For each prime $p$, let $v_{p}$ denote the number of times it divides $\binom{2N}{N}$. Hence $$\binom{2N}{N}=\prod_{p\leq2N}p^{v_{p}}.$$ Let $t=\log_{p}(2N)$ so that $\left\lfloor\frac{2N}{p^{j}}\right\rfloor=0$ when $j\geq t.$ By the lemma, we have $$v_{p}=\left(\left\lfloor\frac{2N}{p}\right\rfloor+\left\lfloor\frac{2N}{p^{2}}\right\rfloor+\cdots\right)-2\left(\left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{N}{p^{2}}\right\rfloor+\cdots\right)=\sum_{i=1}^{t}\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor.$$ Since $\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor=0$ or $\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor=1$ for each $i$ we see that $v_{p}\leq t.$ Hence if $p^{r}$ divides $\binom{2N}{N}$ we must have $p^{r}\leq2N.$  Consequently $$\binom{2N}{N}=\prod_{p\leq2N}p^{v_{p}}\leq\prod_{p\leq2N}(2N)=(2N)^{\pi(2N)}.$$Applying the other lemma we get $\frac{4^{N}}{2N}\leq(2N)^{\pi(2N)}$and upon taking logarithms we find$$N\ln4-\ln(2N)\leq\pi(2N)\ln(2N)$$and hence $$\frac{2N\ln2}{\ln2N}\leq\pi(2N)+1.$$Since this holds for every even integer, and there is at most one prime between even integers we get

$$\frac{x\ln2}{\ln x}-2\leq\pi(x)\ \text{for all } \ x\geq2,$$ as desired.


问:
Bank balance interest using Generating Functions
I have faced an interesting question:

Jack puts $\$4,000 $ in his bank account with yearly interest rate of $5\%$. In the beginning of every year (starting the second year after the first deposit), he puts an additional $\$1,000$ in the same bank account. What is the bank balance after $n$ years?

Now I clearly thought about Generating Functions, but failed to describe what in what regarding the interest rate and the bank balance!
I thought that one year gives a power of $x^{0.05n}$, but I genuinely feel that is truly wrong.

Any help would be appreciated

答:
You can do this with recurrence relation: $a_0=4000$ and $a_n=1.05a_{n-1}+1000$. From here you can solve by using generating functions (if you want to):

Let $f(x)=\sum_{n=0}^\infty a_nx^n$. Then $xf(x)=\sum_{n=1}^\infty a_{n-1}x^n$. So
$$\begin{align*}(1-1.05x)f(x)&=f(x)-1.05xf(x)=a_0+\sum_{n=1}^\infty (a_n-1.05a_{n-1})x^n=\\
&=a_0+\sum_{n=1}^\infty a_nx^n-1.05\sum_{n=1}^\infty a_{n-1}x^n=\\
&=a_0+\sum_{n=1}^\infty 1000x^n=4000+1000\left(\frac1{1-x}-1\right)
\end{align*}$$
Now find $f(x)$ and expand into series to find $a_n$.


问:
Find the relationship between $x_0$ and $y_0$ that ensures $x(t) → 0$ as $t → ∞$
If the roots of the auxiliary equation of a second order linear homogeneous
ODE are $k_1 > 0$ and $−k_2 < 0$ then the solution is
$$x(t) = Ae^{k_1t}+ Be^{−k_2t}.$$

For most choices of initial conditions
$x(0) = x_0$, $\dot x(0) = y_0$
we will have that $x(t) → ±∞$ as $t → ∞$. However, there are some special
initial conditions for which $x(t) → 0$ as $t → ∞$. Find the relationship between
$x_0$ and $y_0$ that ensures this

My working so far is $x_0 = A + B$ and $y_0 = Ak_1 - Bk_2$ and I'm really unsure of how to tackle this question further

答: The second term always converges to zero for $t\to\infty$. The first term always diverges at the same time  if $A\ne 0$. Thus you need $A=0$, elimination of $B$ gives the desired relation.


问:
Giving a formula for an integral and showing whether it can integrated
I have a question which states:

Given that $(r_n)_{n\in\mathbb{N}}$ is a strictly increasing sequence in $\mathbb{R}$ with $r_0 = 0$ and $\lim_{n\to\infty}r_n = 1$ and that the function $f$ is defined by:

$$f: [0,1) \to [0, \infty):f(x) = \begin{cases} (x-r_n)^{-\alpha} & \text{if } r_n < x < r_{n+1} \text{ for } n\in \mathbb{N}\\
+\infty & \text{if } x=r_n \text{ for } n \in \mathbb{N}
\end{cases} $$

Give a formula for $\int_{[0,1)}f(x)dx$ in terms of $(r_n)$ and $\alpha$. Prove this formula. Consider $\alpha <1$ and $\alpha \geq 1$ separately.

Can you choose $\alpha >0$ and $(r_n)$ such that $\int_{[0,1)}f(x)dx < \infty $?

I don't really know where to start with this question. Any help would be appreciated.

答:
Consider

$$f_n(x)=\sum_{k=0}^n(x-r_k)^{-\alpha}\chi_{(r_k,r_{k+1})}(x)$$

and define $\tilde{f} = \sum_{n=0}^\infty f_n$.

Note that $\tilde{f}=f \ \ a.e.$.

If $\alpha \neq 1$, we have(by the Monotone Convergence Theorem)

$$\int f d\mu = \int \tilde{f} d\mu=\sum_{n=0}^\infty \sum_{k=0}^n \frac{(r_{k+1}-r_k)^{1-\alpha}}{1 - \alpha}$$

and if $\alpha=1$, we have

$$\int f d\mu = \int \tilde{f} d\mu=+\infty.$$

If $0 < \alpha < 1$, we have

$$\sum_{k=0}^n (r_{k+1}-r_k)^{1-\alpha} > \sum_{k=0}^n (r_{k+1}-r_k)=r_{n+1} \to 1, \ \text{as} \ \  n \to \infty,$$

so

$$\int f d\mu = \int \tilde{f} d\mu=+\infty.$$

Finally, if $\alpha > 1$, we have

$$\sum_{k=0}^n (r_{k+1}-r_k)^{1-\alpha} > Cn$$ for large $n$, where $C > 0$,

so

$$\int f d\mu = \int \tilde{f} d\mu=-\infty.$$


问:
How to find a positive semidefinite matrix satisfying the following condition?
Consider such a problem:
$$\mathbf A\mathbf x=\mathbf b$$
where the vectors $\mathbf x$ and $\mathbf b$ are known.

I want to find a positive semidefinite matrix $\mathbf A$ satisfying the aforementioned equation.

What conditions do $\mathbf x$ and $\mathbf b$ need to satisfy?

Thanks a lot.

答:
If $x$ and $b$ are vectors, they at least have to satisfy $x \cdot b \geq 0$. You can see this by considering the quadratic form associated with $A$:

$$Q(x) = x^T A x = x^T b = x \cdot b \geq 0,$$

since $A$ is positive semidefinite if and only if the quadratic form is.


答:
There are three cases to consider when doing these limits.

Firstly, fix $x$ at its limit point, so $x=2$ and take $y\to 1$. We get:
$$\lim_{y=1}\frac{\tan(y-1)\sin^2(2y-2)}{(y-1)^2}=0$$

Secondly, fix $y$ at its limit point, so $y=1$ and take $x\to 2$. We get:
$$\lim_{x\to 2}\frac{\tan(0)\sin^2(2-x)}{(x-2)^2}=0$$

Thirdly, let $y(x)$ be any function defined such that $x\to 2 \implies y(x)\to 1$

Then, you take $$\lim_{x\to 2}\frac{\tan(y(x)-1)\sin^2(2y(x)-x)}{(x-2)^2+(y(x)-1)^2}$$
and evaluate this general limit. If all the limits are the same, then that limit is the limit of the multivariate function, if there is a single exception, it has no limit.

In this case, the denominator is screaming at you to take $x-2=r\cos\theta, y-1=r\sin\theta$, as $$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{r\to 0}f(r\cos\theta, r\sin\theta)$$

Title: Evaluate Limit two variables , how to tackle these questions?

Body:
I'm aware this is a beginner question :D but im new to the subject.
I thought about known limits like $\tan(x)/x$ as $x\to0$ but im stuck, I can't find a way to include both known limits because of the denominator... maybe using the fact that $|\sin x|<|x|<|\tan x|$ ? ...
a hint would be much appreciated :D
$$
\displaystyle
\lim\limits_{\substack{x\to 2 \\ y\to 1}}
\frac{\tan(y-1)\sin^2(2 y -  x)}{(x-2)^2 +(y-1)^2}=
$$

furtermore, any tips on how do i tackle these types of questions?


问:
Are the names and symbols for common mathematical operators the same in every language?
Do all human languages that have arithmetic use +, -, ×, ÷, and ^?

How about sin, cos, tan, asin, acos, and atan?

答:
There are human languages that have arithmetic but their users still do not make use of any formal symbols for arithmetic operations (or even numbers).

You mentioned internationalizing an application so two more things may be useful:

1)in some languages it is common to use dot "." in decimals, like in "1.5", while in other languages it is common to use comma ",", like in "1,5".

2) ${\bf tg}$ is in some languages used instead of ${\bf tan}$, and ${\bf arcsin}$ instead of ${\bf asin}$, similarly for other functions.


问:
help integrating
I'm really struggling to integrate this.

$$\int_0^t \sqrt{\cos^2\left(1\right)\sinh^2\left(u\right)+4\sin^2\left(\dfrac{1}{2}\right)\cosh^2\left(\dfrac{u}{2}\right)} \,\mathrm{d}u$$

This is to find the arc length of a plane curve. An online integral calculator gives me such a complicated result with $\ln$ and I don't know how they got to that result.

I also then need to use the result to find the inverse.

答:
So I followed your hint and got

$s(t)=2\sqrt{b}\sinh\left(\dfrac{u}{2}\right)\sqrt{\dfrac{a\sinh^2\left(\frac{u}{2}\right)}{b}+1}+\dfrac{2b\ln\left(\left|\sqrt{a\sinh^2\left(\frac{u}{2}\right)+b}+\sqrt{a}\sinh\left(\frac{u}{2}\right)\right|\right)}{\sqrt{a}}$

After substituting the values of $a$ and $b$ into the equation and then calculating the definite integral from lower bound $0$ and upper bound $t$ I will need to find the inverse.

Any ideas how to go about this with a $ln$ in the result??? I need to make $t$ the subject

Title: help integrating


问:
Solve for time in constant velocity question (easy)
A speeder drives by a cop car moving 50m/s. The cop takes 12 seconds to get things in order and then chases after the speeder, traveling 60 m/s. How much time will it take the cop to catch up to the speeder?

This is what I have so far:

$v_c= 60$ m/s

$x_c = 60(t-12)$

$v_s = 50$ m/s

$x_s = 50t$

I set the two equations equal to each other and I have

$$50t = 60(t-12)$$

$$t=72$$

However the solution says

$$60t = 600+50t$$

$$t=60$$

Why am I incorrect in my thinking? And shouldn't both methods of solving give the same answer?

答:
You need to focus on what's been asked in the question.

It's asked How much time will it take for the cop to catch up to the speeder?, not after what time they meet or cross each other?.

As it's given that the cop takes $12$ seconds to get things in order and he starts after that, we need to find the time after the 12$^{th}$ second. So you have time = $t_{\text{used by you }}-12 = 72-12 = 60$.

The time $t$ you've considered gives the total time after which they meet and you just need to subtract the $12$ seconds the cop took to get things in order in order to find the time taken by the cop to catch the speeder.


问:
Finding function from its solutions
Let's say I want to find a function of which I only know its solutions, i.e. the points where it goes to zero.

$x_0 \: | \: f(x_0) = 0$

$x_1 \: | \: f(x_1) = 0$

$...$

Is there a way to programmatically find the function?

Thanks a lot.

答:
The problem is called interpolation.

You have to specify the type of the function you want. This gives a functional term with parameters. The mathematical problem is transformed to the problem of calculating the parameters.

There is software available for functions often used in sciences and engineering:

TableCurve 2D

TableCurve 3D

Title: Finding function from its solutions


问:
The functional equation $f(x + y) = f(x) + f(y) + xy$

Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the functional equation
$$f(x + y) = f(x) + f(y) + xy\text,$$
for all $x,y \in \mathbb{R}$.

I need hints for this problem.

答:
Define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - \frac { x ^ 2 } 2 $. Then the original functional equation leads to
$$ g ( x + y ) = g ( x ) + g ( y ) \text , $$
which is the famous Cauchy's functional equation.


问:
$\int_0^1 \frac{{f}(x)}{x^p} $ exists and finite $\implies f(0) = 0 $
Need some help with this question please.

Let $f$ be a continuous function and
let the improper inegral
$$\int_0^1 \frac{{f}(x)}{x^p} $$
exist and be finite for any $ p \geq 1 $.

I need to prove that
$$f(0) = 0 $$

In this question, I really wanted to use somehow integration by parts and/or the Fundamental theorem of calculus.
Or even maybe use Lagrange Means value theorem,
but couldn't find a way to se it up.
I'll really appreciate your help on proving this.

答:
Assume $f(0) > 0$. Since $f$ is continuous, we can find $\epsilon > 0$ so that $\forall x \in [0, \epsilon] : f(x) > 0$. Since $[0, \epsilon]$ is compact, we can find $m > 0$ so that $\forall x \in [0, \epsilon] : f(x) > m$. Thus:

$$
\int_0^\epsilon \frac{f(x)}{x^p}dx \ge m \int_0^\epsilon \frac{dx}{x^p}
$$

But the integral on the RHS diverges for all $p \ge 1$.

What's left is to show that this implies that the integral on $[0, 1]$ diverges too, and handle the case $f(0) < 0$. Both should be easy.


问:
What happens to the Stone-Cech compactification if you change “compact Hausdorff” to “T1 compact”?
I have asked a similar question elsewhere:

what happens to the Stone-Cech compactification if you change “compact Hausdorff” to “$T_1$ compact”? Here, I've added $T_1$
as opposed to this. Is $K$ there $T_1$ or at least $T_0$ ? In other words, does this modified thing always exist ?

答:
Have a look at the Wallman extension $wX$ of a $T_1$-space $X$. This is a compact $T_1$-space together with an emdedding $i : X \to wX$ such that

$i(X)$ is dense in $wX$.

Each continuous map $f : X \to Z$ to a compact Hausdorff $Z$ has a continuous extension to $wX$, i.e. there exists a continuous $F : wX \to Z$ such that $f \circ i = f$.

See here or consult

Engelking, Ryszard. "General topology." (1977).


问:
A proof of a suprising limit on Dirichlet's theorem
Dirichlet's theorem states that there are infinitely primes exist, which are a residues class of $r$ of $\pmod{n}$ where $n>2$ or ($p$ is congruent to $r \pmod{n}$),and also there occurs unique numbers $a$ and $b$ which represents a prime number of the form $a+nb$ where $n>0$ and the numbers in these unique collection are relative prime to each other. Eeg there exist infinity many primes numbers of form $6k+1$,$6k-1$ such as $5,7,11,13,19$ and many more.

So my question; is there a formal elementary proof which shows that as $x$ tends to infinity, the proportion of number of primes less than $x$, which are congruent to $r \pmod{n}$, is to number of primes less than $x$ approaches to reciprocal of Euler totient function $\varphi(n)$?

答: Yes, there is an "elementary proof", namely the one given by Selberg in 1949 here on JSTOR. I would be wary for asking about elementary proofs, since most likely you are thinking that "elementary=simple" which is most definitely not true. For example, the elementary proof of the PNT given by Erdos and Selberg is one of the most complicated proofs I've seen, and the original "complex" one using the fact that $\zeta(s)$ is zero free for $\Re(s)=1$ is in my opinion much more simple and straightforward.


问:
Let $f : A \to B$ and $g : B \to C$ be functions, Suppose that $g \circ f$ is bijective and $g$ is bijective. Prove $f$ is bijective.
My work so far is:

for 1-1:
for all $x,y \in A$, if $f(x)=f(y)$, then
since $g \circ f$ is bijective, $g(f(x))=g(f(y))$.
Since we know $g$ is bijective, then
$f(x)=f(y)$

is there a way I can get this down to $x=y$? I'm not sure I'm doing this the right way but this was all I could think of to at least start the problem. I'm not sure how to start solving onto; should I use substitution from knowing that $g \circ f$ and $g$ are surjective to try to get $f(a)=b$ ?

答:
I think you didn't notice but you actually proved that $f$ is $injective$.

if $g(f(x))=g(f(y))$ and since $(g\circ f)$ is $bijective$ $\Rightarrow$ $x=y$

Now let  $b \in B$ and let $c=g(b)$. Now since $g \circ f$ is $bijective$ so $\exists$ $k \in A$ such that $(g \circ f)(k)=c$. (Notice that $c \in C$).

So $g(f(k))=c=g(b)$ and since $g$ is $bijective$ so $f(k)=b$ and hence $f$ is $surjective$.

And we are done!!


问:
A question about the limit of a Lebesgue integration
For arbitrary $r_n\in\mathbb{R}$,show that
$$
\lim_{n\rightarrow\infty} \int_{0}^{+\infty} e^{-x}\sin^{n}(x+r_n)\, dx=0
$$
By Lebesgue Dominated Convergence Theorem, since
$$ |e^{-x}\sin^{n}(x+r_n)|<e^{-x}\in\mathbb{L}^1(0,+\infty)$$ We have
$$
\lim_{n\rightarrow\infty} \int_{0}^{+\infty} e^{-x}\sin^{n}(x+r_n) dx=
\int_{0}^{+\infty} e^{-x}\lim_{n\rightarrow\infty} \left(\sin^{n}(x+r_n)\right) dx
$$
However, the limit of $\lim_{n\rightarrow\infty} \sin^{n}(x+r_n)$ doesn't exist.

答:
The integral is bounded above in absolute value by

$$\int_0^\infty e^{-x}|\sin^n(x+r_n)|\,dx = \sum_{k=0}^{\infty} \int_{k\pi}^{(k+1)\pi}e^{-x}|\sin^n(x+r_n)|\,dx $$ $$\tag 1 \le \sum_{k=0}^{\infty} e^{-\pi k}\int_{k\pi}^{(k+1)\pi}|\sin^n(x+r_n)|\,dx.$$

But note $|\sin^n x|$ is $\pi$-periodic. Thus

$$\int_a^{a+\pi}|\sin^n x|\,dx =\int_0^\pi|\sin^nx|\,dx$$

for any $a.$ Therefore

$$\int_{k\pi}^{(k+1)\pi}|\sin^n(x+r_n)|\,dx = \int_{k\pi+r_n}^{(k+1)\pi+r_n}|\sin^n(y)|\,dy = \int_0^\pi |\sin^n(y)|\,dy$$

for all $k,n.$ Now the last integral $\to 0$ as $n\to \infty$ by the DCT. So the limit of $(1)$ equals

$$\lim_{n\to \infty}\left (\int_0^\pi |\sin^n(y)|\,dy\right)\cdot \sum_{k=0}^{\infty} e^{-\pi k}=0.$$

This gives the desired result.


问:
When $S$, the set of straight lines in $\mathbb{R}^2$, is a subbasis for a topology, what is the topology?
Salam. What are your thoughts on this? Again, please provide as much detail in your explanation as possible...

Let $\mathcal{S}$ be the collection of all straight lines in the plane $\Bbb R^2$. If $\mathcal{S}$ is a subbasis for a topology  $\tau$ on the set $\Bbb R^2$, what is the topology?

These are my thoughts so far:

If $\mathcal{S}$ is a subbasis then the basis it generates is all of the finite intersections of all of the straight lines in $\Bbb R^2$. Let's call this basis $\mathcal{B}$. The topology $\tau$ is generated by basis $\mathcal{B}$.

In order to get an idea of what $\tau$ is let's look at what kind of sets are in $\mathcal{B}$. We know that any straight line is in the basis since any straight line is its intersection with itself. But what do finite intersections of lines look like? Are they just all the points in the plane? If that is so then what is $\tau$?

Thanks.

答: Notice, that the intersection of the straight lines is a singleton, hence all finite intersections make a collection $\mathcal{B}$ of all singletons, which is a basis for the discrete topology in $\mathbb{R}^2$.


问:
Function spaces and domains

Hi, I was wondering if anyone could help me tackle this linear algebra question?

Currently I have shown that λ1 + λ2 = 0 as when using x=x1

f(x)= 1, g(x)=1 and h(x)=0

then λ1*(f(x))+λ2*(g(x))+λ3*(h(x)) = 0

λ1*(1) + λ2*(1)+λ3*(0) = 0

λ1+λ2=0

Now for b) I have assumed that the way to approach this is to replicate part a) but use x=x2 and then again with x=x3.

Which gives me:
λ1+λ3=0   and -λ2+λ3=0

and then great a matrix from those three conditions:

Row 1 = 1 1 0 | 0      Row 2 = 1 0 1 | 0    Row 3 = 0 -1 1 | 0

Which row reduces to give a row of zeros in row three hence creating a free variable which then tells us that they cannot be linearly independent but in fact linearly dependent.

Am I correct in this method or am I mucking up somewhere? Thanks in advance.

答: Your approach is correct. Another way of showing (b) would be to observe that $f-g=h$.

Title: Function spaces and domains


问:
Why is a field $A$ isomorphic to it's fractionnal field $K(A)$?
I am asked to demonstrate a field $A$ is isomorphic to it's fractionnal field $K(A)$ [it's not the notation i commonly use but i see wikipedia uses it so ,]

I tried to use different approaches but it seems i'm not sure of what's best.

I think a bit about the example of reals $R$ , constructing a isomorphism between reals and the reals fractions , seem impossible to me (I mean , in term of standard function , how could we send 0 on 0 , 1 to 1 , and be able to send elements on $\dfrac35$ as exemple , I don't really get it)

Any idea of what theorem would be the best to try to demonstrate that or , how to construct such a isomorphism ?

Francis Benjamin

答:
Define the map from $K(A)$ to $A$ by:
$$ \frac{a}{b} \mapsto ab^{-1} $$

Then show this map is an isomorphism and you are done!

Abit more details:

This map is clearly surjective.

Showing injectivity is essentially is recognizing that $ab^{-1} = a' b'^{-1}$ is the 'cross multiplication' of fractions


答: The solution of $t=60$ is correct. The speeder is $600$m ahead of the cop when he starts at $60$ meters per second. His speed relative to the speeder is $10$ meters per second so it takes $60$ seconds to catch the speeder.

Title: Solve for time in constant velocity question (easy)

Body:
A speeder drives by a cop car moving 50m/s. The cop takes 12 seconds to get things in order and then chases after the speeder, traveling 60 m/s. How much time will it take the cop to catch up to the speeder?

This is what I have so far:

$v_c= 60$ m/s

$x_c = 60(t-12)$

$v_s = 50$ m/s

$x_s = 50t$

I set the two equations equal to each other and I have

$$50t = 60(t-12)$$

$$t=72$$

However the solution says

$$60t = 600+50t$$

$$t=60$$

Why am I incorrect in my thinking? And shouldn't both methods of solving give the same answer?


问:
Prove trigonometric function is Lebesgue integrable
Is there a way to find an upper bound for $\frac{1}{(\cosh(x)-1)^{1/2}}$ that is Lebesgue integrable on $(0,2]$?

I am struggling to find an upper bound for the function which leads me to think there’s isn’t one. I’ve tried to make the denominator smaller but I can’t really replace the $-1$ by an exponential function without making the denominator negative at certain values in the domain.

答:
HINTS:

Note that we have

$$\sqrt{\cosh(x)-1}=\sqrt 2 \sinh(x/2)$$

As $x\to 0$, $\sinh(x)=O(x)$.  Can you conclude now?


答:
$\int_{0}^{1} \frac{f(x)}{x^p}dx=\int_{0}^{\delta}\frac{f(x)}{x^p}dx+\int_{\delta}^{1}\frac{f(x)}{x^p}dx$.

We know that $\int_{\delta}^{1}\frac{f(x)}{x^p}dx$ is bounded for given $\delta>0$.

For near $x=0$, $f(x)$ is almost constant, called $c$, because $f$ is continuous.

For small $\delta>0$,

$\int_{0}^{\delta} \frac{f(x)}{x^p}dx$ is almost $\int_{0}^{\delta} \frac{c}{x^p}dx=c\int_{0}^{\delta} \frac{1}{x^p}dx$

Thus $c$ should be $0$.

Title: $\int_0^1 \frac{{f}(x)}{x^p} $ exists and finite $\implies f(0) = 0 $

Body:
Need some help with this question please.

Let $f$ be a continuous function and
let the improper inegral
$$\int_0^1 \frac{{f}(x)}{x^p} $$
exist and be finite for any $ p \geq 1 $.

I need to prove that
$$f(0) = 0 $$

In this question, I really wanted to use somehow integration by parts and/or the Fundamental theorem of calculus.
Or even maybe use Lagrange Means value theorem,
but couldn't find a way to se it up.
I'll really appreciate your help on proving this.


问:
Matrix Idempotency Questions

Firstly I apologise for the use of an image but I am still unfamiliar with using the math text on here.

So I am currently studying linear algebra at second year level and have been working on this question.

I have proved a) by squaring the matrix and obtaining that it is equal to the original matrix.

I have also shown b) by saying that

$$(I-B)^2= I^2 - 2IB + B^2$$

Which then equates to $I - B$ due to the fact that $I^2=I$ and $B^2=B$ for Idempotency.

I have also shown that for c) the determinants may only equal 0 or 1 by saying:

$det(C^2)= det(C)$

$det(c)* (det(c) - 1)=0$

However I'm now stuck on how to approach questions d) and e).

Any help would be greatly appreciated.

答:
I'll expand my comments as required.

For (d): If $D$ is the matrix of $S$ and $I_n-D$ of $S'$, then the matrix of $S\circ S'$ is $$D(I_n-D)=D-D^2=D-D=0$$
so $S\circ S'$ is the null aplication.

For (e): As the matrix of $T$, $E$, is idempotent, then $E^2=E$. Also the matrix of $T^{9001}$ is $$E^{9001}=E^2 \cdot E^{8999}=E\cdot E^{8999}=E^{9000}=\ldots=E$$
so $T^{9001}=T$.

Title: Matrix Idempotency Questions


问:
Perfect powers in Horadam sequence
The Horadam sequence $\{W_n\}$ is defined by the Binet formula [Using notation from here]:

$$W_n=\frac{A\alpha^n-B\beta^n}{\alpha-\beta}$$

where,

$$A\ =\ b\ -\ a\beta \text{ and } B\ =\ b\ -a\alpha$$

$$\alpha = {p + d \over 2}, \beta = {p - d \over 2}, d = \sqrt {p^2 - 4q}$$

Questions:

Take all $a,b,\alpha,\beta \in \mathbb{Z}$.

Does $W_n$ ever equal a perfect power? Since the Horadam sequence is a generalization of Fibonnacci, Lucas sequences, there are a finite set of perfect power values assumed by $W_n$. Can we say anything about perfect powers in general for $A, B, \alpha, \beta$ that defines $W_n$?

If we take $A = \alpha, B = \beta$ and $GCD(\alpha, \beta) = 1$, is $W_n$ ever a power of $\alpha - \beta$? [Note: This is a special case of Beal's conjecture. So, the answer is likely to be $W_n$ is never a power of $\alpha - \beta$.]

For the second question, I tried $\beta = 1$ and set $W_n = y^m$ and obtained the Nagell-Ljunggren equation:

$$y^m = {\alpha^{n+1} - \beta^{n+1} \over \alpha - \beta} = {\alpha^{n+1} - 1 \over \alpha - 1}$$

with solutions $(\alpha, n+1, y, m) \in \{(3,5,11,2), (7,4,20,2), (18,3,7,3)\}$

What can we say about $\alpha \ne 1, \beta \ne 1$?

答:
If $A=\alpha$ and $B=\beta$ then
$$W_n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}=\sum_{i=0}^n\alpha^i\beta^{n-i}.$$
In particular, for $n=1$ this becomes $W_1=\alpha+\beta$, which is clearly a perfect power for plenty of choices of $\alpha$ and $\beta$. For $n=2$ this becomes
$$W_2=\alpha^2+\alpha\beta+\beta^2,$$
and for example for $\alpha=8$ and $\beta=7$ this becomes
$$W_2=\alpha^2+\alpha\beta+\beta^2=8^2+8\times7+7^2=13^2.$$

Title: Perfect powers in Horadam sequence


问:
Prove that every prime except one has the form $a^2 − b^2$ for some positive integers $a$ and $b$
This is a homework question, so I'm looking for some help to get started with the problem since I cannot think of a good way to do so. I know that we can say the integers a and b can be written as $a = p_1p_2...p_k$ and $b = q_1q_2...q_k$ where $p_i$ and $q_i$ are prime numbers. After this I'm stuck and don't know how to continue.
Any help is appreciated, thank you.

答: Hint: Try showing that every odd number can be written as the difference of two squares. Handle the prime number 2 separately.</p>


答: Note that $a^{2}-b^{2} = (a + b)(a - b)$. Note that every odd number, and thus every odd prime, can be written this way by taking $a-b = 1$. Thus, every prime except $2$ can be written in the form $a^{2}-b^{2}$. $\blacksquare$

Title: Prove that every prime except one has the form $a^2 − b^2$ for some positive integers $a$ and $b$

Body:
This is a homework question, so I'm looking for some help to get started with the problem since I cannot think of a good way to do so. I know that we can say the integers a and b can be written as $a = p_1p_2...p_k$ and $b = q_1q_2...q_k$ where $p_i$ and $q_i$ are prime numbers. After this I'm stuck and don't know how to continue.
Any help is appreciated, thank you.


问:
find $T(x,y,z)$ given the eigenvalues and eigenspaces
I have to find $T \in \mathcal{L}(\mathbb{R}^3)$ (which is linear) given the eigenvalues $\lambda_1=1$ and $\lambda_2=3$ with eigenspaces
$$V(1)=\{(x,x+y,y); x,y \in \mathbb{R}\} \equiv [(1,1,0),(0,1,1)] $$
$$V(3)=\{(0,x,2x); x \in \mathbb{R}\} \equiv [(0,1,2)]$$

My approach was to consider the basis $\beta=\{(1,1,0),(0,1,1),(0,1,2)\}$ and by the equation $T(v)=\lambda{v}$, where $v$ is an eigenvector associated to the eigenvalue $\lambda$ (right?). Thus:
$$T(1,1,0)=\lambda_1(1,1,0)=(1,1,0)\\
T(0,1,1)=\lambda_1(0,1,1)=(0,1,1)\\
T(0,1,2)=\lambda_3(0,1,2)=(0,3,6)$$
writting $(x,y,z) \in \mathbb{R}^3$ in terms of the elements of $\beta$:
$$(x,y,z)=(\alpha_1,\alpha_1+\alpha_2+\alpha_3,\alpha_2+2\alpha_3)$$
it's easy to show that $\alpha_1=x,\;\;\alpha_2=2y-2x-z,\;\;\alpha_3=z+x-y$. Plugging them in the equation above and applying T (which is linear) I got
$$T(x,y,z)=(x,2x+2z-y,4x-4y+5z)$$
I've checked that this transformation has eigenvalues 1 and 3 (it's good!). and the eigenspaces for these eigenvectors are generated by the same vectors ($[(1,1,0),(0,1,1)]$ for $\lambda=1$ and $[(0,1,2)]$ for $\lambda=3$). It seems to be consistent to me, but I wanna know if I made some mistake during my calculations.

答:
Your work looks correct.  In matrices, you could see this as converting to coordinates in the eigenvector basis, in which the transformation matrix is diagonal, and back again to standard coordinates, as follows:

$\begin{pmatrix}
1&0&0 \\ 1&1&1 \\ 0&1&2 \end{pmatrix}\begin{pmatrix}
1&0&0 \\
0&1&0 \\
0&0&3
\end{pmatrix}\begin{pmatrix}
1&0&0 \\
1&1&1 \\
0&1&2
\end{pmatrix}^{-1}=\begin{pmatrix}
1&0&0 \\
2&-1&2 \\
4&-4&5
\end{pmatrix}$


问:
Is the countable union of measure-zero sets zero?
Let $E=\cup_{i=1}^\infty A_i$ where the measure of $A_i$ is zero. How can I conclude that $E$ has measure zero?</p>

答:
It follows by sigma-aditivity.
$$m\bigg(\bigcup_{i\in\mathbb {N}}A_{i}\bigg)\leq\sum_{i\in\mathbb{N}}m(A_{i})=0$$</p>


问:
Every compact set is an image of a compact set under a continuous open surjection from locally compact into Hausdorff space
Firstly, the task:

Let $X$ be locally compact space, $Y$ be a Hausdorff space, $f:X \rightarrow Y$ be a continuous open surjection and let $K$ be a compact set in space $Y$. Prove that there exists a compact set $C$ in space $X$ such that $f(C)=K$.

I think that I have a proof, but it seems to use the axiom of choice. I have tried, but I can't decide whether I really use that axiom or not. I need a proof that doesn't use it. Please, help me prove it without that axiom, or explain me that I haven't really used it.

Sketch of my (incorrect) proof:

Since $f$ is surjective, for each $y \in Y$ there is some $x_{y} \in X$ such that $f(x_y)=y$. Let's choose $C=\{x_y | y \in Y\}$. Now $f$ maps each open covering of $C$ to an open covering of $K$ and that one has a finite subcovering. Now for every $y \in K$ there exists some open set in that subcovering that contains it, so the corresponding open set in the initial covering of $C$ contained $x_y$. The collection of all the corresponding open sets (the sets whose images are in the subcover of $K$) is finite and cover whole $C$, since elements $x_y$ are unique among those that map to a chosen $y \in Y$, so we can conclude that $C$ is compact too.

I doubt that this is correct, since I used only one of four conditions, and I think that it is because I use the axiom of choice.

答:
Your proof isn't correct, indeed. The subtle mistake is that, since $f$ is not (in general) injective, if $y\in f(U)$ you can't argue that the $x_y \in U$.

HINT: Every point of $f^{-1}(K)$ admits a compact neighborhood. Use these compact neighborhoods to get an open covering of $K$. (Keep in mind that these neighbourhoods are subsets of $X$ whereas $K$ is a subset of $Y\dots$).


答: Every odd number $2n+1$ can be written as the sum  of two consecutive numbers: $n,n+1$. The difference between two consecutive numbers is simply $1$. We see that $((n+1)+n)\cdot ((n+1)-n)=(n+1)^2-n^2=2n+1$, so every odd number, to include every odd prime number, can be readily represented as the difference of two squares. Can you finish by showing why the only even prime, $2$, cannot be so represented?


问:
How to prove $(\frac{n+1}{e})^n<n!<e(\frac{n+1}{e})^{n+1}$ without integrating method?
How to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.</p>

答:
Induct - the base case $n = 1$ is trivial. We will show both sides of the equality by looking at ratios; indeed, $\displaystyle\frac{(n+2)^{n+1} e^n}{e^{n+1} (n+1)^n} = \frac{1}{e}\cdot\left(1+\frac{1}{n+1}\right)^{n} \cdot(n+2) < \frac{1}{1+1/(n+1)} \cdot (n+2) = n+1$ since $\left(1 + \frac{1}{n+1}\right)^{n+1} < e.$

Hence, $\left(\frac{n+2}{e}\right)^{n+1} < \left(\frac{n+1}{e}\right)^{n}\cdot (n+1) < n!\cdot (n+1) = (n+1)!$ by hypothesis. For the other side, the ratio is $\displaystyle\frac{1}{e}\cdot \left(1 + \frac{1}{n+1}\right)^{n+1}\cdot(n+2)$; it suffices to show that this is greater than $n+1.$ That is, $\displaystyle \frac{1}{e}\cdot\left(\frac{n+2}{n+1}\right)^{n+1} > \frac{n+1}{n+2},$ which is equivalent to $\left(1+ \frac{1}{n+1}\right)^{n+2} > e.$ To prove this, we use the fact that $\ln (1+h) > \frac{h}{1+h}.$ Then $(n+2)\ln\left(1+\frac{1}{n+1}\right) > (n+2)\frac{1/(n+1)}{(n+2)/(n+1)} = 1,$ whence the desired inequality follows.


问:
$\lim_n nr^n$ if $|r| < 1$
In an exercise I'm asked to study the limit of the following sequence:

$$(nr^n)_{n\in \Bbb N}$$

I'm aware of the following:

If $|r|>1$ then the sequence diverges if $r$ is positive, or it has no limit if $r$ is negative.

If $r=1$ then it also diverges

If $r = -1$ then it has no limit

But I'm not being able to find out what the limit is when $|r| < 1$. I know that if that's the case then $\lim_n r^n = 0$, but I don't know how to do this for this sequence.

答:
Assume $0 < r < 1$ and note that
$$n=n(r + (1-r))^n=nr^n\left(1 + \frac{1-r}{r}\right)^n.$$

Using the binomial theorem, we have

$$nr^n = \frac{n}{\left(1 + \frac{1-r}{r}\right)^n} < \frac{n}{1 + n\frac{(1 - r)}{r} + \frac{n(n-1)}{2}\frac{(1-r)^2}{r^2} }.$$

Therefore, $nr^n \to 0$ as $n \to \infty$.

Title: $\lim_n nr^n$ if $|r| < 1$


问:
Find the relationship between $x_0$ and $y_0$ that ensures $x(t) → 0$ as $t → ∞$
If the roots of the auxiliary equation of a second order linear homogeneous
ODE are $k_1 > 0$ and $−k_2 < 0$ then the solution is
$$x(t) = Ae^{k_1t}+ Be^{−k_2t}.$$

For most choices of initial conditions
$x(0) = x_0$, $\dot x(0) = y_0$
we will have that $x(t) → ±∞$ as $t → ∞$. However, there are some special
initial conditions for which $x(t) → 0$ as $t → ∞$. Find the relationship between
$x_0$ and $y_0$ that ensures this

My working so far is $x_0 = A + B$ and $y_0 = Ak_1 - Bk_2$ and I'm really unsure of how to tackle this question further

答:
If

$x(t) \to 0 \; \text{as} \; x \to \infty, \tag 1$

we have

$A = 0, \tag 2$

lest the term

$Ae^{k_1t} \to \infty \tag 3$

and dominate

$x(t) = Ae^{k_1t} + Be^{-k_2t} \tag 4$

as $t$ increases without bound.  (We note here that

$Be^{-k_2t} \to 0 \; \text{as} \; x \to \infty.) \tag 5$

In light of (2),

$x(0) = x_0 = B \tag 6$

and

$\dot x(0) = y_0 = -Bk_2; \tag 7$

thus the relationship 'twixt $x_0$ and $y_0$ is

$x_0 = B = -\dfrac{y_0}{k_2}, \tag 8$

that is,

$y_0 = -k_2 x_0. \tag 9$

It is also easy to see that this equation forces $A = 0$, for

$x_0 = A + B \Longrightarrow k_1A - k_2B = y_0 = -k_2x_0 = -k_2A - k_2B \Longrightarrow k_1A = -k_2A$
$\Longrightarrow (k_1 + k_2)A \Longrightarrow A = 0, \tag{10}$

since

$k_1 + k_2 > 0; \tag{11}$

thus, (9) implies

$x(t) \to 0 \; \text{as} \; t \to \infty. \tag{12}$


问:
Transition matrix from basis to orthonormal basis
I am given a set $T = \{u_1,\ u_2,\ ... \ ,u_k\}$, and I am another set $T' = \{w_1,\ w_2, \ ...\ ,w_k\} $, where $T'$ is obtained from applying the Gram-Schmidt process to orthogonalise and normalise the vectors in $T$. I am tasked to find the transition matrix $P$ from $T$ to $T'$.

I know what the Gram-Schmidt process does, and I have attempted to express each element of $T'$, $w_i$, as the expression as deduced from the corresponding elements of $T$.

But beyond that, I am completely stuck. I admit that I do not have a good grasp on the concepts of linear algebra at this level. Does anyone have any tips on how I can proceed? Thank you!

答:
First, consider the fact that vectors in both $T$ and $T'$ are from the same subspace. Hence, we can express the vectors in $T$ as a linear combination of the vectors in $T'$. The general formula is given as follows:

$$u_k = (u_k \cdot w_1)w_1 + (u_k \cdot w_2)w_2 + \cdots + (u_k \cdot w_k)w_k$$

You can then express each of the vectors as a column matrix:

\begin{pmatrix}
u_k \cdot w_1 \\
u_k \cdot w_2 \\
\vdots \\
u_k \cdot w_k
\end{pmatrix}

The transition matrix can then be obtained by joining the column matrices together.


问:
Problem on implicit functions
I'm asked in Ted Shifrin's Textbook Multivariable Mathematics to :

Suppose $h: \mathbb{R}^2 \to \mathbb{R}$ is $\mathcal{C^1}$ and $\frac{\partial h}{\partial x_2} \neq 0$. Show that the equation $h \begin{pmatrix}y/x \\z/x \end{pmatrix}=0$ defines $z$  $\\$ (locally) implicitly as a $\mathcal{C^1}$ function $z= \phi \begin{pmatrix}x \\y \end{pmatrix} $,

This is easy enough: I fix a $x \neq 0$, use the first assumption and then apply the implicit function Theorem.

However I'm also asked to show that $$x \ \frac{\partial \phi}{\partial x} + y \ \frac{\partial \phi}{\partial y} = \phi \begin{pmatrix}x \\y \end{pmatrix}$$

Which has me stuck. I thought about trying to show that the equation of the tangent plane at $\begin{pmatrix}x \\y \end{pmatrix}$ is zero at the point $\pmb 0$ but I can't go further than that.
$$\\$$

Edit after Ted Shifrin asked me to.

Since $h$ does change as we wiggle $x_2$ we can fix $x \neq 0$ to obtain a continuous one to one onto function $\mathbb{R} \to \mathbb{R}$, namely: $z/x$, and so we know that if we wiggle $z$ then $h \begin{pmatrix}y/x \\z/x \end{pmatrix}$ changes too (the partial derivative isn't zero) using the previous fact. This, together with the assumptions in the exercise about $h$, let us deduce that there is a $\mathcal{C^1}$ function $\phi \begin{pmatrix}x \\y \end{pmatrix} = z$ .

The equation of the tangent plane of the graph of $\phi$ at $\pmb x$ is:
$$g\begin{pmatrix}v \\ w \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}v-x\\ w-y \end{pmatrix}$$
So, evaluating $g$ at $\pmb 0$ we obtain:

$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}= \phi\begin{pmatrix}x \\ y \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}-x\\ -y \end{pmatrix}$$

$$g\begin{pmatrix}0 \\ 0 \end{pmatrix}+ D\phi\begin{pmatrix}x \\ y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} = \phi\begin{pmatrix}x \\ y \end{pmatrix}$$

But I can't show how $g\begin{pmatrix}0 \\ 0 \end{pmatrix}=0$

答:
If one naively ignores all the possible pitfalls and differentiates the equation $h(y/x, z/x) = 0$ (pardon, but I'm not a fan of the vertical notation), which must be true of all $x, y$ within some neighborhood, one gets:

$$0 = \frac{\partial h(y/x, z/x)}{\partial y} = h_1\frac 1x + h_2\frac 1x\frac{\partial z}{\partial y}\\\frac{\partial z}{\partial y} = -\frac{h_1}{h_2}$$
Where $h_1$ and $h_2$ are the partial derivatives of $h$ with respect to $x_1, x_2$, respectively, and evaluated at $(y/x, z/x)$. And
$$0 = \frac{\partial h(y/x, z/x)}{\partial x} = -h_1\frac y{x^2} + h_2\dfrac{x\frac{\partial z}{\partial x} - z}{x^2}\\z = -\frac{h_1}{h_2}y + x\frac{\partial z}{\partial x}$$
substituting from the previous equation,
$$z = y\frac{\partial z}{\partial y} + x \frac{\partial z}{\partial x}$$

Of course, it needs some clean up and care to make sure all the "i"s are dotted and all the "t"s crossed, but the basic calculation is there.

Title: Problem on implicit functions


问:
How can I deduce a lower hemisphere's boundary's orientation?
Source: Stewart. Calculus: Early Transcendentals (6 edn 2007). p. 1098. §16.8, Exercise #19.

Despite reading this, I don't understand how to deduce the red sentence beneath.

$\text{19}.$ If $S$ is a sphere and $\mathbf{F}$ satisfies the hypotheses of Stokes' Theorem, show that $\iint_S \text{curl} \, \mathbf{F} \cdot d\mathbf{S} = 0 $.

Author's Solution:
Assume $S$ is centered at the origin with radius $a$. Let $U$ and $L$ be the upper and lower hemispheres, respectively, of $S.$ Then by Stokes' Theorem: $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_U (\nabla \times \mathbf{F}) \cdot d\mathbf{U} + \iint_L (\nabla \times \mathbf{F}) \cdot d\mathbf{L} = \oint_{\partial U} \! \mathbf{F}\cdot d\mathbf{r} + \oint_{\partial L} \! \mathbf{F}\cdot d\mathbf{r}  $

But ${\partial U}$ is $x^2 + y^2 = a^2$ oriented counterclockwise  and $\color{red}{{\partial L} \text{ is the same circle oriented clockwise}}$. (Rest of solution omitted)

Criterion A for Orientation of Boundary Curve (from Stewart p. 1093):

If you walk in the positive direction around $\partial S$ and your head points in the direction of $\mathbf{\hat{n}}$, then the surface will always be on your left.

Criterion B (from Thomas' Calculus, 2009 12 edn, p 963):

If the thumb of a right hand points along $\mathbf{\hat{n}}$, then the fingers curl in the direction of $\partial S$.

I know that the orientation of $S$ determines the positive orientation of $\partial S$. Also, the outward unit normals determine any closed surface's positive orientation . Here, the outward unit normals have $z$-component $< 0$ which yields the person in red.

$\color{purple}{\text{By Criterion A, the red person's head is pointing downwards (in the $-z$ direction)}}.$

$\color{purple}{\text{The purple direction is compulsory for the surface to be on that person's left.}}$

By Criterion B, my thumb must point in the $-z$ direction. So my fingers curl in the orange direction.

Why's there a discrepancy between Criteria A and B? Where's the mistake?

$\Large{\text{Supplementary to Prof Shifrin's Answer:}}$

Here's my interpretation: The red person must travel in the purple direction, for the surface to be on her left. Then the purple direction is counterclockwise when viewed from above the $xy$-plane, BUT when viewed from below (as the red person is doing), the purple direction is CLOCKwise.

I had to draw a clock in my updated picture to spot this. Without this clock, I still don't understand why the purple direction can be both clockwise and counterclockwise, depending on from where I look at ahe $xy$-plane. Is there a more intuitive or natural explanation, without drawing clocks?

$\Large{\text{2nd Supplementary to Prof Shifrin's Answer:}}$

I still don't understand how to spot that reflections across the $xy$-plane reserve orientation (but preserve direction), without looking at a sheet of paper from above and below.

$\Large{1.}$ Why did you choose the basis {$j, -i, k$}, instead of the standard basis for $\mathbb{R}^3$?

$\Large{2.}$ Doesn't $-i := (-1,0,0)$ imply that the tip of the vector, lying on the $x$-axis, point towards $-\infty.$ How does it "still point inward"?

答: You're wrong about criterion A. (Fantastic artwork, BTW.) For the lower hemisphere (think disk with downward-pointing normal) to be on your left, you must walk clockwise.</p>


问:
Let $U = \{ p \in {P_4}(F):p(2) = p(5)\}$ find basis of U
Let $U = \{ p \in {P_4}(F):p(2) = p(5)\}$ find basis of
Then extend the basis to a basis of P4.

I don't have any idea where should I start, can anyone please give me a hint?

答:
Let $u \in U$ then $u(x)=n_4x^4+n_3x^3+n_2x^2+n_1x_1+n_0$ we also know that

$u(2)=u(5)$ $\Rightarrow$ $16n_4+8n_3+4n_2+2n_1+n_0=625n_4+125n_3+25n_2+5n_1+n_0$

$609n_4+117n_3+21n_2+3n_1=0$ $\Rightarrow$ $n_1=-203n_4-39n_3-7n_2$

so we can say that

$U=\{p(x)=n_4x^4+n_3x^3+n_2x^2+n_1x_1+n_0:(n_0,-203n_4-39n_3-7n_2,n_2,n_3,n_4)\}$

there are 4 generators as you can see all we have to do is prove that they are linearly independent

$1,-7x+x^2,x^3-39x,x^4-203x$

$c_0+c_1(-7x+x^2)+c_2(x^3-39x)+c_3(x^4-203x)=0$ $\Rightarrow$ $c_0=c_1=c_2=c_3=0$

which means they are linearly independent.If you add x to this set you will have an extended basis


问:
Proving Coprimes - algebra
Let $a,b,c,d$ be integers . Suppose that a$ = 3b+2c$ and that $b$ is odd.

Suppose that $b$ is coprime to $c$. Prove that a is coprime to $b$.

so far....

since $b$ is coprime to $c$ there exists integers $x,y$ such that:
$1=xb+yc$ by Bezuts lemma.

not sure where to go from to show that:
$1= xb+az$ , where $x,z$ are integers.

答:
We try to link $1 = xb + yc$ to the given, $a = 3b+2c$.

Multiplying by $2$ we have:

$$2 = 2xb+ 2yc = 2xb + y(a-3b) = ya + (2x-3y)b$$

By Bezout's Lemma, $\gcd (a,b)$ divides $2$.

Now we just need to prove that $\gcd(a,b) \ne 2$. This can be shown by a condition given by the question that we haven't used yet.

$b$ is odd so $$b=2n+1$$ and b is not divisible by 2.

Title: Proving Coprimes - algebra


问:
How to find integrable upper bound for a hyperbolic function?
How do I find an upper bounds for the function

$$h(x)=\frac{1}{(\cosh(x)-1)^{1/3}}$$

on $(0,2]$? I have tried to use the monotone convergence theorem on intervals the functions $h(x)\chi_{[\epsilon,2]}$ but calculating the integral and it’s limit as $\epsilon \rightarrow 0$ is not viable.

答:
Note that we have

$$\frac1{(\cosh(x)-1)^{1/3}}=\frac1{(2\sinh^2(x/2))^{1/3}}=O\left(\frac1{x^{2/3}}\right)$$

which is Lebesgue integrable on $[0,2]$


答: Hint: Assuming the measure is Lebesgue measure, for each $A_i$, you can find an open set $O_i$ such that $A_i\subseteq O_i$ and the measure of $O_i$ is less than $\varepsilon/2^i$. If not, as long as the measure is subadditive, you can follow Dominic Michaelis' approach.</p>

Title: Is the countable union of measure-zero sets zero?

Body: Let $E=\cup_{i=1}^\infty A_i$ where the measure of $A_i$ is zero. How can I conclude that $E$ has measure zero?</p>


问:
Nowhere dense set characterization
Let $A \subseteq X$ then $A$ is nowhere dense ($int(\overline{A}) = \emptyset$) if and only if $A \subseteq \overline{X-\ \overline{A}}$

I've already done one side, $int(\overline{A})=\emptyset \Leftrightarrow X-\overline{(X-\overline{A})} = \emptyset \Leftrightarrow \overline{X-\overline{A}} = X \Rightarrow A \subseteq X = \overline{X-\overline{A}}$

答: For the other direction, for convenience let $U=\operatorname{int}\operatorname{cl}A$. $U\subseteq\operatorname{cl}A$, so $U\cap X\setminus\operatorname{cl}A=\varnothing$, and therefore $U\cap\operatorname{cl}(X\setminus\operatorname{cl}A)=\varnothing$, since $U$ is open. If $U\ne\varnothing$, let $x\in U$. Then $x\in\operatorname{cl}A$, and $U$ is an open nbhd of $x$, so $U\cap A\ne\varnothing$. Let $y\in U\cap A$; then $y\in A\setminus\operatorname{cl}(X\setminus\operatorname{cl}A)$, so $A\nsubseteq\operatorname{cl}(X\setminus\operatorname{cl}A)$.

Title: Nowhere dense set characterization


问:
Solve $2^{2x} + 9e^{-2x} = 6$ for x using substitution.
This is the equation I have:

$$2^{2x} + 9e^{-2x} = 6$$

I want to solve for x using the substitution method.

I've turned it into

$$4^x+\frac{9}{e^{2x}} - 6 = 0$$

But I do not know what to substitute and how to solve it.

答:
This based on the equation

$$
a^b = e^{b \log a}.
$$

Apply this to the only exponential not with base $e$ in your problem:

$$
\begin{aligned}
2^{2x} + 9e^{-2x} &= 6\\
e^{2x \log 2} + 9e^{-2x} &= 6
\end{aligned}
$$
Now, the only substitution for $x$ I see is $u = e^{2x}.$

$$
\begin{aligned}
u^{\log 2} + \frac{9}{u} &= 6\\
u^{1 + \log 2} + 9 &=6u\\
u^{1 + \log 2} -6u + 9 &=0
\end{aligned}
$$

I don't see any way to solve for $u$, and so for $x$, here.  Something involving the Lambert $W$-function, perhaps.


问:
Find the norm of a continuos linear operator
I know that the formula $$(T u)(t) := u(
\sqrt{t}),\qquad
 0 \le t \le  1,$$ defines a
continuous linear operator $T : L^1[0, 1]  \to L^1[0, 1]$, and I want to determine its
norm.

It's easy to see that $\|T\|\le 2$ but I could'n go further, neither by finding an element which norm is = 2, neither by lowering the upper bound.

答:
We have
$$\|Tu\| = \int_0^1 |Tu(t)| \, \mathrm{d}t = \int_0^1 |u(\sqrt{t})| \, \mathrm{d}t = \int_0^1 |u(s)| 2s \, \mathrm{d}s \leq 2\|u\|.$$
To maximize this, you need a function that has support only close to $1$. So let $u$ be the characteristic function of $[1-\frac{1}{n},1]$. Then
$$\frac{\|Tu\|}{\|u\|} = n\int_{1-\frac{1}{n}}^1 2s \, \mathrm{d}s = n\left(1-\left(1-\frac{1}{n}\right)^2\right) = 2 - \frac{1}{n}.$$
Letting $n \to \infty$ gives $\|T\| = 2$.


问:
Putting Pins in a Grid - UK Senior Maths Challenge

In an $8 \times 2$ grid, you can place $5$ blue pins, $2$ red pins and $1$ yellow pin. The only rule is that the two red pins cannot share an edge. How many different arrangements of the pins can be made?

Looking at the arrangements top to bottom are considered different arrangements, as are using left and right columns.

I think you can have 36 different arrangements for the red pins and 8 for the yellow pin, and since the blue pins are all the same, does this mean that the answer would be 288? I feel like my reasoning has gone wrong somewhere though, so if you could offer any advice, thanks.

答:
Considering all rotation as different arrangements (as clarified).

$2$ red pins, $5$ blue pins, $1$ yellow pin in $8$ x $2$ grid. Cells with red pins cannot share an edge (note, they can share a corner).

Say $8$ x $2$ grid is $2$ columns and $8$ rows. You put one red pin in the left top cell. Then you have $13$ valid cells to place the second red. Similarly for the left top cell. That makes it $13 \times 2$. Now let's come to the second row from top. Please note we have to ignore the first row as any possible (which is only $1$ each in this case for left and right) combinations is already counted. Now you have $2 \times 11$. Similarly keep striking off rows that are done and this gives you

$2 \times (13 + 11 + 9 + 7 + 5 + 3 + 1) = 98$ possibilities for red pins.

This leaves $14$ cells. You can choose $5$ cells for blue pins from it in ${14 \choose 5}$ ways. You are then left with $9$ cells and you can place $1$ yellow pin in $9$ ways.

That gives total of $98 \times {14 \choose 5} \times 9 = 1765764$ arrangements.


问:
Eigenvectors of $A^TA$ and $AA^T$
Let $A$ be a real $m\times n$ matrix, $m\leq n$. Now, what can we say about the eigenvectors of $A^TA$ and $AA^T$?

First of all, the rank of $A\leq \min(m,n)=m$.

Now, $R=A^TA$ is $n\times n$ matrix and $L=AA^T$ is $m\times m$ matrix.

What can we say about their eigenvectors and eigenvalues? (Like, how many eigenvectors corresponding to $0$ eigenvalue)

I know that the eigenvector of L corresponding to a non-zero eigenvalue is an eigenvector of R with the same eigenvalue and vice-versa. So if one has exactly $x$ eigenvectors corresponding to non-zero eigenvalues, then the other one also has exactly $x$ eigenvectors corresponding to those non-zero eigenvalues.  But what about the rest (if any)?

Note: Sorry for the confusion, I meant equivalence only. For every eigenvector (non-zero eigenval.) of $L/R$, there is a corresponding eigenvector for the other.

Please give the reasoning behind the claims too, thank you!

答:
What is true is that if $x$ is an eigenvector of $AA^T$ for nonzero eigenvalue $\lambda$, then $A^T x$ is an eigenvector of $A^T A$ for $\lambda$, and if $y$ is an eigenvector of $A^T A$ for nonzero eigenvalue $\lambda$, then $A y$ is an eigenvector of $A A^T$ for $\lambda$.

There are $r$ linearly independent eigenvectors $u_n$ of $A A^T$ for nonzero eigenvalues, where $r$ is the rank of $A$, and $r$ linearly independent eigenvectors $v_n$ of $A^T A$ for nonzero eigenvalues.  These may be chosen so that $u_n = A v_n$ for each $n$, and the corresponding eigenvalues are the same.  Another $n-r$ linearly independent eigenvectors of $A^T A$ are for eigenvalue $0$, and  $m-r$ linearly independent eigenvectors of $A A^T$ are for eigenvalue $0$.
These form bases of the null spaces of $A$ and $A^T$ respectively.

Title: Eigenvectors of $A^TA$ and $AA^T$


问:
Prove $010010001...0^i1$ is irregular without using pumping lemma
I think that we can prove it either using pigeonhole principle and a contradiction or showing that if it's regular then there's only one way to show it's DFA and since the number of states are finite and we can't make use of loops we have a contradiction.
But I somehow think the latter one is pumping lemma itself so I didn't spend so much efforts on it.</p>

答:
If this is a regular language there exists a DFA for it.
Since in the DFA's numbers of states are finite and number of strings belonging to language are infinite due to pigeonhole principle there exists a state $q_{i,j}$ in which by starting from $q_0$ we can read $w_i=01001...0^i$ and $w_j=01001...0^j$ we can reach that state. Without loss of generality we can assume that $i<j$. For reading $w_{j+1}$ first we have to read $w_j$ by reaching $q_{i,j}$ and then read $0^{j+1}1$ and reach an accepting state $q_{j+1}$. Although we can reach $q_{i,j}$ by reading $w_i$ and then read $0^{j+1}1$ and reach $q_{j+1}$ and finally read $w'=01001...0^i10^{j+1}$. Since $w'$ is not a string accepted by language we have reached a contradiction and hence $L$ is not regular.


问:
What relationship(s) [if any] exist between primorial primes and palindromic primes?
Information on primorial primes are in the following hyperlinks:

MathWorld - Primorial Prime

Wikipedia - Primorial Prime

On the other hand, we have the following hyperlinks providing information on palindromic primes:

MathWorld - Palindromic Prime

Wikipedia - Palindromic Prime

My question at this point would be: What relationship(s) [if any] exist between primorial primes and palindromic primes?

The first few primorial primes are

$$3, 5, 7, 29, 31, 211, 2309, 2311, 30029, 200560490131, 304250263527209$$

The first few palindromic primes in base-10 are:

$$2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, 14741, 15451, 15551, 16061, 16361, 16561, 16661, 17471, 17971, 18181$$ … sequence A002385 in OEIS

From these two lists, the following "conjecture" appears plausible:

$\mathbf{CONJECTURE}$:  In base-10, the only prime numbers that are both primorial primes and palindromic primes are $3, 5$ and $7$.

Is this conjecture known in the literature?  If so, does it have a name (i.e., has it been proposed before)?

答:
There is a big heuristic justification for why it is probably true.  First though, instead of thinking about a primorial prime also being a palindrome, we can simply think of it as a primorial number also being a palindrome and the math still will show that your conjecture is probably true.

So, primorial numbers are of the form $p_n\# \pm 1$, but that form is not easy to deal with as primes are complicated, so we will approximate it with:
$$p_n\# \pm 1 \sim e^n$$
So then let's think about the chance that a random number is a palindrome.  Well, the definition of a palindrome is that the first half of the digits are the same as the reverse of the last half of the digits.  Therefore if a number is about $10^x$ then the odds that it is a palindrome is $10^{x/2}$ because the first half of the digits can be anything, but the last half of the digits have to be match it.  The equation $10^x \to \frac{1}{10^{x/2}}$ is equivalent to $x \to \frac{1}{\sqrt{x}}$ so well be using the $2$nd as it is easier to use.

Now let's consider asking how many primorial numbers are also a palindrome with the number $>10^{11}$, because there are no counterexamples to your conjecture at least up to over $10^{11}$. Well, starting with $n=11$ we get $p_n\# > 10^{11}$, so well write a sum starting with it:
$$(\text{Approximate number of palindrome primorials} > 10^{11}) \approx \sum_{k=11}^\infty \frac{1}{\sqrt{e^x}}$$
That last step might be a little confusing.  Basically, we are adding up the chances that the numbers are counterexamples to see how many (or little) we would expect.

Next well write $\text{(Approximate number of palindrome primorials > 10000)}$ as $C$ for count.  Well also write $\sqrt{e^x}$ as $(\sqrt{e})^x$ to get:

$$C \approx \sum_{k=11}^\infty \frac{1}{(\sqrt{e})^x}$$

By solving the series we get: (Wolfram page here)

$$C \approx \frac{1}{(\sqrt{e} - 1)e^5}$$

Which means $C \approx 0.0103\dots$ which is less than a 1.1% chance!

To summarize what this is saying, is that the chance that besides the small examples you gave there is approximately less than a 1.1% chance that there is another.  This could be dramatically improved by just checking a few more primorials, and remember, this heuristic did not even take into account that the numbers have to be prime!

If you have any questions, just ask!