问:
How to find the side of the triangle?

I need to find the value on the $v_{1}$ side of the triangle, the book's answer is $v_{1} = V cos(\psi) + v'_{1} cos (\theta - \psi)$, but I couldn't understand how you get to that result. I used the cosine law but it didn't work, how can I solve it?

The main exercise is about elastic collisions, in which I need to relate the initial kinetic energy and the final kinetic energy of the particle in the LAB framework :

$$
\frac{T_{1}}{T_{0}}=\frac{m^2}{(m_{1}+m_{2})^2}.S^2
$$

$$
S= \cos(\psi)+\frac{\cos(\theta-\psi)}{(m_{1}/m_{2})}
$$

thereby

$$
T_{0}=\frac{m_{1}u_{1}^2}{2}
$$

$$
T_{1}=\frac{m_{1}v_{1}^2}{2}
$$

$$
\frac{T_{1}}{T_{0}}=\frac{v_{1}^2}{u_{1}^2}
$$

and the book says that by the triangle diagram you get the relationship
$v_{1} = V cos(\psi) + v'_{1} cos (\theta - \psi)$

The figure is the final state of mass $m_{1}$ for the elastic collision of two particles for the case $V<v'_{1}$ for which there is one trajectory.

$V$ is the velocity of the center of mass in the LAB framework.

That is question 9.38 from the book Classical Dynamics of Particles and Systems by Marion, Thornton.

答:
Taking inspiration from Piquito, in order to see the answer clearly,
draw the perpendicular
from the endpoint of $V$
(where $\theta$ is)
to $v_1$.

That divides $v1$ into 2 segments: The first is equal to
$V\cos(\psi)$,
and the second is equal to
$v_1'\cos(\theta-\psi)$. These two segments add up to the expression in your book's answer: $v1=V\cos(\psi)+v_1'\cos(\theta-\psi)$.

Title: How to find the side of the triangle?


问:
Show that no non-trivial open set in $\mathbb{R}^n$ has measure zero in $\mathbb{R}^n$
Show that no non-trivial open set in $\mathbb{R}^n$ has measure zero in $\mathbb{R}^n$.

What does non-trivial mean? (I think it means that the set contains a general subset of $\mathbb{R}^n$, doesn't it?)
Any idea for a rigorous proof?

My proof:
Let $A$ be the non-trivial open set in $\mathbb{R}^n$. It contains a closed rectangle $Q$. Since $A$ is open, there exist an open covering of $Q$, say {$Q_i$}$^{\infty}$, contained in $A$; moreover a finite subcollection, say {$Q_1,...,Q_n$}, covers $Q$ (choose the open covering so that each set has at most boundary points in common). Thus:$$\sum_{i=1}^{n}v(Q_i)\ge v(Q)$$
where $v(Q)$ is the volume of $Q$.
Now let {$A_i$}$^{\infty}$ be a countable collection of rectangles which covers $A$. My attempt is to show that the total volume of the rectangles $A_1,A_2,...$ cannot be made less than $v(Q)$; since {$A_1,A_2,...$} covers $A$, it will also covers {$Q_1,...,Q_n$}; therefore:$$\sum_{i=1}^{\infty}v(A_i)\ge \sum_{i=1}^{n}v(Q_i)$$
so that one ends the proof.

Is this correct?

答:
I guess this exercise is from "Analysis on Manifolds" by James R. Munkres. (Exercise 2. on p.97.)

Let $U$ be a non-empty open set in $\mathbb{R}^n$.

Assume that $U$ has measure zero in $\mathbb{R}^n$.

Since $U$ is open, there is a (closed) rectangle $Q$ which is contained in $U$.

Since $U$ has measure zero in $\mathbb{R}^n$, for $\epsilon=v(Q)$, there is a covering $\text{Int}\,Q_1,\text{Int}\,Q_2,\dots$ of $U$ by countably many open rectangles such that $$\sum_{i=1}^{\infty}v(Q_i)<\epsilon. \text{(See Theorem 11.1(c) on p.91.)}$$
Since $Q$ is compact, there is a finite subcollection $\text{Int}\,Q_{i_1},\dots,\text{Int}\,Q_{i_k}$ that covers $Q$.

So, the finite collection of (closed) rectangles $Q_{i_1},\dots,Q_{i_k}$ covers $Q$.

By Corollary 10.5 on p.88, $$v(Q)\leq\sum_{j=1}^{k}v(Q_{i_j})\leq\sum_{i=1}^{\infty}v(Q_i)<\epsilon=v(Q).$$
This is a contradiction.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
How to calculate $\int \frac{x^3}{\sqrt{9-x^2}}$
According to my textbook this is : $\frac{\sqrt{9-x^2}^3}{3} -9 \sqrt{9-x^2}$. However my solution is completely different. Here's my work:

$ x = 3\sin \theta, dx = 3 \cos\theta d\theta$

$$\int \frac{27\cos^3 \theta}{\sqrt{9-9\sin^2\theta}} = \int \frac{27\cos^3\theta}{3\cos^2\theta} = 9\int\cos^2 \theta$$

$$9\left(\int \frac{1}{2} d\theta+ \int \cos 2\theta\right) = 9\left(\frac{1}{2}\theta + \frac{\sin 2\theta}{2}\right)$$

Solving for $\theta$ yields:

$$\frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{4}\sin\left(2\arcsin\left(\frac{x}{3}\right)\right)$$

What's wrong with my solution?

答:
Substitute $u^2=9-x^2$ and integrate to obtain:
$$-9u + \frac{u^3}{3} +C.$$
(Observe that $2u\ du=-2x\ dx.$)


问:
Inner product defined by summation
I am trying to prove that $V$ is a finite dimension space over C and $(u_1,...,u_n)$ is a basis for $V$. Show that $\langle\,,\rangle$ : $V × V → C,$ defined by $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} b_iu_i\rangle=\sum_{i=1}^{n} a_i\overline {b_i}$

I tried to show the 4 properties of an inner product:

for additivity: $\langle\sum_{i=1}^{n} a_iu_i+\sum_{i=1}^{n} c_iu_i,\sum_{i=1}^{n} b_iu_i\rangle$ = $\sum_{i=1}^{n} (a+c)_i\overline {b_i} = \sum_{i=1}^{n} a_i\overline {b_i}+\sum_{i=1}^{n} c_i\overline {b_i} = \langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} b_iu_i\rangle +\langle\sum_{i=1}^{n} c_iu_i, \sum_{i=1}^{n} b_iu_i\rangle$

Then for homogeneity and conjugate symmetry, it's the same way as above, it's just properties of summations and scalars

For positivity and defineness, $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} a_iu_i\rangle = \sum_{i=1}^{n} a_i\overline {a_i} = \sum_{i=1}^{n} |a_i|^2 \geq0$

and if  $\langle\sum_{i=1}^{n} a_iu_i, \sum_{i=1}^{n} a_iu_i\rangle = 0, \sum_{i=1}^{n} |a_i|^2 =0$ and since $|a_i|^2 \geq0$ , $a_i=0$ and each  $\sum_{i=1}^{n} a_iu_i =0$

My question is, is my outline correct? If so, why did my proof did not mention anything about a basis and linear independence which is mentioned in the assumption? Where should those conditions about a basis appear in the proof?

答: Those conditions were used to make $\langle \cdot,\cdot \rangle$ a well-defined function. More concrete: as $(u_1,\dots,u_n)$ is a basis for $V$, the function $\phi : \mathbb C^n \to V$ given by $\phi(a_1,\dots,a_n) = a_1u_1 + \cdots + a_nu_n$ is a linear isomorphism (why?) and then we define $\langle \cdot,\cdot \rangle : V \times V \to \mathbb C$ by $\langle v_1,v_2 \rangle = \langle \phi^{-1}(v_1),\phi^{-1}(v_2) \rangle_{\mathbb C^n}$ where $\langle \cdot,\cdot \rangle_{\mathbb C^n}$ is the standard inner product in $\mathbb C^n$.

Title: Inner product defined by summation


问:
Support of a $L^2$ function as a distribution.
For a
$$f \in L^2(\mathbb{R})=L^2(\mathbb{R})^*
\subset D^*$$
so we can regard it as a distribution.

So we have support set of f as a distribution.
$$supp f=\{x\in \mathbb{R}: \text{for any open neighborhood U of x , restrictions of f on U is not }0\}.$$

Actually $f \in L^2$  as a function we have
$$Supp f=\overline{\{x\in \mathbb{R}:f(x) \neq 0\}}$$
But they are different.
For example ,$f(x)=1$,when x is rational,otherwise $f=0$.
For f as a distribution we have $supp f$ is empty,but $Supp f$ is the $\mathbb{R}$.

Here is my question:
For a nonnegative function $f\in L^2$ , do we have $f>0 \quad a.e$ on its support set $supp f$.

Any suggestion will be appreciated.

答:
Look at $$f=\sum_{n\ge 1}\sum_{k =0}^{2^n} 1_{\displaystyle [\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}}]} \in L^2[0,1]$$

$f\ge 1$ on a neighborhood of each $k/2^n$ so $$supp(f)= [0,1]$$ but $f$ vanishes on the set with non-zero measure $$[0,1] - \bigcup_{k,n}(\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}})$$

($f>0$ a.e. on its support means that $\mu(f^{-1}(0)\cap supp(f)) = 0$ right ?)


问:
Lagrange's Multiplier for $f(x,y,z) = (xyz)^2$
Given a function:

$$f(x,y,z) = (xyz)^2$$

And the set: $$S = \{(x,y,z) : x^2 + y^2 + z^2  = a^2\} $$

Does $f$ have a maxima on $S$?

(Is this argument correct? : Yes, because f has a maxima and minima in S if S is compact and f is continuous. But here S is compact so we have a maxima)

Also, how do I tell if a function has maxima or minima on a set if the set is not compact?

I tried using the lagrange multiplier.

$f_x = 2xy^2z^2$, $f_y = 2yx^2z^2$,  $f_z = 2zx^2y^2$, $f_\lambda = x^2 + y^2 + z^2 - a^2$

$$f_x + \lambda g_x = 0$$
$$f_y + \lambda g_y = 0$$
$$f_z + \lambda g_z = 0$$
$$f_\lambda = 0$$

However on solving this system of equations using substitution, I am unable to find a solution for (x,y,z). I did find one point though (0,0,0), for $a = 0$,  Can I set $a$ to be $0$?

$$2xy^2z^2 = \lambda2x$$
$$2yx^2z^2 = \lambda2y$$
$$2zx^2y^2 = \lambda2z$$

So suppose $x = 0$ and $\lambda \not = 0$, we have $\lambda2y = 0$, so $y = 0$, similarly $z = 0$, $(0,0,0)$ satisfies $f_\lambda = 0$ for $a = 0$, so this is one point.

However, evaluating for other points, I am unable to simplify.

For $x \not = 0$, we have: $y^2z^2 = \lambda$

But how do I proceed from now?

Can I say by symmetry, if any one of $x$, $y$ or $z$ is $0$ then they all have to be $0$?

If so, then we have all non zero :

$$x^2z^2 = x^2y^2 = y^2z^2$$

which means $$ x^2 = y^2 = z^2$$

And then substituting this into constraint, we get 6 points of the form

$$( \pm\ a/\sqrt3, \pm\ a/\sqrt3, \pm\ a/\sqrt3)$$

Is this correct?

答:
Define : $\mathcal{L}(x,y,z,\lambda) = (xyz)^2 - \lambda\cdot(x^2 + y^2 + z^2 - a^2)$ for any $\lambda \in \mathbb{R}$ , and for any $x, y, z \in \mathbb{R}$ .

So, $~\mathcal{L}_x = 2xy^2z^2 - 2\lambda x~$ , $~\mathcal{L}_x = 2x^2yz^2 - 2\lambda y~$ , and $~\mathcal{L}_x = 2x^2y^2z - 2\lambda z~$ . Also, we have : $~\mathcal{L}_\lambda = a^2 - (x^2 + y^2 + z^2)~$ .

Equating these to $0$ , we get : $\lambda = x^2y^2 = y^2z^2 = z^2x^2~$ , and $~~a^2 = x^2 + y^2 + z^2$ .

Now, if any of $x, y, z$ is $0$ , then $f$ won't be maximized for sure, in fact it'll get minimized then. So , multiplying the first $3$ constraints by $z^2, x^2 , y^2$ respectively , we get $\lambda x^2 = \lambda y^2 = \lambda z^2$ , which implies that $f$ is maximized when $x^2 = y^2 = z^2 = k~$ (Say) .

Thus, by Lagrange Multiplier Theorem, using the 4th constraint, we get that $f$ is maximized when $a^2 = x^2 + y^2 + z^2 = 3k$ , i.e. $~x^2 = y^2 = z^2 = \frac{a^2}{3}$ .

Therefore, $f$ is maximized when $~(x,y,z) = \left(\pm\dfrac{| a |}{3} , \pm\dfrac{| a |}{3} , \pm\dfrac{| a |}{3} \right)$ , and the maximum value of will be $\dfrac{a^6}{27}$ .

In fact, you don't need a strong result like Lagrange Multiplier in order to solve this problem. Observe that by simple AM-GM Inequality, we have : $$(xyz)^2 = x^2\cdot y^2 \cdot z^2 \leqslant \left[\dfrac{x^2 + y^2 + z^2}{3}\right]^3 = \left(\dfrac{a^2}{3}\right)^3 = \dfrac{a^6}{27}$$ and the maximum is attained when $x^2 = y^2 = z^2 = \dfrac{a^2}{3}$ , which is certainly same as the result produced using Lagrange Multiplier. Hope it helps.


问:
Existence of analytic function with certain conditions

$1$. Does there exists an analytic function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(z)=z$ for all $|z|=1$ and $f(z)=z^2$ for all $|z|=2$.

$2$.There exists an analytic function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(0)=1,f(4i)=i$ and for all $z_j$ such that $1 < |z_j| < 3, j=1,2$ we have
$|f(z_1)-f(z_2)| \leq |z_1-z_2|^{\frac{\pi}{3}}$

For $2$, I get that $f$ is constant on the annulus. Then by identity theorem, I can conclude that f cannot exist. Am I right?

I don't know how to proceed $1$

答: Your answer for second part is correct. The first part is much simpler: $f(z)-z$ is analytic on $\mathbb C$ and its zeros have a limit point. Hence $f(z)=z$ for all $z$.


问:
How to show that this polynomial has all it's roots in in the disk $|z|<3$
This question was asked in my analysis assignment and I am unable to solve it.So, I am looking for help here.

Show that the polynomial $z^3 -z^2 +4z+5=0 $ has all its roots in the disk |z|<3.

I thought of using Rouché's theorem but I am not able to choose a $g(z)$ which would work. So, I would be unable to provide anything as attempt as this is the starting point of the answer if question has to be solved using Rouché's theorem

Can you please help me with finding $g(z)$?

I hope Rouché's theorem would work in this question.

答: Take $f(z)=z^3$ and $\varepsilon(z)=-z^2+4z+5$. Then, $z^3-z^2+4z+5=f(z)+\varepsilon(z)$ and, if $|z|=3$,\begin{align}\bigl|\varepsilon(z)\bigr|&=|-z^2+4z+5|\\&\leqslant3^2+4\times3+5\\&=26\\&<27\\&=|z^3|\\&=\bigl|f(z)\bigr|.\end{align}So, $z^3-z^2+4z+5$ has as many zeros on the disk $D(0,3)$ as $f$, which has $3$ zeros there (if we count them with their multiplicities).


问:
Solve the following integral $\iiint \left(x+y+z\right)\left(x-y-z\right)\left(x+y-z\right)dV$
Hello there I am looking how to solve this integral

$$\iiint \left(x+y+z\right)\left(x-y-z\right)\left(x+y-z\right)dV$$

using change of variable under the following planes:

$x\:+\:y\:+\:z\:=\:0$

$x\:+\:y\:-\:z\:=\:0$

$x\:-\:y\:-\:z\:=\:0$

Please give some advices to do solve it or any hints you have in mind...

Thanks

答:
Multiply out the polynomial and integrate each term to find:

$$\frac{1}{6} x y z \left(2 x^2-2 y^2-3 y z-2 z^2\right)$$


答: No theorem is needed for this. If $|z| \geq 3$ we get $|z^{3}-z^{2}| \geq |z|^{2} (|z|-1)\geq 9(|z|-1)$. But $|4z+5| \leq 5+4|z| <9(|z|-1) \leq |z^{3}-z^{2}|$.  Hence, the equation cannot hold if $|z| \geq 3$.

Title: How to show that this polynomial has all it's roots in in the disk $|z|<3$

Body:
This question was asked in my analysis assignment and I am unable to solve it.So, I am looking for help here.

Show that the polynomial $z^3 -z^2 +4z+5=0 $ has all its roots in the disk |z|<3.

I thought of using Rouché's theorem but I am not able to choose a $g(z)$ which would work. So, I would be unable to provide anything as attempt as this is the starting point of the answer if question has to be solved using Rouché's theorem

Can you please help me with finding $g(z)$?

I hope Rouché's theorem would work in this question.


问:
For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even
I want to know for what valuyes this function is even

I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!

答:
Hint: Show that a polynomial is even if and only if the coefficients of all the odd powers are 0.

This is a complete classification of even polynomial functions, which is more general than the question you're asking, but a good concept for you to be aware of.


问:
Subset of Permutations of distinct symbols, with positional count constraint/
I am working a personal project, and I need to derive subsets of permutations of unique elements.

For example:  I have a set of 5 unique elements {A, B, C, D, E}.   There are 60 permutations tuples of size 3 of this: {A,B,C}, {A,B,D}...{E,C,D}, {E,D,C}.  What I want is to select tuples from this list such that the count of each symbol in the same position across all tuples is constrained as 0 < A# < B# < C# < D# < E# the count of each symbol is the same in each position.

E.g.

```<code>{A, B, C}
{A, B, D}
⋮
{E, C, D}
{E, D, C}
</code>```

For the full set of tuples, there are 12 of each symbol in each position within the tuple, or specifically for what is explicitly written above there are quantity A2 and E2 in position 0, B2 C1 D1 in position 1, and C2 and D2 in position 2, so this does not meet the criteria.

With the example of 5 permute 3, as there are only 60 permutation tuples I was able to create such a desired subset by brute force, finding a collection of 20 permutation tuples resulting in counts of A2, B3, C4, D5, E6 in each position.

My question is:   Is there a known algorithmic way of selecting subsets with this type of criteria, and is there a predictive formula to calculate how many valid solutions can be derived?  I will need this to be for quantities of symbols up to 15 and up to tuple size 5 so brute force is sub-optimal.

For completeness, the solution subset I found for 5 symbols permut 3 is the following set:

```<code>AEC
AED
BCE
BDE
BED
CBE
CDB
CDE
CED
DBC
DBE
DCE
DEB
DEC
EAC
EAD
ECA
ECD
EDA
EDB
</code>```

Edit:  Using logic inspired by Ross Millikan, I verified that 4 permut 3 has no solution, as the permut domain is too small.  I did find via brute force a solution for 6 permut 3: A3,B4,C5,D6,E7,F8 for a total of 33 tuples:

```<code>ABC
ABE
ACD
BCF
BDF
BEF
BFE
CAD
CED
CEF
CFD
CFE
DCE
DCF
DEC
DEF
DFC
DFE
ECF
EDB
EDC
EDF
EFB
EFC
EFD
FAD
FAE
FBA
FBE
FDA
FDB
FEA
FEB
</code>```

答:
Not an answer, but I hope helpful and too long for a comment.

There are a number of constraints that you can use to guide the search.  The big one is that the sum of the counts in each position must match the number of tuples.  For your example, you have $20$ tuples and $2+3+4+5+6=20$  Among the $60$ permutations you have $E$ at each slot in $12$ of them.  If you were to demand $E\#=12$ you would need $36$ permutations to satisfy that because there are $12\ E$s in each position, which is all the permutations that include $E$.  There are $24$ non-E slots which are equally distributed among the other letters, violating your constraint that there be less $A$s than $B$s and so on.

The number of permutations is the sum of the counts because you have one letter in each position of each permutation.  If you demanded $E\#=11$ you would need at least $33$ permutations (more if some do not contain $E$ at all).  The sum of the other four counts would be at least $22$.  That means $D\#$ is at least $7$, so there are at least four $D$s in one position matched with $E$s in the same position.  However there are only three choices for the third position, so this will not work.  The same argument works for $E\#=10$.  You have a hope with $E\#=9, 27$ total permutations with $D\#=6, C\#=5, B\#=4, A\#=3$  Three of the $D$s in each column go with $E$s in each of the other two.  That gives you $18$ permutations including $DE$ and  one of each of the other letters and uses two of the other letters in each column.  If we focus on the ones with $E$ in the first column we have
E,D,(A,B,C)

E,(A,B,C),D

for six and we need three more which can only be
EAC
ECB
EBA

and we have an equal number of $A,B,C$ so this fails.

I don't think it is easy


问:
Looking for spectrum of an operator
Let's define the following Hilbert space on the field $\mathbb{C}$:
$$ \ell^2(\mathbb{Z}) = \{(...,x_{-1},x_0,x_1,...): \sum_{n \in \mathbb{Z}} |x_n|^2 < \infty   \}  $$
with the inner product $\langle x,y \rangle = \sum_{n \in \mathbb{Z}} x_n \overline{y_n}$ and define operator $T \in L(\ell^2(\mathbb{Z}))$ by
$$T(...,x_{-1},x_0,x_1,...) = (...,\frac{x_{-2} + x_0}{2},\frac{x_{-1} + x_1}{2},\frac{x_{0} + x_2}{2},...). $$

I would like to classify the spectrum of this operator.

First of all, this operator is self-adjoint, $\lVert T \rVert = 1$ so from few properties of spectra of such operators we know that:

$\sigma(T)$ is real,

$\sigma(T) \subset [-1,1]$

$r_{\sigma}(T) = \lVert T \rVert$ hence at least one of $-1,1$ belong to spectrum,

residual spectrum of this operator is empty.

I wanted now to check what is the continuous spectrum and the point spectrum of this operator, since then $\sigma(T) = \sigma_P(T) \cup \sigma_C(T)$.

Let's fix $\lambda \in \sigma_P(T)$. We look for such $x \neq 0$ that
$$Tx - \lambda x = 0.$$
I noticed the following: suppose that there exists (a finite) $N \in \mathbb{Z}$ such that $N$ is the smallest integer satisfying $x_n \neq 0$. Then, from the definition of $T$ we get
$$\frac{x_{N-2} + x_{N}}{2} = \lambda x_{N-1}$$
and since by assumption $x_{N-2} = 0 = x_{N-1}$ hence
$$\frac{0 + x_N}{2} = \lambda 0$$
which means that $x_N = 0$; contradiction.

Similarly, if there exists (a finite) $N \in \mathbb{Z}$ being the largest integer satisfying $x_N \neq 0$ then
$$\frac{x_N + x_{N+2}}{2} = \lambda x_{N+1}$$
and since $x_{N+1} = 0 = x_{N+2}$ hence $x_{N} = 0$; contradiction.

This means that we cannot have a situation, where $x$ is equal $0$ starting from some point.

Unfortunately here I'm stuck. I would appreciate if anyone could share some thoughts here.

Thanks!

答:
The solution is straightforward but the details tedious.

Suppose such a (real) eigenvalue $\lambda$ exists. Then $x$ satisfies the difference equation
$x_{k+2}=2\lambda x_{k+1}-x_k$ in both 'directions' (positive & negative $k$).

The roots of the difference equation are ${ 1\over 2}(2\lambda \pm \sqrt{4\lambda^2-4})$, or $\lambda \pm \sqrt{\lambda^2-1}$.

Since $|\lambda| \le 1$ we see that the real part of the roots is $\lambda$.

If $(x_0,x_1) \ne0$, then if $\lambda \ge 0$ then $x_k \not\to 0$ and if $\lambda <0$ then $x_{-k} \not\to 0$, and so we must have
$(x_0,x_1) =0$ and so $x=0$.

In particular, no such $\lambda$ can exist. Hence $T$ has no point spectrum.

I claim that the continuous spectrum is $[-1,1]$. Since we know the spectrum is closed, it is sufficient to show that a dense subset of $[-1,1]$ is in the continuous spectrum.

It is straightforward to show that if $\lambda$ is not in the point or residual spectrum then it is in the continuous spectrum iff for any $\epsilon>0$ there is a unit vector $x$ such that $\|(T-\lambda I)v\| < \epsilon$.

Let $T_n$ be the $n\times n$ dimensional tridiagonal Toepliz matrix that 'corresponds' to $T$.
From https://mathoverflow.net/a/188489/31729 we see that the eigenvalues are $\lambda_{n,k} = \cos ({k \pi \over n+1})$, with $k=1,...,n$. Let $v_{n,k}$ be a corresponding unit eigenvector.

It is straightforward to see that $\lambda_{n,k}$ are dense in $[-1,1]$.

Very roughly speaking, the idea is to 'stack' a finite number of the $v_{n,k}$ eigenvectors together so that we can find a unit $v$ with $\|(T-\lambda I)v\| < \epsilon$. Then we see that the $\lambda_{n,k}$ are in the continuous spectrum which finishes the proof.

Define the permutation $P$ on $\mathbb{R}^n$ by $P e_k = e_{n-k+1}$, note that $P^2=I$ and it is straightforward to check that if $T_n v_{n,k} = \lambda_{n,k} v_{n,k}$ then $T_n Pv_{n,k} = \lambda_{n,k} Pv_{n,k}$. In particular, 'reversing' the eigenvector $v_{n,k}$ gives another eigenvector.

Define the linear injection $\sigma_k:\mathbb{R}^n \to l_2$ by $\sigma_k(e_j) = e_{k+j-1}$ where the $e_k$ are the usual basis vectors in their respective spaces.

Note that $T \sigma_k(x) = \sigma_k(T_n x) + {1 \over 2} x_1 e_{k-1}+ {1 \over 2} x_n e_{k+n}$ and in particular,
$(T -\lambda I)\sigma_k(v_{n,k}) = {1 \over 2} [v_{n,k}]_1 e_{k-1}+ {1 \over 2} [v_{n,v}]_n e_{k+n}$ and
\begin{eqnarray}
(T -\lambda I) \sigma_{k+n+1}(-Pv_{n,k}) &=&  {1 \over 2} [-Pv_{n,k}]_1 e_{k+n}+ {1 \over 2} [-Pv_{n,k}]_n e_{k+2n+1} \\
&=& - {1 \over 2} [v_{n,k}]_n e_{k+n}- {1 \over 2} [v_{n,k}]_1 e_{k+2n+1}
\end{eqnarray}
Adding together gives
$(T -\lambda I) (\sigma_k(v_{n,k})+ \sigma_{k+n+1}(-Pv_{n,k})) = {1 \over 2} [v_{n,k}]_1 e_{k-1} - {1 \over 2} [v_{n,k}]_1 e_{k+2n+1}$.

Let $v_m = \sum_{i=0}^m \sigma_{k+i(n+1)} ( (-P)^{i} v_{n,k})$ (that is, $m+1$ copies of $v_{n,k}$ or $-P v_{n,k}$ stacked together but separated by a $0$). It is not too hard to see (assuming I have not made an error) that
$(T -\lambda I) v_m = {1 \over 2} [v_{n,k}]_1 e_{k-1} + {1 \over 2} (-1)^m [v_{n,k}]_{l_m} e_{k+n+ m(n+1)}$,
where $l_m$ is $1$ if $m$ is odd and $n$ is $m$ is even.
Note that $\|v_m\| = \sqrt{m+1}$, and we see that
$\lim_{m \to \infty} {1 \over \sqrt{m+1}} \|(T -\lambda I) v_m\| = 0$.

Title: Looking for spectrum of an operator


问:
Compute the integral $\int_C (z^2-1)^\frac{1}{2} dz$ where $R>1$
Let C be the circle of Radius $R>1$, centered at the origin, in the complex plane. Compute the integral

$\int_C (z^2-1)^\frac{1}{2} dz$

where we employ a branch of the integrand defined by a straight branch cut connecting $z=1$ and $z=-1$, and $(z^2-1)^\frac{1}{2} > 0 $ on the line $y=0$, $x>1$. Note that the singularities are not isolated in this case. One way to do this integral is to expand the integrand in a Laurent series valid for $|{z}|> 1$ and integrate term by term

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\oint_{\verts{z}\ =\ R\ >\ 1}\,\,\,
\pars{z^{2} - 1}^{1/2}\,\,\dd z}\ =\
\overbrace{\oint_{\verts{z}\ =\ R\ >\ 1}\,\,\root{z + 1}\root{z - 1}
\,\,\dd z}^{\substack{\ds{Both\ root\ are\ de\!fined\ as} \\[1mm]
\ds{Principal\ Branchs}}}
\\[5mm] = &\
-\int_{-R}^{-1}\root{-x - 1}\expo{\ic\pi/2}\root{-x + 1}\expo{\ic\pi/2}
\,\dd x
\\[2mm] &\
-\int_{-1}^{1}\root{x + 1}\root{-x + 1}\expo{\ic\pi/2}\,\dd x
\\[2mm] &\
-\int_{1}^{-1}\root{x + 1}\root{-x + 1}\expo{-\ic\pi/2}\,\dd x
\\[2mm] &\
-\int_{-1}^{-R}\root{-x - 1}\expo{-\ic\pi/2}\root{-x + 1}\expo{-\ic\pi/2}
\,\dd x
\\[5mm] = &\
-2\ic\int_{-1}^{1}\root{1 - x^{2}}\,\dd x =
-4\ic\int_{0}^{\pi/2}\cos^{2}\pars{\theta}\,\dd\theta
\\[5mm] = &\
\bbx{-\pi\,\ic} \\ &
\end{align}


问:
Consider the element $2 + \sqrt{-5}$ of $\Bbb Z[\sqrt{-5}] = \{a+b\sqrt{-5}\}$. Who are the associates?
Just to see if I got it right.

$\Bbb Z [ \sqrt{-5}]$ has only 1 and -1 as a unit as seen here:

units of $\mathbb Z[\sqrt{-5}]$

I konow that. Let $a, b \in D$, $a$ and $b$ are said to be associated if $a\mid b$ and $b\mid a$. But as the only units are $1$ and $-1$, then there is no $ \neq b$ such that $a = ub$, as seen here:

Prove that $2$, $3$, $1+ \sqrt{-5}$, and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$.

So $2 +  \sqrt{ -5} $ has no associates.

答:
It sounds like you have the right idea, but you have omitted a lot of details. Let me fill them in.

Let $a$ be an associate of $2 + \sqrt{-5}$ in $\Bbb{Z}[\sqrt{-5}]$. Then we find that there exist $b,c\in\Bbb{Z}[\sqrt{-5}]$ such that $(2 + \sqrt{-5})b = a$ and $ac = 2 + \sqrt{-5}$. Combining these equations and rearranging, we find that
$$
(2 + \sqrt{-5})(bc - 1) = 0.
$$

Now, $2 + \sqrt{-5}$ is not a zero divisor in $\Bbb{Z}[\sqrt{-5}]$, so that we must have $bc = 1.$ Thus, $b$ and $c$ are units, so that $b,c\in\{1,-1\}.$ If $bc = 1,$ we must have $b = c,$ so that the only associates of $2 + \sqrt{-5}$ in $\Bbb{Z}[\sqrt{-5}]$ are $\pm(2 + \sqrt{-5}).$


问:
Does $(1-(1-p)^n)^\frac{n}{2} =o(1)$ when $p < \frac{1}{2}$?
Does $(1-(1-p)^n)^\frac{n}{2} =o(1)$ when $p < \frac{1}{2}$? Given the context where this question arises, I believe it should. But I'm having trouble showing it. I've done the following:

\begin{align*}
    (1-(1-p)^n)^\frac{n}{2} &\leq (1-e^{-np})^{\frac{n}{2}}\\
    &\leq (e^{-e^{-np}})^\frac{n}{2}\\
    &= \big(\frac{1}{e^{e^{-np}}}\big)^\frac{n}{2}
\end{align*}

But I don't think these approximations will work because I believe what's inside the brackets goes to 1 as $n \rightarrow \infty$.

Body:
When $0<p<\frac{1}{2}$
$$(1-(1-p)^n)^\frac{n}{2} = {\Large e}^{\frac{n}{2} \ln(1-(1-p)^n)} \sim {\Large e}^{-\frac{n}{2} (1-p)^n} \to 1$$


问:
What is the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm?

What is the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm? Round your decimal answer to the nearest hundredth.

How do you find this?

答:
If the triangle has (integer) sides $a,b,c$ then as given $a+b+c = 15$. Now let's use imranfat's suggestion of the triangle inequality: $$a+b > c$$ $$a+c > b$$ $$b+c >a.$$

Since we want to minimize area and all our data is in terms of side lengths I think Heron's Formula is our best bet: let $s = \frac{a+b+c}{2} = 15/2 =7.5$ then the area $A$ of our triangle is $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(7.5-a)(7.5-b)(7.5-c)}.$$

From looking at the above one can see to make the area small we essentially want some integer, say $a$, closest to $7.5$, in particular $7$ is good because it leaves $0.5$ which scales down our area. Any integer small leaves a whole number multiply which increases the area and we can't go bigger than $7$ by the triangle inequalities and the bound on perimeter. All of which should motivate the choice of $(7,7,1)$ as our triple for $(a,b,c)$ which gives minimal area (for intuition this is a very skinny isosceles triangle). Doing the calculation and rounding gives $A \approx 3.49$.

[Note this approach isn't the most elegant, and we could be more exact by using the tools of calculus, but from a purely high-school geometry point of view it's a workable way of thinking.]


Title: For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even

Body:
I want to know for what valuyes this function is even

I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!

Body:
$f(-x)=(-x)^6 + a(-x)^3 -2(-x)^3 -2(-x)^2 + 1 = x^6 -ax^3 + 2x^3 - 2x^2 +1 $

$= x^6 +(-a+2)x^3 -2x^2 +1$

So for the two polynomials to be equal we need $2-a=a-2$ Hence, $2a=4, a=2$


问:
Why does this shortcut to the section formula work?
If a straight line $ax+by+c=0$ divides the line segment joining the points $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $k:1$, the standard way of finding $k$ is to use the section formula and input the point of intersection in the equation of the straight line. However, my textbook gives the following shortcut but doesn't prove it:

$$\frac{AP}{PB} = k = - \frac{ax_1+by_1+c}{ax_2+by_2+c}$$

If $k>0$, then $P$ divides $AB$ internally else externally.

Can someone please explain to me why this method works?

答:
Suppose I give you a line $L: ax + by + c = 0$ and a point $P(x_1, y_1)$

Then

$$d(P,L) = \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}$$

Where $d(P,L)$ is the signed perpendicular distance of point P from line L

So essentially, when you take the ratio of $d(P_1, L)$ to $d(P_2, L)$ you get

$$r = \frac{d(P_1, L)}{d(P_2, L)} = -\frac{ax_1 + by_1 + c}{ax_2 + by_2 + c}$$

The negative sign is because the point in the interior will be negative for one of the lines and positive for the other line - hence to maintain the sign convention of internal ratios being positive, the negative sign will appear

Now, if we say that from points $P_1$ and $P_2$, we drop perpendiculars to the line $L$ at $H_1$, $H_2$. Also, let $L$ intersect $P_1P_2$ at $X$

Then, the triangles $P_1H_1X$ and $P_2H_2X$ are similar

Therefore, we have

$$r = \frac{d(P_1, L)}{d(P_2, L)} = \frac{P_1H_1}{P_2H_2} = \frac{P_1X}{P_2X} = k$$


问:
About the substitution in the integration
$$\int_0^1 2t \cdot \sqrt{1 + 4t^2}dt$$

I see the answer is $\frac{5\sqrt{5} -1}{6}$ but I'm not sure how we got here. Can someone remind me how to do this? My book gets right to this answer but I'm not sure what steps there were done and I haven't done integration in quite some time. Can someone show me the steps?

答:
Substitute $u^2=1+4t^2$ and use $u\ du=4t\ dt$ so that the integral becomes
$$\frac{1}{2}\int_1^{\sqrt{5}}u^2\ du.$$


问:
Show that, $\displaystyle (m+n)I_{m,n}~=~\sin \dfrac{n\pi}{2}-mJ_{m-1,n-1}$
Let $m$ and $n$ be positive integers. Given,
$$\displaystyle I_{m,n}~=~\int_0^{\frac{\pi}{2}} \cos (nx) (\sin x)^{m}~ \mathrm{d}x$$
$$\displaystyle J_{m,n}~=~\int_0^{\frac{\pi}{2}} \sin (nx) (\sin x)^m ~\mathrm{d}x$$
Show that,
$$\displaystyle (m+n)I_{m,n}~=~\sin \dfrac{n\pi}{2}-mJ_{m-1,n-1}$$

I have been trying this problem for a long time. First I tried to apply by parts but failed.
I need some help to start off. A rigorous and wordy proof is needed.

答:
Hint

$$I_{m,n} = \int_0^{\pi/2} \cos(nx) (\sin x)^mdx$$

$$\implies I_{m,n} = \int_0^{\pi/2} \cos\left(\frac{n\pi}{2} - nx\right) (\cos x)^mdx$$

I used the property that $$\int_a^bf(x)dx = \int_a^b f(a+b-x)dx$$

Can you proceed?


问:
Complete theory with no atomic model
I'm stuck on the following problem. This is an old qualifying exam and a new homework question. So I need just a hint to think on it.

Let $T$ be complete theory in a countable language $\mathcal{L}$ with no atomic models. Let $\mathfrak{A}\vDash T$ be countable model. Prove that there is a countable model $\mathfrak{B}\vDash T$ such that neither $\mathfrak{A}$ nor $\mathfrak{B}$ can be elementarily embedded into the other.

One can deduce $T$ has no prime model, and the isolated types in $S_n(T)$ is not dense for some $n$. Also since $\mathfrak{A}$ is not atomic, it realizes a non-principal type $p$. So we can construct a model $\mathfrak{B}$ omitting $p$. However, this gives only that $\mathfrak{A}$ cannot be elementarily embedded into $\mathfrak{B}$. How can improve this idea? Or is there another way to prove this?

答:
EDIT: This misses an important subtlety, see Alex Kruckman's answer below. I'll delete this once it's unaccepted.

Besides the non-principal type $p$ realized in $\mathfrak{A}$, there also must be a type $q$ (neccessarily non-principal) which is not realized in $\mathfrak{A}$ since $\mathfrak{A}$ is countable. Now we'll be done if we can whip up a model $\mathfrak{B}$ which realizes $q$ but omits $p$ (if it's not countable, take a countable elementary substructure containing a witness to $q$). This isn't something that the usual statement of the omitting types theorem will do for us, but the proof of the omitting types theorem extends to solve this problem too (we combine omitting $p$ with realizing $q$).

Title: Complete theory with no atomic model


问:
Zariski's Lemma
I'm looking for a proof for this proposition:
"Let $L/K$ be a field extension, where $L$ is a finitely generated $K$ algebra. Then $L/K$ is a finite field extension."

My attempt: In "Field and Galois Theory" written by Morandi, there is a proposition saying that a field extension $K/F$ is finite if and only if $K$ is finitely generated and algebraic over $F$.

Let $K[a_1,...,a_n]/I\cong L$ for some $a_i\in L$ and $I$ is an ideal of $L$. I tried to show that $\{a_i+I\}$ are algebraic over $K$

答:
$L=K[a_1,\ldots,a_n]$ is a field.

If $L/K$ is algebraic then $L$ is a quotient of $K[x_1,\ldots,x_n]/(f_1(x_1),\ldots,f_n(x_n))$ which is a $\prod_j \deg(f_j)$ vector space so $L$ is a finite dimensional vector space (where $f_j\in K[x]$ is the minimal polynomial of $a_j$).

If $L/K$ is not algebraic then take a transcendental basis $a_{i_1},\ldots,a_{i_m}$.

Let $t=a_{i_m}$, $F=K(a_{i_1},\ldots,a_{i_{m-1}})$,

$L$ is an algebraic extension of $F(t)$, it is a quotient of $F(t)[x_1,\ldots,x_n]/(f_1(x_1),\ldots,f_n(x_n))$ where $f_j$ is now the monic $F(t)$-minimal polynomial of $a_j$. The coefficients of $f_j$ are $u_{j,l}(t)/v_{j,l}(t)\in Frac(F[t])$, let $$h(t)= \prod_{j,l} v_{j,l}(t)\in F[t]$$ Then $$L= F[t,h(t)^{-1}][a_1,\ldots,a_n]$$ is integral over $F[t,h(t)^{-1}]$. Take an irreducible polynomial $g(t)\in F[t]$ not dividing $h(t)$,

$g(t)^{-1}$ is not integral over $F[t,h(t)^{-1}]$ so $g(t)^{-1}\not \in L$  which is a contradiction.

Title: Zariski's Lemma


问:
$f$ has a fixed point if $\begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align}$

Let $E=[a,b]\times[c,d]$ be a subset in $\Bbb R^2$, and $f:E\to E$ be a function satisfying $$\begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align} $$
Prove that there is a point $x_0\in E$ such that $f(x_0)=x_0$.

I know the case when the contraction constant is $0\le c<1$, but this is a case when $c=1$. Thanks for any suggestion.

After reading the comments below, I think I have some idea.
Choose $x_0 \in E$, and define $x_{n+1}=f(x_n)$ for $n\in \Bbb N$.
Then $\{x_n\}$ is a sequence in $E$, which is compact, $\{x_n\}$ has a convergent subsequence, say $x_{n_k}\to x$.
By assumption, $f$ is continuous on $E$, so $\lim_\limits{k\rightarrow\infty}f(x_{n_k})=f(\lim_\limits{k\rightarrow\infty}x_{n_k})=f(x)$. On the other hand, $\lim_\limits{k\rightarrow\infty}f(x_{n_k})=\lim_\limits{k\rightarrow\infty}x_{n_{k+1}}=x$. Hence, we get $x=f(x)$.

答:
HINT:

Consider $x_0$ such that $d(x_0, f(x_0))$ is smallest.


问:
Projecting a grid on a 2d plane
Unsure if this is the right network to post this, but maybe someone can help...

I'm working on a personal project where, with the help of AI, I want to automatically estimate the perspective of a given image.

I've been able to programatically estimate the position of the horizon, as seen below. This is based on how "quickly" cars farther away shrink in the image.

Not sure the above is useful, but now I want to draw a grid on the plane. Something like:

This is where it's getting too complicated for me. I have no idea how to solve this. Not the drawing per se, but the mathematics behind it.

I first thought I needed to find the plane equation based on the coordinates of the bottom left and bottom right corners of each box, since these are the ones intersecting the plane (the tires touching the road).

I did manage to find the equation ax + by + cy + d, but I am not sure I am on the right track. Looking for some guidance if anyone has any ideas.

EDIT:

Just to clarify the unknowns:

Distance to any object or horizon.

Physical size of objects (in meters or otherwise)

Height or angle of camera

答:
Or just calculate whatever lane direction is needed with Cos(direction) * Distance = y and Sin(direction) * Distance = x . North is 0 degrees direction while East is positive 90 degrees direction and West is negative 90 degrees direction. The coordinates are "forwards", polar style, from the origin coordinates and are netted with the origin coordinates for the overall x-y result. Or directly "forward" each lane direction from each beginning lane position.

An "inverse" between two pairs of coordinates of the same lane gives the lane direction. That's InverseTan((x2 - x1) / (y2 - y1)) . If (x2 - x1) is negative then that is West otherwise it's East. If (y2 - y1) is negative then that is South otherwise it's North.

Title: Projecting a grid on a 2d plane


答:
Having even values is different to being an even function. For example, the function $\operatorname{f} : x \mapsto x$ has many even values, e.g. $2 \mapsto 2$ and $4 \mapsto 4$. However, $\operatorname{f}$ is not an even function since $\operatorname{f}(-x) \neq x$ for all $x \neq 0$.

In the example you give, the function cannot be even if it contains odd powers of $x$. Consider

\begin{array}{ccc}
\operatorname{f}(-x) &\equiv& (-x)^6 + a(-x)^3 - 2(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& (-x)^6 + (a-2)(-x)^3 - 2(-x)^2 + 1 \\
&\equiv& x^6 +(2-a)x^3 -2x^2 + 1
\end{array}

Hence, $\operatorname{f}(x) \equiv \operatorname{f}(-x)$ if and only if $a-2=2-a$, i.e. $a=2$. Notice that $a=2$ if and only if the $x^3$, i.e. the odd powered term, vanishes.


问:
What is the distribution of the combination of two indepenently and normally distributed random variables?
Let $X$ and $Y$ be two independent random variables such that $X \sim \mathcal{N}(\mu_1, \sigma_1)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2)$. In addition, let $Z=X^2+XY$ be the combination of these two random variables, then what is the PDF or CDF of $Z$?

Honestly, I know what to do with $Z=X + Y$ or $Z=XY$ since $X$ and $Y$ are independent. However, things are different here because both $X^2$ and $XY$ depend on $X$, and unfortunatly, I have totally no idea  about how to deal with the dependency.

答:
What you need is to get the distribution function of $Z=X^2$.

Knowing that $X\sim f_X(x)$, if $z>0$, then

$F_Z(z) = P(Z\le z) = P(X^2 \le z) = P(-\sqrt{z}\le X\le \sqrt{z}) = F(\sqrt{z}) - F(-\sqrt{z}).$

Taking derivative, we have

$f_Z(z) = \frac{f_X(\sqrt{z}) + f_X(-\sqrt{z})}{2\sqrt{z}}.$

If $z\le 0$, then $F_Z(z) = P(Z\le z) = P(X^2\le z) = 0$, we have
$f_Z(z) = 0$.

Taking $f_X(x) = N(\mu,\sigma^2)$ in to the above formula we get the desire result.


问:
simplifying an asymptotic expression
I have this expression in a statistics book, namely $nh(f(x) +o(1)+O_p(1/\sqrt{nh}))$. Where $f$ is a density function. Now, this expression is equal to $nhf(x)\{1+o_p(1)\}$. Note, that  $n\to \infty$, $h\to 0$, and $nh\to \infty$. I do appreciate any comment on how to get this final expression. Thanks in advance.</p>

答:
We are given an expression

$$E=nh\left(f(x)+o(1)+O_p\left(\sqrt{1/nh}\right)\right).$$

Provided that $f(x) \neq 0$ (a necessary assumption not given in your statement, but hopefully present in context), we obtain

$$E=nhf(x)\left(1+o(1)+O_p\left(f(x)^{-1} \sqrt{1/nh}\right)\right).$$

We are taking the limit independently of $x$, so the constant $f(x)^{-1}$ may be absorbed into the $O$-notation:

$$E= nhf(x)\left(1+o(1)+O_p\left(\sqrt{1/nh}\right)\right).$$

Finally, we use that $O_p(\sqrt{1/nh}) =o_p(1)$ as $nh \to \infty$, to get
$$E=nhf(x)\left(1+o(1)+o_p(1)\right)=nhf(x)\left(1+o_p(1)\right),$$
in which we've used that $o(1)=o_p(1)$.

Title: simplifying an asymptotic expression


答:
Following orangeskid's hints, the proof is straight forward:

Let $g:E\rightarrow[0,\infty)$ be defined by $g(x)=d(x,f(x))$. Clearly
$g$ is continuous. Since $E$ is compact, $g$ attains it minimum. Choose $x_0\in E$ such that $g(x_0)=\min g(E):=\xi$. If $x_{0}\neq f(x_{0})$, then
$g(f(x_{0}))=d(f(x_{0}),$$f(f(x_{0})))<d(x_{0},f(x_{0}))=\xi$, a
contradiction.

Title: $f$ has a fixed point if $\begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align}$

Body:
Let $E=[a,b]\times[c,d]$ be a subset in $\Bbb R^2$, and $f:E\to E$ be a function satisfying $$\begin{align} \|f(x)-f(y)\|<\|x-y\|\text{ for all }x,y \in E, x\ne y \ . \end{align} $$
Prove that there is a point $x_0\in E$ such that $f(x_0)=x_0$.

I know the case when the contraction constant is $0\le c<1$, but this is a case when $c=1$. Thanks for any suggestion.

After reading the comments below, I think I have some idea.
Choose $x_0 \in E$, and define $x_{n+1}=f(x_n)$ for $n\in \Bbb N$.
Then $\{x_n\}$ is a sequence in $E$, which is compact, $\{x_n\}$ has a convergent subsequence, say $x_{n_k}\to x$.
By assumption, $f$ is continuous on $E$, so $\lim_\limits{k\rightarrow\infty}f(x_{n_k})=f(\lim_\limits{k\rightarrow\infty}x_{n_k})=f(x)$. On the other hand, $\lim_\limits{k\rightarrow\infty}f(x_{n_k})=\lim_\limits{k\rightarrow\infty}x_{n_{k+1}}=x$. Hence, we get $x=f(x)$.


问:
upper bound of $\sum_{k=1}^n k$
$\sum_{k=1}^n k$

We know that

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$

Now how do I show that it is not bounded?

答:
For showing unboundedness is the same as showing the expression is greater than any any natural number for sufficiently big $n$

Let $a_n \ = \dfrac{n(n+1)}{2}$

clearly $ a_n > n^2/2$

so for all $n>\sqrt{2M} \\a_n > M$

and as the set of naturals is unbounded we can always find such $n$ so we are done

Title: upper bound of $\sum_{k=1}^n k$


问:
Finding all integrands
I am trying to characterize all the functions that satisfy the following integrability property

$$
\int_{-\infty}^\infty x f(x)e^{-x^2/2}dx=\int_{-\infty}^\infty f'(x)e^{-x^2/2}dx.
$$

Attempt: if $f$ is even, then the left side is always 0, similarly the right side will be 0 only since the derivative of an even function is odd, so $f'(x)e^{-x^2/2}$ is odd. So the class of all even functions does the trick, is this the best I can do? Additionally, if I just combine the integrals into 1 I get that

$$ \int_R (xf(x)-f'(x))e^{-x^2/2}dx=0$$
which means we need $xf(x)-f'$ to be odd. Is this more general than just the class of all even $f$?

答:
The formula holds for any continuously differentiable function $f$ satisfying the following subgaussian growth condition:

$$ \lim_{\left|x\right|\to\infty} f(x)e^{-x^2/2} = 0. $$

This is very easy to prove. Indeed, by the fundamental theorem of calculus,

$$ \int_{a}^{b} (x f(x) - f'(x)) e^{-x^2/2} \, \mathrm{d}x
= \left[ -f(x) e^{-x^2/2} \right]_{a}^{b}
= f(a)e^{-a^2/2} - f(b) e^{-b^2/2}. $$

Then letting $a \to -\infty$ and $b \to \infty$ proves the desired claim.

Title: Finding all integrands


问:
If $G$ is a group with $N$ normal and $H_1$, $H_2$ subgroups of $G$ with $H_1$ a normal subgroup of $H_2$, show $NH_1$ is a normal subgroup of $NH_2$.
I was unsure how to prove this property which seems necessary for the solution to a problem I did recently. Since $N$ is normal, it seems pretty clear that $NH_1 = H_1N$ and $NH_2 = H_2N$ so they are indeed subgroups of $G$. I tried to prove normality directly by taking $nh \in NH_2$ and was able to show that $$nh NH_1 = N nh H_1 = Nn H_1 h$$ using the fact that $H_1$ is normal in $H_2$ and so $h H_1 = H_1 h$. Then, it seems like this problem reduces to showing that $nH_1 = H_1n$ which isn't immediately obvious. In fact, I'm not sure if that's even true.

My latest attempt uses the fact that $NH_1 = H_1N$ and $Nn = N$ so that we can find some $n' \in N$ so that $n'H_1 = H_1n$ which would allow us to conclude $$nh NH_1 = N nh H_1 = n H_1 h = Nn' H_1 h = NH_1nh$$ which is what we wanted but I had some trouble showing that this idea actually worked.

Thank you in advance for your help!

答:
Since $NH_1$ is a subgroup, we have (refer here) $$NH_1=H_1N.$$
So, $$nhNH_1=nNhH_1=nNH_1h=NH_1h=H_1Nh=H_1Nnh=NH_1nh.$$


问:
Prove or disprove the following state $\lim_{m\to\infty}||A^m||=0$ then $\rho(A)\leq1$
Prove or disprove the following state $\lim_{m\to\infty}||A^m||=0$ then $\rho(A)\leq1$,

I know that $\rho(A)\leq ||A||$ for every induced norm and my intution is that the statment is true but not sure how to prove it.

Any hints please?

答: If there is an eigen value $\lambda$ with $|\lambda |>1$ the there is  vecor $x \neq 0$ such that $Ax=\lambda x$. This gives $A^{m}x=\lambda ^{m}x$. So $A^{m}y=\lambda ^{m}y$ where $y=\frac x {\|x\|}$ is a unit vector. Now $\|A^{m}\| \geq \|A^{m}y\|=|\lambda ^{m}| \to  \infty$, a contradiction.


问:
Using the snake lemma to prove that if $N$ and $M/N$ are finitely generated, then so is $M$.
Let $M$ be a $R$-module and $N$ be a submodule. Suppose that both $N$ and $M/N$ are finitely generated. I want to prove that so is $M$, using the snake lemma.

(Following If $M$ and $N/M$ are finitely generated, then so is $N$.)

My idea is as follows:

We clearly two exact sequences:
  $$0\to R^{\oplus n}\to R^{\oplus n}\oplus R^{\oplus m}\to R^{\oplus m}\to 0$$
  and
  $$0\to N \to M\to M/N\to 0.$$
  We also have surjective morphisms $R^{\oplus n}\to N$ and $R^{\oplus m}\to M/N$. As soon as we have any $R$-module morphism $R^{\oplus n}\oplus R^{\oplus m}\to M$ that makes the desired diagram commute, the Snake lemma will imply that it is surjective and then we are done.

In order to create this morphism, I think we can sum
  $$R^{\oplus n}\to N\to M$$
  and a lift of
  $$R^{\oplus m}\to M/N$$
  to $M$. This certainly makes the left side of the diagram commute.

So, I have two doubts:

1) Does this make the right side of the diagram commute? That is, the square containing $R^{\oplus n}\oplus R^{\oplus m}$, $R^{\oplus m}$, $M$ and $M/N$.

2) How can I prove the existence of such a lift?

Since I'm in the beginning of P. Aluffi's Algebra Chapter 0, I would appreciate if the answer does not contain heavy category theory.

答:
Just to add, for (2), the existence of such a lift is the same as proving the “projectivity” of free modules. (Once we know what a projective module is.)

In other words, the existence of such a lift is just another interpretation of saying that a free $R$-module is projective. The proof of this fact is what @Maxime Ramzi pointed out in the wonderful answer.


问:
Weak convergence of some sequences of function $u_{n,m}$ in $H^1$
Suppose $B_1$ is a unit ball in $\mathbb{R}^n$ and $u_n$ is a sequence of function in $H^1(B_1)$. It is well-known that if $|u_n|_{H^1(B_1)}$ is uniformly bounded, then $u_n$ converges weakly to $u_0$, up to a subsequence.

Now suppose we have a sequence of function $u_{n,m}$ in $H^1(B_1)$. If $|u_{n,m}|_{H^1(B_1)}$ is uniformly bounded in both $n$ and $m$, I believe that we also have the fact that $u_{n,m}$ is weakly convergent up to some subsequences. Here we do not relabelled the subsequence.

I would like to ask that is the weak convergence uniform in $n$ and $m$? Is it true that

$$ \lim_{n \to \infty}\lim_{m \to \infty} u_{n,m} =  \lim_{m \to \infty}\lim_{n \to \infty} u_{n,m}? $$
The above limit is taken as weak sense in $H^1$. Thank you so much!

答:
Even if you replace $H^1$ by $\mathbb R$, this does not seem to be true.
We pick
$$
a_{n,m}
=
\begin{cases}42 & \text{for } n > m, \\ 23 & \text{for } m \ge n.\end{cases}$$
Note that we do not need to select subsequences, since we have convergence for $n \to \infty$ and for $m \to \infty$ separately. Similarly, the iterated limits exist, but are not equal.


问:
Convexity of root
I need some help proving that
$$
\sqrt{x^T \Sigma x},
$$
with $\Sigma$ being a positive definite matrix, is convex. I already found some questions about this topic, indicating that one should use a conic representation but I did not understand these approaches, I was hoping maybe someone has another idea or could explain it better (maybe with a link to a resp. paper or book).

答:
There are at least two ways to prove this:

$(x,y) \mapsto x^\top \Sigma \, y$ is a scalar product and your function is the corresponding norm.

You have $\sqrt{x^\top \Sigma x} = \| \Sigma^{1/2} x\|$, i.e., the combination of a convex function and a linear function.

Title: Convexity of root


问:
A problem in polynomials to show that $a_{n−1}$ and $a_n$ must be coprime
Let P(x) = $a_0x^n$ + · · · + $a_{n−1}x$ + $a_n$ be a polynomial with integer coefficients. Suppose the
equation P(x) = 0 has n distinct integer roots which are pairwisely coprime. Then prove
that $a_{n−1}$ and $a_n$ must be coprime.

This was asked in an examination and I couldn't solve it.
Please solve this problem.

Can anyone solve it...I hope people can see my post

答:
Edit Embarrassingly, I overlooked the concept behind Mark Bennet's comment.  This response therefore assumes that $a_0 = \pm 1.$

Hint

First, see Vieta's formulas.

Then, assume that $p$ is a prime number that divides both $a_n$ and $a_{(n-1)}.$

This means that $p$ divides some root of the equation.  This means that $p$ divides every term that comprises $a_{(n-1)}$ but one term.  What about this outstanding term.  Can $p$ also divide this outstanding term?


问:
Does $Y_t^-\geq a+bM_t^-$ and integrability of $M_t^-$ imply that $Y_t^-$ is integrable?
Consider a martingale $(M_t)_{t \in T}$. Then $M_t$ is integrable $\forall t \in T.$ So $M_t^-$ is also integrable $\forall t \in T$.

Let $a \in \mathbb R$ and $b\geq0$.

Assume $Y_t^-\geq a+bM_t^-$

Does this imply that $Y_t^-$ is integrable or can the integral of $Y_t^-$ still be $\infty$?

答: Take $M_t=0$ and $a=0$. The hypothesis becomes $Y_t^{-} \geq 0$ which is always true. So you can get a counter-example by taking any positive random variable $Z$ with infinite expectation and taking $Y_t =-Z$.


问:
Use Laplace transform to solve $f'(t) = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t')$
Consider the following differential equation for $f$
$$
\frac{df}{dt} = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t')
$$
where we have some initial condition $f(0) = f_0$, the function $g(t)$ is known and $a>0$ is a parameter, and $H(t)$ is the Heaviside step function.

Assuming that $t-a>0$, we can ignore the Heaviside function that appears in the RHS.

I would like to solve the above using a Laplace transform, but the upper limit on the integral makes the RHS of the DE not exactly a convolution. I can write the RHS of the above as (assuming $t-a>0$)
$$
H(t-a) \int_{0}^{t-a} dt'\; g(t') f(t-t') = \int_{0}^{t} ds\; g(t') H(t - a - t') f(t-t') = \int_{0}^{t} dt'\; \left[ H(t' - a) f(t) \right] g(t-t')
$$
So the above can be written as a convolution of the function $g(t)$ and $e_{a}(t) := H(t - a) f(t)$. This means that taking the Laplace transform of my DE ends up in
$$
s F(s) - f(0) = G(s) E_a(s)
$$
where $\mathcal{L}\{ f \}(s) = F(s)$, and $\mathcal{L}\{ g \}(s) = G(s)$, and $\mathcal{L}\{ e_a \}(s) = E_a(s)$ (where $\mathcal{L}$ is the usual one-sided Laplace transform)

Using the above, I cannot isolate for $F(s)$ in my DE (since $E_a(s)$ does not depend easily on $F(s)$).

How do I solve this differential equation with the Laplace transform?

答:
Upon a first look, one notices immediately that the derivative is zero in the region $t<a$ and is also continuous at $t=a$. We immediately conclude that $f$ assumes the piecewise constant form

$$f(t)=\begin{Bmatrix}f_0 &~,~t\leq a\\f_0+\int_{a}^th(x)dx&~,~t>a\end{Bmatrix}$$

One computes directly from this form that $\frac{df}{dt}=H(t-a)h(t)$, so $h(t)$ is to be identified with the convolution-like integral on the RHS of the ODE to be solved. Now we consider the ODE in the region $t>a$ and integrate it with a Laplace weight

$$\int_{a}^{\infty}\frac{df}{dt}e^{-st}dt=\int_{a}^{\infty}dt~ e^{-st}\int_{0}^{t-a}dt'g(t')f(t-t')$$
The left-hand side evaluates to
$$LHS=s\int_{a}^{\infty}f(t)e^{-st}dt-f(a)e^{-sa}\equiv s F_a(s)-f_0e^{-sa}$$
In the right-hand side we change the order of integration carefully and after a shift in $t$ we obtain
$$RHS=\int_{0}^{\infty}dt'g(t')\int_{t'+a}^{\infty}f(t-t')e^{-st}dt=\int_{0}^{\infty}g(t')e^{-st'}\int_{a}^{\infty}f(t)e^{-st}=G(s)F_a(s)$$
Equating the two and rearranging we obtain that
$$F_a(s)=\frac{f_0 e^{-sa}}{s-G(s)}$$
and thus the Laplace transform of the function in the entire domain  is given by
$$F(s)=f_0\frac{1-e^{-sa}}{s}+\frac{f_0 e^{-sa}}{s-G(s)}$$
Of course this form could have been derived directly from the very first equation. Note that solving this equation by a Laplace transform may not pick up various solutions that may happen to not have a well-defined Laplace transform.

Finally, this reproduces the solution for the case where $g(t)=b=const.$, where the problem can be solved without using a Laplace transform:

$$f'(t)=b\int_{a}^t f(t')dt'\Rightarrow \int_a^t dt'f(t')=A(e^{\sqrt{b}t}+e^{\sqrt{b}(2a-t)})\Rightarrow f(t)=\frac{f_0}{2}\sinh((t-a)\sqrt{b})$$
whose Laplace transform reads (taking into account the constant part as well)
$$F(s)=f_0\frac{1-e^{-sa}}{s}+f_0\frac{e^{-sa}}{s-\frac{b}{s}}$$
which directly confirms the result of the method presented here.


问:
If Function is a set of ordered pairs then why do we call it as a rule?
I am confused here with the vocabulary we use while defining functions. If function f is a set of ordered pairs, then what is meant by a function f is a Rule which which produce new elelments out of some given elements.Why do we have two definitions of function.</p>

答:
When the concept of function $f$ from a set $A$ into a set $B$ is introduced for the first time, it is usually described as a rule that converts each element of $A$ into an element of $B$. This is just an idea of what a function is. It cannot be a definition unless we then define the meaning of the word “rule”.

So, we have the formal definition of a function $f$ from a set $A$ into a set $B$ as a subset of $A\times B$ such that, for each $a\in A$, there is one and only $b$ in $B$ such that $(a,b)\in f$. And then we introduce the notation $f(a)$ (with $a\in A$), which means that only element $b\in B$ such that $(a,b)\in f$. This definition may be hard to understand at first, but it is a real definition (that is, it is not just an attempt to explain the meaning of the word).


答:
$$f'(t)=H(t-a)\int_{0}^{t-a} g(s) f(t-s)\mathrm{d}s =\int_{0}^\infty
g(s)H\big((t-a)-s\big) f(t-s)\mathrm{d}s.$$

Note that for $t<a$, $f'(t)=0$ and hence $f(t)=f(0)$ for
$0\leq t\leq a$. The Laplace transform of $H(t-a) f(t)$ is given by

\begin{align*}
  \mathcal{L}\big\{H(t-a) f(t)\big\}(s)&=
%
\int_{0}^\infty e^{-sx}H(x-a) f(x)\mathrm{d}s
%
=F(s)-\int_{0}^a e^{-sx}f(x)\mathrm{d}s\\
%
&=F(s)+\frac1s e^{-sx}f(x)\biggm|_0^a-\frac1s\int_{0}^a
e^{-sx}f'(x)\mathrm{d}s=
%
F(s)-\frac1s f(0)(1-e^{-as}).
%
\end{align*}

So,

$$F(s)=f(0)\frac{1-\frac1s(1-e^{-as})}{s-G(s)}.$$

Title: Use Laplace transform to solve $f'(t) = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t')$

Body:
Consider the following differential equation for $f$
$$
\frac{df}{dt} = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t')
$$
where we have some initial condition $f(0) = f_0$, the function $g(t)$ is known and $a>0$ is a parameter, and $H(t)$ is the Heaviside step function.

Assuming that $t-a>0$, we can ignore the Heaviside function that appears in the RHS.

I would like to solve the above using a Laplace transform, but the upper limit on the integral makes the RHS of the DE not exactly a convolution. I can write the RHS of the above as (assuming $t-a>0$)
$$
H(t-a) \int_{0}^{t-a} dt'\; g(t') f(t-t') = \int_{0}^{t} ds\; g(t') H(t - a - t') f(t-t') = \int_{0}^{t} dt'\; \left[ H(t' - a) f(t) \right] g(t-t')
$$
So the above can be written as a convolution of the function $g(t)$ and $e_{a}(t) := H(t - a) f(t)$. This means that taking the Laplace transform of my DE ends up in
$$
s F(s) - f(0) = G(s) E_a(s)
$$
where $\mathcal{L}\{ f \}(s) = F(s)$, and $\mathcal{L}\{ g \}(s) = G(s)$, and $\mathcal{L}\{ e_a \}(s) = E_a(s)$ (where $\mathcal{L}$ is the usual one-sided Laplace transform)

Using the above, I cannot isolate for $F(s)$ in my DE (since $E_a(s)$ does not depend easily on $F(s)$).

How do I solve this differential equation with the Laplace transform?


答:
If $c$ is a positive number, then there is $m_0 \in \mathbb N$ such that

$$||A^m|| \le c$$

for $m >m_0.$ Hence

$$||A^m||^{1/m} \le c^{1/m}$$

for $m >m_0.$ With $ m \to \infty$ we get

$$\rho(A) \le 1.$$

Remark: this holds for any bounded linear operator $A: V \to V$ on a normed linear space $V$.

Body:
Prove or disprove the following state $\lim_{m\to\infty}||A^m||=0$ then $\rho(A)\leq1$,

I know that $\rho(A)\leq ||A||$ for every induced norm and my intution is that the statment is true but not sure how to prove it.

Any hints please?


答:
The former is a rigorous definition for a function in axiomatic set theory, whereas the latter is the intuitive understanding of functions. You use terms like "rule" and "produce new elements" but these do not have a wholly rigorous definition. In math, we take for granted only basic assumptions (axioms) and logical inference rules. From these, we formally define every object we are interested in. Specifically, almost all modern math uses the axioms of set theory, under which every mathematical object is a set. In fact, even ordered pairs can be defined as specific sets. From here, we can define functions $f: A \longrightarrow B$ as a special subset $f \subseteq A \times B$ where $\forall a \exists b (a, b) \in f$ and $(a, b)\in f \text{ and } (a, b') \in f \implies b = b'$. We write $f(a)=b$ for $(a, b) \in f$.

We can continue further to make set theoretic definitions for most of the objects of modern math. By doing so, we can put math on a rigorous and uniform foundation.

Title: If Function is a set of ordered pairs then why do we call it as a rule?

Body: I am confused here with the vocabulary we use while defining functions. If function f is a set of ordered pairs, then what is meant by a function f is a Rule which which produce new elelments out of some given elements.Why do we have two definitions of function.</p>


问:
To what number $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}$ converges
This question is from Ponnusamy and Silvermann's complex variables with applications Ch- 12 , subsection Infinite products.

Question: prove that $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}=2/3$.

I factored numerator and denominator to $\frac{(n-1) \times n^2+n+1 } { (n+1) \times n^2 -n+1}$ . But I think it's not possible to cancel denominantor and numerator in a way to get a pattren of cancelling due to factors $n^2-n+1$ and $n^2+n+1$ . So, I am struck on how should this be approached and unable to find any idea.

Kindly guide on how this should be approached.

Thanks!

答: Hint: The first five terms in the product $$\prod_{n = 2}^{\infty}\dfrac{n^3+1}{n^3-1} = \prod_{n = 2}^{\infty}\dfrac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)}$$ are $\dfrac{3 \cdot 3}{1 \cdot 7}$, $\dfrac{4 \cdot 7}{2 \cdot 13}$, $\dfrac{5 \cdot 13}{3 \cdot 21}$, $\dfrac{6 \cdot 21}{4 \cdot 31}$, and $\dfrac{7 \cdot 31}{5 \cdot 43}$. When you multiply these out, a lot of terms cancel. In general, when you multiply the first $N$ terms, a lot of stuff cancels, and you can simplify what's left. Then take the limit as $N \to \infty$.


问:
Locally nilpotent derivations closed under isomorphism?
I have been struggling to prove or find a counterexample to the following:

Fix a field $k$ of characteristic $0$. Let $R:= k[x_1,...,x_n]$ and $A$ be the ring of differential operators on $R$ (so $A$ is isomorphic to the $n$th Weyl algebra; in particular, it is just a ring of polynomials in the operators $\hat{x_1}, ..., \hat{x_n}, \partial_1,...,\partial_n$, where $\hat{x_i}$ is the multiplication by $x_i$ operator and $\partial_i$ is the ith partial derivative operator). We say an operator $D \in A$ is locally nilpotent if for all $f \in k[x_1,...,x_n]$ there exists a positive integer $n$ so that $D^n(f) = 0$. Let $\phi: A \to A$ be an automorphism of $A$. If $D$ is a locally nilpotent $k$-derivation in $A$, is the same true of $\phi(D)$? If this is true, does the statement hold in general when $R$ is a finitely generated integral domain over $k$?

Any thoughts would be appreciated.

答: No, this is not true - in fact, automorphisms of $A$ need not even send derivations to derivations. Let $n=1$ so that the Weyl algebra is $k\langle x,\partial\rangle/(\partial x- x\partial=1)$. Then $\partial$ is locally nilpotent: if $p$ is a polynomial of degree $n$, $\partial^{n+1}p=0$, while $x$ acts injectively on $k[x]$. But there is an automorphism sending $\partial\mapsto x$ and $x\mapsto -\partial$.


答:
$\textbf{Hint}$: The fraction can be simplified down to

$$\frac{1+\frac{1}{n}-\frac{1}{n+1}}{1+\frac{1}{n-1}-\frac{1}{n}}$$

which does have terms that cancel out consecutively. What is the formula for the $N$th partial product?

Title: To what number $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}$ converges

Body:
This question is from Ponnusamy and Silvermann's complex variables with applications Ch- 12 , subsection Infinite products.

Question: prove that $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}=2/3$.

I factored numerator and denominator to $\frac{(n-1) \times n^2+n+1 } { (n+1) \times n^2 -n+1}$ . But I think it's not possible to cancel denominantor and numerator in a way to get a pattren of cancelling due to factors $n^2-n+1$ and $n^2+n+1$ . So, I am struck on how should this be approached and unable to find any idea.

Kindly guide on how this should be approached.

Thanks!


问:
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$
$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

Can we write it as following

$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$

Let's see what happens:-

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$

In the same way for further terms, we will get $0$

Let's also confirm for general term

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$

So the whole expression $E$ will be zero

But actual answer is $1$

Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$

Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?

I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.

But I got the following counter thought:-

$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$

As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$

$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$

$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$

Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.

But surprisingly $\dfrac{1}{3}$ is the correct answer.

I am feeling very confused in these two things. Please help me.

答:
I will be giving a $\varepsilon -\mathcal{N}$ proof

notice

$$\begin{align}\frac{n}{1+n^2} + \frac{n}{2+n^2} + \cdots + \frac{n}{n+n^2} - 1=
n\Big( \frac{1}{1+n^2} - \frac{1}{n^2} +   \cdots +  \frac{1}{n+n^2} - \frac{1}{n^2}   \Big) = h_n\end{align}$$

Claim $\dfrac{n}{n+n^2} > \dfrac{r}{r+n^2}$

So

$|h_n| < \Big|n\cdot(\dfrac{1}{n^2}\cdot\dfrac{n^2}{n+n^2})\Big|=\dfrac{1}{n+1}<\dfrac{1}{n}$

so $\forall n > \dfrac{1}{\varepsilon}$
$|h_n| < \varepsilon$


问:
Find all integer solutions for the equation $|5x^2 - y^2| = 4$
In a paper that I wrote as an undergraduate student, I conjectured that the only integer solutions to the equation
$$|5x^2 - y^2| = 4$$
occur when $x$ is a Fibonacci number and $y$ is a Lucas number. I was able to prove that when $x$ was a Fibonacci number there existed a Lucas number $y$ such that $|5x^2 - y^2| = 4$. This is easily shown with Cassini's Identity $$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$

The challenge is this ... prove (or disprove) that these are the ONLY solutions.

By the way, this is how I generated the Diophantine equation.
$$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$
$$F_{n-1}(F_{n}+F_{n-1}) - F_{n}^2 = (-1)^n$$
$$F_n^2 - F_{n-1}F_n-F_{n-1}^2+(-1)^n=0$$
because $F_n \gt \frac{F_{n-1}}{2}$
$$F_n=\frac{F_{n-1} + \sqrt{F_{n-1}^2-4((-1)^n-F_{n-1}^2)}}{2}=\frac{F_{n-1} + \sqrt{5F_{n-1}^2+4((-1)^{n+1})}}{2}$$
Letting $y= \pm \sqrt{5F_{n-1}^2+4((-1)^{n+1})}$ and $x=F_{n-1}$ we have
$$y= \pm \sqrt{5x^2+4((-1)^{n+1})}$$
$$y^2= 5x^2 \pm 4$$
$$|5x^2 - y^2| = 4$$

Body:
If no enough basic knowledge, we can also directly caculate.

If $(x_0,y_0),(x_1,y_1)$ solved $x^2-5y^2 = \pm 4$

$ 1 = \frac{x_0^2-5y_0^2}{x_1^2-5y_1^2}$

$ = \frac{x_0+\sqrt{5}y_0}{x_1+\sqrt{5}y_1} \cdot \frac{x_0-\sqrt{5}y_0}{x_1-\sqrt{5}y_1} $

$ = \frac{(x_0+\sqrt{5}y_0)(x_1-\sqrt{5}y_1)}{x_1^2-5y_1^2} \cdot \frac{(x_0-\sqrt{5}y_0)(x_1+\sqrt{5}y_1)}{x_1^2-5y_1^2} $

$ = \frac{(x_0x_1-5y_0y_1)+\sqrt{5}(y_0x_1-x_0y_1)}{4} \cdot \frac{(x_0x_1-5y_0y_1)-\sqrt{5}(y_0x_1-x_0y_1)}{4} $

$ = (\frac{x_0x_1-5y_0y_1}{4})^2+\sqrt{5}(\frac{y_0x_1-x_0y_1}{4})^2$

if $ y_0x_1-x_0y_1 = 0 \pmod 4 $ then $ (\frac{x_0x_1-5y_0y_1}{4},\frac{y_0x_1-x_0y_1}{4})$ satisfy $x^2 - 5y^2 = 1$ which is pell's equation

so we can group $(x,y)$ into different group, (if exist)

$(4k_0,4k_1+2),(4k_0+2,4k_1)$

$(4k_0+1,4k_1+1),(4k_0+3,4k_1+3)$

$(4k_0+1,4k_1+3),(4k_0+3,4k_1+1)$

$(4k_0+2,4k_1+2),(4k_0+4,4k_1+4)$

Note that $(4k_0+2,4k_1+2),(4k_0+4,4k_1+4)$ if exist, it can be same group with above group. When $(4k_0+3,4k_1+3)$ be the same group with $(4k_0+2,4k_1+2)$ and $(4k_0+1,4k_1+3)$ be the same group with $(4k_0+2,4k_1+2)$, but $(4k_0+1,4k_1+3)$ and $(4k_0+3,4k_1+3)$ can never be in same group, so some group may have no answer. So I split it out.

The base answer for $x^2-5y^2=1$ is $(9,4)$

so if $(x,y)$ is the answer for $x^2-5y^2=-4$ , ($x^2-5y^2=4$ can be solved same way)

the next answer in its group will be get from $(x+\sqrt{5}y)(9+4\sqrt{5}) = (9x+20y) + \sqrt{5}(9y+4x)$,which is $(9x+20y,9y+4x)$

the previous answer in its group will be get from $(x+\sqrt{5}y)(9+4\sqrt{5})^{-1} = (9x-20y) + \sqrt{5}(9y-4x)$,which is $(9x-20y,9y-4x)$

we only care about $x>0,y>0$, In each group we have the the smallest positive solutions.

so the smallest answer satisfy $9x-20y \le 0$ or $9y-4x \le 0$

$\frac{x}{y} \le \frac{20}{9}$ or $\frac{x}{y} \ge \frac{9}{4}$

$\sqrt{5- \frac{4}{y^2}} \le \frac{20}{9}$ or $\sqrt{5- \frac{4}{y^2}} \ge \frac{9}{4}$

$0.894427 \le y \le 8.04984 $

so we can just test $y$ from 1 to 8, and we will find 3 base answer for $x^2-5y^2=-4$ , $(x,y) = (1,1),(4,2),(11,5)$

if $x^2-5y^2=4$,the above ineqution become $\sqrt{5 + \frac{4}{y^2}} \le \frac{20}{9}$ or $\sqrt{5 + \frac{4}{y^2}} \ge \frac{9}{4}$

$0<y \le 8$

so we can just test $y$ from 1 to 8, and we will find 3 base answer for $x^2-5y^2=4$ , $(x,y) = (3,1),(7,3),(18,8)$

all other answers can be generate by $(x+\sqrt{5}y)(9+4\sqrt{5})^n$ , $n$ can be positive or negative integer


问:
How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$
I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
But I've not been successful in doing so, what I tried is
$$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx=\int _0^1\left(x\right)'\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
$$\overset{\operatorname{IBP}}=\ln \left(\zeta (2)\right)+\int _0^1\frac{\ln \left(1-x\right)}{\operatorname{Li}_2\left(x\right)}\:dx$$
I also tried using identities for the dilogarithm but they dont do much.

I'm not sure what to do now, any help will be very well regarded, thank you.

答:
I do not know if there is a closed form.

Being myself quite lazy, I should compose Taylor series  starting from
$$\text{Li}_2(x)=\sum_{n=1}^\infty \frac {x^n}{n^2}$$ and being very patient (this is cheating ! I used a CAS); write
$$\log \left(\operatorname{Li}_2\left(x\right)\right)=\log(x)+\sum_{n=1}^\infty a_n x^n$$ where, unfortunately, all $a_n$'s are positive. The coefficients make the sequence
$$\left\{\frac{1}{4},\frac{23}{288},\frac{23}{576},\frac{50119}{2073600},\frac{33769
   }{2073600},\frac{257986487}{21946982400},\frac{130265243}{14631321600},\cdots\right\}$$ which decrease quite fast; halas, nothing being recognized by $OEIS$.

This gives
$$\int_0^1\log \left(\operatorname{Li}_2\left(x\right)\right)\,dx=-1+\sum_{n=1}^\infty \frac{a_n}{n+1}$$ Unfortunately, the convergence is not very fast
$$\left(
\begin{array}{cc}
p & 1+\sum_{n=1}^p \frac{a_n}{n+1} \\
 10 & -0.826294 \\
 20 & -0.824783 \\
 30 & -0.824456 \\
 40 & -0.824334 \\
 50 & -0.824276 \\
 60 & -0.824243 \\
 70 & -0.824223 \\
 80 & -0.824210 \\
 90 & -0.824201 \\
 100 & -0.824194 \\
\dots & \dots \\
\infty &-0.824166
\end{array}
\right)$$

However, it is much faster than the same work done using your last expression.


问:
Is my method of solving equation correct?
The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$

Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!

答:
When you went from :
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
to
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
You incorrectly assumed that
$$a^3 + b^3 = (a+b)^3 = \biggl(\frac{32}{(a+b)^2}\biggr)$$
Therefore you cannot say that
$$a^3 + b^3 = \frac{32}{2^2}$$


问:
Upper Triangular Matrix with zeros up
I have a linear algebra homework, but honestly I'vee been trying to solve this problem for some days without success:

Suppose T $\in$ $M_{5 \times 5}$ (K) is a triangulable matrix. Be $\beta = \{ \vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4}, \vec{v_5} \}$ a base of $K^5$ such that:

$$[T]_\beta = \begin{pmatrix}
4 & * & a & b & c \\
0 & 4 & d & e & f \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$

And I must prove that exists a base $\beta'$ such that:

$$[T]_{\beta'} = \begin{pmatrix}
4 & * & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0 \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$

How can I prove it?

Thoughts: If a=b=c=d=e=f=0 means that there is nothing in the coordinates corresponding to the vectors related to 4, but I don't know how to prove that there is a basis with vectors that satisfy this.

答:
The generalized eigenspaces $\ker(T-4)^2$ and $\ker(T-5)^3$ intersect at only the zero vector.

In general, if $(T-\lambda)^nv=0=(T-\mu)^mv$ with $\lambda\ne\mu$ and $n,m\ge1$, then by writing the gcd $I$ of the coprime factors $(T-\lambda)^n$, $(T-\mu)^m$, as a Bezout sum of them, $$v=p_1(T)(T-\lambda)^nv+p_2(T)(T-\mu)^mv=0$$

Hence it is possible to find a spanning set $\{w_1,w_2\}$ for $\ker(T-4)^2$ and another $\{w_3,w_4,w_5\}$ for $\ker(T-5)^3$, such that together form a basis $\beta'$ for $K^5$.

The coefficients $a,\ldots,f=0$ since generalized eigenspaces are $T$-invariant. For example, suppose $Tw_3=\sum_{i=1}^5\alpha_iw_i$, then $$0=(T-5)^3Tw_3=\alpha_1(T-5)^3w_1+\alpha_2(T-5)^3w_2$$ $$\therefore\ (T-5)^3(\alpha_1w_1+\alpha_2w_2)=0$$ which forces $\alpha_1=0=\alpha_2$ by the above.

Title: Upper Triangular Matrix with zeros up


问:
Proving that no term is an initial segment of another term.

The Question:

If $s$ and $t$ are strings, we say that $s$ is an initial segment of $t$ if there is a nonempty string $u$ such that $t = su$, where $su$ is the string $s$ followed by the string $u$. For example, $KUMQ$ is an initial segment of $KUMQUAT$ and $+24$ is an initial segment of $+24u - v$.

The task at hand is to prove, by induction on the complexity of $s$, that if $s$ and $t$ are terms, then $s$ is not an initial segment of $t$.

[Suggestion: The base case, when $s$ is either a variable or a constant symbol, should be easy. Then suppose that $s$ is an initial segment of $t$ and $s := f t_1 t_2 t_3\dots t_n$, where you know that each $t_i$ is not an initial segment of any other term. Look for a contradiction.]

What I have so far:

For bookkeeping, let us first use a definition for terms. A term of $L$ is a nonempty finite string $t$ of symbols from $L$ such that either:

$t$ is a variable

$t$ is a constant symbol, or

$t := f t_1 t_2 t_3\dots t_n$, where $f$ is an n-ary function symbol of $L$ and each of the $t_i$ terms is a term of $L$. This is a recursive definition.

Base Case:

We begin our inductive proof with the base case, as you would expect. What this looks like, is actually, considering the case where $s$ is a variable or a constant symbol.

Suppose that $s$ is a variable. Then, $t := su$, where $u$ is non-empty, can only mean that $t$ is a variable term followed by some other sequence $\in L$. However, since there is no function symbol preceding $s$, this is not possible, since $t$ too is a term. Therefore, if $s$ is a variable, and $t$ is a term, $s$ can not be an initial segment of $t$.

The case for when $s$ is a constant symbol is identical to the case above.

Inductive Hypothesis:

Let the statement of theorem we are trying to prove be that if $s$ and $t$ are terms, then $s$ is not an initial segment of $t$. We define the induction property $P$ such that $P$ holds for term $s$ if $\forall i$, $s$ is not an initial segment of $t_i$ (where $t_i$ is a term) i.e. including a non-empty string $u$, $t_i := su$ is not possible, and conversely, $t_i$ is not an initial segment in $s$.

Inductive Step:

The inductive step of a proof by induction on complexity of a term takes the following form: Assume that $s$ is a term by virtue of clause (3) in the definition of terms above. Also assume that the statement of the theorem is true when applied to the terms $t_1, t_2, \dots, t_n$. Suppose, towards contradiction, that $s$ takes the form $f t_1, t_2, \dots, t_n u$ and is indeed an initial segment for $t$. This implies that $t$ must, by definition of an initial segment, must begin with $f t'_1, t'_2, \dots, t'_n$.

Since we said that:

$$t = su$$

It must be that:

$$f t'_1, t'_2, \dots, t'_n = f t_1, t_2, \dots, t_n u$$

We can then get rid of the $n$-ary function $f$ to achieve the following:

$$t'_1, t'_2, \dots, t'_n = t_1, t_2, \dots, t_n u$$

How to proceed..?

I got some leads from this answer while I was writing this post, but it seems like my answer is missing something and the aforesaid didn't really affirm my understanding or confidence in the correctness of my answer so far. Moreover, I'm looking for a bit more intuitive, and dare I say, simple-worded explanation of how to carry the aforesaid induction forward.

答:
Found this proof by author Kunen in ... don't know at the moment, but will try to find out later.

Induction on length of term $t$.
We prove simultaneously:

No proper initial segment of a term is a term

If $t$ has first symbol $f$ of arity $n$ then unique terms $t_1,\dots,t_n$ exist with $t=ft_1\cdots t_n$.

In this context variables and constants have arity $0$. We assume that both bullets hold for terms that have smaller length than term $t=ft_1\cdots t_n$. Then it is enough to prove that they also hold for $t$.

Let $t'$ be any term which is an initial segment (possibly not proper) of term $t=ft_1\cdots t_n$. Terms are non-empty strings hence $t'$ starts with $f$. So we must have $t'=ft_1'\cdots t_n'$ where every $t_i'$ denotes a term. Then we must have $t_1=t_1'$ because otherwise one would be a proper initial segment of the other (contradicting induction hypothese first bullet). Likewise we can prove $t_i=t_i'$ applying induction on $i$: if $t_j=t_j'$ for every $j<i$ then $t_i$ and $t_i'$ start with the same symbol of $t$ so $t_i=t_i'$ because otherwise one would be a proper initial segment of the other.

Now we conclude that $t=t'$ and have established both bullets to be true for $t$.

Nice illustration of the profit of "proving more than you want at first hand" by induction proofs. Your hypothese is stronger.


问:
Bipartite graph with weight function such that the sum of weight of neighbours is 1
Suppose $G=(A,B)$ is a bipartite graph with bipartition $(A,B)$ with weight function $w: E(G)\rightarrow \mathbb{R}_{\geq 0}$. Suppose $G$ has the property that for any vertex $v$, $\sum_{u \in N(v)}w(uv)=1$. We can prove that $G$ must have a perfect matching. But I also want to prove the following, there exists perfect matchings, $M_1,...,M_k$ and non-negative numbers $c_1,...,c_k$ such that for every $e \in E(G)$, $\sum_{i: e\in M_i}c_i = w(e)$ and further $\sum_{i=1}^{k}c_i=1$

答:
Note that if there is some constant $C > 0$ such that for every vertex $v$, $\sum_{u \in N(v)}w(uv) = C$, then there is a perfect matching in $G$, since after multiplying the weights by $1/C$, we're back to the $C = 1$ case.

This observation allows us to solve the problem greedily. Take a perfect matching $M_1$ in $G$ and let $c_1$ be $\min\{w(e) : e \in M\}$. Now we update the graph by decreasing the weight of the edges in $M_1$ by $c_1$ and removing the edges from $G$ whose weight have become zero.

Since $M$ was a perfect matching, in this new graph we have that for every vertex $v$, $\sum_{u \in N(v)}w(uv) = 1 - c_1$. Now we can iterate the above to get another perfect matching $M_2$, and so on. Note that in each iteration we delete at least one edge of $G$, so in a finite number of steps we reach a state where $G$ has no more edges, or equivalently, where every edge of the original graph has zero weight. Since at the end of each iteration $\sum_{u \in N(v)}w(uv) = 1 - (c_1 + \dots + c_k) = C$ was constant for all $v$, this must mean that $C = 0$, so that $\sum c_i = 1$. Also, the final weight $w'(e)$ of an edge is just $w(e) - \sum_{i: e \in M_i}c_i$. Since $w'(e) = 0$, this gives $w(e) = \sum_{i: e \in M_i}c_i$, as desired.


问:
find domain of given question. f(x,y)=arcsin( √x- √y/ √x+ √y)

Find domain of given question.
$$f(x,y)= \arcsin \frac{ \sqrt x-  \sqrt y}{ \sqrt x+ \sqrt y}$$

Body: $\{(x,y)|x\geq 0, y\geq 0\}-\{(0,0)\}.$


问:
Solve differential equation using 'variation of parameters'
Given the differential equation $L \ u = f$, with $$L \ u = a_2(t) \frac{d^2u}{dt^2}+a_1(t)\frac{du}{dt}+a_0(t)u$$
with $a_i(t)$ sufficiently smooth and $a_2(t) \neq 0$ for every $t$.
Suppose that $u_0$ is a solution to the homogenenous equation $L \ u=0$, with $u_0(t)\neq0$ for every $t$.
 How can I then use the variation of parameters method by Lagrange to find the solution $u =u_0v$ ? And how can I find the general solution if the equation isn't homogeneous?

Update: I've substituted $u=u_0v$ into the equation and found a new $2^{nd}$ order differential equation of the form
$$f = \frac{d^2v}{dt^2}(a_2(t)u_0)+\frac{dv}{dt}(2a_2(t)u_0'+a_1(t)u_0)+va_0(t)u_0$$ which I'm supposed to solve for $w$ using $w=v'$. But I don't seem to be able to proceed as I'm not sure how to solve the system
$$\begin{equation}
\begin{split}
w =&\ v'\\
w'a_2(t)u_0+w(2a_2(t)u_0'+a_1(t)u_0)+v(a_0(t)u_0)=&\ f\ \\
\end{split}
\end{equation}$$

答:
Here is how we can solve the homogeneous equation $ L u = 0$. Once we have both solutions of this equation, we can use the method of variation of parameters to find a solution to $ L u = f $.

First we calculate derivatives of the solution $ u = u_0 v $ that you suggested:
$$ u' = u_0' v + u_0 v' $$
$$ u'' = u_0'' v + 2 u_0' v' + u_0 v'' $$
Next, because $ u = u_0 v $ is a solution to the homogeneous equation $ a_2(t) u'' + a_1(t) u' + a_0(t) u = 0 $ we have:
$$ a_2(t) (u_0'' v + 2 u_0' v' + u_0 v'') + a_1(t) (u_0' v + u_0 v') + a_0(t) u_0 v = 0 $$
We can re-arrange the terms and the equation in terms of $ v $ and its derivatives:
$$ a_2(t) u_0 v'' + (2 a_2(t) u_0'  + a_1(t) u_0) v' + (a_2(t) u_0'' + a_1(t) u_0' + a_0(t) u_0) v = 0 $$
Now, because $ u_0 $ satisfies the equation $ L u = 0 $ , the coefficient of $ v $ is zero, and the equation becomes:
$$ a_2(t) u_0 v'' + (2 a_2(t) u_0'  + a_1(t) u_0) v' = 0 $$
This looks like a second order equation, but it can be solved as a first order equation, if we take $ w = v' $ :
$$ a_2(t) u_0 w' + (2 a_2(t) u_0'  + a_1(t) u_0) w = 0 $$
From here, we solve this equation for $ w $, calculate the integral of $ w $ to find $ v $, and multiply $ v $ by $ u_0 $ to find the solution $ u $.


问:
Sum of squares of eigenvalues
This came up in a sample final of mine, but I couldn't figure out a way to solve it.

For any $n \times n$ complex-valued matrix $A$, define $N(A) = \sum_{i,j=1}^n a_{ij}\bar{a_{ij}}.$ Let $A$ be a matrix with the eigenvalues $\{ \lambda_1, ..., \lambda_k \}$. Show that $N(A) \geq \sum_{i=1}^k \lambda_{i}^2.$

I managed to show that equality holds when $A$ is self adjoint, but I was not able to figure out anything further from this. I also tried an approach where $A = UTU^*$, where $U$ is unitary and $T$ is upper-triangular, but again I hit a dead-end. Could anyone help with this? Thank you!

答:
Let $M_n(\mathbb{C})$ denote the set of all $n\times n$ complex matrices. Also, define

$$ \langle A, B \rangle = \sum_{i,j} a_{ij}\overline{b_{ij}} = \operatorname{Tr}(A B^*) $$

for $A, B \in M_n(\mathbb{C})$. Then it is clear that $\langle \cdot, \cdot \rangle$ defines a Hermitian inner product on $M_n(\mathbb{C})$. So, if $\lambda_1,\dots,\lambda_n$ are eigenvalues of $A \in M_n(\mathbb{C})$, counted with multiplicity, then by the Cauchy-Schwarz inequality,

$$ \Biggl| \sum_{j=1}^{n} \lambda_j^2 \Biggr| = \left| \operatorname{Tr}(A^2) \right| =  \left| \langle A, A^* \rangle \right| \leq \left|\langle A,A \rangle\right|^{1/2}\left|\langle A^*,A^* \rangle\right|^{1/2} = \operatorname{Tr}(AA^*). $$

Title: Sum of squares of eigenvalues


答:
Your idea to try $A = UTU^*$ is good.

Since in $N(A) \geq \sum_{i=1}^k \lambda_{i}^2$ you treat the eigenvalues as members of an ordered field, I think you assume that they are real. We will prove a slightly more general $N(A) \geq \sum_{i=1}^k |\lambda_{i}|^2$ where eigenvalues are allowed to be complex (which may have been your intention all along).

First note that $N(A) = {\rm tr}(AA^*)$. Using your substitution and the cyclic property of the trace $N(A) = {\rm tr}(UTU^*UT^*U^*) = {\rm tr}(TT^*)$. Therefore

$$
N(A) = \sum_{i \leq j} |t_{ij}|^2 = \sum_{i=0}^k |\lambda_i|^2 + \sum_{i \lt j} |t_{ij}|^2 \geq \sum_{i=0}^k |\lambda_i|^2
$$

where we used the fact that diagonal entries of a triangular matrix are its eigenvalues (which in turn follows from properties of the determinant).

Title: Sum of squares of eigenvalues

Body:
This came up in a sample final of mine, but I couldn't figure out a way to solve it.

For any $n \times n$ complex-valued matrix $A$, define $N(A) = \sum_{i,j=1}^n a_{ij}\bar{a_{ij}}.$ Let $A$ be a matrix with the eigenvalues $\{ \lambda_1, ..., \lambda_k \}$. Show that $N(A) \geq \sum_{i=1}^k \lambda_{i}^2.$

I managed to show that equality holds when $A$ is self adjoint, but I was not able to figure out anything further from this. I also tried an approach where $A = UTU^*$, where $U$ is unitary and $T$ is upper-triangular, but again I hit a dead-end. Could anyone help with this? Thank you!


答:
You can correct the error that Ryan points out. Instead, note that
$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 8 - 6ab.$$
So,
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
becomes
$$a^5 + b^5 + 5ab(8 - 6ab) + 20a^2b^2 = 32.$$
Using $a^5 + b^5 = 32$, then expanding and collecting like terms,
$$40a^2b^2 - 10ab = 0 \implies ab = 0 \text{ or } ab = 4.$$
So, now we know that
$$(r - a)(r - b) = r^2 - (a + b)r + ab$$
equals either $r^2 - 2r + 4$, or $r^2 - 2r$, depending on the value of $ab$. Note that the former has no real roots (though it does have complex solutions, but that's whole other kettle of fish), but the latter does. The two numbers would have to be the roots of the former polynomial: $0$ and $2$. We must therefore have, putting the numbers in ascending order,
$$\sqrt[5]{16 - \sqrt{x}} = 0 \quad \sqrt[5]{16 + \sqrt{x}} = 2.$$
It's not hard to see that $\sqrt{x} = 16 \implies x = 256$.

Note: this is the same answer as your method. The reason is, because as it turned out $b = 0$, we actually had $(a + b)^3 = a^3 + b^3$ for that particular $a$ and $b$.

Title: Is my method of solving equation correct?

Body:
The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$

Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!


问:
Showing non-differentiability of a function at $0$

Starting from a function $f:t\mapsto f(t)$ such that:

$f(0)=0\tag{1}$

$\liminf\limits_{t\to\infty} \sqrt{t}f\left(\dfrac{1}{t}\right)=-\infty\tag{2}$

$\limsup\limits_{t\to\infty} \sqrt{t}f\left(\dfrac{1}{t}\right)=+\infty\tag{3}$

setting:

$f_l(t)=\liminf\limits_{\delta\to0} \dfrac{f(t+\delta)-f(t)}{\delta}$;

$f^u(t)=\limsup\limits_{\delta\to0} \dfrac{f(t+\delta)-f(t)}{\delta}$.

I have to show that:

$f^u(0)=+\infty$;

$f_l(0)=-\infty$.

I succeeded in showing 1. by the following steps:
\begin{align}
f^u(0)=\limsup\limits_{t\to\infty} \dfrac{f\left(\dfrac{1}{t}\right)-f(0)}{\dfrac{1}{t}}&\geq\limsup_{t\to\infty}\sqrt{t}f\left(\dfrac{1}{t}\right)\\&=+\infty
\end{align}
with the inequality coming from $(1)$ plus the fact that $\dfrac{f\left(\dfrac{1}{t}\right)}{\dfrac{1}{t}}=tf\left(\dfrac{1}{t}\right)\geq\sqrt{t}f\left(\dfrac{1}{t}\right)$ for $t\geq1$ and the last equality following from $(3)$.

However, I cannot show 2. (I have tried to show it by the same token used for showing 1., but I did not succeed). Could you please give me some hint?

答: $\lim \inf_{t \to 0} \frac {f(t)} t=\lim \inf_{t \to \infty} t f(\frac  1 t)\leq \lim \inf_{t \to \infty} \sqrt t f(\frac  1 t)=-\infty$. [The inequality follows from the fact that $f(\frac 1 t)<0, t>1$ implies $tf(\frac 1 t) \leq \sqrt t f(\frac1  t)$].


问:
Mulitplicativity of $\Vert \cdot\Vert_{L^\infty}$
Let $(X,\mathscr{A},\mu)$ be some measure space and let $L^\infty(X)$ denote $L^\infty(X,\mathscr{A},\mu)$.
Furthermore let $f\in L^\infty(X)\cap L^1(X,\mathscr{A},\mu)$ i.e. $\int_X \vert f\vert \ d \mu <\infty$ and $\operatorname{ess sup}_X \vert f \vert <\infty$. For some $k\in \mathbb{R}^+$  is the following true:

$\Vert f^k\Vert_{L^\infty(X)} \le \Vert f\Vert_{L^\infty(X)}^k$? Where $\Vert g\Vert_{L^\infty(X)}:= \operatorname{ess sup}_X \vert g\vert$ for any $g\in L^\infty(X)$.

答:
I think I might have figured out all the details thanks to all of you.
So for future reference, I'll share the answer.
Let $f\in L^\infty(X)$, then we may assume without loss of generality that $\Vert f\Vert_{L^\infty}=1$. Thus, for every $\epsilon >0$ there exists some set $N\subset X$ with measure zero, such that
$$ \sup_{x\in X\setminus N} \vert f(x) \vert -\epsilon \le 1 \implies \sup_{x\in X\setminus N}\vert f(x)\vert \le 1+\epsilon $$
Therefore we get that for any $k\in \mathbb{R}^+$
$$ (\sup_{x\in X\setminus N} \vert f(x)\vert )^k \le (1+\epsilon)^k $$
and since $\epsilon >0$ has been chosen arbirtrarily we have that $(\sup_{x\in X\setminus N} \vert f(x)\vert)^k \le 1$. Since we know that $(\sup_{x\in X\setminus N} \vert f(x)\vert)^k \ge \sup_{x\in X\setminus N}\vert f(x)\vert ^k$, we obtain that $\Vert f^k\Vert_{L^\infty}\le 1.$

Note: I do not claim the correctness of this proof


问:
Duality pairing between finite signed measures and bounded continuous functions
Let $E$ be a metric space, $C_b(E)$ denote the set of real-valued bounded continuous functions on $E$ and $\mathcal M(E)$ denote the set of finite signed measures on $\mathcal B(E)$.

How can we show that $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E)$$ is a duality pairing between $C_b(E)$ and $\mathcal M(E)$? Do we need impose further restrictions on $E$ (e.g. completeness and/or separability) and/or $\mathcal M(E)$ (e.g. regularity of the measures)?

We need to show that \begin{align}\forall f\in C_b(E)\setminus\{0\}&:\exists\mu\in\mathcal M(E)&:\langle f,\mu\rangle\ne0\tag1;\\\forall\mu\in\mathcal M(E)\setminus\{0\}&:\exists f\in C_b(E)&:\langle f,\mu\rangle\ne0\tag2.\end{align}

$(1)$ Should be easy. If $f\in C_b(E)\setminus\{0\}$, there is a $x\in E$ with $f(x)\ne 0$. Now we can choose $\mu$ to be the Dirac measure $\delta_x$ concentrated at $x$ and obtain $\langle f,\mu\rangle=f(x)\ne0$.

But how can we show $(2)$?

答:
To put this result into context, I will show how to deduce it from the big theorems. (This is not nearly as elementary as Kavi's proof.)

Unfortunately, different authors use different notation / terminology, so let me set up a few definitions first.

Definition. A finitely additive signed measure (or charge) is a function $\mu : \mathcal A \to [-\infty,\infty]$ with the following properties:

$\mu(\varnothing) = 0$;

$\mu$ assumes at most one of the values $-\infty$ and $+\infty$;

$\mu$ is finitely additive.

Comparted to signed measures, the requirement of $\sigma$-additivity is weakened to finite additivity.

If $\Omega$ is a topological space, let $\mathcal A_\Omega$ and $\mathcal B_\Omega$ denote respectively the algebra and the $\sigma$-algebra generated by the open (or closed) sets of $\Omega$. Then $\mathcal A_\Omega \subseteq \mathcal B_\Omega$, and $\mathcal B_\Omega$ is the $\sigma$-algebra generated by $\mathcal A_\Omega$, so every measure on $B_\Omega$ is uniquely determined by its values on $A_\Omega$.

Definition. Let $\Omega$ be a topological space. We say that a finitely additive signed measure $\mu$ on $\mathcal A_\Omega$ is regular if for every $A \in \mathcal A_\Omega$ one has
\begin{align*}
\mu(A) &= \sup\{\mu(F) \, : \, F \subseteq A\ \text{closed}\} \\[1ex]
&= \inf\,\{\mu(V) \: : \: V \supseteq A \ \text{open}\}.
\end{align*}
Note. Different authors use different notions of regularity. In particular, sometimes the closed sets $F \subseteq A$ are replaced by compact sets.

We will use the following well-known results:

Theorem. Let $\Omega$ be a normal Hausdorff space. Then $C_b(\Omega)'$ is isometrically isomorphic to the space $rba(\Omega)$ of all regular, finitely additive signed measures of bounded variation on $\mathcal A_\Omega$, equipped with the total variation norm.

See [DS58, Theorem IV.6.2 (p.262)] or [AB06, Theorem 14.10 (p.495)].

For general topological spaces, see this question on MathOverflow.

Lemma. Every finite signed Borel measure on a metric space is regular.

See [DS58, Exercise III.9.22 (p.170)] or [AB06, Theorem 12.5 (p.436)], among others.

(Side note: this is not true for the other notion of regularity, with compact sets instead of closed sets, as can be seen from this answer on MathOverflow.)

It follows that the space $\mathcal M(\Omega)$ of finite signed Borel measures is a subspace of $rba(\Omega) \cong C_b(\Omega)'$. To complete the proof, note that every normed space $X$ separates points on every subspace of $X'$: if $\varphi(x) = 0$ for all $x \in X$, then $\varphi = 0$.

(More generally, for a bilinear pairing $\langle E , F \rangle$ to be non-degenerate, so that it is a proper dual pairing, it is necessary and sufficient that the induced maps $E \to F^*$ and $F \to E^*$ are injective.)

References.

[DS58] Nelson Dunford, Jacob T. Schwartz, Linear Operators, Part I: General Theory, Interscience, 1958.

[AB06] Charalambos D. Aliprantis, Kim C. Border, Infinite Dimensional Analysis, A Hitchhiker's Guide, Third Edition, Springer, 2006.


问:
Where does the following formula come from
Where does the following formula come from:
$\sum_{i=1}^n i \cdot  2^i $
$= ( − 1)2^{n+1}  + 2$

Body:
Sum of GP:
$$\sum_{i=1}^{n} {x^i}=x \frac{x^n-1}{x-1}$$
D.w.r.t. $x$ both sides to get
$$\sum_{i=1}^{n} i ~x^{i-1}=\frac{nx^{n+1}-(1+n)x^n+1}{(x-1)^2}$$
Put $x=2$ to get
$$\sum_{i=1}^{n}i ~2^i =2^{n+1}(n-1)+2$$


问:
If $G$ is a group with $N$ normal and $H_1$, $H_2$ subgroups of $G$ with $H_1$ a normal subgroup of $H_2$, show $NH_1$ is a normal subgroup of $NH_2$.
I was unsure how to prove this property which seems necessary for the solution to a problem I did recently. Since $N$ is normal, it seems pretty clear that $NH_1 = H_1N$ and $NH_2 = H_2N$ so they are indeed subgroups of $G$. I tried to prove normality directly by taking $nh \in NH_2$ and was able to show that $$nh NH_1 = N nh H_1 = Nn H_1 h$$ using the fact that $H_1$ is normal in $H_2$ and so $h H_1 = H_1 h$. Then, it seems like this problem reduces to showing that $nH_1 = H_1n$ which isn't immediately obvious. In fact, I'm not sure if that's even true.

My latest attempt uses the fact that $NH_1 = H_1N$ and $Nn = N$ so that we can find some $n' \in N$ so that $n'H_1 = H_1n$ which would allow us to conclude $$nh NH_1 = N nh H_1 = n H_1 h = Nn' H_1 h = NH_1nh$$ which is what we wanted but I had some trouble showing that this idea actually worked.

Thank you in advance for your help!

答:
Hp:  $N\trianglelefteq G$, $H_1,H_2\leq G$, $H_1 \trianglelefteq H_2$

Ts: $NH_1 \trianglelefteq NH_2$

Note that, since $N\trianglelefteq G$ and $H_1,H_2 \leq G$, we have $NH_1,NH_2 \leq G$.

Take $x$ in $NH_1 \implies \exists n_1 \in N,h_1\in H_1 :x=n_1h_1$ and consider $y$ in $NH_2$, so $\exists n_2 \in N,h_2\in H_2: y=n_2h_2$. We have to show that $NH_1$ is normal in $NH_2$ (which means that $y^{-1}xy\in NH_1$), so

$$(n_2h_2)^{-1}(n_1h_1)(n_2h_2)=h_2^{-1}(n_2^{-1}n_1)h_1n_2h_2= h_2^{-1}n_3 h_1n_2h_2 = \underset {N\trianglelefteq G,H_1\trianglelefteq H_2}{n_4(h_2^{-1}h_1h_2)n_5}=n_4\bar h_1n_5=\underset{N\trianglelefteq G}{n_6\bar h_1} \in NH_1 \implies NH_1\trianglelefteq NH_2$$


问:
What's wrong with my boundary

Evaluate the triple integral $$\iiint_E x^8 e^y \,dV$$ where $E$ is bounded by the parabolic cylinder $z=4−y^2$ and the planes $z=0,x=2$, and $x=−2$.

I used boundary dxdzdy(BADCFE):
B: 2
A: -2
D: 4
C: 0
F: sqrt(4-z)
E: 0

the answer is 2500 something?
how can I get the right boundary

答:
Specifically on your boundary -

it is a parabolic cylinder and the parabola ($y^2 = 4 - z$) is in $YZ$ plane. Values of $x$ is bound by the planes $x = -2, x = 2$.

So your bounds can be -

$-2 \leq x \leq 2$

$- \sqrt{4-z} \leq y \leq \sqrt{4-z}$

$0 \leq z \leq 4$ (bound below by the plane $z = 0$)

And if you are using above bounds, the order of integral should have $dy$ first and then $dz$. $dx$ can be pretty much in any spot - keep it first or last, does not matter.

If you go with order $dz$ then $dy$, the bounds of $y, z$ will be -

$0 \leq z \leq 4-y^2$

$-2 \leq y \leq 2$

Title: What's wrong with my boundary