问:
What does it mean that a Jacobian matrix has full rank in an optimisation problem?
I learned that when we want to find candidate points for our optimisation problem of

$$\underset{x}{max} \quad f(\mathbf{X}) \text{ s.t. } g_j(\mathbf{X}) = c_j$$

where $\mathbf{X}$ is a point in multidimensional space, i.e. a vector, we find them through solving $\frac{\partial L}{\partial x_i}$, where $L$ is our Lagrangian,

$$L(\mathbf{X}, \Lambda) = f(\mathbf{X}) - \sum_{j=1}^{m}\lambda_j \,g_j(\mathbf{X})$$

and verifying that the rank of the jacobian matrix, $J_g(\mathbf{X})$ is $m$. What does it mean when the jacobian matrix has not full rank, $m$?

I know from linear algebra, that it means columns (or rows) are linearly dependent. However, I fail to deduct from that what this actually means for the optimisation problem.

Partially I understand that less-than-full-rank means that a constraint has similar first derivatives, however, whata does that mean for the optimisation problem?

Slide on which I base my question on:
Slide

答:
Let's assume the constraint functions $g_1, \ldots, g_m$ are smooth ($C^\infty$), though this can be relaxed.

The rows of the Jacobian $J_g(X)$ are the gradients of the constraints at $X$; let me write them as $\nabla g_1(X), \ldots, \nabla g_m(X)$.

If these are linearly independent (that is, if the Jacobian has full row rank), then they remain linearly independent in a neighborhood of $X$. You can see this by considering the determinant of $J_g(X) J_g(X)^\top$: it is nonzero at $X$, and it is continuous as a function of $X$; therefore, it remains nonzero in a neighborhood of $X$.

It is a fact from differential geometry that---under the conditions described above---the set of solutions to the constraints (locally) is a smooth manifold, that is, a smooth surface: at $X$, this surface has a well-defined tangent space (a linearization) and a normal space (the orthogonal complement of the tangent space with respect to some inner product).

The normal space is spanned exactly by the gradients of the constraints. In other words: vectors of the form $\sum_{j=1}^m \lambda_j \nabla g_j(X)$ are vectors normal to the surface, and all vectors normal to the surface can be written (uniquely) in that way.

This gives a nice interpretation to the Lagrangian formalism: $X$ is stationary exactly if the gradient of $f$ at $X$ is in the normal space to the surface. Intuitively, this should make sense.

When the gradients are not linearly independent (that is, when the Jacobian does not have full row rank), the above discussion does not go through: we cannot be certain that the search space (the set of solutions of the constraints) looks like a smooth surface around $X$. It might look weird (have kinks, angles, cusps, ...). It is then much harder to say precisely what is going on. The simplest results from typical introductory courses in optimization simply avoid those complications by assuming linear independence of the gradients. That is not to say that things necessarily go bad when you don't have linear independence, but they could; and at any rate, the situation requires more care.


问:
Comparison of rank and eigenvalues of matrix AB and BA
I was unable to solve this question of linear algebra and hence I am posting it here.

Let A and B be  n$\times $ n matrices over $\mathbb{C}$.Then which of the following are true:

A  AB and BA have same set of Eigenvalues.

B If AB and BA have same set of eigenvalues then BA= AB.

C If $A^{-1}$ exists then AB and BA  are similar

D The rank of AB is always equal as rank of BA.

I have done (C) .  Tried some matrices for D, but couldn't contradict it.  Fr A,B I don't have any idea so, your help is required.

Please guide me!

答:
$AB$ and $BA$ have same set of Eigenvalues

True.

See here

If $AB$ and $BA$ have same set of eigenvalues then $BA$=$AB$.

False.

From (A), the if part in (B) is trivially true for any two matrices. So you just need to pick any two non-commuting matrices as counter-example. Google Image has a good number of those. (But funny you have to search examples of commutative matrices.)

The rank of $AB$ is always equal as rank of $BA$.

False.

The easiest way to spot why this is false is to consider matrices as linear operators. If $A$ has non-trivial nullspace, and $B$ happens to map into that, yet A still maps out of the nullspace of $B$...

For example, consider

$$
A=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
1 & 0 & 0
\end{bmatrix},
B=\begin{bmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}
$$

In $\mathbb{R}^3, A$ only keeps 1st coordinate and throws it to 3rd. $B$ keeps 2nd & 3rd coordinate. Going through $A$ first and then $B$, you will end with 1st coordinate on 3rd slot. In other words, rank 1. Going through $B$ first and then $A$, since you lose 1st coordinate right away, you end with zero. In other words, rank 0.

ContentLicense: CC BY-SA 4.0


问:
If $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, show f is quadratic polynomial.
Suppose the function f is analytic in the punctured plane $z!=0$ (it means we excluded the zero) and satisfies the above condition, $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, then show f is quadratic polynomial.

I think that if we multiply $z$ on the both side then we get $g(z)=zf(z)$ goes to 0 when z is going to 0. therefore, g is analytic function. Then what we should do to prove this?

答:
First, since $\lim_{z \to 0} z f(z) = 0$, we see that $f$ has a removable singularity at $z=0$. Hence we may as well assume that $f$ is analytic everywhere, and so has an entire  power series expansion at $z=0$.

Second, if $|z| \ge 1$, we have $\frac{1}{\sqrt{|z|}} \le |z|^2$, hence $|f(z)| \le 2 |z|^2$, for $|z|\ge 1$.

Third, if we choose $R \ge 1$, and suppose $|z| = R$, then Cauchy's estimate gives $|f^{(k)}(0)| \le \frac{2 R^2}{R^k}$. Letting $R \to \infty$, it follows that if $k>2$, then $f^{(k)}(0) = 0$. Consequently $f(z) = f(0) + f'(0) z + \frac{1}{2} f''(0) z^2$.

ContentLicense: CC BY-SA 3.0


问:
Asymptotic formula for Mean of Sum of Power of Divisors $\frac{ \sum_{i=1}^{n}\sigma_v(i)}{n}\sim\frac{n^v\zeta(v+1)}{v+1}$

Question:

Define the sum of $v$-powers of divisor $\sigma_v(n)=\sum_{d|n}d^v$ for $v \in \mathbb{R}$.

Prove that for all $v>0$,
$$\frac{ \sum_{i=1}^{n}\sigma_v(i)}{n}\sim\frac{n^v\zeta(v+1)}{v+1}$$
where $\zeta$ is the Riemann zeta function.

Attempts:

By using the convolution $\sigma_1=\mathbb{1}*\text{Id}$,
(where $\mathbb{1}(n)=1$, $\text{Id}(n)=n$)

we have $\displaystyle \sum_{i=1}^n\sigma_v(i)=\sum_{d=1}^{n}\sum_{k\leq \frac{n}{d}}k^v$.

When $v=1$, we know the sum $\displaystyle \sum_{k\leq \frac{n}{d}}k = \sum_{k=1}^{[\frac{n}{d}]}k=\frac{1}{2}([\frac{n}{d}]+1)[\frac{n}{d}]$,

then by bounding $\frac{n}{d}-1<[\frac{n}{d}]\leq\frac{n}{d}$,

we can show $\displaystyle \sum_{i=1}^n\sigma_1(i)=\frac{n^2}{2}\sum_{d=1}^n\frac{1}{d^2}+f(n)$ for some $|f(n)|<n\log{n}$ , when $n\geq 4$.

The "2" in $d^2$, $n^2$, $\frac{1}{2}$ comes from the rather easy formula of sum of positive integers.

However, when $v\neq 1$, the term becomes the sum of power of integers and I cannot easily find a formula for it.

The closest I have heard of is Faulhaber's_formula with Bernoulli numbers as coefficients, but the coefficients are hard to handle and the formula seems to be for integer power only.

Another way I think of is by inequality. Power-Mean Inequality looks usable, but it does not have the ratio $\displaystyle\frac{1}{v+1}$ so it does not seems to be the way.

The third way is possibly constructing the general case from the result of $\sigma_1$, but I don't know how to either.

Any help is very much appreciated.

答:
Using techniques like Riemann-Stieltjes integration, we can show that

$$
\sum_{k\le n}{1\over k^s}=\zeta(s)+\mathcal O\left(1\over n^{s-1}\right)\quad \Re(s)>1\tag1
$$
$$
\sum_{k\le n}\frac1k=\log n+\gamma+\mathcal O\left(\frac1n\right)\tag2
$$
$$
\sum_{k\le n}k^a={k^{a+1}\over a+1}+\mathcal O(k^a)\quad a\ge0\tag3
$$

and further works on this problem solely depend on these identities.

First, let's exchange the order of summation

$$
\begin{aligned}
\frac1n\sum_{k\le n}\sigma_v(k)
&=\frac1n\sum_{k\le n}\sum_{d|k}d^v
=\frac1n\sum_{qd\le n}d^v \\
&=\frac1n\sum_{q\le n}\color{green}{\sum_{d\le n/q}d^v}
\end{aligned}
$$

To continue, we apply (3) to the green part:

$$
\begin{aligned}
\frac1n\sum_{k\le n}\sigma_v(k)
&=\frac1n\sum_{q\le n}\left[{n^{v+1}\over(v+1)q^{v+1}}+\mathcal O\left(n^v\over q^v\right)\right] \\
&={n^v\over v+1}\color{orange}{\sum_{q\le n}{1\over n^{v+1}}}+\mathcal O\left(n^{v-1}\color{blue}{\sum_{q\le n}{1\over q^v}}\right)
\end{aligned}
$$

Since $v>0$, we can apply (1) to the orange sum. Because it is not known whether $v>1$, it is better for us to consider each different situations for the blue sum:

For $v>1$, we know that the blue series converges, so it becomes $\mathcal O(n^{v-1})$.

For $v=1$, we can apply (2) to get $\mathcal O(n^{v-1}\log n)$.

For $0<v<1$, we can use (3) to obtain $\mathcal O(1)$.

In each of the aforesaid situations, we observe that the error term would not exceed $n^v$, leaving us this asymptotic relation:

$$
\frac1n\sum_{k\le n}\sigma_v(k)\sim{n^v\zeta(v+1)\over v+1}
$$


问:
Determinants and Characteristic Polynomial of Matrices

Let $A \in M_{mxn}(F)$ and $B \in M_{nxm}(F)$, show that
$$x^ndet|xI_m-AB| = x^mdet|x I-BA|.$$ Deduce that if n=m, then AB and BA have the same characteristic polynomial.

The second part should be easy to prove: if n=m, then $$x^ndet|xI-AB| = x^ndet|x I-BA| \implies det|xI-AB| = det|xI-BA|$$ which means the characteristic polynomial is the same since $det|xI-AB|$ or $det|xI-BA|$ gives the exact characteristic polynomial of AB and BA respectively.

For the first part, I am not sure how to proceed with the left side being determinant of mxm matrix while right side being determinant of nxn matrix.

答:
Consider the matrices
$\mathcal C$, $\mathcal D$ below,
compute $\mathcal C\mathcal D$ and $\mathcal D\mathcal C$
$$
\begin{aligned}
\mathcal C&=\begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix}\ ,\\
\mathcal D&=\begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix}\ ,\\[3mm]
%
\mathcal C\mathcal D &=
\begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix}
\begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix}
=
\begin{bmatrix} xI_m-AB & \\ *& xI_n\end{bmatrix}\ ,\\
%
\mathcal D\mathcal C &=
\begin{bmatrix} xI_m & xA\\ B&xI_n\end{bmatrix}
\begin{bmatrix} I_m & -A\\ & I_n\end{bmatrix}
=
\begin{bmatrix} xI_m & \\ * & xI_n-BA\end{bmatrix}\ ,
\end{aligned}
$$
(empty entries are zero block matrices, star entries do not matter any more,)
and use finally the multiplicativity of the determinant
$$
\det(\mathcal C\mathcal D)
=\det(\mathcal C)\det(\mathcal D)
=\det(\mathcal D)\det(\mathcal C)
=\det(\mathcal D\mathcal C)
\ ,
$$
and the relations
$$
\det\begin{bmatrix}S&\\*&T\end{bmatrix}=\det(S)\det(T)\ ,\qquad
\det(xI_m)=x^m\det(I_m)=x^m\ .
$$


问:
In how many ways can a cinema hall with 100 seats be filled with 100 people such that certain conditions are satisfed?
In how many ways can a cinema hall with 100 seats be filled with 100 people such that certain conditions are satisfed?

Conditions:

40 seats are on the first floor

60 seats are on the ground floor

10 of the 100 people will only be seated on the first floor and 15 others will only be seated on the ground floor

My solution:

Select 10 seats from the first floor and find the arrangements for the 10 people ( $^{40}C_{10} \times10!$ ), then similarly select 15 seats from the ground floor and find arrangements for the 15 people ($^{60}C_{15}\times15!$), and then arrange the rest in the remaining seats(75!): $$^{40}C_{10} \times10!\times^{60}C_{15}\times15!\times75!$$
But this is incorrect, why is this incorrect and how to solve this?

答:
Your answer is correct.

Here is another approach to the problem:

We know that we must reserve $10$ seats on the first floor for the $10$ people who will only sit on the first floor and $15$ seats on the ground floor for the $15$ people who will only sit on the ground floor.  Therefore, there are $40 - 10 = 30$ seats left on the first floor.  We must choose which $30$ of the remaining $100 - 10 - 15 = 75$ people will sit on the first floor.  We can do this in $\binom{75}{30}$ ways.  The remaining $75 - 30 = 45$ people must take the remaining $60 - 15 = 45$ seats on the ground floor.  The $40$ people on the first floor can be seated in the $40$ seats on the ground floor in $40!$ ways.  The $60$ people on the ground floor can be seated in the $60$ seats on the ground floor in $60!$ ways.  Hence, there are
$$\binom{75}{30}40!60!$$
admissible seating arrangements, which agrees with your answer.

The stated answer
$$\binom{75}{30} = \frac{75!}{30!45!}$$
is the number of ways we can select which $30$ of the $75$ people who have not indicated that they will only sit on a particular floor will be seated on the first floor.  It is not the answer to the question you stated.


问:
Proof by strong induction for odd and even sequence
I have a proof by strong induction question (I won't ask the full question since someone in a previous course got in trouble for it). We basically have to prove, for a recurring sequence, that when $n$ is an odd the result of $S(n)$ is odd and when $n$ is an even number $S(n)$ is even.

I know that for strong induction we have to show multiple basis cases but with the induction step I am not sure in this case whether we have to to do it in two parts one showing that it's odd and one showing that it's even. Could anyone point me to the right direction?

答:
After you have split into cases for $n$ even or $n$ odd, you tell us that you must prove something about $S(n)$ using strong induction? Without knowing the question it is hard to tell.

Perhaps you might need to induct on even and odd terms separately.

For example,
if we have some recurrence relationship between $a_{n}$ and $a_{n+2}$ for some sequence $S$ then we might use induction to get information about $a_{n+4}$ and in the case of strong induction to get information about $a_{n+2}$ from $\{a_{2},a_{4},...,a_{n}\}$.

Does this help at all?


问:
Proving a geometric inequality $BE+CF>EF$
In the figure $D$ is the midpoint of $BC$ and $\Delta EDF$ is right triangle.Prove that $BE+CF>EF$

I tried triangle inequality $$BE+ED>BD$$
$$CD+FC>DF$$
Add:$$BE+FC>DF-ED$$

I am stuck now!

答:
Rotate the figure by $180^\circ$ around $D$.

Now the desired inequality becomes $BE + BF' > EF$.

However $EF = EF'$ by congruent triangles $\triangle EDF' \cong \triangle EDF$.

Finish off using triangle inequality.


问:
How many values of positive integers 'm' exist such that, $m^4 - 4m^3 + 22m^2 - 36m + 18$ is a perfect square.
I found this question in a math olympiad book. I figured that,

$m^4 - 4m^3 + 22m^2 - 36m + 18$ = $(m-1)^2(m^2 - 2m + 17) + 1$

But I don't know if I'm solving it the right way, and if yes, I don't know how to proceed.

答:
Note that we have:
$$
(x^2 -2x + 9)^2
=x^4 - 4x^3 + 22x^2 - 36x + 81\ ,
$$
and the previous square number is not far away.


问:
PDEs in biology
I am student who mostly heard lectures on partial differential equations and homogenization. But I really like the idea of working in biology or with biologists - but (with my lack of overview) it seems to me it's either statistics or some ODE problems.
Is there anything more PDE releated in biology?</p>

答:
The book referenced by @Henrik Finsberg is OK if you'd like to learn some basic facts about PDEs (and about their applications in biology). If you feel already confident about PDEs, try to get a hold of

James D. Murray, Mathematical Biology, Vol. 2

Benoît Perthame, Transport equations in biology

Robert S. Cantrell, Chris Cosner, Spatial Ecology via Reaction-Diffusion Equations

Akira Okubo, Simon A. Levin, Diffusion and Ecological Problems, Modern Perspectives

Title: PDEs in biology


问:
Limit of piecewise function and find a

I calculated the limit for 1 ^ - and found it infinite. But for 1 ^ + I can't find a when the 2nd function equals infinity. What solution should I do?

Title: Limit of piecewise function and find a

Body:
Right limit at $x=1$ is $$RL=\lim_{h\to 0} [2(1+h)-4+3^\frac{1+h}{1-(1+h)^2}$$
$$RL=-2+\lim_{h \to 0}3^{\frac{(1+h)}{1-(1+2h)}}=-2+\lim_{h\to 0}3^{\frac{1+h}{-2h}}=-2$$
Left limit at $x=1$ is
$$LL=\frac{8}{\pi}\tan^{-1}a $$
Finally, $LL=RL$ gives $a=-1$.


问:
$f\cap(A\times Y)$ is a function
I have the following exercise:

If $f:X\rightarrow Y$ is a function and $A\subset X$ is a set. It is true that $f\cap(A\times Y)$ is a function?

At first I thought this was the restriction function, but I checked the definition of the restriction of a function and it does not match this one. I thought that since f is a function, and $f \subset X\times Y$ then that was enough to see that $f\cap(A\times Y)$ is a function, but I think it is not enough, could you please help me. Thank you.

答:
$f\cap(A\times Y)$ is indeed a function:

Its elements are elements of $X \times Y$.

Each element is paired with just one element of $Y$.

However it is not a function from $X$ to $Y$. It is a function from $A$ to $Y$.

Title: $f\cap(A\times Y)$ is a function


问:
Reference for optimization on Riemannian manifolds
Can anyone recommend me a fairly good textbook on optimization on manifolds?
I only managed to find this: Convex Functions and Optimization Methods on Riemannian Manifolds

It seems a bit old, nothing against it obviously the content seems ok to me, but maybe there's something more recent out there.

Ideally I'd like something like Numerical Optimization
By Jorge Nocedal, Stephen Wright but translated on manifolds.

But any reference you can suggest in general is fine.

答:
Adding to Mark L. Stone's answer, I wrote a textbook for Riemannian optimization which may address the needs you describe:

An introduction to optimization on smooth manifolds

http://www.nicolasboumal.net/book

It does not assume any prerequisites in geometry. It addresses certain aspects of practical importance for numerical implementations, such as numerical pitfalls and recommended parameter values.


问:
How do I solve $\frac{d^2y}{dx^2} = (1+(\frac{dy}{dx})^2)^{3/2}$?
$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$

My progress: I have used substituon $u = \frac{dy}{dx}$ and arrived at $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ equation with $c$ - constant. After that I was unsure on whether it is allowed to consider separate cases for $u$ or not: maybe for some values of $x$, $u$ will have positive sign, on other occasions negative sign. In fact, any help related to such sign issues would be welcomed.

答:
Substitute

$$ u = \frac{dy}{dx}$$

This is a first order homogeneous differential equation nonlinear. Because

$$f(\frac{dy}{dx},\frac{d^2y}{dx^2})$$

alone and the square is risen to third order about a polynomial with a constant and the first order differential in seconder order.

Our new differential equation with $u$ is:

$$\frac{du}{dx}=(1+u^2)^\frac{3}{2}$$

This can be solve by the very famous method of serepation of constants found everywhere in the literature for differential equation solution methodology. It is as simple as integrating:

$$\frac{du}{(1+u^2)^\frac{3}{2}}=dx$$

$$\int\frac{du}{(1+u^2)^\frac{3}{2}}=\frac{u}{(1+u^2)^\frac{1}{2}}=x+c$$

Now we have x(u) but wanted x and u=y'. So we need to invert the equation for u(x). Squaring the equation and regrouping for u gives

$$u(x)=(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}= \frac{dy}{dx}$$

This equation is first order and already integrable.

$$(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}dx= dy$$

$$ y(x)= -(1-(x+c)^2))^\frac{1}{2} + C $$

Only this two integration step give us the allowance for two integration constants.

We can than rearrange the equation to the general representation of a unit circle:

$$1=(y+c_1)^2+(x+c_2)^2$$

that is shifted away from the origin in the plane. This is indeed a hard restriction we did had with the differential equation at the start. To cover all $ℝ^2$ we need to cover the space of ${c_1,c_2}$ in $ℝ^2$.

Only in the second step of this solution path there is a discussion necessary about complexes solution of the square root. For the second step to be real

$1-(x+c)^2)\geq0$ and $x+c\geq0$

are required. Can that be fullfilled for all $c\in ℝ$? That can be.

The question of interest, how can a nonlinear differential equation be so selective for a small part of values and can also be solvable for rather bigger number of solutions tuples $(x,y)$? Now the answer is the nonlinear differential equation is for the constants only. It does not matter on the tuples $(x,y)$ and in this case this is $ℝ^2$ each. In the case we leave $ℝ^2$ it gets not so restricted. The restriction arises from $ℝ^2$. We can define metrics on $ℝ^2$ and that is the point of investigation making this nonlinear differential equation of first order in the first derivative of $y$ so interesting. The metrics can be positive definite and therefore impose well defined orders.

It is necessary to formulate the equation with the square root for that result.


问:
A problem about method of moments (MOM) and find the distribution of a event

An urn contains $\theta$ black balls and $N-\theta$ white balls. A sample of $n$ balls is to be selected without replacement. Let $Y$ denote the number of black balls in the sample. Show that $(N/n)Y$ is the method-of-moments (MOM) estimators of $\theta$.

My approach: I know that:

Method of moment (MOM): Choose as estimates values of the parameters that are solutions of the equations $\mu_{k}'=m_{k}'$, for $k=1,2,\ldots,t$ where $t$ is the number of parameters to be estimated and $\mu_{k}'=\mathbb{E}[Y^{k}]$ is the $k$-th moment of a random variable and $\displaystyle m_{k}'=\frac{1}{n}\sum_{i=1}^{n}Y_{i}^{k}$ is the corresponding $k$-th sample moment.

Now, 1) I need to know of the distribution of random variable $Y$, so I think that our sole observation  $Y$ is hypergeometric, but I don't sure about it. How can I check the first part?

I think here $t=1$, so we need to solve $\mu_{1}'=m_{1}'$. I think that part is easy if a can find the distribution for $Y$.

答:
Considering the Hypergeometric distribution,

$$\mu=\frac{\theta n}{N}$$

thus

$$\theta=\frac{N}{n}\mu$$

thus the MOM estimator of $\theta$ is

$$\hat{\theta}_{MM}=\frac{N}{n}\overline{Y}_n$$

because it has been obtained substituting the first moment with the first sample moment


问:
How do I calculate the equation of a circle given 3 complex numbers?
Given three complex values (for example, $2i, 4, i+3$), how would you calculate the equation of the circle that contains those three points?  I know it has something to do with the cross ratio of the three points and $z$ and the fact that the cross ratio is a real number, but I don't know what to do with that information.</p>

答:
Might as well flesh out Theo's comments a bit...

If you have four points $z_i,\quad i=1\dots4$, in the following configuration:

the cross ratio of these four points is

$$\Delta=\frac{z_1-z_3}{z_1-z_4}\left(\frac{z_2-z_3}{z_2-z_4}\right)^{-1}$$

If $\Delta$ is real, this means that

$$\arg\left(\frac{z_1-z_3}{z_1-z_4}\right)-\arg\left(\frac{z_2-z_3}{z_2-z_4}\right)=0$$

Interpreted geometrically, the angles $\angle z_3 z_1 z_4$ and $\angle z_3 z_2 z_4$ are congruent, and must thus be inscribed angles in a circle; i.e. all four points are concyclic.

If we let $z_4$ be a variable point $z$ tracing the circle and $\Delta$ be a varying parameter, we obtain an equation for the circle through $z_1,z_2,z_3$:

$$\Delta=\frac{z_1-z_3}{z_1-z}\left(\frac{z_2-z_3}{z_2-z}\right)^{-1}$$

Solving for $z=x+iy$ gives

$$z=\frac{z_2(z_1-z_3)-z_1(z_2-z_3)\Delta}{z_1-z_3-(z_2-z_3)\Delta}$$

which as Theo says is a Möbius transformation. Taking real and imaginary parts and eliminating $\Delta$ should yield a Cartesian equation. (Obtaining the center and radius of this circle is a bit messy and is left as an exercise in algebraic manipulation.)


问:
If you randomly call $5$ students and each call is independent, what is the probability that no student is called twice?
I am thinking about this question and I don't really know how to tackle it.

The problem is:

The class has $30$ students. If you want to call 5 students, each one is independent of the others and randomly with equal probability, what is the probability that none are repeated?

I am thinking as $$\frac{30\cdot 29\cdot 28 \cdot  27\cdot 26}{30^5},$$ but I am not sure it is correct. Thank you so much!

答:
Think about it as a bowl with 30 blue marbles (all the students). Whenever a student is called, we replace a blue marble with a red marble (so this is a student we have already called). Now what is the probability that we don't grab a red marble after 5 times?

So the first attempt, P(blue) is obviously one.
The second attempt, we have P(blue) = $\frac{29}{30}$.
The third, fourth and fifth attempt, we have respectively:

$\frac{28}{30}$, $\frac{27}{30}$, $\frac{26}{30}$

Multiplying these

$1 \times \frac{29}{30} \times \frac{28}{30} \times \frac{27}{30} \times \frac{26}{30}$

we get $\approx$ $70.4$%...

The same will be:

$(\frac{1}{30})^5 \times (30 \times 29 \times 28 \times 27 \times 26) $

So yes, your answer is correct as well.


问:
How to apply Change of Variable theorem with Probability Integral Transform?
I have a function which outputs samples and the density of a random variable on $(-\infty, \infty)$. On the samples, I apply the Gaussian CDF to get samples on [0,1]. Now, I would like to transform the density accordingly.

My idea was to use the Change of Variables theorem. If $G(y)$ is the Gaussian CDF and $f(x)$ my function, I think it should look like this:

$g(x) = f(G^{-1}(x)|\det(\dfrac{\partial G^{-1}(x)}{\partial x})|$

Is this correct?

If so, how do I find the determinant of the inverse Gaussian CDF? I believe another form to express this is

$g(x) = f(z)|\det(\dfrac{\partial G(z)}{\partial z})|^{-1}$, for $z=G^{-1}(x)$

but then I'm still not sure how to find the determinant term.

答:
OK, I think I figured it out:

Let's say $x$ is the random variable on $(- \infty, \infty)$ and $f(x)$ it's PDF. Lets call the Gaussian CDF $H(x)$, and $z:=H(x)$. I am interested in the PDF of z, called g(z).

Given the Change of Variables Formula, we get

$g(z) = f(H^{-1}(z))|\det \dfrac{\partial H^{-1}(z)}{\partial z}|$

and with $z=H(x)$

$g(z) = f(x) | \det \dfrac{\partial H(x)}{\partial x}|^{-1}$

and since the derivative of the Gaussian CDF is the Gaussian PDF, I think $\dfrac{\partial H(x)}{\partial x}$ just becomes a matrix with $h_1(x), ..., h_n(x)$ on the diagonal. The determinant of that is $\prod_{i=1}^n h_i(x)$. We can then take the log:

$\log g(z) = \log f(x) - \log \prod_{i=1}^n h_i(x)$

which I can calculate using my original function and the Gaussian PDF.


问:
What is the name of this inequality of sigma and product?
Does this inequality have a name so I could search for it's prove?

$$for \sum_{i=1}^{n} t_i=1 $$
We have:
$$
    (x_1^{t_1})  ... (x_n^{t_n}) <= (t_1x_1) +...+ (t_nx_n)
$$

答: Well I did some more searches and apparently this is a form of AM–GM inequality. So it can be proved by Jensen's inequality.</p>


问:
In $\Delta ABC$ , bisectors of $\angle A, \angle B$ meet at incentre $I$. The extension of $AI$ meets the circumcircle of $\Delta ABC$ at $D$.

In $\Delta ABC$ , bisectors of $\angle A, \angle B$ meet at incentre $I$. The extension of $AI$ meets the circumcircle of $\Delta ABC$ at $D$. Let $P$ be the foot of the perpendicular from $B$ to $AD$, and $Q$ be a point on the extension of $AD$ such that $ID = IQ$. Find $\frac{(BQ * IB)}{(BP * ID)}$.

What I Tried: Here is a picture :-

I was struggling to make a good picture out of this, and I am now totally stumped to solve the problem. A good amount of information is not given, for example no lengths of any sides of the triangles or circles are given, and they expect me to find the value of $\frac{(BQ * IB)}{(BP * ID)}$ . Second is that I think the problem looks very complicated so I cannot understand how I should start solving it.

Can anyone help me? Thank You.

答:
It suffices to prove that $DI = DB$.

Note that $\angle IBC = \angle IBA$ and $\angle BAI = \angle CAI = \angle CBD$.

Hence $\angle DBI = \angle DBC + \angle CBI = \angle IAB + \angle IBA = \angle DIB$.

This proves $DI = DB = DQ$. By Thales' Theorem (or otherwise), $IB \perp BQ$.

Now consider the areas of $\triangle DBI$ and $\triangle QBI$ and you will have the ratio you need.


问:
Show that image of simplex contains its barycenter
Let $T^n$ be a standard $n$-simplex and $f_\varepsilon \colon T^n \to T^n$ be a family of continuous mappings, $\varepsilon > 0$. Furthermore $f_\varepsilon \to \mathop{\mathrm{id}}\nolimits_{T^n}$ in $C(T^n,T^n)$ when $\varepsilon \to +0$. How to show that for small $\varepsilon$ there is an $x(\varepsilon) \in  T^n$ such that $f_\varepsilon(x(\varepsilon)) = x^\ast$ where $x^\ast$ is the barycenter of $T^n$?</p>

答: I have taken the liberty to interpret the converge as a homotopy from $f_1$ to $f_0=$ Id$_X$. Let $X=T^n$ and assume that the barycenter $b$ is not in the image $f_t(T^n)$ for positive $t$. The boundary of $T^n$ is homeomorphic to $S^{n-1}$. We can restrict $f_t$ to this boundary to get a map $g_t:S^{n-1}\to T^n-{b}$ since $g_0=$Id and $g_t(S^{n-1})\not\ni b$. Now $g_0$ is the embedding of $S^{n-1}$ as the boundary and induces the identity on $\tilde H_{n-1}(S^{n-1})\approx\Bbb Z$ since $S^{n-1}\xrightarrow{g_0}T^n-\{b\}\ \xrightarrow{r}S^{n-1}$ (where $r$ is the deformation retraction onto the boundary) is the identity and induces the identity on homology $\tilde H_{n-1}(S^{n-1})\xrightarrow{g_{t*}}\tilde H_{n-1}(T^n-\{b\})\xrightarrow{r_*}\tilde H_{n-1}(S^{n-1})$. But $g_t$ is null-homotopic for $t>0$ since it factors as $S^{n-1}\hookrightarrow T^n\xrightarrow{f_t}T^n-{b}$, so it induces the zero map on homology.</p>


答:
Yes this is correct.

$30 \cdot 29 \cdot 28 \cdot 27 \cdot 26$ or $^{30}P_5$ is the number of ways to permute $5$ students out of $30$ without repetition. $30^5$ is the total number of ways, so the probability is:

$$\frac{30 \cdot 29 \cdot 28 \cdot 27 \cdot 26}{30^5} \approx 0.704.$$

Title: If you randomly call $5$ students and each call is independent, what is the probability that no student is called twice?

Body:
I am thinking about this question and I don't really know how to tackle it.

The problem is:

The class has $30$ students. If you want to call 5 students, each one is independent of the others and randomly with equal probability, what is the probability that none are repeated?

I am thinking as $$\frac{30\cdot 29\cdot 28 \cdot  27\cdot 26}{30^5},$$ but I am not sure it is correct. Thank you so much!


问:
How is the cross product an orientation dependend pseudo-vector, since the right hand rule uniquely determines its direction?
The geometric definition of the cross product $a\times b$ of two vectors (arrows) $a$ and $b$ seems unambiguous, in particular its direction is uniquely determined by the right hand rule.

So why is the cross product called a pseudo vector (or pseudo form) whose direction depends on the orientation of the vector space? There must be a connection between the geometric definition and the formula $a\times b=*(a\wedge b)$ I found, where the dependency on the orientation is explicit, musn't it?

答:
In $\mathbb{R}^3$ with the canonical basis $B=(e_1,e_2,e_3)$, there exist two different orientations. The classical one is saying $B$ is positively oriented, the other is saying $B$ is negatively oriented.

The cross product of two linearly independant vectors $u$ and $v$ is defined to be the unique vector $w$, orthogonal to $u$ and $v$, such that $\|w\| = \|u\|\|v\||\sin(u,v)|$ and $(u,v,w)$ is a positively oriented basis. Positively oriented means that $\det_B(u,v,w) >0$ if $B$ is positively oriented, $<0$ otherwise.

The second point is where the orientation of $\mathbb{R}^3$ matters. If you chose $B$ to be positively oriented, then $e_1 \times e_2 = e_3$ (right hand rule).
If you say that $B$ is negatively oriented, then $e_1\times e_2 = -e_3$.
In general, the change of orientation induces a multiplication by $(-1)$ on the cross product.

In a general orientable riemannian manifold, a metric $g$ induces the Hodge operator $\star$, which can define similar things. It closely depend on the volume form induced by the metric $g$ and thus, on the orientation of the manifold.


问:
Non-linear recurrence relation $\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{1}{2n^2}$
Problem:
I need to find all sequences $\{a_n\}_{n=1}^\infty$ that satisfy the following non-linear recurrence relation:
$$\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{1}{2n^2}$$

What would the general approach be, if there is any? I couldn't find any certain literature on that topic, so pointing me in some direction would be really helpful!

Motivation:
As you probably recognize, this is namely the difference of arctangents formula:
$$arctan(a) - arctan(b) = arctan\left(\frac{a - b}{1 + ab}\right)$$

I was initially confronted with the following series:
$$\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right)$$
and my initial intuition was that it's a telescopic series. So the approach was namely the named above - to find a sequence $a_n$ such that I can present $arctan\left(\frac{1}{2n^2}\right) = arctan(a_n) - arctan(a_{n+1})$ and thus the general solution for the value of the series would be:
$$\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right) = arctan(a_1) - \lim_{N\to\infty}arctan(a_{N+1})$$
Of course, I searched the web for this certain problem and people were suggesting various options for $a_n$, the most common of which were:

$a_n = \frac{1}{2n-1}$

$a_n = 1 - 2n$

Both satisfy the mentioned recurrence relation and lead to a satisfying result for the series, namely $\sum_{n=1}^\infty arctan\left(\frac{1}{2n^2}\right) = \frac{\pi}{4}$

Anyways, I was simply not satisfied with the explanation that this series requires some "clever rewriting" as a telescopic one. I wanted to build some stronger intuition as to how to find such rewritings, when I have the idea that a series might be telescopic. I just couldn't resist wondering whether these are for some reason the only two possible sequences, whether there are more possible ones or they are maybe infinitely many? Obviously, all boils down to generally solving the non-linear recurrence relation.

My progress:

I would like also share my approach to this problem, which only leads to the solution $a_n = 1 - 2n$ since I'm making a restriction at the beginning in order to make any reasonable progress. What I'm basically doing is expressing:
$$\frac{a_n - a_{n+1}}{1 + a_na_{n+1}} = \frac{d}{2dn^2}$$

And artificially building the following system (which, naturally, loses solutions, so I'm not happy...):
$$\begin{array}{|l}a_n - a_{n+1} = d \\ 1 + a_na_{n+1} = 2dn^2\end{array}$$

After some transformations and case distinction, one is able to obtain $a_n = 1 - 2n$. But the first solution I mentioned simply doesn't fit this form, so I need something more general as an approach to the recurrence relation.

Many thanks in advance!

答:
This kind of recurrence relations are quite difficult to solve. If my memory is good (not too sure at my age !), I think I saw them solve using complexes.

Anyway, taking into account what you had as solution, let us try for
$$\frac{A_n - A_{n+1}}{1 + A_n\,A_{n+1}} -\frac{1}{2n^2}=0$$ Reduce to common denominator

$$2 n^2 A_n-2 n^2 A_{n+1}-A_n A_{n+1}-1=0$$

Now, assume that $A_n=\frac{a+bn}{1+cn}$; replace and reduce to same denomiantors everywhere to arrive at
$$-\left(a^2+a b+c+1\right)-n \left(2 a b+b^2+c^2+2 c\right)+n^2 \left(2 a   c-b^2-2 b-c^2\right)=0$$ Now, we need to cancel all these coefficients.

The first one gives $c=-a^2-ab-1$. Replace to obtain
$$\left(a^2+1\right)(a+b+1) \big[ (a+b-1) n+
   (a+b+1) n^2\big]=0$$ So $b=-a-1$ and $c=a-1$

So
$$A_n=\frac{a-(a+1) n}{(a-1) n+1}$$


答:
Draw $\triangle DEG \cong \triangle DEB$ ($DG = DB, EG = BE$)

Then $\angle GDF = 90^0 - \angle EDG = 90^0 - \angle BDE = \angle FDC$

Also, $ DG = DB = DC$.

So $\triangle DFG \cong \triangle DFC$ (by S-A-S, $DF$ is common).

So $GE + GF \gt EF \implies BE + CF \gt EF$.

Title: Proving a geometric inequality $BE+CF>EF$

Body:
In the figure $D$ is the midpoint of $BC$ and $\Delta EDF$ is right triangle.Prove that $BE+CF>EF$

I tried triangle inequality $$BE+ED>BD$$
$$CD+FC>DF$$
Add:$$BE+FC>DF-ED$$

I am stuck now!


问:
How to solve this trigonometric equation further?
Find the sum of the roots of the equation
$$
\sin (2\cos x - \sqrt{3}) = 0
$$

belonging to the segment

$$
A_m = [2m\pi; ~ \frac{\pi}{2} + (2m+2)\pi]
$$

where $m = -4$. Round the number to two decimal places if necessary.

my incomplete solution:

$$2\cos x - \sqrt 3 = k\pi$$

$$k = 0 \text{ fits}$$
$$k = 1$$

$$ 2\cos x = pi + \sqrt3 = 3.14 + 1.73> 2$$ - not suitable. For
larger $k$, it is even worse ...

$$k = -1$$
$$2\cos x = -3.14 + 1.73 = -1.41$$ Hmm Suitable unfortunately.
Smaller ones won't fit

This means that there are 2 options

$$cos x = 0$$ $$x = \pi / 2 + 2n\pi$$ The segment $$[-8\pi, \pi / 2 - 6\pi]$$ 2
roots hit $$-8\pi + \pi / 2$$ and $$-6\pi + \pi / 2$$

$$\cos x = -0.705$$ (coincidence with the root of 2 is random here) But
approximately you can say $$x = + -2\pi / 3 + 2n\pi$$

I have a problem with subsequent calculations, it lies in the fact that I do not understand how and what to calculate

I would be grateful if you could solve this problem to the end with an explanation

答:
based on the data given by the user Yves Daoust, I solved this problem

5 roots fall into the specified interval from the graph.

the interval [-8pi, pi / 2 - 6pi] is divided into two: [-8pi, - 6pi]
and [-6pi, - 6pi + pi / 2].

The first interval will have 4 roots:

$$x = pi / 6 - 8pi$$

$$x = -pi / 6 - 6pi$$

$$x = -7pi + arccos ((pi-sqrt3) / 2)$$

$$x = -7pi - arccos ((pi-sqrt3) / 2)$$

The second interval [-6pi, - 6pi + pi / 2] corresponds on the unit
circle to the first coordinate quarter, so $$x = pi / 6 - 6pi$$ will get
here.

The sum of these five roots is $$pi (-8-6-7-7-6 + 1/6) = - pi * 203/6 =
> -106, 29.$$

to check the result


问:
Proof of 'stronger' Hensel's Lemma
I'm trying to understand the proof of the following statement:

Let $f\in\mathbb{Z}_p[X]$ be a polynomial with coefficients in $\mathbb{Z}_p$. Suppose there is a $p$-adic integer $\alpha_1$ such that $$|f(\alpha_1)|_p<|f'(\alpha_1)|_p^2.$$ Then there is a unique $p$-adic integer $\alpha$ such that $f(\alpha)=0$ and $\alpha\equiv\alpha_1\;\text{mod }p^{\nu_p(f(\alpha_1))-\nu_p(f'(\alpha_1))}.$

To prove it in the case $p\neq 2$, we write $f(\alpha_1)=p^j\xi$ and $f(\alpha_1)=p^k\gamma$, with $0\leq 2k< j$ and $\nu_p(\xi)=\nu_p(\gamma)=0$, and we look for $\alpha_2=\alpha_1+b_2 p^{j-k}$  such that $f(\alpha_2)\equiv 0\;\text{mod }p^{j+1}$.

Using the Taylor expansion we have $$f(\alpha_2)\equiv f(\alpha_1)+f'(\alpha_1)b_2p^{j-k}\text{ mod }p^{j+1}=p^j\xi+b_2\gamma p^j\text{ mod }p^{j+1}.$$ Imposing $f(\alpha_2)\equiv 0\text{ mod }p^{j+1}$, we get $$0\equiv \xi+b_2\gamma \text{ mod }p,$$ which can be solved in $b_2$. Then we construct the following sequence ($\alpha_n$) by induction.

My question is about the claim. To apply induction we need that $$|f(\alpha_2)|_p<|f'(\alpha_2)|_p^2.$$ How can this condition be checked?

Thanks in advance.

答:
Write
$$f(\alpha_n+x)=a_0+a_1x+a_2x^2+\cdots $$
where $a_0=f(\alpha_n)$, $a_1=f'(\alpha_n)$.
If $\alpha_{n+1}=\alpha_m+h$, then
$$ f(\alpha_{n+1}+x)=b_0+b_1x+\cdots$$
where
$$b_0=a_0+a_1h+a_2h^2+\cdots$$
and
$$b_1=a_1+2a_2h+\cdots$$

We choose our $h$ such that $a_0+a_1h$ becomes small, namely as $0<|a_0|_p<|a_1^2|_p\le 1$, our pick $h\approx -\frac{a_0}{a_1}$ will make $|h|_p=\frac{|a_0|_p}{|a_1|_p}<|a_1|_p$, hence for all $k\ge2$, we have $|a_kh^k|_p\le |h^2|_p<|a_1h|_p$ so that $$|b_0|_p=|a_0+a_1h|_p<|a_0|_p.$$
On the other hand, at least for $2\le k<p$, we have $|ka_kh^{k-1}|_p\le |h|_p^{k-1}\le |h|_p<|a_1|_p$; and for $k\ge p>2$, we also get $|ka_kh^{k-1}|_p\le |h|_p<|a_1|_p$ because certainly $h^{k-1}$ is more divisible by $p$ than $k$. Thus
$$|b_1|_p=|a_1|_p.$$

In other words, when using Newton's method in $p$-adics ($p>2$), surprisingly only the function value gets smaller while the derivative keeps its size.

Title: Proof of 'stronger' Hensel's Lemma


问:
A fair six sided dice is rolled 4 times
I have this problem I am working on about permutations. I just want to know if what I am doing is correct.

A fair six-sided die is tossed four times and the numbers shown are recorded in a sequence. How many different sequences are there?

So by the rules of permutations, I used the formula:

$$P(n,r) = \frac{n!}{(n-r)!}$$
$$P(6,4) = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1}$$

which gave me the answer $360$. Is this correct? Another thought is saying since a dice has six sides, it only has six permutations.

答:
No, this is not correct.

You are assuming that the digits cannot repeat, when in fact they can. There are $6$ possibilities for the first digit of the sequence, $6$ possibilities for the second, and $6$ possibilities for the third and fourth. The order of the digits matters.

Hence there are $6^4 = 1296$ different sequences.

Title: A fair six sided dice is rolled 4 times


问:
How to prove that if an entire function has uncountably many zeroes then it must be constant
I am solving assignment questions of complex analysis which might not be discussed due to pendamic and I was unable to solve this particular question.

Let f be an entire function that has uncountably many zeroes .  Prove that it is constant.

I was unable to get an idea on which result to use although I am good in complex analysis. So, It's my humble request to you tp kindly guide me.

I thought  of $\prod_{r_i  varies in\in \mathbb{R}} (z-r_i) $ , and I don't know why it must be constant.

答:
Analytic functions which are non-constant are such that their zeroes are isolated, that is, for every zero $z_0$ of $f$ there is a neighborhood $U$ about $z_0$ where $f(z)\not=0$ except if $z=z_0$. We say that the set of zeroes is totally disconnected.

It remains to show that an uncountable set cannot be totally disconnected.


问:
How can this sum be rewritten like this?
How can this sum be rewritten like this

$1 + 1 + 2 + 2 + · · · + l + l + (l + 1) + (l + 1) + l + ε$

$= 2\sum\limits_{i=1}^{l+1}i+l+ε$

$= (l+1)(l+2)+l+ε $

First of all I dont even understand how we can go from the first step to the second. To make this a bit more clear for myself I tried plugging in numbers into the second step like this.
But I'm pretty its wrong-

$2\sum\limits_{i=1}^{l+1}i+l+ε$

$=2\sum\limits_{i=1}^{l+1}1+l+ε  + 2 + l + ε $

...

答:
You seem to be including the $l$ and $\varepsilon$ inside the sum as if it were $$\sum \left( i + l + \varepsilon \right)$$

But this is not the situation.

The first line has two copies of every number from $1$ to $l+1$, and a lone copy of $l + \varepsilon$. To put this into a summation form, we can write $$\left(\sum_{i=1}^{l+1} 2i \right) + l+ \epsilon$$

Using summation properties, we can factor out the $2$ from the sum:
$$2 \left( \sum_{i=1}^{l+1} i \right) + l+ \epsilon$$

This is how the second line should read.

Can you get to the third line by evaluating the sum?


答: Hint: If for each $z\in S\subset\Bbb C$ we find $r_z>0$ such that the open ball $B_{r_z}(z)$contains no point of $S$ other than $z$, then the balls $B_{r_z/2}(z)$ are pairwise disjoint and each contains a point with rational coordinates.

Title: How to prove that if an entire function has uncountably many zeroes then it must be constant

Body:
I am solving assignment questions of complex analysis which might not be discussed due to pendamic and I was unable to solve this particular question.

Let f be an entire function that has uncountably many zeroes .  Prove that it is constant.

I was unable to get an idea on which result to use although I am good in complex analysis. So, It's my humble request to you tp kindly guide me.

I thought  of $\prod_{r_i  varies in\in \mathbb{R}} (z-r_i) $ , and I don't know why it must be constant.


Title: How can this sum be rewritten like this?

Body:
How can this sum be rewritten like this

$1 + 1 + 2 + 2 + · · · + l + l + (l + 1) + (l + 1) + l + ε$

$= 2\sum\limits_{i=1}^{l+1}i+l+ε$

$= (l+1)(l+2)+l+ε $

First of all I dont even understand how we can go from the first step to the second. To make this a bit more clear for myself I tried plugging in numbers into the second step like this.
But I'm pretty its wrong-

$2\sum\limits_{i=1}^{l+1}i+l+ε$

$=2\sum\limits_{i=1}^{l+1}1+l+ε  + 2 + l + ε $

...

Body:
Using the sum of first $n$ numbers formula $(1+2+...+ n = \frac {n(n+1)}{2})$,

$1+2+...+(l+1)= \sum\limits_{i=1}^{l+1}i = $
${\frac {(l+1)(l+2)}{2}} \implies 2\sum\limits_{i=1}^{l+1}i=(l+1)(l+2)$ so that

$2(1+2+...+(l+1))=1+1+2+2+...+(l+1)+(l+1)=(l+1)(l+2)$ so that

$1+1+2+2+...+(l+1)+(l+1)+l+\epsilon=(l+1)(l+2)+l+\epsilon$


问:
Find the sum of he series to infinity : $\tan^{-1} (\frac 12) + \tan^{-1} (\frac 18) + \tan^{-1} (\frac{1}{18})...$
The series is

$$\tan^{-1} (\frac{1}{(2)(1^2)}) + \tan^{-1} (\frac{1}{(2)(2^2)})+ \tan^{-1} (\frac{1}{(2)(3^2)}).....$$

I know the series will telescope, but I don’t know how it would work. These problems usually reduce to $\tan^{-1} a- \tan^{-1} b$, but that’s not happening here

答: You can prove by induction the $n$th partial sum is $\arctan\frac{n-1}{n}$, so the series has limit $\arctan1=\frac{\pi}{4}$. So your telescoping strategy works viz. $a=\frac{k}{k+1},\,b=\frac{k-1}{k}$.


问:
When does $d$ mean a differential and when does it mean a derivative?
In a lot of books that I've read about calculus the authors will sometimes use the operator $d$ to mean a differential (an "infinitesimal" change), and sometimes they'll use it to mean an actual derivative. For example they'll write: $d(x_1x_2)=dx_1x_2+ x_1dx_2$ or they'll write $d x^2 = 2x$ when if they at least write it as $d x^2 = 2xdx$ it would make more sense as a derivative notation.

When should one interpret this $d$ operator as a "small change" or conversely as a derivative? This mix-up gets even more confusing when manipulating differential equations.

Thanks.

答:
The notation $dx^2=2x$ is wrong and might confuses you.

The abstract derivative as a limit of differential quotient comes first. The following notation is all the same: $$
\lim \limits_{x\to a} \frac{f(x)-f(a)}{x-a}\equiv f'(a)\equiv \left.\frac{dy}{dx}\right\vert _{a}.$$ However, since a derivative is computed by the quotient, we might design a new notation of differential that looks like a fraction but the action when the numerator and the denominator come together is differentiation.

The new notation $dy$ is defined to be a function of the point $a$ and a small increment $\Delta x$, $$dy:=f'(a)\cdot \Delta x.$$ You can use $dx$ instead of $\Delta x$ if you wish, though. From this definition and linear approximation $f(a+\Delta x)\approx f(a)+f'(a)\cdot \Delta x$, we have $$dy\approx f(a+\Delta x)- f(a)$$

You see that $dy$ is now viewed as a change in the value of $y$ (also $f$) at the point $a$ after an increment of $\Delta$ as how we designed it should be.

To sum up, the differentials $dy$ or $dx$ is to be viewed as a small change while $\frac{dy}{dx}$ is to be viewed as a derivative.


问:
Determine the nature of this series?
How do I determine if this series converges ? $\overset{\infty}{\sum}\frac{\ln n!}{n^3}$. Cauchy and d'Alembert have proved inconclusive.

Title: Determine the nature of this series?

Body: Hint : $$\ln(n!)=\sum_{k=1}^n \ln(k) \leq n \ln(n)$$


问:
Why do we need 2 equations to solve 2 variables, 3 equations to solve 3 variables, etc...?
The question put another way: why does solving a system of two equations with 2 variables give an exact answer?

I understand that using graphing, I will get either 2 parallel lines, which means 'no solutions' or intersecting lines, which means '1 solution'. But I'm wondering why this happens in terms of algebra, not graphing. I.E. I am asking for a proof/illustration.

答:
A system of $n$ (for simplicity: homogenous) equations in $m$ unknowns can be viewed as a linear map from $m$-dimensional space to $n$-dimensional space, and the solutions are the preimiage of a point under this map.
Now to make a statement about solvability and dimension of solution you need an important result from linear algebra:

If $f\colon V\to W$ is a linear map, then $\dim V=\dim\ker f+\dim \operatorname{im} f$.

To get there, you need of course a bit of linear algebra and the notions of kernel, image, dimension, etc.


问:
Making sense of $x'=Ax\implies x(t)=e^{Jt}x(0)$ when $A$ is nondiagonalizable

How do I make sense of $x'=Ax$ is solved by $x(t)=e^{Jt}x(0)$ when $A$ is nondiagonalizable.

When we have an $n\times n$ system of ODE's, $\dfrac{dx}{dt}=Ax$ and if $A=VD_AV^{-1}$ is diagonalizable.

Substitute $x=Vy$,

$$
\dfrac{dx}{dt}=Ax\implies V\dfrac{dy}{dt}=AVy\implies \dfrac{dy}{dt}=V^{-1}AVy=D_Ay\\
\begin{bmatrix}y_1'\\\vdots\\y_n'\end{bmatrix}=\begin{bmatrix}\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\lambda_n\end{bmatrix}\begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}\implies y_i(t)=c_ie^{\lambda_it}\\
x(t)=Vy(t)=\begin{bmatrix}v_1&\cdots&v_n\end{bmatrix}\begin{bmatrix}c_1e^{\lambda_1t}\\\vdots\\c_ne^{\lambda_nt}\end{bmatrix}\implies x(t)=c_1v_1e^{\lambda_1t}+\cdots+c_nv_ne^{\lambda_nt}\\
x(0)=V\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=Vc\implies c=V^{-1}x(0)
$$
$$
x(t)=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}c=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}V^{-1}x(0)=e^{At}x(0)
$$

where $e^{At}=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}V^{-1}$

I think it makes sense why we define $e^{At}=Ve^{D_At}V^{-1}$ the way it is and how it comes into picture and we can also prove the Taylor series expansion is the same thing.

When $A$ is nondiagonalizable $A$ can be decomposed into $A=XJX^{-1}\implies J=X^{-1}AX$ is called the Jordan canonical form and $X$ contains generalized eigenvectors $x_k$ such that $Ax_k=λ_kx_k$ , or $Ax_k=λ_kx_k+x_{k−1}$ and $J=X^{-1}AX=\begin{bmatrix}J_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&J_s\end{bmatrix}=J_1\oplus\cdots\oplus J_s$ where $J_i=\begin{bmatrix}\lambda_1&1&0&\cdots&0\\0&\lambda_2&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda_i\end{bmatrix}$

Substitute $x=Az$

$$
x'=Ax\implies Bz'=ABz\\
z'=B^{-1}ABz=Jz\\
\begin{bmatrix}z'_1\\\vdots\\z'_n\end{bmatrix}=\begin{bmatrix}J_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&J_s\end{bmatrix}\begin{bmatrix}y_1\\\vdots\\z_n\end{bmatrix}
$$
I don't think it goes anywhere with it !

In this case how do I make sense of the fact that $x(t)=e^{Jt}x(0)$ solves $x'=Ax$ similar to the one we obtained above ?

答:
The solution of the equation

$$
x'(t) = A x(t)
$$

is

$$
x(t) = e^{t A } x(0)
$$
wether or not $A$ is diagonalizable. This can be easily seen differentiating the solution. (Added: I give a proof below). Note that $e^{tA}$ and $A$ commute with each other so there is no problem with that.

The Jordan canonical form is a way to compute the matrix exponential. But that is not needed here.

If there is an invertible matrix $V$ such that

$$
A = V D V^{-1}
$$

then

$$
(t A)^n = t^n V D^n V^{-1}
$$

and so

$$
\exp{(tA) } = V e^{tD} V^{-1}.
$$

Note $D$ can be whatever for this to hold. The point is that when $D$ is in Jordan block form the computation of $e^{tD}$ is greatly simplified.

Added

I will show here that the exponential solves the differential equation. In fact I will show something slightly more general, namely:

Let $B(t)$ be a differentiable matrix valued function. Provided, $B'(t)$ commutes with $B(t)$ (for all $t$ in the needed interval) then we have

$$
\frac{d}{dt} e^{B(t)} = B'(t) e^{B(t)}
$$

Proof. -

\begin{align}
\frac{d}{dt} [B(t)]^n &= B'(t) [B(t)]^{n-1} + B(t) B'(t) [B(t)]^{n-2} +\ldots \\
& + \ldots  [B(t)]^{n-1} B'(t) \\
&= n B'(t) [B(t)]^{n-1}
\end{align}

where the last equality stems from commutativity. Hence

\begin{align}
\frac{d}{dt} e^{B(t)} &=  \frac{d}{dt}\sum_{n=0}^{\infty}  \frac{[B(t)]^n}{n!} \\
&= \sum_{n=0}^{\infty}  \frac{n B'(t) [B(t)]^{n-1}}{n!} \\
&= B'(t) \sum_{n=1}^{\infty}  \frac{[B(t)]^{n-1}}{(n-1)!} \\
&= B'(t) e^{B(t)}.
\end{align}

Now to prove the claim simply consider $B(t) = t A$, $\square.$


问:
Two forms of $\big(A^{-1} + B^{-1}\big)^{-1}$
I am trying to calculate $\big(A^{-1} + B^{-1}\big)^{-1}$ and I find that there are two ways to get the answers:

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(A^{-1}(I + AB^{-1})\big)^{-1}
 = \big(A^{-1}(B+A)B^{-1}\big)^{-1} = B(A+B)^{-1}A$$

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(B^{-1}(BA^{-1} + I)\big)^{-1}
 = \big(B^{-1}(B + A)A^{-1}\big)^{-1}
 = A(A+B)^{-1}B$$
So my question is that: is there a way to prove that
$$B(A+B)^{-1}A = A(A+B)^{-1}B$$
without proving that they are both equal to the inverse of $(A^{-1}+B^{-1})$ ?

Body:
$$\begin{split}
B(A+B)^{-1}A&=B(A+B)^{-1}(A+B-B)\\&= B-B(A+B)^{-1}B\\&=B-(A+B-A)(A+B)^{-1}B
\\&= B - B + A(A+B)^{-1}B\\& = A(A+B)^{-1}B
\end{split}$$


问:
Use laws of set theory to show equality
We have sets $X,Y,Z$. By laws of set theory we are to show that $$(X∪Y∪Z)\triangle(X∩Z∩Z)=(X∪Y∪Z)\backslash(X∩Z∩Z),$$
where $\triangle$ denotes symmetric difference.

I started by using the following identity for sets $A, B$:
$$A\triangle B=(A-B)∪(B-A)$$

By applying this to our case, the left side of the equation yields
$$(X∪Y∪Z)∪(X∩Z∩Z)-(X∪Y∪Z)∩(X∪Y∪Z).$$

By then applying the absorbation law, this can be rewritten as
$$(X∪Y∪Z) - (X∩Z∩Z),$$
given that I haven't made any mistakes. From my understanding this equals $(X∪Y∪Z)\backslash(X∩Z∩Z)$, which we were supposed to show.
The problem is from a discrete mathematics class and since we haven't really covered symmetric difference, I would appreciate some input.

答: Your solution contains a gap in the first step. You need to first show that $A\Delta B = (A\cup B)\backslash (A\cap B)$. Apply this to the LHS and use the absorption law and we are done. Note that the backslash symbol ($\backslash$) stands for the set minus ($-$) that you are familiar with.

Title: Use laws of set theory to show equality


问:
a,b are positive integers. If $21ab^2$ and $15ab$ are perfect squares, the minimum values of a+b is?
I tried by splitting $21ab^2$ and $15ab$ and equating them to distinct integers $m^2$ and $n^2$.

$2100 + 10a + b^2 = m^2$

$1500 + 10a + b = n^2$

Subtracting,

$900 + b^2 - b = (m + n)(m-n)$

After that I am not able to proceed further...

For reference the answer of the question is $56$.

答:
If $21ab^2$ and $15ab$ are perfect squares, then so is their product

$$ 21\cdot 15\cdot a^2b^3 = 3^2\cdot a^2b^2\cdot5\cdot7\cdot b$$

For $5\cdot7\cdot b$ to be perfect square, min $b=5\cdot7=35$.

Also $15ab = 3\cdot5\cdot a\cdot5\cdot7$ is a perfect square $\Rightarrow $ min $a=3\cdot7=21$

Check that for these values $21ab^2 = 21^2b^2$ is indeed a perfect square.


问:
Where and what is the mistake?
A bag contains 2 green balls and 3 red balls. two balls are drawn at random. Find the probability that the balls  are red and green?

Solution- A there are 2 green balls and 3 red balls, two balls taken from this can be

G G

G R

R G

R R

As all the balls are identical

So probability of getting a red and green = 2/4 = 1/2.

I know this is absolutely wrong but someone please explain correctly what is the mistake here. I cant see what is wrong in this.

The correct solution is 2c1 * 3c1 / 5c2 = 3/5

答:
Re your comment:

"but even if there are equal red and green balls it does not come like that. why is it that"

Firstly, you need to be clear whether the balls are drawn with replacement or without replacement. From the context, they are drawn w/o replacement. Taking two balls of each color, let us work out probabilities directly, and understand why this happens.

$P(RR) = \frac 2 4 \frac 1 3 = \frac 1 6,$
$P(BB) = \frac 2 4 \frac 1 3 = \frac 1 6,$
$P(RB) = \frac 2 4 \frac 2 3 = \frac 1 3,$
$P(BR) = \frac 2 4 \frac 2 3 = \frac 1 3,$

So you can clearly see why the probabilities aren't identical.

The shorter way to compute P(one of each color) is $\frac {^2C_1\cdot^2C_1}{ ^4C_2} = \frac2 3$

Title: Where and what is the mistake?


问:
How many values of positive integers 'm' exist such that, $m^4 - 4m^3 + 22m^2 - 36m + 18$ is a perfect square.
I found this question in a math olympiad book. I figured that,

$m^4 - 4m^3 + 22m^2 - 36m + 18$ = $(m-1)^2(m^2 - 2m + 17) + 1$

But I don't know if I'm solving it the right way, and if yes, I don't know how to proceed.

答:
A different take on @dan_fulea's observation:

Let $b=m^2 -2m + 9$. Note that $b>0$.

If $m^4 - 4m^3 + 22m^2 - 36m + 18=a^2$, then $63=b^2-a^2=(b-a)(b+a)$.

The only solutions with $b>0$ and $m \in \mathbb N$ are $(b,m)=(8,1)$ and $(b,m)=(12,3)$.

Bottom line: $m=1$ or $m=3$.


问:
Integration of some floor functions
Can anyone please answer the following questions ?

1) $\int\left \lfloor{x}\right \rfloor dx$

2) $\int$ $ \left \lfloor{\sin(x)}\right \rfloor $ $dx$

3) $\int_0^2$ $\left \lfloor{x^2+x-1}\right \rfloor$ $dx$

4) $\int_o^\pi$ $\left \lfloor{x(1+\sin(\pi x)}\right \rfloor$

Also can anyone please make me understand the way in which to proceed in these types of sums?

$\left \lfloor{x}\right \rfloor$ is the floor function

Thanks

答:
$1)$ The area under the graph of $\lfloor{x}\rfloor$ is $0+1+2 +\cdots+(x-1)$ so $$\int{\lfloor{x}\rfloor} \, dx =\frac{x(x-1)}{2}$$ $2)$ The graph of $\lfloor{\sin(x)}\rfloor$ alternates between $0$ and $-1$ every $\pi$ values, so every $2\pi$ values the integral will $-1$. So $$\int\big\lfloor{\sin(x)}\big\rfloor \, dx = -\lfloor{\frac{x}{2\pi}}\rfloor-\frac{\operatorname{sgn}\big((x\bmod{2\pi}) - \pi\big)+1}{2}$$
$3)$ $$\int_{0}^{2}\Big\lfloor\left(x^{2}+x-1\right)\Big\rfloor \, dx\approx1.726$$
$4)$ $$\int_0^\pi  \big\lfloor x(1+\sin(\pi x)\big\rfloor \, dx\approx4.530$$

Title: Integration of some floor functions


问:
Proving the following sequence is bounded (Proof Verification)
I want to prove that the following sequence is bounded,

$a_{n+1}=\sqrt{2+a_n}$, $a_1=\sqrt{2}.$

We can prove that it is monotonically increasing by induction. (I'll skip this part and treat it as an assumption.)

Assume $\exists m \in \mathbb{N}$ such that $M\leq a_m < M^2-2$ for $M>2$, the second inequality gives $a_{m+1}<M$ (contradiction because the sequence is increasing.) Then assume $\exists m \in \mathbb{N}$ such that $M^2-2\leq a_m < (M^2-2)^2-2.$ This gives $a_{m+1}<M^2-2$ (also a contradiction.) Repeating the procedure over and over we find that there is no $a_m$ starting from $M$ up to infinity.
Therefore the sequence is bounded.

Is the logic sound? Also is there a better way to write the proof?

答:
The hypothesis $$(\exists m\in \Bbb N ) : M\le a_m<M^2-2$$

implies a contradiction $(a_{m+1}<a_m)$, hence, this hypothesis is false.

An other way :
The sequence is increasing and if it converges to $ L $, then
$L=\sqrt{L+2} \; or \; L=2 $ and $$(\forall n\in \Bbb N)\; a_n<2$$

this can be proven easily by induction :
$$a_{n+1}<\sqrt{2+2}=2$$


答:
All of your numbers ($M$, $M^2-2$, $\left(M^2-2\right)^2-2$, …) are greater than $2$. So, you have not considered all possible cases.

Note that, of $f(x)=\sqrt{2+x}$ and if $0<x<2$, then$$f(x)-x=\sqrt{2+x}-x=\frac{2+x-x^2}{\sqrt{2+x}+x}=\frac{(x+1)(2-x)}{\sqrt{2+x}+x}>0$$and that therefore $f(x)>x$. Also, $f(x)<2$, since$$2-f(x)=2-\sqrt{2+x}=\frac{2-x}{2+\sqrt{2+x}}$$and therefore $2>f(x)$. So, since $a_1=\sqrt2$, $a_1<a_2<2$. Since, $a_2<2$, $a_2<a_3<2$ and so on…

Title: Proving the following sequence is bounded (Proof Verification)

Body:
I want to prove that the following sequence is bounded,

$a_{n+1}=\sqrt{2+a_n}$, $a_1=\sqrt{2}.$

We can prove that it is monotonically increasing by induction. (I'll skip this part and treat it as an assumption.)

Assume $\exists m \in \mathbb{N}$ such that $M\leq a_m < M^2-2$ for $M>2$, the second inequality gives $a_{m+1}<M$ (contradiction because the sequence is increasing.) Then assume $\exists m \in \mathbb{N}$ such that $M^2-2\leq a_m < (M^2-2)^2-2.$ This gives $a_{m+1}<M^2-2$ (also a contradiction.) Repeating the procedure over and over we find that there is no $a_m$ starting from $M$ up to infinity.
Therefore the sequence is bounded.

Is the logic sound? Also is there a better way to write the proof?


答:
Taking nested roots of positive numbers leads one to think that $0$ is a lower bound for $(a_{n})$ so we do the following:

Since $a_{1} > 0$, suppose $a_{n} > 0$ for some $n$.

Then $a_{n+1} = \sqrt{(2+a_{n})} > 0$. By induction, $a_{n} > 0$ for all $n$. I assumed a limit existed and saw that it would be $2$ if it did. This made me consider $2$ being an upper bound.

Since $a_{1} < 2$, suppose $a_{n} < 2$ for some $n$.

Then $a_{n+1} = \sqrt{(2+a_{n}}) < \sqrt{(2+2})=\sqrt 4 = 2$. By induction, $a_{n} < 2$ for all $n$.

Hence $0<a_{n}<2$ for all $n$ and $(a_{n})$ is bounded.

To prove monotonicity, check $a_{1}$ and $a_{2}$, and we see it is the case that $a_{1} \leq a_{2}$.

Suppose for some $n$ that $a_{n} \leq a_{n+1}$.

Then we must show that $a_{n+1} \leq a_{n+2}$.

That is, $a_{n+1} \leq a_{n+2} \Longleftrightarrow\sqrt{2+a_{n}} \leq \sqrt{2+a_{n+1}} \Longleftrightarrow 2+a_{n} \leq 2+ a_{n+1} \Longleftrightarrow
a_{n} \leq a_{n+1}$

Since $(a_{n})$ is bounded and monotonic, the sequence converges.


问:
Power sets of sets with nonempty intersection (Set Theory)
I started learning about Set Theory and I ran into the following unproven lemma:
There exist sets $A$ and $B$ with $A\cap B=\emptyset$ such that the power set of $A$ is equal to the power set of $A\setminus B$.

The statement is: P(A) = P(A/B) $\Leftrightarrow$  $A$ and $B$ are disjoint.

I tried every possible pair of sets I could come-up with, yet I cannot find a concrete example.
I would really appreciate it if someone could point me in the right direction.
Thanks

答:
Are you sure that this is supposed to be correct.

When $A$ and $B$ are sets with $A\cap B\neq\emptyset$, then there is $x\in A$ with $x\in B$.

Then $\{x\}\in\mathcal{P}(A)$. But $\{x\}\notin\mathcal{P}(A\setminus B)$. Since $\{x\}$ can not be a subset of $A\setminus B$.

Suppose otherwise, and $\{x\}\subseteq A\setminus B$. Then $x\in A\setminus B$, but $x\in A$ and $x\in B$. So $x\notin A\setminus B$. Which is a contradiction.

Edit: On your update.

When the statement is to show that $\mathcal{P}(A)=\mathcal{P}(A\setminus B)\Leftrightarrow~ A\cap B=\emptyset$, then we can prove this as follows:

$\Rightarrow$.

Let $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.

Suppose $A\cap B\neq\emptyset$. Then the proof given above can be applied here, to yield a contradiction.
You might want to work out the differences, or details yourself.

$\Leftarrow$:

Let $A\cap B=\emptyset$. Show $\mathcal{P}(A)=\mathcal{P}(A\setminus B)$.

This is trivial, as $A\cap B=\emptyset$ implies that $A\setminus B=A$.

Indeed:

$A\setminus B\subseteq A$ holds always.

For: $A\subseteq A\setminus B$ let $x\in A$. Since $A$ and $B$ are disjoint, we have that $x\notin B$. So $x\in A\setminus B$.

So we have shown that $A=A\setminus B$ (under the assumption that $A\cap B=\emptyset$.)


答:
You only need that $A+B$ is invertible.

If $A$ and $B$ are both invertible, then as you demonstrated, it follows that $A^{-1}+B^{-1}$ is also invertible and $$B(A+B)^{-1}A = (A^{-1}+B^{-1})^{-1} =A(A+B)^{-1}B$$

If $A$ or $B$ is not invertible, then since $\sigma(A)$ and $\sigma(B)$ are finite, there exists $r> 0$ such that for all $\varepsilon \in \langle0,r\rangle$ we have that $A_\varepsilon := A-\varepsilon I$ and $B_\varepsilon := B+\varepsilon I$ are both invertible. Now from the first case it follows that
$$B_\varepsilon(A+B)^{-1}A_\varepsilon = B_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}A_\varepsilon = A_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}B_\varepsilon = A_\varepsilon(A+B)^{-1}B_\varepsilon.$$
Letting $\varepsilon \to 0$ from continuity of matrix operations we get $B(A+B)^{-1}A =A(A+B)^{-1}B$.

Title: Two forms of $\big(A^{-1} + B^{-1}\big)^{-1}$

Body:
I am trying to calculate $\big(A^{-1} + B^{-1}\big)^{-1}$ and I find that there are two ways to get the answers:

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(A^{-1}(I + AB^{-1})\big)^{-1}
 = \big(A^{-1}(B+A)B^{-1}\big)^{-1} = B(A+B)^{-1}A$$

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(B^{-1}(BA^{-1} + I)\big)^{-1}
 = \big(B^{-1}(B + A)A^{-1}\big)^{-1}
 = A(A+B)^{-1}B$$
So my question is that: is there a way to prove that
$$B(A+B)^{-1}A = A(A+B)^{-1}B$$
without proving that they are both equal to the inverse of $(A^{-1}+B^{-1})$ ?


答:
Your proof uses the phrase "Repeating the procedure over and over". This is not a finite proof, thus not valid. Also you seem to use proof by contradiction where I don't see necessary.

Here's my proof which uses induction:

We will show that the sequence $(a_n)$ is bounded by $2$. It is easy to see that $(a_n)$ is a sequence of positive reals, so we instead check, by induction, if  $a_n<2$ for all $n\in\mathbb{N}$. The basis step is easy, since $a_1=\sqrt{2}<2$. Now let $n\in\mathbb{N}$ and assume $a_n<2$. Consider $$a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=\sqrt{4}=2.$$ Thus, the inductive step is proved. By Principle of Mathematical Induction, we have that $a_n<2$ for all $n\in\mathbb{N}$. Thus, the sequence is bounded.


问:
Combining differential equations
Can anyone see how to combine the following 3 equations

$$\dot r^2-\dot\theta^2=-\theta^2$$
$$\theta\ddot \theta-2\dot \theta^2=2(\dot r^2-\dot \theta^2)$$
$$\dot r=a \theta^2$$
to get $$\theta^2=r^2+cr+d$$
where $c,d$ are constants?

I've tried various ways of substituting one into another but I just end up with very high orders of derivatives, solving of which I don't think should be necessary.

What I did was (2)-(1) then substitute in (3), giving
$$(a-1)\theta^2=\theta\ddot\theta-\dot\theta^2$$
So I need the RHS to be $$(a-1)r^2+Ar+B$$ for some constants $A,B$? But how??

or perhaps start differently?

答:
I think I have found the solution. You only need the first equation and the third one.

First, note that

$$\frac{d\theta}{dr}=\frac{\dot\theta}{\dot r} \; .$$

Then, divide the first equation by ${\dot r}^2$ to get

$$1-\frac{\dot\theta^2}{\dot r^2}=-\frac{\theta^2}{\dot r^2}$$

Using the third equation and rearranging, you can transform this to

$$1+\frac{1}{a^2\theta^2}=\left(\frac{d\theta}{dr}\right)^2$$

Take the square root and you can easily integrate by separation of variables to give the sought result.

Title: Combining differential equations


答:
Here is an intuition in a $2\times2$ setting:

Say your system is $\dot{y} = Ay$, with $A$ a $2\times 2$ matrix which cannot be diagonalized. $A$ can be put in Jordan normal form, so there is a change of basis matrix $P$ such that $A = PJP^{-1}$. Let's say $J = \begin{aligned} \begin{bmatrix}2 & 1\\\ 0 & 2\end{bmatrix} \end{aligned}$ for the sake of the example.

Then, $\dot{y} = Ay \implies $ if we set $x = P^{-1}y$, then $\dot{x} = P^{-1}\dot{y} = P^{-1}A(Px) = Jx$. So you have the reformulated system:

$\dot{x} = \displaystyle\binom{\dot{x_1}}{\dot{x_2}} = \begin{aligned} \begin{bmatrix}2 & 1\\\ 0 & 2\end{bmatrix} \end{aligned}\displaystyle\binom{x_1}{x_2}$.

Now, if you solve this system by hand, starting with the second row you get:

$x_2(t) = x_2(0) e^{2t}$.

So $\dot{x_1} = 2x_1 + x_2(0) e^{2t} \implies x_1(t) = x_1(0) e^{2t} + x_2(0) te^{2t}$.

So... what you end up getting is: $\displaystyle\binom{x_1}{x_2} = \begin{aligned} \begin{bmatrix}e^{2t} & te^{2t} \\ 0 & e^{2t}\end{bmatrix} \end{aligned}\displaystyle\binom{x_1(0)}{x_2(0)}$, where the matrix in the middle is exactly $e^{Jt}$.

Basically, the jordan normal form is simply another way to encode your system. Which you can always solve by starting backwards and working your way up the diagonal.

Now, when you switch back to your original basis, everything works out fine because, if $A = PJP^{-1}$, then $e^{At} = Pe^{Jt}P^{-1}$ (you can verify this by simply writing out the expression for $e^{At}$ and then factoring out $P$ and $P^{-1}$ on both sides).

Title: Making sense of $x'=Ax\implies x(t)=e^{Jt}x(0)$ when $A$ is nondiagonalizable

Body:
How do I make sense of $x'=Ax$ is solved by $x(t)=e^{Jt}x(0)$ when $A$ is nondiagonalizable.

When we have an $n\times n$ system of ODE's, $\dfrac{dx}{dt}=Ax$ and if $A=VD_AV^{-1}$ is diagonalizable.

Substitute $x=Vy$,

$$
\dfrac{dx}{dt}=Ax\implies V\dfrac{dy}{dt}=AVy\implies \dfrac{dy}{dt}=V^{-1}AVy=D_Ay\\
\begin{bmatrix}y_1'\\\vdots\\y_n'\end{bmatrix}=\begin{bmatrix}\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\lambda_n\end{bmatrix}\begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}\implies y_i(t)=c_ie^{\lambda_it}\\
x(t)=Vy(t)=\begin{bmatrix}v_1&\cdots&v_n\end{bmatrix}\begin{bmatrix}c_1e^{\lambda_1t}\\\vdots\\c_ne^{\lambda_nt}\end{bmatrix}\implies x(t)=c_1v_1e^{\lambda_1t}+\cdots+c_nv_ne^{\lambda_nt}\\
x(0)=V\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=Vc\implies c=V^{-1}x(0)
$$
$$
x(t)=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}c=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}V^{-1}x(0)=e^{At}x(0)
$$

where $e^{At}=V\begin{bmatrix}e^{\lambda_1t}&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&e^{\lambda_nt}\end{bmatrix}V^{-1}$

I think it makes sense why we define $e^{At}=Ve^{D_At}V^{-1}$ the way it is and how it comes into picture and we can also prove the Taylor series expansion is the same thing.

When $A$ is nondiagonalizable $A$ can be decomposed into $A=XJX^{-1}\implies J=X^{-1}AX$ is called the Jordan canonical form and $X$ contains generalized eigenvectors $x_k$ such that $Ax_k=λ_kx_k$ , or $Ax_k=λ_kx_k+x_{k−1}$ and $J=X^{-1}AX=\begin{bmatrix}J_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&J_s\end{bmatrix}=J_1\oplus\cdots\oplus J_s$ where $J_i=\begin{bmatrix}\lambda_1&1&0&\cdots&0\\0&\lambda_2&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda_i\end{bmatrix}$

Substitute $x=Az$

$$
x'=Ax\implies Bz'=ABz\\
z'=B^{-1}ABz=Jz\\
\begin{bmatrix}z'_1\\\vdots\\z'_n\end{bmatrix}=\begin{bmatrix}J_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&J_s\end{bmatrix}\begin{bmatrix}y_1\\\vdots\\z_n\end{bmatrix}
$$
I don't think it goes anywhere with it !

In this case how do I make sense of the fact that $x(t)=e^{Jt}x(0)$ solves $x'=Ax$ similar to the one we obtained above ?


答:
Nope. The logic is flawed, because you start from an unjustified assumption that your sequence is increasing.

It actually is increasing, but in some interval only, and you didn't show what the interval is nor the starting point belongs to the interval.

The interval is $(-2,2)$ and the starting point does belong to it, hence the sequence is increasing and, as a result, it is bounded – but your reasoning does not prove it.


问:
help with Harmonic Sequence inequality proof by using induction of H2^n <= n+1
I tried to prove Harmonic sequence inequality by using induction. I couldn't solve it.I'll glad for some enlightenments or any help.
Let:
$$\sum_{n=1}^n \frac{1}{n} = H_n$$
Prove:
$$H_{2^n} <= 1+n$$

Body:
$$H_{2^n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\ldots +\frac{1}{2^n-1}+\frac{1}{2^n}$$
$$H_{2^n}\leq 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\ldots+\frac{1}{2^n}$$

that is for each $2^{j}\leq k<2^{j-1}$ bound $\frac{1}{k}\leq\frac{1}{2^j}$
$$H_{2^n}\leq 1+2\cdot\frac{1}{2}+4\cdot\frac{1}{4}+8\cdot\frac{1}{8}+\ldots+2^{n-1}\cdot\frac{1}{2^{n-1}}+\frac{1}{2^n}=n+\frac{1}{2^n}$$


答:
Let $x_n$ be the successive values of $x$ where $f(x)$ takes an integer value (they can be obtained by solving $f(x)\in\mathbb Z$ and sorting). As $\left\lfloor f(x)\right\rfloor$ is piecewise constant, the antiderivative is piecewise linear, continuous, with a slope equal to $f(x_k)$.

If we start integrating at $x_0$, up to some $x_n$, the integral is the sum

$$\int_{x_0}^{x_n}\left\lfloor f(x)\right\rfloor dx=\sum_{k=0}^{n-1}f(x_k)(x_{k+1}-x_k).$$

And if we continue integrating to some $x$ that does not exceed $x_{n+1}$,

$$\int_{x_0}^x\left\lfloor f(x)\right\rfloor dx=\sum_{k=0}^{n-1}f(x_k)(x_{k+1}-x_k)+f(x_n)(x-x_n).$$

Title: Integration of some floor functions

Body:
Can anyone please answer the following questions ?

1) $\int\left \lfloor{x}\right \rfloor dx$

2) $\int$ $ \left \lfloor{\sin(x)}\right \rfloor $ $dx$

3) $\int_0^2$ $\left \lfloor{x^2+x-1}\right \rfloor$ $dx$

4) $\int_o^\pi$ $\left \lfloor{x(1+\sin(\pi x)}\right \rfloor$

Also can anyone please make me understand the way in which to proceed in these types of sums?

$\left \lfloor{x}\right \rfloor$ is the floor function

Thanks


问:
What if the heat equation was of order n instead of order 2?
I am trying to understand the heat equation in one dimension, $$\partial_t u +\partial_x^2 u = 0$$
I've been introduce to the idea that as time goes to infinity that the function $u$ moves towards the average, and eventually flattens out to a horizontal line. I am wondering if $$\partial_t u +\partial_x^n u = 0$$

would have similar effects or descriptions. In the case $n=1$ it seems to me like, max's and min's which have 0 derivative with respect to $x$ would also have 0 derivative with respect to $t$, points at which the derivative is positive such as before a max,  would decrease to the previous min. Similarily points which are negative such as after a max, would increase to the previous max. So the end result would be like a Heaviside function, for initial conditions like $u(x,t)=sin(x)$.

What about $n=3,4$ for the initial condition case of $u(x,t)=sin(x)$ then things are similar but opposite, in the $n=4$ case the points which are not fixed with time, tend towards infinity as time increased. in the $n=3$ case the points which are moving towards the previous max, will move towards the next min, and similar for other cases.

My question is first, does this make sense, or where am I going wrong? Second is there anywhere I can go to find out more about the equation $\partial_t u +\partial_x^n u = 0$? Lastly, is there an analogue to "moving towards the average" for the $n$th order heat equation, like "moving towards the average of the average" for the 3rd order heat equation?

P.S. is the Laplacian of the Laplacian the $n=4$ scenario?

答:
Where you say "would decrease to the previous min" it is important to remember that a derivative is a local rate of change, so that a decrease right now need not continue. For your $u_t+u_x = 0$ case with $u(x,0) = \sin(x)$, you don't get Heaviside, in fact a solution is $\sin(x-t)$.

A second point involves boundary conditions. PDE's usually have many solutions that have different behavior depending on the boundaries. For $u_t-u_{xx} = 0$ with $u(x,0) = \sin(x)$, one solution is $e^{-t}\sin(x)$, which satisfies the boundary conditions $u(0,t) = u(\pi,t) = 0$. But there are many other solutions that you can find in many books on differential equations, one solution for each case $u(0,t) = u(b,t) = 0$,
one for each choice of $b>0$ all having the same initial value $u(x,0) = \sin(x)$ for $0 <x < b$. Look in the section on Fourier series.


问:
Problems solveing an ODE
I'm having problems solving the following nonlinear second-order differential equation:
$$y''=Ay\left(1+y'^2\right)^{3/2} + \left(1+y'^2\right)^{2}$$
I've tried the traditional methods when the independent variable is missing but couldn't solve it anyways. Every help is welcome.

答:
Assume $A\neq0$ for the key cases:

Let $u=y'$ ,

Then $u\dfrac{du}{dy}=Ay(1+u^2)^\frac{3}{2}+(1+u^2)^2$

Let $v=1+u^2$ ,

Then $\dfrac{dv}{dy}=2u\dfrac{du}{dy}$

$\therefore\dfrac{1}{2}\dfrac{dv}{dy}=Ayv^\frac{3}{2}+v^2$

$\left(y+\dfrac{\sqrt v}{A}\right)\dfrac{dy}{dv}=\dfrac{1}{2Av^\frac{3}{2}}$

This belongs to an Abel equation of the second kind.

Let $w=y+\dfrac{\sqrt v}{A}$ ,

Then $y=w-\dfrac{\sqrt v}{A}$

$\dfrac{dy}{dv}=\dfrac{dw}{dv}-\dfrac{1}{2A\sqrt v}$

$\therefore w\left(\dfrac{dw}{dv}-\dfrac{1}{2A\sqrt v}\right)=\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dv}-\dfrac{w}{2A\sqrt v}=\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dv}=\dfrac{w}{2A\sqrt v}+\dfrac{1}{2Av^\frac{3}{2}}$

Let $x=\dfrac{\sqrt v}{A}$ ,

Then $\dfrac{dw}{dv}=\dfrac{dw}{dx}\dfrac{dx}{dv}=\dfrac{1}{2A\sqrt v}\dfrac{dw}{dx}$

$\therefore\dfrac{w}{2A\sqrt v}\dfrac{dw}{dx}=\dfrac{w}{2A\sqrt v}+\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dx}=w+\dfrac{1}{v}$

$w\dfrac{dw}{dx}=w+\dfrac{1}{A^2x^2}$

This belongs to an Abel equation of the second kind in the canonical form.

Title: Problems solveing an ODE


问:
Does a simple group of order 60 has a cyclic subgroup of order 6
I was unable to solve the following question in abstract algebra quiz, so I am posting it here.

Let G be a simple group  of order 60. Then prove that it doesn't have a cyclic subgroup of order 6.

Using sylow theorems and that group is simple , I got No. of sylow 5 subgroups are 6 and number of Sylow 3 subgroup are 4 or 10 and number if sylow 2 subgroups are 15 or 3 or 5.

But I noticed that in any combinations sum of elements in not 60.

Can you please tell what mistake I am making?

Another thought i had was that by cauchy theorem there exists an element og=f both 2 and 3 so they will generate a subgroup of order 2 and 3 respectively say A and B . SO, why doesnot there exists a cyclic subgroup  of order 6 (A $\times$ B)?

Please illuminate on this.

答:
Any simple group of order $60$ must be isomorphic to $A_5$. [refer here]

Suppose $A_5$ has a cyclic subgroup of order $6$, this means that $A_5$ has an element of order $6$. But element of order $6$ in $S_5$ must be in the form $(ab)(cde)$ which is an odd permutation. Hence $A_5$ does not have any cyclic subgroup of order $6$.


问:
How should I evaluate the flux?
I have the following vector field

$F= \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{i} + \frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{j} + \frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{k}$

I need to calculate the flux, $\iint_{S} F \cdot \vec{n}\ \text{dS}$, through the surface $S: x^2+y^2+z^2 = 16$ (sphere). As $F$ has a singularity at origin, I can't use the Divergent Theorem (right?). So I try to calculate using the surface integral of a vector field, $\iint_{S} F \cdot \vec{n}\ \text{dS} = \iint_{D} F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})\ \text{dA}$.

I've used the parametric representation $\mathbf{r}(\phi,\theta) = (4 \sin\phi\cos\theta, 4 \sin\phi\sin\theta, 4 \cos\phi)$. When I perform $F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})$, I'm getting 0. With this result, my flux would be 0, but according to my teacher answer, the flux is $44\pi$. What I'm doing wrong?

答:
There is no real need to use divergence theorem here. The polar coordinate representation should do.

\begin{gather*}
Transferring\ to\ polar\ coordinates,\ we\ see\ that\ \\
F( r) =\frac{11}{r^{2}} \ \\
Since\ the\ field\ is\ spherically\ symmetric,\ \\
we\ can\ directly\ calculate\ the\ flux\ as\ \\
\oint F( r) dr=\int 4\pi r^{2} F( r) dr=\int 44\pi dr=176\pi \\
\\
\end{gather*}
I am not sure whether your final answer is correct, because that would mean that the radius of the sphere should be 1 in that case.

Title: How should I evaluate the flux?


问:
Length of root strings is at most 4
Let $\Phi$ be a root system. In his Introduction to Lie algebras and Representation Theory, J. Humphreys proves that if
$$\beta-p\alpha,\dots,\beta,\dots,\beta+q\alpha$$
is the $\alpha$-root string through $\beta$, then $$\frac{2(\beta,\alpha)}{(\alpha,\alpha)}=p-q.$$
Then, he states that it follows "at once" that root strings are of length at most four.

I dont get why. Of course, we know that $-4\le 2(\beta,\alpha)/(\alpha,\alpha)\le 4$, but then length of the root string is $p+q+1$, not $p-q$...

答:
Here is a slight refinement of Sid's argument. Using Humphreys'  notation we know that $-3 \le \langle \alpha,\beta \rangle \le 3$ for all roots $\alpha,\beta$. Hence, in particular, for the $\alpha$-string through $\beta$ of length $q+r+1$ given by $\beta-r\alpha,\ldots ,\beta+q\alpha$ we find
$$
\langle \beta+q\alpha,\alpha\rangle \le 3\mbox{ and} -3\le \langle \beta-r\alpha,\alpha\rangle.
$$
But $\langle \alpha,\alpha\rangle=2$ therefore implies
$$
2q\le 3- \langle \beta ,\alpha\rangle  \mbox{ and } 2r\le 3+\langle \beta,\alpha\rangle.
$$
Thus together we find $q+r+1\le 4$.

Title: Length of root strings is at most 4


问:
if $n>5$ is prime,prove $(n-1)|(n-2)!$
Could you give a hint how to prove this?

if $n>5$ is prime, prove $(n-1)|(n-2)!$

答:
This is not the same proof as before, I have changed the proof to the following:

If $n > 5$ is prime then $n-1$ is composite as there are no consecutive primes in this case. Hence, $n-1 = ab$ such that $1 < a < b < n-2 < n-1$ or $n-1 = p^{2}$ where $p$ is prime. For the first case this means that $a,b \in \{2,...n-2\}$ so that $ab|(n-2)!$ and $(n-1)|(n-2)!$. If $n-1 = p^{2}$, in order to get both $p$'s in the expansion of $(n-2)!$, we need to get $(n-2) \geq 2p = 2\sqrt{n-1}$ since if we consider the set $\{1,2,...,p,...2p\}$, this is the smallest set that when one takes the product, we have that $p^{2}$ divides the result. The previous inequality is exactly true for all $n>7$, but more specifically when $n$ prime.
Therefore in either case, $(n-1)|(n-2)!$.

Title: if $n>5$ is prime,prove $(n-1)|(n-2)!$


问:
How to integrate $\int_0^\infty e^{-ty^2} \sin t dt$
My book suggests that I do some sort of limiting

$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t$

But I'm not getting anywhere.

答:
Starting with the initial integral.
You can integrate it by parts.
$$\int_0^A e^{-ty^2} \sin t d t=-\int_0^A e^{-ty^2}  d \cos(t)=-\left(e^{-ty^2}\cos(t)\bigg|_0^A  -(-y^2)\int_0^A \cos(t)e^{-ty^2}  dt \right)$$
$$\int_0^A e^{-ty^2} \sin t d t=-e^{-ty^2}\cos(t)\bigg|_0^A  -y^2\int_0^A e^{-ty^2}  d\sin(t)=-e^{-ty^2}\cos(t)\bigg|_0^A -y^2\left(e^{-ty^2}\sin(t)\bigg|_0^A  -(-y^2)\int_0^A \sin(t)e^{-ty^2}  dt \right) $$
$$\int_0^A e^{-ty^2} \sin t dt=1-e^{-Ay^2}(\cos(A)+y^2\sin(A))-y^4\int_0^A e^{-ty^2} \sin t dt$$
$$(1+y^4)\int_0^A e^{-ty^2} \sin t dt=1-e^{-Ay^2}(\cos(A)+y^2\sin(A))$$
$$\int_0^A e^{-ty^2} \sin t dt=\frac{1-e^{-Ay^2}(\cos(A)+y^2\sin(A))}{(1+y^4)}$$
And finally takong the limit one can obtain the answer:
$$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t=\frac{\lim_{A \to \infty} (1-e^{-Ay^2}(\cos(A)+y^2\sin(A))) }{1+y^4}=\frac{1}{1+y^4}$$</p>


问:
Finding basis of a span in $\mathbb R^4$
Let
$$S=\left\{(x_{1}, x_{2}, x_{3}, x_{4})\in \mathbb{R}^{4} : x_{1}=x_{2},\ x_{3}^{2}+x_{4}^{2}=1\right\}.$$

I have found that one vector will be $(1\ 1\ 0\ 0)$ because $x_1=x_2$ but since $x_{3}=\pm \sqrt{1-x_{4}^{2}}$ I am not sure how the find other vectors that will be basis for $[S]$.

答:
Note that $S\subseteq\mathbb R^4$ is a subset of the $3$-dimensional subspace $U\subseteq\mathbb R^3$ given by $x_1=x_2$ and hence for the linear span $\langle S\rangle$ we also have $\langle S\rangle\subseteq U$.

Now observe that the three vectors
$$
v_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix},\
v_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\
v_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}
$$
are elements of $S$ and linearly independent. Hence $\langle S\rangle$ is at least of dimension $3$, but as a subspace of $U$ it is also at most of dimension $3$.

Thus, we must have $\langle S\rangle = U$ and the vectors $v_1, v_2, v_3$ form a basis.


问:
To find the limiting value of the function
[The following image here. ][1]

*My answer came out to be 1 but doesn't seem so through graph *
What must be right? 1 or 0
[1]: https://i.stack.imgur.com/dPDr7.jpg

答:
Hint : When $h$ tends to $0$, $$\frac{\ln^2(1+h)\sin\left( \frac{1}{h}\right)}{h} \sim \frac{h^2\sin\left( \frac{1}{h}\right)}{h} = h \sin\left( \frac{1}{h}\right)$$
Now you can use a squeeze argument.


问:
Formal way to evaluate this limit
I've some doubt about the formal way to evaluate the following limit:

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}$$

I know that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}$$

my calculus teacher says that we can't evaluate the limit by pieces, so we can't (in general) say that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{\left(e^{-1}\right)^{n}}=0$$ even if in this case it is the right answer.

So I'm asking a formal way to solve this, my approach is:

Say that the limit of a product is the product of the limits, so I can say that

$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2}{\lim_{n\to\ +\infty}\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{-n}}=0$$

Can I do this (I've used the continuity of the exponential function) or am I still doing the limit by pieces when I've substituted $ln\left(1-\frac{1}{n}\right)$ with $-\frac{1}{n}$?

Furthermore, can I use the theorem of the product even if there is the indeterminate form $0\cdot(+\infty$)?

Thanks for your time.

答:
Late answer since it was tagged as duplicate.

An elementary way without any $e$ or logarithm could be:

\begin{eqnarray*}\left(1-\frac{1}{n}\right)^{n^2}
& = & \frac 1{\left(1+\frac{1}{n-1}\right)^{n^2}}\\
& \stackrel{\text{binomial formula}}{\leq} & \frac 1{\frac{n^2}{n-1}}\\
& < & \frac 1 n \stackrel{n\to \infty}{\longrightarrow} 0
\end{eqnarray*}

Title: Formal way to evaluate this limit


问:
Proof of series representation of inverse of sum of matrices
Is there a proof that $(A+B)^{-1} = A^{-1} - A^{-1}BA^{-1} + A^{-1}BA^{-1}BA^{-1} - \cdots$?

答:
There is no proof because of the following counter example:

Take

$A=\left[\begin{matrix}
1&0\\
0&0\\
\end{matrix}\right]$

$B=\left[\begin{matrix}
0&0\\
0&1\\
\end{matrix}\right]$

We have $A+B= \left[\begin{matrix}
1&0\\
0&1\\
\end{matrix}\right]$ $\Rightarrow$ $(A+B)^{-1}= \left[\begin{matrix}
1&0\\
0&1\\
\end{matrix}\right]$

But $A$ is not invertible, so $A^{-1}$ is not defined.


问:
Deduce that $H$ has no elements of finite order other than the identity element.
Given operation $*$ defined on the set $G$, where $G =\{(a, b)\mid a, b \in \mathbb{Q}\}$ with $a$ and $b$ not both
zero, $(a, b) * (c, d) = (ac + 3bd, ad + bc)$.

Prove that subset $H = \{(a, 0)\mid a \in \mathbb{Q} \land a\neq 0\}$ is a subgroup of $G$. Find $(a, 0)^r$
for $r \in \mathbb{Z^+}$, where $(a, 0) \in H$ and deduce that $H$ has no elements of finite order other than the
identity element.

Attempt

I found identity element of $G$ as $(1,0)$ to prove $H$ is a subgroup of $G$ I took $(a,0),(b,0)\in H$ then I proved $(a,0)*(b,0)^{-1} \in H$ so $H$ is a subgroup of $G$

Next part, $(a, 0)^r=(a,0)*(a,0)*\dots*(a,0)=(a^r,0)$ so then,

$$(a, 0)^r=(a^r,0)=(1, 0).$$

This implies $a^r=1$ so $a$ can be $-1$ or $1$

If $a=1$ then $(1,0)$ we can neglect it since it is an identity element so we have another element $(-1,0) $ it is also have a finite order but the question is said to deduce there is no elements of finite order other than the identity element.

Is there anything wrong my steps?

Thank you!

答:
You're correct.

It is a subgroup.

I will use the one-step subgroup test. (Note that I have to show $\varnothing\neq H\subseteq G$.)

Observe that $1\in \Bbb Q$ and $1\neq 0$, so $(1,0)\in H$. Hence $H\neq\varnothing$.

Let $(a,0)\in H$. Then $a\neq 0$, so $a$ and $0$ are not both zero but are both rational. Hence $(a,0)\in G$. Hence $H\subseteq G$.

Let $A=(a,0), B=(b,0)\in H$. Then

$$\begin{align}
AB^{-1}&=(a,0)*(b,0)^{-1}\\
&=(a,0)*(b^{-1}, 0)\\
&=(ab^{-1}+3(0)(0), a(0)+(0)b^{-1})\\
&=(ab^{-1}, 0),
\end{align}$$

which is in $H$ since $ab^{-1}\in \Bbb Q\setminus \{0\}$ as $a,b\in\Bbb Q\setminus \{0\}$.

Thus $(H,*)\le (G,*)$.

The element $(-1,0)\in H$ has order two.

Indeed, if $r\in \Bbb Z^+$, then

$$\begin{align}
(a,0)^r&=\underbrace{(a,0)*\dots*(a,0)}_{r\text{ times}}\\
&=(a^r, 0),
\end{align}$$

since the second argument of $(a,0)$ is zero, meaning $3(0)(0)=a(0)=(0)a=0$.

But $a\neq 0$ by definition of $H$, so, since also $a\in \Bbb Q$, if

$$(a,0)^r=(a^r,0)=e_G=(1,0),$$

then $a^r=1$, meaning $a=\pm 1$ if $\pm r>0$, depending on whether $r$ is odd or even.

This can be checked directly.

Indeed,

$$\begin{align}
(-1,0)^2&=(-1,0)*(-1,0)\\
&=((-1)(-1)+3(0)(0), -1(0)+0(-1))\\
&=(1,0)
\end{align}$$

and $(-1,0)\in H$.


问:
Prove about a integral with the special integral Beta $\mathbb{B}(\alpha,\beta)$.
How can I prove this identity:
\begin{eqnarray*}
\int_{0}^{1}y^{a}\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}dy=\frac{\Gamma(\alpha+a)\Gamma(\alpha+\beta)}{\Gamma({\alpha})\Gamma({\alpha+\beta+a})}
\end{eqnarray*}
for all $a$ such that $a+\alpha>0$ and $\alpha,\beta>0$.

My attempt: This problem it comes from the context of a probability and distribution theory problem that I am trying to solve. Where the random variable has a beta distribution.
I know, that $$\mathbb{B}(\alpha,\beta)=\int_{0}^{1}y^{\alpha-1}(1-y)^{\beta-1}dy=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$
and can proved that $$\int_{0}^{1}y\left[\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}\right]dy=\frac{\alpha}{\alpha+\beta}$$
but $\alpha \in \mathbb{R}$ so I can't use induction on $\alpha$.
How can I prove that identity?

答:
If we were to move the Beta function outside, we would expect to obtain

$$
\int_0^1y^\alpha{y^{\alpha-1}(1-y)^{\beta-1}\over B(\alpha,\beta)}\mathrm dy={1\over B(\alpha,\beta)}\int_0^1y^{2\alpha-1}(1-y)^{\beta-1}\mathrm dy={B(2\alpha,\beta)\over B(\alpha,\beta)}
$$

Now, apply the Gamma function representation for the Beta function, so that

$$
\int_0^1y^\alpha{y^{\alpha-1}(1-y)^{\beta-1}\over B(\alpha,\beta)}\mathrm dy={\Gamma(2\alpha)\Gamma(\alpha+\beta)\over\Gamma(2\alpha+\beta)\Gamma(\alpha)}
$$


问:
Starting point for solving this integral
I have to calculate a pretty ugly integral which is given by:
$$
\psi(x,t)=\int_{-\infty}^{\infty}\exp\left(-\frac{(p-p_0)^2}{4\,\Delta p^2} - \frac{i}{\hbar}(p-p_0)x_0+\frac{i}{\hbar}(px-\frac{p^2}{2m}t)\right)\mathrm{d}p
$$

It's a solution to the Schrodinger equation and $x_0,p_0$ and $\Delta p$ are constants.

I tried to complete the square on this one, but for me it doesn't seem that the integral gets easier by that. I would be really grateful if someone could give me a good starting point in solving this integral. Maybe completing the square is a good way to do it, if this the case, I will try to do it again with that method.

答:
Hint:

The exponent is a quadratic polynomial with some complex coefficients, so the integrand has the form

$$e^{ap^2+2bp+c}$$ where $a$ is real and negative.

You can get rid of $a$ by a scaling of the variable, and of the real part of $b$ by shifting the variable. The constant coefficient can be factored away. You end-up with an integral for the form

$$\int_{-\infty}^\infty e^{-p^2}(\cos\omega p+i\sin\omega p)\,dp$$ which is a classical one. (The imaginary part vanishes.)

https://www.wolframalpha.com/input/?i=integrate+e%5E%28-p%5E2%29+cos%28omega+p%29+from+-+inifinity+to+infinity


答:
The integral can be rewritten as

$$\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_0^1y^{2a-1}(1-y)^{b-1}dy=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\frac{\Gamma(2a)\Gamma(b)}{\Gamma(2a+b)}=\frac{\Gamma(a+b)\Gamma(2a)}{\Gamma(a)\Gamma(2a+b)}$$

Title: Prove about a integral with the special integral Beta $\mathbb{B}(\alpha,\beta)$.

Body:
How can I prove this identity:
\begin{eqnarray*}
\int_{0}^{1}y^{a}\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}dy=\frac{\Gamma(\alpha+a)\Gamma(\alpha+\beta)}{\Gamma({\alpha})\Gamma({\alpha+\beta+a})}
\end{eqnarray*}
for all $a$ such that $a+\alpha>0$ and $\alpha,\beta>0$.

My attempt: This problem it comes from the context of a probability and distribution theory problem that I am trying to solve. Where the random variable has a beta distribution.
I know, that $$\mathbb{B}(\alpha,\beta)=\int_{0}^{1}y^{\alpha-1}(1-y)^{\beta-1}dy=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$
and can proved that $$\int_{0}^{1}y\left[\frac{y^{\alpha-1}(1-y)^{\beta-1}}{\mathbb{B}(\alpha,\beta)}\right]dy=\frac{\alpha}{\alpha+\beta}$$
but $\alpha \in \mathbb{R}$ so I can't use induction on $\alpha$.
How can I prove that identity?


问:
Why is this matrix not diagonalizable over $\mathbb{Q}$
Consider  this matrix ( I don't know what mistake i am making in mathjax, I have thought a lot) ( Its a 3 * 3 matrix with 1 st row 2 ,2, 1 )

$$\begin{pmatrix} 2 & 2 & 1  \\
   0 & 2 & -1   \\
 0 & 0 & 3   \\
 \end{pmatrix}$$

Question was asked in my algebra quiz that is this diagonalizable over $\mathbb{Q}$?

Eigenvalues of A are 2,2,3 $\in  \mathbb{Q}$ . So, why it is not diagonizable.

I seriosly have no idea!

Can you please tell why it is not diagonizable.

答:
$A$ is not diagonalizable over $\Bbb Q$ because it is not diagonalizable at all (i.e. not diagonalizable over $\Bbb C$). It suffices to note that even though $2$ is an eigenvalue with algebraic multiplicity $2$, the geometric multiplicity (dimension of the eigenspace) is
$$
\dim \ker (A - 2 I) = 1.
$$


问:
Do the pairing axiom, subset axiom and union axiom imply replacement axiom?
I am studying Set theory from Daniel Cunningham's textbook. This is how Cunningham states replacement axiom:

Let $\psi (x,y)$ be a formula. For every set $A$, if for each $x\in A$, there is a unique $y$ such that $\psi (x,y)$, then there is a a set $S$ that consists of all the elements $y$ such that $\psi(x,y)$ for some $x\in A$.

Suppose that we assume the hypothesis of the previous axiom. Then for each $x\in A $, there is a unique $y$ such that $\psi (x,y)$ holds. Then $P_{x}=\{ x, y \}$ is a set by the pairing axiom. Then $C=\bigcup \{ P_x : x \in A \}$ is a set by union axiom. Then $C\setminus A$ is the set $S$ that is provided by the replacement axiom (obviously we assume the subset axiom).

So, is the replacement axiom really needed or am I incorrect?

答:
In your argument, you are forming the set
$$
\{ P_x : x \in A\}.
$$
To prove that this set exists (i.e., a set such that for each $x \in A$ there is a unique element in the set of the form $\{x, y\}$ where $\psi(x, y)$ holds) you need the replacement axiom, in general.


问:
PDEs in biology
I am student who mostly heard lectures on partial differential equations and homogenization. But I really like the idea of working in biology or with biologists - but (with my lack of overview) it seems to me it's either statistics or some ODE problems.
Is there anything more PDE releated in biology?</p>

答:
Mathematicians often look to other fields to gain inspiration for some of their research (the mother of all examples might be vector calculus and electromagnetism), so this is not a bad question.  Lots of mathematical modelling in biology is very approximate and many models are entirely useless to the average experimentalist. An example of this are the PDEs e.g. "governing" neural transfer, where $u$ is the presynaptic firing rate and $v$ is a depolarization field :

$$\left(\frac{\partial}{\partial t} + (1-\Delta)\right)u(x,y,t) = C(x,y,t)v,$$
coupled to
$$\frac{\partial^2}{\partial t^2} v + \frac{\partial}{\partial t}v + v = K(x,y,t)u.$$

In my view, more interesting PDEs arise in the study of biological membranes. For instance, for vesicles with a phospholipid bilayer, Helfrich found (based on his work in liquid crystals) that a possible energy for the membrane is
$$E[u] = \kappa\int_\Sigma H^2 - K dS.$$
The absolute minimizer among surfaces of genus 1 for this energy is the Clifford torus. This problem provides interesting mathematical insight and matches biological findings very accurately.

One can also consider the energy for liquid crystals whose molecules follow a director field n:

$$\int |\nabla_s n|^2 = \int (div_s n)^2 + |n \times \nabla_s \times n|^2 + |n\cdot \nabla_s \times n|^2$$
where the $s$ always denotes ``surface'' and $\nabla_s \times$ is surface curl.

Title: PDEs in biology


答:
A nonabelian simple group cannot contain a proper subgroup of index less than $5$. ($*$)

(See here, for example,  for a proof.)

So the number of Sylow $3$-subgroups cannot be $4$, and hence it must be $10$.

Let $P \in {\rm Syl}_3(G)$. Then $|N_G(P)| = 6$, and to prove that $G$ has no cyclic subgroup of order $6$, we need to prove that $N_G(P)$ is not cyclic.

So suppose it is, and let $g \in N_G(P)$ have order $2$. Let $H = C_G(g)$.
Since $g \in H$ and $H$must also contain a Sylow $2$-subgroup of $G$, we have $|H| \ge 12$, and then by $(*)$ we must have $|H| = 12$.

Since $N_G(P) \le H$, $H$ has at most two Sylow $3$-subgroups, and so by Sylow it has only one, but then $P \lhd H$, contradiction.

Title: Does a simple group of order 60 has a cyclic subgroup of order 6

Body:
I was unable to solve the following question in abstract algebra quiz, so I am posting it here.

Let G be a simple group  of order 60. Then prove that it doesn't have a cyclic subgroup of order 6.

Using sylow theorems and that group is simple , I got No. of sylow 5 subgroups are 6 and number of Sylow 3 subgroup are 4 or 10 and number if sylow 2 subgroups are 15 or 3 or 5.

But I noticed that in any combinations sum of elements in not 60.

Can you please tell what mistake I am making?

Another thought i had was that by cauchy theorem there exists an element og=f both 2 and 3 so they will generate a subgroup of order 2 and 3 respectively say A and B . SO, why doesnot there exists a cyclic subgroup  of order 6 (A $\times$ B)?

Please illuminate on this.


问:
Probability Question: Rolling two dice with re-roll
I came across the following dice-roll problem:

You roll two dice consecutively. Rolling a 6, then a 6 wins you \$100.

Rolling a 6, then not a 6 loses \$x

For all other cases you, redo the dice rolling.

When should you play the game?

If X is the value amount I receive by playing this game; that event A: (6,6); event B: (6, not 6); I calculated the expected value E[X] as follows:

E[X] = P(A)*100 - P(B)*x = $\frac{100}{36} - \frac{5x}{36}$. E[X] is positive when $5x < 100$. I should therefore play the game when $x< \$20$.

Am I missing something by not taking into account the "re-roll" information? Does that change my probabilities, or can it just be treated as the following independent re-running of the game?

答: No, you can just assume the first roll is $6$ because you reroll anything else.  It becomes a single roll with $+100$ on a $6$ and $-20$ on anything else, which is fair.