问:
Problem in "proof" that nonempty perfect set without rationals cannot exist.
Q: (Baby Rudin Chap $2$ Prob $18$) Is there a nonempty perfect set in $\mathbb R$ which contains no rational number?

Proof (which is obviously wrong since there are counterexamples):

Consider that such a set exists. It follows that it must have a irrational point. By the definition of perfect set, this point must be a limit point. By definition of limit point, we get that a rational number must be part of this set (since there is a rational between any two numbers). This contradicts the property of the set. Hence, no such set can exist.

What is wrong with this "proof"? I must be missing some basic definition.

答:
The problem is here:

By definition of limit point, we get that a rational number must be part of this set (since there is a rational between any two numbers).

If $x$ is a limit point of $A\subseteq\mathbb{R}$ then there must be a sequence $(y_i)_{i\in\mathbb{N}}$ of points in $A$ with limit $x$ ... but all those points could be irrational! That is, we could "skip over" each rational.

It might help to consider the picture with "rational" and "irrational" switched: there are irrationals between any two numbers, but $$0,{1\over 2}, {3\over 4}, {5\over 6}, ...$$ is a sequence containing no irrationals which converges to the rational $1$. It's an easy exercise to "flip this around" and get a sequence of irrationals converging to an irrational.


问:
Where did I go wrong differentiating arccos(4/x)?
I've been trying to differentiate $y=\arccos(4/x)$ and ended up with the answer $\frac{dy}{dx} = \frac{4}{x\sqrt{x^2-16}}$ however the solution to this question is marked as $\frac{dy}{dx}= \frac{4}{|x|\sqrt{x^2-16}}$. I opted to use implicit differentiation and honestly have no idea where the $|x|$ could of come from. Any help would be greatly appreciated.

答:
Using implicit differentiation,

$$\begin{align*}
y &= \arccos \frac4x\\
\cos y &= \frac 4x
\end{align*}$$

Differentiating both sides with respect to $x$,

$$\begin{align*}
-\sin y\cdot \frac{dy}{dx} &= -\frac 4{x^2}\\
\sqrt{1-\cos^2y}\cdot \frac{dy}{dx} &= \frac 4{x^2}\\
\sqrt{1-\frac{16}{x^2}}\cdot \frac{dy}{dx} &= \frac{4}{x^2}\\
\frac{\sqrt{x^2-16}}{\sqrt{{x^2}}}\cdot\frac{dy}{dx} &= \frac{4}{x^2}
\end{align*}$$

For $x$ that may be positive or negative, $\sqrt{x^2}$ is really $|x|$, and is different from $x$ when $x$ is negative.

$$\begin{align*}
\frac{\sqrt{x^2-16}}{|x|}\cdot\frac{dy}{dx} &= \frac{4}{x^2}\\
\frac{dy}{dx} &= \frac{4}{|x|\sqrt{x^2-16}}
\end{align*}$$

ContentLicense: CC BY-SA 4.0


问:
Skorohod topology extended to whole real line
Results about the standard Skorohod topology on the space $D([0,\infty))$ of cadlag functions from $[0,\infty)$ to $\mathbb{R}^d$, can easily be found in many classic texts, such as Billingsley's Convergence of Probability Measures. Completeness, separability, ... are well explained.

However, I am really struggling to find a reference for the Skorohod topology on the space $D(\mathbb{R})$ of cadlag functions from $\mathbb{R}$ to $\mathbb{R}^d$. Does anybody know of any reference?

答:
The only explicit treatment I could find is in a diploma thesis:

https://www.ruhr-uni-bochum.de/imperia/md/images/mathematik/folder/lehrstuhl-xii/vogel_diplomarbeit_2005.pdf

From what I understand the construction of the Skorokhod metric on $D(\mathbb{R})$ is very similar to the one on $D([0,\infty))$.

ContentLicense: CC BY-SA 3.0


问:
Inequality Involving Complex Norm
I am trying to prove, that for any real $x$ and any real $t \neq 0$ that
$$\bigg{|}\frac{e^{itx}-1}{t}\bigg{|} \leq |x| $$
Graphically it seems to be true but I cant quite get it formally.

答: Use $\frac{e^{itx}-1}{itx}=e^{itx/2}\operatorname{sinc}\tfrac{tx}{2}$ and $|\operatorname{sinc}z|\le1$ for $z\in\Bbb R$.

Title: Inequality Involving Complex Norm


问:
Dual of holomorphic functions (with the $L^1$ topology)
Let $\Omega$ be a connected domain of the complex plane, and let $E$ be the vector space of integrable holomorphic functions on $\Omega$. Then it can be checked that $E$ is a closed subspace of $L^1(\Omega)$. My question (which I'm sure has a well-known answer, but I couldn't find it) is : what is the dual of $E$ ? More precisely, does it identify with a nice (i.e. explicit) subspace of $L^\infty(\Omega)$ ?

Note that since we do not endow $E$ with the usual topology of holomorphic functions (uniform convergence on compacts) this has a priori nothing to do with hyperfunctions.

答:
The key term is Bergman space. The space you introduced is usually denoted  $A^1(\Omega)$ or $L^1_a(\Omega)$. When $\Omega$ is a disk, the dual of $A^1(\Omega)$ can be identified with the space of Bloch functions on the disk. This is proved in the book Bergman spaces by Duren and Schuster. (By the way, in the reflexive range $1<p<\infty$ we have $(A^p)^*=A^q$ on the disk, with $p^{-1}+q^{-1}=1$.)

If $\Omega$ is simply-connected and has smooth boundary ($C^{1,\alpha}$ with $\alpha>0$), then the conformal map of $\Omega$ onto the disk has derivative bounded from $0$ and $\infty$. By the change of variable formula, the map induces isomorphism between Bergman spaces on $\Omega$ and on the disk. Thus,   the problem reduces to the case of the disk; the dual is again the Bloch space.

For a general simply-connected domain $\Omega$, the identification of the dual of $A^p(\Omega)$ is difficult; it is related to a major open problem in complex function theory (Brennan's conjecture). As a starting point, see the paper The dual of a Bergman space on simply connected domains by Hedenmalm.


问:
How can I show that $\lim_{n\rightarrow \infty} ((-n^x)/e^n)) = 0$
How can I show that $$\lim_{n\rightarrow \infty} ((-n^x)/e^n)) = 0$$ , where n is a natural number and $x>0$?

I wanted to use l'hôspital, but it doesn't work or rather not defined, if n is a natural number.

答:
$$n^x=e^{x\ln (n)},$$

and thus $$\frac{n^x}{e^n}=e^{x(\ln(n)-n)}.$$
Using the fact that $$\ln(n)-n\underset{n\to \infty }{\longrightarrow }-\infty $$ gives the wished result.


问:
Notations for complex logarithms
Are there any specific notations for complex logarithms in different branch cuts? The principal branch is denoted as $\operatorname{Log}z$. But how do I write a logarithm which is sliced on the negative imaginary axis?

Edit: If there are none, how would you write it? Maybe something like $\log_w^{\arg}$? This looks not good in my opinion that's why I need your advice.

答: I do not think that there is any widely used notation. However, you can distinguish different branches by specifying the domain of the function. This means that, if the branch you are dealing with has domain $U$, then you can write $\log_U$ or explicitely say that $\log : U \rightarrow \mathbb{C}$. For instance, the principal branch of $\log$ is obtained by using $U=\mathbb{C} \setminus \mathbb{R}_{\leq 0}$ as domain.

Title: Notations for complex logarithms


问:
In a queue of $n$ people, what is the average number of spaces in between two people?

There are $n$ people in a queue. You randomly select two of them. What is the expected number of people standing in between them?

I noticed that the question can be reformulated in the following way. Let $X$ and $Y$ be discrete random variables that can take values between $1$ and $n$. What is $\mathbb{E}(|X-Y|-1 \text{ given that $X\neq Y$})$?

Unfortunately, this didn't seem to get me anywhere. I also considered what the probability was for small values of $n$. Let $Z$ be the number of people standing in between the two people. For $n=2$, $Z=0$ trivially. For $n=3$, there is a two-thirds chance that the people are standing next to each other, and a one-third chance that they are on either side of the queue. So $E(Z)=\frac{2}{3}\cdot0+\frac{1}{3}\cdot1=\frac{1}{3}$. I also calculated it for $n=4$, but stopped after that point since the calculations were getting unwieldy.

Could anyone hint at what approach I could use instead?

答:
Update: Just realized there's a simple way: pick any three people at random, there are $n \choose 3$ ways. Fix the two people at both ends, the total number of cases are the same as the number of people in between them. Therefore the expected value is $\frac{n \choose 3}{n \choose 2} = \frac{n-2}{3}.$

By symmetry
$$E(|X-Y|-1 \mid X \ne Y) = E[X-Y-1|X>Y]=\frac{\sum_{y=1}^{n-1} {n-y \choose 2} }{n \choose 2} = \frac{n \choose 3}{n \choose 2} = \frac{n-2}{3}$$

where we used the fact that if $Y=y$, total "distance" when $X$ runs from $y+1$ to $n$ is $n-y \choose 2$.

A baby example:
$$\begin{array}{c}
  & X=1 & 2 &3 &4 &5 & \rm Total\\
Y=1 &&0 &1 &2 & 3 & 6\\
2 &&&0&1&2 & 3\\
3 &&&&0&1 & 1\\
4 &&&&&0  & 0\\
\rm Total &&&&&0  & 10\\
\end{array}$$


问:
Lagrange multipliers - missing critical points?

Given $S = \left \{ (x,y,z) : z^2-xy=1\right \}$

a. Prove that there is at least one point which is closest to $(0, 0, 0)$ and find the distance.

b. In section a you should have found several "critical points". show that those which are not a global minimum are also not a local extremum.

My problem is that I only get global minimum points in section a:

Defining $f = || \ ||^2=x^2+y^2+z^2$ and using Lagrange multipliers theorem we know that for any local extremum $p$ s.t $\nabla F(p) \neq 0$, we have $\nabla F(p) = \lambda \nabla f(p)$.

So, $\lambda (2x_0, 2y_0, 2z_0) = (-y_0, -x_0, 2z_0)$, thus $\lambda = 1$, and $x_0=y_0=0$

So we get 2 solutions $(0, 0, 1), (0, 0, -1)$.

Which are both a global minimum.

Can anyone explain what am I missing?

答:
$\lambda (2x, 2y, 2z) = (-y, -x, 2z)$

$2\lambda x = -y$

$2\lambda y = -x$

$\lambda z = z$

$z^2 - xy = 1$

So we also have,

$z = 0$ as a possible solution from third.

From first, $\lambda = - \frac{y}{2x}$

Plugging into second, $y^2 = x^2 \implies y = \pm x$

So given the constraint, $x = -1, y = 1, z = 0$ and $x = 1, y = -1, z = 0$ are other critical points to test.


问:
Definition of metastability
I was reading Terence Tao's blog post on analogies between soft and hard analysis when I saw that the soft analysis statement "$x_n$ is convergent" corresponds to the hard analysis statement "$x_n$ is metastable".

What does "metastable" mean? The Wikipedia article on that term explains it only in the context of physical systems:

Metastability describes the behaviour of certain physical systems that can exist in long lived states that are less stable than the system's most stable state.

I understand that a sequence $x_n$ can be viewed as a (discrete-time) dynamical system, but I need a precise mathematical definition of this property that sequences can have.

答:
Metastability has many interpretations, depending upon which field you are working in. In dynamical systems/Markov chain literature, metastability has been associated with states (or regions in phase space) where a typical trajectory will spend a lot of time, but that state (or region of phase space) is not an attractor (or a sink or so on).

To give you a practical example: Properties of protein molecules can be understood in terms of conformations of such molecules: i.e. millions of different ways in which the atoms can arrange themselves. Very few of such conformations are actually seen in reality, and those states are called metastable states.

Here's a paper that talks more about it:http://biocomputing.mi.fu-berlin.de/publications/ScHu03.pdf

Precise mathematical definition is hard to come by, but imagine partitioning your phase-space into n different subsets. Such a partition can be done in infinite many ways, but a metastable partition would follow certain rules. For example if the phase space subsets are $B_1$,$B_2$,...$B_n$, where $X=B_1 \cup B_2 ... \cup B_n$, $X$ being the state space, each $B_i$ metastable and $F$ the system propagator in time, then you require that at time step k, the probabilities satisfy:

$P(x_{k+1}\in B_j | x_k\in B_i) \ll 1, $ if $i \neq j$ And

$P(x_{k+1}\in B_j | x_k\in B_i) \approx 1, $ if $i = j$

where $x_{x+1}=F(x_k)$

Title: Definition of metastability


答:
Your attempt seems a little messy, so I hope you don't mind that I present an alternative approach to an inductive proof.

You want to show $\displaystyle (1+2+...+n)^2 = 1^3 + 2^3 +... + n^3$

Assume, for some $k$ that $\displaystyle (1+2+...+k)^2 = 1^3 + 2^3 +... + k^3$ as the inductive hypothesis and call the expression on the left hand side $A$.

Now, let $\displaystyle B = (1+2+...+k+ (k+1))^2$

By the difference of squares identity, $\displaystyle B-A \\= (k+1)(2(1+2+...+k) + (k+1)) = (k+1)(2(\frac 12)(k)(k+1)+(k+1)) = (k+1)(k+1)(k+1) = (k+1)^3$

Hence $\displaystyle B = A + (k+1)^3$, and using the inductive hypothesis (and showing the base case), the proof is complete.

Sometimes, using the summation notation tends to cloud the thought process.

Title: Nicomachus theorem proof - what did I do wrong?

Body:
The exercise asked me to proof

$1^3=1$

$2^3=3+5$

$3^3=7+9+11$

$...$

I formulate the equation as

(1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$

Proof base case:

$$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$

Proof (a+1) case:

(2) $$(a+1)^3=a^3+3a^2+3a+1$$
(3) $$(a+1)^3=\sum_{i=0}^{(a+1)-1}((a+1)-1)(a+1)+1+2i=\sum_{i=0}^{a}(a)(a+1)+1+2i$$
factor out last summation term, so number of iterations match (1):

$$\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+(a)(a+1)+1+2a=\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+a^2+3a+1$$

rearrange the summation term, so part of equation match (1):
$$\sum_{i=0}^{a-1}[(a)(a-1)+1+2i]+\sum_{i=0}^{a-1}(2a)+a^2+3a+1$$
substitute summation with (1):

(4) $$a^3+\sum_{i=0}^{a-1}(2a)+a^2+3a+1=a^3+(2a)(a-1)+a^2+3a+1=a^3+3a^2+a+1$$

From (2) I got $a^3+3a^2+3a+1$ but from (3) & (4) I got $a^3+3a^2+a+1$

My result with induction is off by 2a. Can someone please help me to understand where I did wrong?


问:
Show that $\sum_{k=0}^n \binom{2n}{2k} = 2^{2n-1}$
I have played around with this question for a bit, but my answer doesn't feel right. I'm not happy with how I converted the binomial. it feels a bit hand-wavy. Is there any way I can make this proof better?

Let's say that we have two teams that are both size $n$. Let's call one team the Blue team and the other team Red. We are going to send a mixture of Blue and Red players to an event (they can also send no one). There are $\binom{2n}{a}$ ways we can select players, but if we want to only send an even sum of players then $a$ must be even. Lets rewrite it as $\binom{2n}{2k}$. Considering all possible $k$-values tells us that the total ways to send an even sum of randomly composed Blue and Red players is
$$\sum_{k=0}^n \binom{2n}{2k}$$
Since both teams have $n$ players, we know that there are $2^{2n}$ subgroups that could be sent. But again we only want to consider even groups so we divide by 2 to get rid of all odd groups. Thus we have

$$2^{2n}/2 = 2^{2n-1}$$

Both methods give the same result.

Body:
I find the algebraic approach crystal clear:
$$
2^{2n} = 2^{2n}+ 0 = (1+1)^{2n} + (1-1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} + \sum_{k=0}^{2n} (-1)^k\binom{2n}{k} = 2 \sum_{k=0\\k \text{ even}}^{2n} \binom{2n}{k}
$$


问:
Graph $|z+2-i|=2$
I want to graph the complex equation $|z+2-i|=2$
However I have no idea how to proceed. I'm also unsure of what to do with the modulus.
How is this supposed to look?

答: Hint: $|z-z_0|=2$ iff the distance from $z$ to $z_0$ is $2$.

Title: Graph $|z+2-i|=2$


答:
With $f(t) = e^{ixt}$, the mean value theorem gives
$|f(t)-f(0)| \le \max_{z \in [0,t]} |f'(z)| |t-0|$.

Since $|f'(z)| = |x|$ we have $|f(t)-f(0)| = |e^{ixt} -1| \le |x||t|$

Title: Inequality Involving Complex Norm

Body:
I am trying to prove, that for any real $x$ and any real $t \neq 0$ that
$$\bigg{|}\frac{e^{itx}-1}{t}\bigg{|} \leq |x| $$
Graphically it seems to be true but I cant quite get it formally.


问:
Proof of number of equivalence relations on a set.
Consider a set $S$ with cardinality $s$. Prove that the number of equivalence relations on the set $S$ with exactly 5 distinct equivalence classes is:

$$
\frac{1}{5!}\sum^5_{i=0}(-1)^i{5\choose i}(5-i)^s.
$$

I know that there exists a bijection between equivalence relations on S and the number of partitions on that set. Since there are $s$ elements I must work out how many partitions there are of $s$. However, I am told there are $5$ equivalence classes so I understand there are ${5\choose i}$ ways. However I am confused as to where the rest comes from?

答:
If there are $s$ elements, and they can each can be put into one of $5$ equivalence classes, then we get $s^5$ allocations.  But we have some significant overcounting.

By this method, there may be some classes with no members, and this will not do.  To make sure that we exclude those cases we need to apply inclusion-exclusion.

${5\choose 0} 5^s - {5\choose 1} 4^s + {5\choose 2} 3^s - {5\choose 3} 2^s + {5\choose 4} 1^s$

We also have a different sort of overcounting.  Class 1 is not fundamentally different from class 2, etc.  So, far we have treated them differently.

We need to divide by the number of permutations of the 5 classes.

$\frac {{5\choose 0} 5^s - {5\choose 1} 4^s + {5\choose 2} 3^s - {5\choose 3} 2^s + {5\choose 4} 1^s}{5!}$

Which is the same as your formula above.


问:
Is it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?
Help would be very appreciated.

It it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?

答:
Consider the linear transformation $T : \Bbb R^2 \to \Bbb R^2$ so that:
$$
T(x, y) = (y, 0)
$$

We have:
$$
\ker T = \{(x, 0) : x \in \Bbb R\} = \operatorname{im} T
$$

I arrived at this example by noticing that such a linear transformation must have $T^2 = 0$.


答:
Let $z = x+yi$, then

$$\begin{align*}
|x+yi + 2 - i| &= 2\\
\sqrt{(x+2)^2 + (y-1)^2} &= 2\\
(x+2)^2 + (y-1)^2 &= 2^2
\end{align*}$$

which becomes the formula of a circle on the $xy$-plane.

Title: Graph $|z+2-i|=2$

Body:
I want to graph the complex equation $|z+2-i|=2$
However I have no idea how to proceed. I'm also unsure of what to do with the modulus.
How is this supposed to look?


问:
Math school online?
As a hobbyist programmer with little to no math skills (no science skills either, thank you religious schools) I am finding math encroaching on my life more and more often. I have always laughed and said that I'm just not good at math because my brain doesn't work like that.

My younger brother challenged me recently claiming that it wasn't a lack of ability that kept higher math out of my hands but a lack of interest. Because my current realm of studies is really starting to delve into 3D modeling, programming for web, mobile and desktop, I feel that a stronger math skill set would benefit me immensely.

But I don't know where to start...

I'm wondering if there is a math version similar to: http://www.code.org/khan-academy

Some place where I can go and start challenging my brain, increasing my math as often as I have time? Some place online? I would also appreciate any tips for books or other resources that would be recommended for someone wanting to ignite an interest in math.

To give a background, my highest level of math in school was pre-algebra in 7th grade. After that the schools my parents sent me to didn't teach higher math (nor evolution) so when I hit college, I had to get a tutor to pass (with a C) the College Level Algebra... I barely did it and it was so long ago I don't remember anything I learned.

I hope this is the right place to ask this... I stumbled across math.stackexchange by accident. Whether this question gets answered or not I will start browsing around and seeing what all there is...

答:
The obvious answer is the math section of Khan Academy!

More advanced courses can be found here and here, a couple of nice ones on analysis and functional analysis by Joel Feinstein here and some brilliant ones on linear algebra/systems/optimisation by Stephen Boyd here.

It is also worthwhile to check for courses here and (in the future) here.

See also the answers to this question and generally those to questions with the tag (self-learning).

Title: Math school online?


问:
Which cases are not counted here?

Q)Let 15 toys be distributed among 3 children subject to the condition that any child can take any number of toys. Find the required number of ways to do this if the toys are distinct.

This question can be easily answered as there are 3 children for each toy to go to, hence $3^{15}=14348907$
is the correct answer. However, if we look at it through different(wrong) way, the first child can get any one of 15 toys, the second one can get can get remaining 14, the third 13 making it total of $15*14*13=2730$ which is very low as compared to other and I can't figure out why, as all children are counted for?Which cases are not counted here?

答:
$15 \times 14 \times 13$ would be the correct answer if each child only gets to pick one toy, and the other toys are left unclaimed.

Which is to say: the cases covered by the $2\,730$ answer aren't among the cases covered by the $14\,348\,907$ answer, and vice versa.  They count completely disjoint scenarios!

Actually -- I'm not sure $3^{15}$ is the correct answer either!  The children could all think that certain toys just aren't any fun and leave them behind, which would make the true result $4^{15} = 1\,073\,741\,824$, and the cases that covers include both the case in which every toy is claimed, and the case in which each child claims exactly one toy.

Title: Which cases are not counted here?


问:
Show that $\int_0^{2 \pi}\vert rie^{it}\vert dt=2 \pi r$
I am struggling with an normally simple integral calculation. I wanna show:
$$\int_0^{2 \pi}\vert rie^{it}\vert dt=2 \pi r$$
I tried the following:
$\int_0^{2\pi}\vert rie^{it}\vert dt=\vert ri \vert \int_0^{2\pi}\vert e^{it}\vert dt = $r$ *(\vert e^{(2 \pi i)} \vert- \vert e^{0} \vert)=0$...

Where is my mistake? Thanks for your help.

答:
$$|rie^{it}|=|r|\times|ie^{it}|=r\times1=r$$
so you have:
$$\int_0^{2\pi}rdt=2\pi r$$
this is because $e^{it}$ represents a unit circle in the complex domain, so the modulus i.e. the radius from origin of a complex point, is always 1


问:
Continuous function and induced homomorphism
We know that given any continuous map $f : X \to Y$, there exists an induced homomorphism $f_* : \pi_1(X, x) \to \pi_1(Y, f(x))$.

Is the converse true? That is, given any homomorphism $\phi: \pi_1(X, x) \to \pi_1(Y, y)$, is there a continuous map $f : X \to Y$ such that $\phi = f_*$ and $f(x) = y$?

If not, are there conditions for $X$ and $Y$ for the converse to be true?

答:
Here is a really simple example.

Take the real projective plane $\mathbb{R}P^2$ and attach a 3-cell to obtain $\mathbb{R}P^3$. For cellular reasons the inclusion $\mathbb{R}P^2\hookrightarrow\mathbb{R}P^3$ induces an isomorphism $\pi_1\mathbb{R}P^2\xrightarrow\cong\pi_1\mathbb{R}P^3$, both groups being isomorphic to $\mathbb{Z}_2$. However this isomorphism cannot be induced by a map $f:\mathbb{R}P^3\rightarrow\mathbb{R}P^2$ which points in the opposite direction.

For suppose that $f$ exists and is non-trivial on $\pi_1$. Then $f$ induces an isomorphism on $H_1$ by the Hurewitz Theorem, and so also an isomorphism on $H^1(-;\mathbb{Z}_2)$ by the Universal Coefficient Theorem. Now, if $x\in H^1(\mathbb{R}P^3;\mathbb{Z}_2)\cong\mathbb{Z}_2$ is a generator, then $x^3$ generates $H^3(\mathbb{R}P^3;\mathbb{Z}_2)\cong\mathbb{Z}_2$. But if $y\in H^1(\mathbb{R}P^2;\mathbb{Z}_2)$ is a generator, then $f^*y=x$ by the above, giving $x^3=(f^*y)^3=f^*(y^3)=0$ for dimensional reasons. This is a contradiction.


问:
How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$.
How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$ . Justify your answer.

Hello, could someone explain to me how to find the invertible elements of a set as large as $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$

I thought about putting together the set and trying each one but it is very big and it can take forever

答:
I will explain how to find the invertible elements of $\mathbb{Z}_n$.

An element $a$ is invertible in $\mathbb{Z}_n$ if and only if therte exists $b\in\mathbb{Z}_n$ such that $a\cdot b=1$, in other words $a\cdot b\equiv 1\pmod{n}$.

$0$ is clearly not invertible.

Observe that if $a$ is not relatively prime with $n$, then this is impossible, as $a\cdot b$ will always be divisible by $\gcd(a,n)$ so this would mean $\gcd(a,n)|1$.

So all elements not relatively prime woth $n$ are not invertible. What about the elements of $\mathbb{Z}_n$ which are relatively prime to $n$? Well let $a$ be one of them. Take the set $\{a,2a,3a,...,(n-1)a\}$. Suppose $ia\equiv ja$ for some $i$ and $j$. This means $a\cdot|i-j|$ is divisible by $n$ so because $a$ is relatively prime to $n$, then $|i-j|$ is divisible by $n$, but $n>|i=j|$ so $|i-j|=0$ so $i=j$.

Thus, all elements of $\{a,2a,3a,...,(n-1)a\}$ are distinct$\pmod{n}$ and there are no elements divisible by $n$ in the above set (because $a\neq 0$ and $a$ and $n$ are coprime), so in the above set there must be an element $\equiv 1\pmod{n}$, thus $a$ is invertible.

So the invertible elements in $\mathbb{Z}_n$ are all the numbers smaller than $n$, which are coprime to $n$. The number of such elements is $\varphi(n)$


答:
For $k$ to be invertible in $\mathbb Z_n$ means there is an integer $a$ so that $ak \equiv 1 \pmod n$.  And $ak\equiv 1\pmod n$ means there is an integer $b$ so that $ak = 1+bn$ which means $ak -bn = 1$.

By Bezouts Lemma that is possible if and only if $k,n$ are relatively prime.

So find all the numbers $k$ so that $k$ and $990$ are relatively prime.

As $990 = 2\cdot 3^2\cdot 5\cdot 11$ this will mean find all the integers less than $990$ that are not a mulitple of $2,3,5$ or $11$.

.....

There is a function that counts the number of elements that are relatively prime to $n$ and less than $n$.  It is called the Euler Totient function.

$\phi(n) = $#of integers, $k: k< n; \gcd(k,n) = 1$.

There are some formulas I won't prove but:

If $n = mp$ and $\gcd(m,p) = 1$ then $\phi(n) = \phi(m)\phi(n)$.

If $p$ is prime $\phi(p) = p-1$.

If $n =p^k$ and $p$ is prime $\phi(p^k)= p^{k-1}(p-1)$

From those we can conclude if If $n = \prod p_i^{k_i}$ then

$\phi(n) = \prod[p_i^{k_i-1}(p_k-1)] = n\prod\limits_{p\text{ prime};p|n}(1-\frac 1p)$.

So there are $\phi(990) = \phi(3^2)\phi(2)\phi(5)\phi(11) = 3^1\cdot 2\cdot 1\cdot 4\cdot 10 = 240$ such elements.

Or .... $990\prod (1-\frac 1p) = 990(1-\frac 12)(1-\frac 13)(1-\frac 15)(1-\frac 1{11})=$

$990\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$

$445\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$

$330\cdot\frac 45\cdot\frac {10}{11}=$

$264\cdot\frac {10}{11}= 260$

...or ....

$\require{cancel}$

$990\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$

$990\cdot \frac 1{\cancel 2}\cdot\frac {\cancel 2}3\cdot\frac 4{\cancel 5}\cdot\frac {\cancel{10}2}{11}=$

$990 \cdot \frac 8{33}=$

$30\cdot 8 = 240$.

Title: How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$.

Body:
How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$ . Justify your answer.

Hello, could someone explain to me how to find the invertible elements of a set as large as $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$

I thought about putting together the set and trying each one but it is very big and it can take forever


问:
If $\int_{0}^{1} f(x) \,dx=0$, then $f(x)=0$ for $0\leq x \leq 1$. True or False.
I need help to check if my analysis is right with exercise.

If $\int_{0}^{1} f(x) \,dx=0$, then $f(x)=0$ for $0\leq x \leq 1$. True or False.

I think this is true because if $f(x)=0$ for $0\leq x \leq 1$ then we have:

$$\int_{0}^{1} 0 \,dx = 0 (1-0) = 0(1)=0$$

I know that the defined integral could be $0$ as well if the limits of integration are equal $\int_{a}^{a} f(x) \,dx=0$ or if the limits of integration are equal but with opposite signs and if the function $f(x)$ is odd $\int_{-a}^{a} f(x) \,dx=0$. But the here the hypothesis says that the limits of integration are $1$ and $0$. So for me, this statement is true following the analysis I exposed before.

答: Actually, it is false: $\int_0^1x-\frac12\,\mathrm dx=0$. The fact that $\int_0^10\,\mathrm dx=0$ is irrelevant.


答:
Let $\varphi(n) = \{a\in \mathbb{Z}_n : a \text{ is invertible}\} = \{a\in \mathbb{Z}_n : \gcd(a,n) = 1\}.$

If $n=p_1^{e_1} \cdots p_s^{e_s}$ is the prime factorization of $n$, then (one can prove that) $$\varphi(n) = n\prod_{k=1}^{s}\Big(1-\dfrac{1}{p_k}\Big).$$

For $n=990,$ we have that $n=2\cdot3^2\cdot5\cdot11$, so $$\varphi(990)=990\cdot\Big(1-\dfrac{1}{2}\Big)\cdot\Big(1-\dfrac{1}{3}\Big)\cdot\Big(1-\dfrac{1}{5}\Big)\cdot\Big(1-\dfrac{1}{11}\Big)=240.$$


答:
If $f(x)$ is continuous, there exists an $x$ in the interval such that $f(x) = 0$ by the mean value theorem.

But, as stated the proposition is not true.

Body:
I need help to check if my analysis is right with exercise.

If $\int_{0}^{1} f(x) \,dx=0$, then $f(x)=0$ for $0\leq x \leq 1$. True or False.

I think this is true because if $f(x)=0$ for $0\leq x \leq 1$ then we have:

$$\int_{0}^{1} 0 \,dx = 0 (1-0) = 0(1)=0$$

I know that the defined integral could be $0$ as well if the limits of integration are equal $\int_{a}^{a} f(x) \,dx=0$ or if the limits of integration are equal but with opposite signs and if the function $f(x)$ is odd $\int_{-a}^{a} f(x) \,dx=0$. But the here the hypothesis says that the limits of integration are $1$ and $0$. So for me, this statement is true following the analysis I exposed before.


问:
Bounding sum of weighted exponential
I would like to upper bound $\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n)$, where $P_{i}$ are probability measure such that $\sum_{i}^{n} P_{i} = 1$ and $\beta > 0$ is a constant. So basically it's a convex combination, which I could easily upper bound with Hölder's inequality that $$\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq \max_{i} \exp(-\beta \cdot P_{i} \cdot n)$$

Is there any tighter upper bound for the sum like this ? Any idea or just reference would help !

答:
Since there is no other information available on the $P_i$ and on $\beta$, here is a modest approach where no one-size-fits-all bounds can be expected, rather bounds which fit special bodies.

First observe that OP's bound is actually tight if all $P_i = \frac{1}{n}$. Hence it only makes sense to look for improved bounds for a rather broad distribution of probabilities. In this case, denote with $P_{*}$  the smallest $P_i$,  we expect to have $P_*$ which is considerably lower than $\frac{1}{n}$.

Let us distinguish cases for $\beta$. Starting with very small $\beta$, we can use $e^{-x} < 1-x + x^2/2$ and obtain
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq 1 -\beta \cdot n\sum_{i}^{n} P_{i}^2 + \frac{\beta^2 n^2}{2}\sum_{i}^{n} P_{i}^3
$$
By Cauchy-Schwarz, $n\sum_{i}^{n} P_{i}^2 \ge 1$, and the obvious $\sum_{i}^{n} P_{i}^3 \le 1$, so we have
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq 1 -\beta + \frac{\beta^2 n^2}{2}
$$
Compare to OP's bound, and ignore terms of quadratic or higher order in $\beta$, we have indeed that
$$
 1 -\beta \leq 1 -\beta P_{*} n \simeq \exp(-\beta \cdot P_{*} \cdot n) = \max_{i} \exp(-\beta \cdot P_{i} \cdot n)
$$
since $ P_{*} < \frac{1}{n}$ was where we started. So this bound improves on OP's bound.

For not so small $\beta$, consider the following:

Consider the function $f(x) = x e^{- ax}$. It has a maximum of $\frac{e^{-1}}{a}$ at $x = \frac{1}{a}$. Let $a = \beta n$,  then if  $\beta > \frac{1}{n}$, this is a good upper bound. It is valid in any case, however depending on the $P_i$ if may not be attained.

So you have
$$
\sum_{i}^{n} P_{i} \cdot \exp(-\beta \cdot P_{i} \cdot n) \leq \sum_{i}^{n} \frac{e^{-1}}{\beta n} = \frac{e^{-1}}{\beta}
$$

We compare to OP's bound and ask how good this is. We require
$$
\frac{e^{-1}}{\beta} = \frac{e^{-1} P_{*} \cdot n}{\beta P_{*} \cdot n} \leq\!\! ?  \; \exp(-\beta \cdot P_{*} \cdot n) = \max_{i} \exp(-\beta \cdot P_{i} \cdot n)
$$
Understand $q = \beta \cdot P_{*} \cdot n$ as dependent on $\beta$ , then we require
$$
e^{-1} P_{*} \cdot n \leq\!\! ? \; q e^{-q} \le e^{-1}
$$
So this is favourable (since $P_{*} < \frac{1}{n}$)  if  $q$ of Order 1, i.e. $\beta \simeq  \frac{1}{ P_{*} \cdot n} > 1$.

If indeed $P_{*} << \frac{1}{n}$, then we have $q << 1$ and therefore approximately
$$
e^{-1} P_{*} \cdot n \leq\!\! ? \; q  = \beta \cdot P_{*} \cdot n
$$
i.e. $\frac{1}{P_{*} \cdot n} >> \beta > e^{-1}$ would then be the only requirement (where $\frac{1}{P_{*} \cdot n}$ is considered a very large number, so the left constraint is not severe).

From the discussion, it is also clear that the  ranges for $\beta$ obtained right now match the $\beta$-case which we just discuss.

Title: Bounding sum of weighted exponential


问:
Heat equation using separation of variables
I'm trying to solve a PDE using separation of variables.
The heat eq. is given by:
$$u_t = k(u_{rr}+ \frac{2}{r} u_r)$$
The boundary conditions are given by:
$$    \lim_{r \to 0} \frac{\partial u}{\partial r} = 0; \quad \text{for} \quad t>0. $$
and
$$    u(a,t) = u_{0}; \quad \text{for} \quad t>0. $$
where $u_0$ is a constant.
The initial condition is given by:
$$    u(r,0) = 0; \quad \text{for} \quad 0<r<a.$$.

Attempt at solution:
Let:  $u = R(r)T(t)$ be the solution to the PDE.
From this, we have:
$$\frac{\partial u}{\partial t} = R(r)T'(t)$$
$$\frac{\partial u}{\partial r} = R'(r)T(t)$$
$$\frac{\partial^2 u}{\partial r^2} = R''(r)T(t)$$
BY substituting back into the PDE, we have:
$$\frac{T'(t)}{T(t)}=k(\frac{R''(r)}{R(r)} + \frac{2}{r}\frac{R'(r)}{R(r)})= -\lambda^2; \quad \text{where} \quad \lambda > 0.$$
Here, $\lambda$ is the separation constant.
Then, we will get 2 ODEs given by:
$$T'(t)+ \lambda T(t)=0$$
$$k(rR''(r)+2R'(r)+\lambda rR(r))= 0$$
The solution to the first ODE is
$$ T = Ae^{- \lambda t}$$
How do I proceed from here and apply the BCs/ICs?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets $\ds{\varphi\pars{r,t} \equiv \on{u}\pars{r,t} - u_{0}}$ such that

$\left\{\begin{array}{rcl}
\ds{\varphi_{t}} & \ds{=} & \ds{k\pars{\varphi_{rr} + {2 \over r}\,\varphi_{r}}}
\\[2mm]
&&\left.\begin{array}{rcl}
\ds{\varphi_{r}\pars{0,t}} & \ds{=} & \ds{0}
\\[1mm]
\ds{\varphi\pars{a,t}} & \ds{=} & \ds{0}
\end{array}\right\}
\substack{\ds{Boundary} \\[0.5mm] \ds{Conditions}}
\\[2mm]
\ds{\varphi\pars{r,0}} & \ds{=} & \ds{-u_{0}}\
\pars{~Initial\ Condition~}
\end{array}\right.$

Set $\ds{\varphi\pars{r,t} = \on{A}\pars{t}\on{B}\pars{r}}$. Then,
\begin{align}
\on{A}_{t}\on{B} & =
k\pars{\on{A}\on{B}_{rr} + {2 \over r}\on{A}\on{B}_{r}}
\\[2mm]
{\on{A}_{t} \over \on{A}} & =
k\pars{{\on{B}_{rr} \over \on{B}}+ {2 \over r}{\on{B}_{r} \over \on{B}}} = -q^{2}\,,\quad q \in \mathbb{R}
\\[2mm]
\on{A} &= -q^{2}\on{A}\,,\quad
k\on{B}_{rr} + {2k \over r}\on{B}_{r} + q^{2}\on{B} = 0
\end{align}
Lets $\ds{\on{B} \propto {\expo{\alpha r} \over r}}$ which yields $\ds{\alpha = \pm \ic q/\root{k}}$. In order to satisfy the boundary condition at $\ds{r = 0}$, $\ds{\on{B} \propto {\sin\pars{qr/\root{k}}}/r}$. At $\ds{r = a}$, $\ds{\sin\pars{qa/\root{k}} = 0}$ $\ds{\implies q_{n} = n\,{\pi\root{k} \over a}\,,\quad n = 1,2,3,\ldots}$ The general solution becomes
\begin{align}
\varphi\pars{r,t} & =
\sum_{n = 1}^{\infty}\on{A}_{n}\pars{0}\exp\pars{-q_{n}^{2}\,t}
{\sin\pars{q_{n}r/\root{k}} \over r}
\\[5mm]
-u_{0} & =
\sum_{n = 1}^{\infty}\on{A}_{n}\pars{0}
{\sin\pars{q_{n}r/\root{k}} \over r}
\\[5mm]
&-u_{0}\
\overbrace{\int_{0}^{a}r\sin\pars{{q_{n} \over \root{k}}\,r}
\dd r}
^{\ds{-\,{\pars{-1}^{n}a^{2} \over n\pi}}}
\\[2mm] & =
\sum_{m = 1}^{\infty}\on{A}_{m}\pars{0}\
\underbrace{\int_{0}^{a}\sin\pars{{q_{m} \over \root{k}}\,r}
\sin\pars{{q_{m} \over \root{k}}\,r}\,\dd r}
_{\ds{{a \over 2}\,\delta_{mn}}}
\end{align}
$\ds{\implies \on{A}_{n}\pars{0} =
{2\pars{-1}^{n}au_{0} \over n\pi}}$.

Finally
$\ds{\pars{~\mbox{with}\
q_{n} = n\,{\pi\root{k} \over a}~}}$,
$$
\bbx{\on{u}\pars{r,t} = u_{0} +
{2au_{0} \over \pi}\,{1 \over r}\sum_{n = 1}^{\infty}
{\pars{-1}^{n} \over n}\exp\pars{-q_{n}^{2}\,t}
\sin\pars{{q_{n} \over \root{k}}\,r}}\\
$$


问:
Where are there stable equilibrium?
I am trying to find where there are stable of equilibrium points of this system such that $x,y \not= 0  $ in the region of the parameter space, with parameters $a$, $b$, and $c$ all being positive constants.

The system is:
$$x' = x(1-x)-\frac{axy}{x+c}$$
$$y'= by(1-\frac{y}{x})$$

I am only concerned in the first quadrant of the system, namely where $x,y > 0$.

My initial thoughts  were to find all the equilibrium points allowed by my conditions, linearize the system, then analyze the signs of the eigenvalues of the equilibrium points in the Jacobican matrix.
I started by trying to identify all the equilibrium points. Since $y \not= 0$, it is clear that the any equilibrium points must be $x=y$. Yet, when I try to solve $x'=0$, it becomes very messy, and plugging in my values for $x$ back into the Jacobian matrix would make it even worse.

It started to become clear to me that I must not be doing something correctly, or that I am missing something important. Is there a better and more practice way to find where in $abc$-space I get stable equilibrium points? Maybe by finding a Liapunov function?

答:
As soon as you have $y=x$, any equilibrium satisfying your requirements is a positive solution of $(1-x)(x+c)-ax=0$, that is the quadratic equation $x^2+(a+c-1)x-c=0$. Since $c>0$, it follows that you have one positive root. You can write this root in terms of $a$ and $c$. Then you compute the Jacobian of your system,
$$J = \begin{bmatrix} \frac{a\,x\,y}{{\left(c+x\right)}^2}-\frac{a\,y}{c+x}-2\,x+1 & -\frac{a\,x}{c+x}\\ \frac{b\,y^2}{x^2} & \frac{b\,\left(x-2\,y\right)}{x} \end{bmatrix},$$
and substitute the obtained solution as $x$ and $y$. You obtain the linearization $2\times 2$ matrix where all elements are functions of $a$, $b$, and $c$. Then you compute its deterministic polynomial and check when all coefficients are positive. That gives you the desired stability condition. It looks a bit messy to put the equations here, but a CAS should help.

Title: Where are there stable equilibrium?


答:
There are simpler ways to prove this relation.

The sum $\displaystyle S(n)=\sum\limits_{i=0}^n i=\frac{n(n+1)}{2}$ is well known and easy to prove, for instance

$\begin{align}2S(n)&=\bigg(1+2+\cdots+(n-1)+n\bigg)+\bigg(n+(n-1)+\cdots+2+1\bigg)\\&=\bigg((n+1)+(n+1)+\cdots+(n+1)\bigg)\\&=n(n+1)\end{align}$

Then notice $(a-1)a+1$ does not depend on the summation index $i$, thus we can factor it out of the sum and multiply by the number of terms.

$\begin{align}\sum\limits_{i=0}^{a-1}((a-1)a+1+2i)&=((a-1)a+1)\times a+2\sum\limits_{i=0}^{a-1}i\\&=(a^3-a^2+a)+2S(a-1)\\&=(a^3-a^2+a)+(a-1)a\\&=a^3\end{align}$

Now to come back to your solution, it is a bit confusing because in (2) you start from your result, then go back to your starting point and try to match both...

In any case, this method involves also factorizing constant values out of the sum so the previous method is more straightforward in this regard.

$\begin{align}(a+1)^3&=(a^3)+(3a^2+3a+1)
\\&=\sum\limits_{i=0}^{a-1}((a-1)a+1+2i)+(3a^2+3a+1)&\text{induction hypothesis P(a)}
\\&=\sum\limits_{i=0}^{a-1}((a+1)a+1+2i)-\underbrace{\sum\limits_{i=0}^{a-1}(2a)}_{-2a^2}+(3a^2+3a+1)&\text{shift formula by 2a, remove the excess}
\\&=\sum\limits_{i=0}^{a-1}((a+1)a+1+2i)+(a^2+3a+1)&\text{simplify constant sum}
\\&=\sum\limits_{i=0}^{a}((a+1)a+1+2i)-\underbrace{((a+1)a+1+2a)}_{a^2+3a+1}+(a^2+3a+1)&\text{add/remove last term of sum}
\\&=\sum\limits_{i=0}^{a}((a+1)a+1+2i)&\text{after cancelling terms P(a+1) is verified}\end{align}$

You pretty much got all of these ideas, so mostly it is a question of organizing your calculations in a way that is less prone to mistakes (negative sign forgotten or missed term, ...).


问:
How do I integrate this function? I've tried every method I can think of, help!
I want to calculate the value of the integral $\int_{40}^{64} y^2(100y-4000)^{1/2} dy$

Body:
$$y^2(100y-4000)^{1/2}dy=10y^2(y-40)^{1/2}dy$$
now you could try letting $u=y-40,du=dy$ so:
$$10(u+40)^2u^{1/2}du=10u^{1/2}(u^2+80u+40^2)$$
so your integral becomes:
$$10\int_0^{24}\left(u^{5/2}+80u^{3/2}+40^2u^{1/2}\right)du$$
now this should be easier


问:
Discuss Series of funtion is Uniformly convergent or not
Discuss the uniform convergence of this series over $\mathbb R$: $$\sum_{n=1}^\infty\frac{(-1)^n x^n}{n(1+x^n)}.$$
For $0<x\leq 1$, $\frac{x^n}{n(1+x^n)}$ is monotonically decreasing and coverges to zero as $n$ tends to $\infty$.

Similarly for $x > 1$, $\frac{x^n}{n(1+x^n)}$ monotonically decreses for $x$ and converges to zero.

In both case they are bounded uniformly between $0$ and $1$, and $\sum_n{(-1)^n}$ oscillates finately between $0,1$ as $n$ is even or odd, by Abel Test we conclude that series is Uniformly convergence over $(0,+\infty)$.

Doubt what about when $x \leq 0$, I am confused as I am getting alternating series i.e. $$\frac{-x}{1+x}$ + $\frac{x^2}{2(1+x^2)}$-$\frac{x^3}{3(1+x^³)}\dots$$
Here as $x$ tends to $-1$, denominator turns $0$. Please help.

答:
The series is not defined at $x = -1$. Furthermore we can show that it even fails to converge uniformly on $(-1,0]$ when this point is excluded.

Let $y = -x$ for $x \in (-1,0]$ and consider the equivalent series $\displaystyle \sum_{n=1}^\infty\frac{y^n}{n(1+(-1)^ny^n)}$ for $y \in [0,1)$.

We have $y_n = 1 - 1/n \in [0,1)$ for all $n$ and

$$\sup_{y \in [0,1)}\left|\sum_{k=n+1}^\infty\frac{y^k}{k(1+(-1)^ky^k)} \right| \geqslant \sup_{y \in [0,1)}\sum_{k=n+1}^{2n}\frac{y^k}{k(1+(-1)^ky^k)} \geqslant \sup_{y \in [0,1)}n \cdot \frac{y^{2n}}{2n(1+y^n)} \\ \geqslant n \cdot \frac{(1-1/n)^{2n}}{2n(1+(1-1/n)^n)} =  \frac{(1-1/n)^{2n}}{2(1+(1-1/n)^n)} \underset{n \to \infty}\longrightarrow \frac{e^{-2}}{2(1+e^{-1})} \neq 0,$$

Thus, the series fails to converge uniformly by violation of  the necessary condition

$$\lim_{n \to \infty}\sup_{y \in [0,1)}\left|\sum_{k=n+1}^\infty\frac{y^n}{k(1+(-1)^ky^k)} \right|= 0$$


答:
By the Fundamental Theorem of Calculus, if $f$ is continuous and defined on $[0,1]$, then defining $F$ by,

$$F(x)=\int_0^xf(s)ds$$

we have that $F'(x)=f(x)$, for all $x$ in $(0,1)$. As a consequence,

$$\int_0^1f(s)ds=F(1)-F(0)$$.

If we take $F(1)=F(0)$ we will have $\int_0^1f(s)ds=0$. For instance, take $F(x)=x(x-1)$, then $F(0)=F(1)=0$ and $f(x)=F'(x)=2x-1$. In this case, $f$ is not null on $[0,1]$.

Title: If $\int_{0}^{1} f(x) \,dx=0$, then $f(x)=0$ for $0\leq x \leq 1$. True or False.


问:
Eigenvalues and eigenvectors linear algebra conceptual questions

If $A$ is a square matrix and $P$ is an invertible matrix each of whose columns is an eigenvector of $A$, then: $P^{-1}AP$ is a diagonal matrix. True or false.

I am not sure of this question but I think that it must be could be false because if $P^{-1}AP = D$, then $AP = PD$, and wouldn't these yield different results?

Let $T: V \to V$ be a linear transformation and let $B$ be a basis for $V$. Then $T$ and $[T]B$ have the same eigenvalues, but may have different vectors. True or False

I believe this answer is true because $T$ relative to the basis $B$ for $V$ shouldn't have different eigenvalues than $T$ right?

A square matrix $A$ is diagonalizable if and only if for each eigenvalue $c$ of $A$ the algebraic multiplicity of $c$ is equal to the dimension of the eigenspace of $A$ corresponding to $c$. True or False.

I believe this answer to be true because if the multiplicity of $c$ is one, then there would be one eigenvector for $c$, and therefore the eigenspace of $A$ would have the same dimension as $c$ which is one correct?

The dimension of the eigenspace of a matrix $A$ corresponding to an eigenvalue $c$ is equal to the rank of $A - cI$. True or false

This one I am stumped on, I believe it is true but I am not sure

true or false. A square matrix $A$ is singular if and only if $0$ is an eigenvalue of $A$.

I know that this is true if there is a determinant of $0$, but is it also for an eigenvalue? I believe it is but wanted to check.

Please answer any or all of these questions, any help is very appreciated.

答:
Question 1: The statement is true. As you said, $P^{-1}AP = D$ is equivalent to
$$
AP = PD.
$$
If $D$ is diagonal, then what is the relationship between the columns of $P$ and the columns of $PD$? What is the relationship between the columns of $P$ and the columns of $AP$?

Question 2: Indeed, the statement is true.

Question 3: Indeed, the statement is true. However, your justification for the statement doesn't make any sense.

Question 4: The statement is false. The eigenspace is the nullspace of $A - cI$, but the rank of $A - cI$ is the dimension of the column space of $A - cI$. For example, with $c = 1$ and the diagonal matrix
$$
A = \pmatrix{1\\&2\\&&3},
$$
we find that the rank of $A - cI$ is $2$, but the dimension of the associated eigenspace is $1$.

Question 5: Indeed, the statement is true.


问:
Find the Maclaurin series of $f(x) = x^7\sin(4x^3)$
Find the Maclaurin series of $f(x) = x^7\sin(4x^3)$

Can you folks show me some different ways to do this one? Is the best way to do it just to start taking derivatives of $f(x)$, plugging in zero and using the general formula? I keep getting this wrong. The answer is supposed to be:

$$\sum_{n=0}^\infty \frac{(-1)^n4^{2n+1}x^7x^{6n+3}}{(2n+1)!}$$

Body: No, that's not the best way. The best way is\begin{align}x^7\sin(4x^3)&=x^7\sum_{n=0}^\infty\frac{(-1)^n(4x^3)^{2n+1}}{(2n+1)!}\\&=x^7\sum_{n=0}^\infty\frac{(-1)^n4^{2n+1}x^{6n+3}}{(2n+1)!}\\&=\sum_{n=0}^\infty\frac{(-1)^n4^{2n+1}x^{6n+10}}{(2n+1)!}.\end{align}


问:
Uniqueness of Pseudo inverse
Given $X\succeq0$, we can write the SVD decomposition as:
\begin{align}
X&= U \begin{bmatrix}\Omega&0\\0&0 \end{bmatrix}U^T,
\end{align}
where $U$ is an orthogonal matrix and $\Omega\succ0$.

My question is whether
\begin{align}
Y&= U \begin{bmatrix}\Omega^{-1} &0\\0&0 \end{bmatrix}U^T,
\end{align}
is the unique Moore-Penrose pseudo inverse of $X$.

答:
I assume that you are referring to the Moore-Penrose peudoinverse and that $\Omega \succ 0$ means that $\Omega$ is positive definite, which implies that $\Omega$ is symmetric.

Yes, your formula for the psuedoinverse is correct, and (as is generally the case) the MP pseudoinverse is unique.

Title: Uniqueness of Pseudo inverse


问:
Prove that the existence of a bridge is an invariant
An invariant is a property $P$ that is shared by all isomorphic graphs. In other words, a property $P$ is an invariant provided that whenever $G_1$ and $G_2$ are isomorphic graphs, if $G_1$ satisfies $P$, then $G_2$ also satisfies $P$.

A bridge is an edge $e$ whose removal yields a disconnected graph.

答:
Illustration:  Below I generated a random connected graph (the blue graph) and highlighted it's bridges in orange.  Then I relabeled the vertices of the graph to form the red graph below highlighted it's bridges in orange too.

It's quite clear in the above example that bridges map to bridges and non-bridges map to non-bridges.  The graph's structure has not changed.  This is all isomorphism is, relabelling the vertices of the graph.

To prove that the existence of a bridge is a graph invariant formally, we should:

Let $G=(V,E)$ be an arbitrary graph with a bridge $ij \in E$.  Let $\varphi$ denote a bijection with domain $V$.  Show that $\varphi(i)\varphi(j)$ is a bridge in $\varphi(G)$.

Repeat the above for when $ij$ is a non-bridge.

Note: this proves the stronger (but unsurprising) result that the number of bridges is preserved under isomorphism.


问:
About absolute continuity of sigma finite measures
Let $\mu_1,\mu_2$ be sigma finite measures. Prove that $\mu_1\ll \mu_2$ and $\mu_2\ll \mu_1$ $\iff \exists f$, where $f$ is strictly positive such that $\nu(A)=\int_A fd\mu$ for all measurable $A$.

It is the strictly positive nature of $f$ which I am having trouble with. Let $\mu_1\ll \mu_2$ and $\mu_2\ll \mu_1$. I know that there exist $f_1,f_2\geq 0$ s.t. $\nu(A)=\int_A f_1d\mu$ and $\mu(A)=\int_A f_2d\nu$ but am not sure how that would lead me to a strictly positive density function. Any suggestions?

答:
By Radon Nikodym there is a nonnegative function $f$ so that $\nu(A)=\int_A f d\mu$ for all measurable $A$. Consider the measurable set $S=\{x:f(x)=0\}$. Then $\nu(S)=\int_S f(x) d\mu=0$. Therefore $\mu(S)=0$.

Define $$\tilde f(x)=\begin{cases}f(x)&\text{ if } x\not \in S\\
24 &\text{ if }x\in S.\end{cases}$$

Then for any measurable $A$ we have that  $$\nu(A)=\int_A f(x)d\mu=\int_A \tilde f(x)d\mu,$$

because changing a function on a set of measure zero doesn't change integral.


问:
Gluing vector bundles on irreducible components
Here is a question that I thought it was obvious but more I thought about it less obvious it became. Given vector bundles on irreducible components of a scheme that match up on intersections, is it possible to glue it to a vector bundle on the whole scheme? A simple example is like this. Given a projective $R[s,t]/(s)$ module $P_1$ and a projective $R[s,t]/(t)$ module $P_2$ that are isomorphic at $t=s=0$. How to construct a projective module on $R[s,t]/(st)$ as a gluing?

答:
Let $P_0 = P_1 \otimes_{R[t]} R = P_2 \otimes_{R[s]} R$. Then define $P$ from the exact sequence
$$
0 \to P \to P_1 \oplus P_2 \to P_0 \to 0,
$$
where $P_i$ are considered as $R[s,t]/(st)$-modules via the natural morphisms $R[s,t]/(st) \to R[s,t]/(s)$, $R[s,t]/(st) \to R[s,t]/(t)$ and $R[s,t]/(st) \to R[s,t]/(s,t)$. Then $P$ is projective.

Indeed, the claim is local, so we may assume that $P_1$ and $P_2$ (and hence $P_0$ as well) are free. Then the defining sequence of $P$ is isomorphic to the direct sum of several copies of the standard exact sequence
$$
0 \to R[s,t]/(st) \to R[s,t]/(s) \oplus R[s,t]/(t) \to R[s,t]/(s,t) \to 0,
$$
hence $P$ is a direct sum of free modules.

The case of a more general gluing is completely analogous.


问:
Where did I go wrong differentiating arccos(4/x)?
I've been trying to differentiate $y=\arccos(4/x)$ and ended up with the answer $\frac{dy}{dx} = \frac{4}{x\sqrt{x^2-16}}$ however the solution to this question is marked as $\frac{dy}{dx}= \frac{4}{|x|\sqrt{x^2-16}}$. I opted to use implicit differentiation and honestly have no idea where the $|x|$ could of come from. Any help would be greatly appreciated.

答:
The given function $y=\arccos\dfrac{4}{x}$ is not defined as real in $|x|<4$

$$ \text{for } x> 4 , y'=\frac{4}{x\sqrt{x^2-16}},$$
$$ \text{for } x< 4 , y'= \pi - \frac{4}{|x|\sqrt{x^2-16}}.$$


问:
Find values of $a$ for which the integral converges
For my homework, I need to find the values of $a$ for which the following integral converges.  $$\int_0^{1/e}\frac{x^a}{\ln{x}} \,dx$$
I've tried using integration by parts, but only got (seemingly) more complex integrands of of it. It'd be a major help if anyone could give me a hint.

答:
With the substitution $$t=-\ln(x)\;\; or\;\; x=e^{-t}$$

your integral will have the same nature as

$$I=\int_1^{+\infty}f(t)dt=\int_1^{+\infty}\frac{e^{-(1+a)t}}{t}dt$$

$$ (1+a)<0 \implies \lim_{t\to+\infty}f(t)=+\infty$$
$$\implies I\; diverges$$

$$(1+a)=0 \implies f(t)=\frac 1t$$
$$\implies I \; diverges$$

$$(1+a)>0\implies \lim_{t\to+\infty}t^2f(t)=0$$
$$\implies I \;\; converges$$

We conclude that $$\boxed{I\;\; converges \;\;\iff a>-1}$$


问:
Is it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?
Help would be very appreciated.

It it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?

答:
Since $ \dim \operatorname{ker} T + \dim \operatorname{im} T = n$, we must have $n=2k$.

Now, take any basis of $V$, say $v_i$

Define $T(v_i) = v_{k+i}$ for $i=1$ to $k$, and $T(v_i) = 0 $ for $i=k+1 $ to $2k$. Extend this out as a linear functional to $V$. Then , $\operatorname{im} V = \operatorname{ker} V$.


问:
Find median of a trapezoid if diagonals, which intersect at right angles, are 3 and 4.
I wrote down similar triangles thing but don't know what to do next. Any help would be appreciated.</p>

答:
What we are given.

$AC = 4\\
BD = 3\\
AC\perp BD$

$\triangle ABE\sim \triangle CDE$

$AE = \alpha EC\\
BE = \alpha ED\\
AB = \alpha CD$

Because the similar triangles share proportionality.

$AE+EC = AC = 4\\
\alpha EC +EC = 4\\
(1+\alpha) EC = 4$

Similarly, $(1+\alpha) ED = 3$

By Pythagoras $EC^2 + ED^2 = CD^2$

This is enough information to find $CD$ in terms of $\alpha.$  And by the similarity rules find $AB$

The median equals $\frac {1}{2} (AB + CD).$

Hopefully, the $\alpha$'s cancel.

Or, you could consider the degenerate case where $\alpha = 0$ and then decide if this indeed describes the general case.


问:
How is this true in case of proportionality?
If $a \propto t$ (directly proportional), then $a = Kt$.

If $a = 6Kt$ , then also $a \propto t$.

Why are these two results same?

Displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained is?

This is the question from where my doubt arose.

答:
If $a$ is directly proportional to $t$, then  $A = Kt$, for some constant $K$ (which we call the constant of proportionality).

If $a = 6t$, then observe that $a$ is directly proportional to $t$, with $K = 6$.

[Perhaps you are confused as to how mathematical definitions work? The if in a definition is typically an if and only if, which can certainly be confusing at first!
]


问:
In a compact metric space, a sequence with a certain property is convergent
The question:

Let $(K,d)$ be a compact metric space, if $\{x_n\}$ is a sequence such that $\{d(x_n,x)\}$ converges for all $x \in K$, then $\{x_n\}$ is a convergent sequence.

My attempt:

Notice that we can define for each $n$, $f_n(x)=d(x_n,x)$, this is an isometry and thus these are equicontinuous functions.

Also, each converging sequence gives us the fact that these functions are pointwise bounded.

Thus we can us Arzela-Ascoli Therom to show that in fact the sequence of functions $\{f_n\}$ has a unifromly convergent subsequence.

Now if I can show that their limit is indeed of the form $f(x)=d(a,x)$ for some $a\in K$, then I can show that $a$ is the limit of the sequence, but I don't know how to do it or even if it is true.

答:
If $(x_n)_{n\in\Bbb N}$ diverges, it has subsequences $(x_{n_k})_{k\in\Bbb N}$ and $(x_{m_k})_{k\in\Bbb N}$ with distinct limits, $l_1$ and $l_2$ respectively. The fact that it has a convergent subsequence whose limit is some $l_1$ follows from the compacity of $K$. Since the sequence diverges, there is some $\varepsilon>0$ such $d(x_k,l_1)\geqslant\varepsilon$ for infinitely many $k$'s. And from those $x_k$'s you can extract another convergent subsequence, again by the compacity of $K$.

But then the sequence $\bigl(d(x_n,l_1)\bigr)_{n\in\Bbb N}$ diverges, since it has a subsequence whose limit is $0$ and another subsequence whose limit is $d(l_1,l_2)$.


问:
Limit of an expression with exponential function
I was trying to understand how we can approximate exp.

One example is:

$$ \exp(t) = \sum_{i=0}^\infty t^i/i! $$

however, why is the following true:

$$\lim_{x\to \infty}\exp \left ({\frac{t^2}{2!} +\frac{t^3}{3!\cdot x^{.5}} + \frac{t^4}{4!\cdot x^{1.5}} + \dots } \right) = \exp(\frac{t^2}{2})$$

答:
(Hint) First, as experimentX suggest, take the log of both sides to get rid of $\exp$ (you can pull $\exp$ out of the limit because $\exp$ is a continuous function).  Now since $t$ is fixed, you can treat it as constant in the limit.
What happens to a term like $\displaystyle \frac{t^3}{3!x^{.5}}$ as $x \to \infty$?  </p>


问:
$\lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$
$\lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$ so i've thought of simplifying into smaller fractions as such $$\lim_{n\to \infty}(\frac{1}{n^2(n+1)}+\frac{2^2\sqrt{2}}{n^2(n+1)}+\dots +\frac{n^2\sqrt[n]{n}}{n^2(n+1)})$$ and as all of the terms we  get k/something that approaches infinity, most of the terms approach 0 andon the last one the n^2 cancel out, and we still get all terms to be 0, so the answer is 0? is this logic right? i doubt it's that easy

答: @BenGrossmann's comment explains why this doesn't work. Instead use Stolz–Cesàro to write your limit as$$\lim_{n\to\infty}\frac{n^{2+1/n}}{n(3n-1)}=\frac13\color{blue}{\lim_{n\to\infty}n^{1/n}}.$$The blue limit is famously $1$ (see e.g here), making the original limit $\tfrac13$.


问:
Having trouble with showing a basic property of the gradient of a function.
Consider functions of the type

$$f(ax) = a^k f(x)$$

where $f: \mathbb{R}^n \to \mathbb{R}$

A simple example is $f(x) = x^k$.

Then I claim,

$$\nabla f(ax)^T(a \delta_i) = a^k \dfrac{\partial f}{\partial x_i}
 (x)$$

where $\delta_i$ is the vector of all zeroes except for $1$ at the ith position.

Cannot show this. I always have a $a^{k+1}$ on the right hand side. I suspect there might be a typo? Can someone please verify.

答:
There is no typo. For your example you get $\nabla f(x) = k x^{k-1}$, so $\nabla f(ax)a = k(ax)^{k-1}a = a^{k}(kx^{k-1}) = a^k \nabla f(x)$

You apparently made the mistake to go from $$f(ax) = a^k f(x)$$ to$$\nabla f(ax) = a^k \nabla f(x).$$
That is not true, because you think that you take  the derivative with respect to $x$ whereas you actually take the derivative w.r.t. $ax$ on the left hand side. To compute $\frac{d}{dx} f(ax)$ you need the chain rule:
$f(ax) = f(h(x))$ with $h(x) = ax$, so $\frac{df(ax)}{dx} = \frac{df(h(x))}{dh}\frac{dh}{dx} = \nabla f \cdot a$.


问:
The definition of $\wedge$ using $\lor$ and $\neg$
We have that $\neg(A\wedge B)\iff \neg A\lor\neg B.$ How can I get the definition of $\wedge$ using $\lor$ and $\neg?$

I am new to logic and don't know how to start. Any help would be appreciated. Thank you in advance!

答: $A \land B = \lnot(\lnot A \lor\lnot B)$


问:
In a queue of $n$ people, what is the average number of spaces in between two people?

There are $n$ people in a queue. You randomly select two of them. What is the expected number of people standing in between them?

I noticed that the question can be reformulated in the following way. Let $X$ and $Y$ be discrete random variables that can take values between $1$ and $n$. What is $\mathbb{E}(|X-Y|-1 \text{ given that $X\neq Y$})$?

Unfortunately, this didn't seem to get me anywhere. I also considered what the probability was for small values of $n$. Let $Z$ be the number of people standing in between the two people. For $n=2$, $Z=0$ trivially. For $n=3$, there is a two-thirds chance that the people are standing next to each other, and a one-third chance that they are on either side of the queue. So $E(Z)=\frac{2}{3}\cdot0+\frac{1}{3}\cdot1=\frac{1}{3}$. I also calculated it for $n=4$, but stopped after that point since the calculations were getting unwieldy.

Could anyone hint at what approach I could use instead?

答:
The sum of the number of spaces between two people is
$$\sum_{i=1}^{n-1} \sum_{j=i+1}^n (j-i-1) = \sum_{i=1}^{n-1} \binom{n-i}{2} = \sum_{k=0}^{n-2} \binom{k+1}{2} = \binom{n}{3},$$
so the average is
$$\frac{\binom{n}{3}}{\binom{n}{2}} = \frac{n-2}{3}.$$


问:
Pigeonhole Principle ( 65 particles, 2x2x2 box, distance between 2 no greater than 1.7)
Question:
Imagine a small box with 2x2x2 dimensions which contains 65 particles floating around. Show that at any point, there are at least 9 particles with a distance no larger than 1.7 between any 2 of them.

My attempt:
I understand I must divide the box into 64 even sectors and place a particle in each.
Using the pigeonhole principle,

pigeons: particles

pigeonholes: sectors

I am just struggling to prove that there are 9 particles (with distance between any two) no greater than 1.7 apart.

Any help would be appreciated.

答: Hint: If you take $8$ unit cubes within the cube of dimension $2 \times 2 \times 2$ and take $8$ groups of $8$ particles, the maximum distance that you can maintain between each of those groups is $\sqrt6$ and no two subgroups of those $8$ particles can be farther than $\sqrt3$ from each another without violating distance of $\sqrt3$ with another group. Now the $65$th particle will be at maximum distance of $\sqrt3$ at least from one of the groups. You still need to fill in the gaps with your working.


问:
Help Understanding This Linear Algebra
I am working through a paper here and trying to understand all of the linear algebra involved, but I get stuck going from equations 5-6 to equation 7. I will skip the bold matrix notation because everything is a matrix/vector

Equation 5

$$
p(y|X, \bar{X}, \bar{f}) = \mathcal{N}(y|K_{NM}K_{M}^{-1}\bar{f}, \Lambda + \sigma^2I)
$$

Equation 6

$$
p(\bar{f}|\bar{X}) = \mathcal{N}(\bar{f}|0, K_{M})
$$

Equation 7

The paper says "we find the posterior distribution over pseudo targets $\bar{f}$ using Bayes' rule on (5) and (6)."

$$
p(\bar{f}|\mathcal{D}, \bar{X}) = \mathcal{N}(\bar{f}|K_MQ^{-1}_MK_{MN}(\Lambda + \sigma^2I)^{-1}y, K_MQ_M^{-1}K_M)
$$

where

$$
Q_M = K_M + K_{MN}(\Lambda + \sigma^2I)^{-1}K_{MN}
$$

By saying that they used Bayes' rule, they are saying that they are doing the following...

$$
p(\bar{f}|\mathcal{D}, \bar{X}) = \frac{p(y|X, \bar{X}, \bar{f})p(\bar{f}|\bar{X})}{p(y)}
$$

But I do not see where they can get the denominator from as it is not given. I tried multiplying the likelihood by the prior in the numerator using these Gaussian identities, but I go something totally different, not even close to the definition of $p(\bar{f}|\mathcal{D}, \bar{X})$. How can you combine the equations 5 and 6 with Bayes' rule to get equation 7?

EDIT: WIP to get details final derivation correct

Point 1

I am now very close to getting the full thing, but there is one part that is eluding me, which is probably something trivial. For one, the proportions have different x values, so I think there needs to be some subscript to set the variables apart, like so...

$$ \exp(a(x_1-b)^2) \exp(cx_2^2) \propto \exp((a+c)x_2^2 - 2abx_1).$$

which would mean that $1 \rightarrow 2$ could be represented as...

$$
\begin{align}
 & \propto \exp(-\frac{1}{2} (y - L \mu)^T \Sigma^{-1}(y - L \mu))\exp(-\frac{1}{2}\mu^T K^{-1} \mu) \\
&\propto \exp(-\frac{1}{2}(\underbrace{-2\mu^T L^T \Sigma^{-1}y + \mu^T(\Sigma^{-1}+K^{-1})\mu}_\text{Terms that do not include $\mu$ are factorized away})) \\
\end{align}
$$

if it is the case that...

$$
\begin{aligned}
a &= \Sigma^{-1} \\
b &= \mu^TL^T \\
c &= K^{-1} \\
x_1 &= y \\
x_2 &= \mu \\
\end{aligned}
$$

But I do not see how we would end up with $L^T\Sigma^{-1}L + K^{-1}$ inside of the parentheses at the end.

Point 2

I think the final line of the derivation is missing an inverse at the end, is it correct?

$$
\propto \exp(-\frac{1}{2}\left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T \Sigma^{-1} y\right)^T (L^T\Sigma^{-1}L+K^{-1})^{\color{red}{-1}} \left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y\right)).
$$

答:
Preamble

I shall use the common Bayesian statistics trick used in figuring out the posteriors - work with PDF kernels and figure out the normalization constants later. I will also make an intensive use of the proportionality arithmetic, $\propto$, which simplifies the derivations.

The main logic of $\propto$ relation is expressed as:

$$ C_1\exp(C_2 + h(x)) \propto C_3\exp(C_4 + h(x)) \propto \exp(h(x))$$

for any constants $C_i$ and $h(\cdot)$ is some function of $x$.

Obviously $f(x) \propto h(x)$ implies $h(x) \propto f(x)$.

The following relations are relatively not obvious but will be useful below:

$$ \exp(a(x-b)^2) \propto \exp(ax^2 - 2ab x) \propto \exp(ax(x-2b)),$$
$$ \exp(a(x-b)^2) \exp(cx^2) \propto \exp((a+c)x^2 - 2ab x).$$

Tedious algebra

So, we have normally distributed data $y|\mu \sim \mathcal{N}(L \mu, \Sigma), \mu\in\mathbb{R}^n$ and normally distributed mean parameter $\mu \sim \mathcal{N}(0, K)$. The posterior density of $\mu$ is proportional to the product of PDF kernels of the two:

\begin{align}
  \overline{p}(\mu|y) \quad & \propto \exp(-\frac{1}{2} (y - L \mu)^T \Sigma^{-1}(y - L \mu))\exp(-\frac{1}{2}\mu^T K^{-1} \mu) \\
&\propto \exp(-\frac{1}{2}(\underbrace{-2\mu^T L^T \Sigma^{-1}y + \mu^T(L^T \Sigma^{-1}L+K^{-1})\mu}_\text{Terms that do not include $\mu$ are factorized away})) \\
&\propto \exp(-\frac{1}{2} \underbrace{\mu^T(L^T\Sigma^{-1}L+K^{-1})}_\text{Factorized inside $\exp$} \left(-2 (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y + \mu\right))\\
&\propto \exp(-\frac{1}{2}\left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T \Sigma^{-1} y\right)^T (L^T\Sigma^{-1}L+K^{-1}) \left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y\right)).
\end{align}

Careful examination of the last line reveals the PDF kernel of a normal rv

$$\mu|y \sim \mathcal{N}((L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y, (L^T\Sigma^{-1}L+K^{-1})^{-1}). $$

Using the Woodbury identity one can rewrite the posterior covariance with an eye on simplifying it in the context of the paper:

$$
(L^T\Sigma^{-1}L+K^{-1})^{-1} = K - K L^T(\Sigma + L K L^T)^{-1} L K
$$

In the context of the paper,

\begin{align}
\mu &= \overline{f}, \\  \Sigma &= \Lambda + \sigma^2 I, \\  K &= K_M,\\
L &= K_{NM}K_M^{-1}, \\ KL^T &= K_{MN}, \\  L K L^T &= K_{NM} K_M^{-1} K_{MN},
\end{align}

and the posterior covariance matrix equals

\begin{align}
&K_M - K_{MN} (\Lambda + \sigma^2I + K_{NM} K_M^{-1} K_{MN} )^{-1} K_{NM} \\
=& K_M (K_M^{-1} - K_M^{-1} K_{MN} (\Lambda + \sigma^2I + K_{NM} K_M^{-1} K_{MN} )^{-1} K_{NM} K_M^{-1} ) K_M\\
=& K_M (K_M + K_{MN} (\Lambda + \sigma^2 I)^{-1}K_{NM})^{-1} K_M \\
= & K_M Q_M^{-1}K_M.
\end{align}

(Woodbury identity is used again to get the third line.)

The mean of the posterior is obtained in a similar fashion.

Title: Help Understanding This Linear Algebra


问:
Computing the limit of an integral of a function series
I am trying to figure out how to compute the limit

$$\lim_{t \rightarrow \infty} \int_{-1}^1 \frac{\cos^2(t^3x^{10})}{tx^2 + 1} \, dx.$$

If I exchange the limit and integral (when is this allowed?) then I get the integral of 0.  This seems too easy.  What should I be looking for?

Body:
Note that

$$\left|\int_{-1}^1 \frac{\cos^2(t^3x^{10})}{tx^2 + 1} \, dx\right|\leqslant\int_{-1}^1 \frac{|\cos^2(t^3x^{10})|}{tx^2 + 1} \, dx \leqslant 2 \int_0^1\frac{1}{tx^2 + 1} \, dx = \frac{2}{\sqrt{t}}\int_0^{\sqrt{t}} \frac{du}{1+u^2}$$

Try to finish from here.


问:
Math school online?
As a hobbyist programmer with little to no math skills (no science skills either, thank you religious schools) I am finding math encroaching on my life more and more often. I have always laughed and said that I'm just not good at math because my brain doesn't work like that.

My younger brother challenged me recently claiming that it wasn't a lack of ability that kept higher math out of my hands but a lack of interest. Because my current realm of studies is really starting to delve into 3D modeling, programming for web, mobile and desktop, I feel that a stronger math skill set would benefit me immensely.

But I don't know where to start...

I'm wondering if there is a math version similar to: http://www.code.org/khan-academy

Some place where I can go and start challenging my brain, increasing my math as often as I have time? Some place online? I would also appreciate any tips for books or other resources that would be recommended for someone wanting to ignite an interest in math.

To give a background, my highest level of math in school was pre-algebra in 7th grade. After that the schools my parents sent me to didn't teach higher math (nor evolution) so when I hit college, I had to get a tutor to pass (with a C) the College Level Algebra... I barely did it and it was so long ago I don't remember anything I learned.

I hope this is the right place to ask this... I stumbled across math.stackexchange by accident. Whether this question gets answered or not I will start browsing around and seeing what all there is...

答:
If you've enjoyed the Khan Academy for coding, then check out it offerings with respect to math! That's my "starting point" recommendation. (Math:...that's where Sal Khan got going, before branching off into other areas.)

See also Paul's Notes: Click on course notes: you'll see a drop down menu: Algebra, Calculus I, II, III, Linear Algebra, etc. Many students find the notes (tutorials) with exercises and examples really helpful.

You might want to investigate OpenStudy where you can interact with other "online" students studying topics you're currently studying. It's a wonderful way to "hold yourself accountable"  (peer pressure/participating in study groups/etc) to maintain some sort of progress, and it's an opportunity to simply learn from and support other students working independently, as you plan to do.

Lastly, you can "enroll" here at Mathematics.SE: post questions when you get stuck, to get feedback on problems you've completed, search through questions that are similar to those you might have, and interact more casually asking informal questions in the Chat room.  The Math.SE community includes a lot of self-studying students who ask, and sometimes answer, questions regularly.

Title: Math school online?


问:
Solve Matrix Equation $\sqrt{x^TAx} - x^\top\theta = 0$
let $x,\theta \in \mathbb{R}^n$, $A\in \mathbb{R}^{n\times n}$ Positive semi-definite and symmetrtic. How to solve (for $x \neq 0$) the following equation?

$$\sqrt{x^TAx} - x^\top\theta = 0$$

答:
The equation can have an infinity of solutions. Try a simple trivial case. Choose $n=2$ and $A=I_2$. Then the general form of $x^T$ is $(x_1\ x_2)$. So the equation becomes $$\sqrt{x_1^2+x_2^2}-x_1\theta_1-x_2\theta_2=0$$
Moving the last two terms from the left to the right, and squaring the equation, you get:
$$x_1^2+x_2^2=x_1^2\theta_1^2+x_2^2\theta_2^2+2x_1x_2\theta_1\theta_2$$
This is the equation of a conic (degenerate). In this case any point on the conic is a solution. Simpler way to see that, if $x$ is a solution, $ax$ is a solution as well, with $a>0$,


问:
How to establish (formally) equality of sets?
The axiom of extensionality establishes that "if two sets have the same elements, then they are identical", but not the second implication (the one I will use to prove that two sets are equal).

If we try to solve this by defining "A and B are equal sets iff they have the same elements", then why do we need to establish the axiom of extensionality?

答:
The axiom of extensionality is:

$\forall A\ \forall B: (A=B\Leftrightarrow \forall C: (C\in A\Leftrightarrow C\in B))$

It already includes both directions. There is nothing to prove here.


问:
If $np + c \sqrt{npq}$ is an integer, does there exits infinitely many $n_1<n_2<n_3\dots$ such that $n_ip + c \sqrt{n_ipq}$ is an integer?
Consider the following situtation: given $p \in (0,1)$, $q:=1- p$ and and some $k<n \in \mathbb N$ we define
$$
c := \frac{k - np}{\sqrt{npq}}.
$$

I would like to motivate that there exists an infinite sequence $n <
 n_1 < n_2 < \dots$ such that  $$ c = \frac{k_i - n_i p}{\sqrt{n_i
> pq}}. $$ for some interger $k_i$, or equivalently, such that $$ n_i p
> + c \sqrt{n_i pq} $$ is an integer.

I got stuch on this, so any help would be appreciated!

答:
Three cases:

$np\in\Bbb N$: If $np+\sqrt{npq}$ is an integer, $n m^2 p+\sqrt{n m^2 pq}=m(nmp+\sqrt{npq})$ is also an integer provided $m\in\Bbb N$. Therefore, choosing $n_i=(i+1)^2\times n$ results in $n<n_1<n_2<\dots$ and, for all $i$, $n_ip+\sqrt{n_ipq}\in\Bbb Z$.

$np\in\Bbb Q$: Let $p=\frac{a}{b}$.

\begin{aligned}
np+\sqrt{npq}&=n\frac{a}{b}+\sqrt{n\frac{a}{b}\left(1-\frac{a}{b}\right)}\\
&=\frac{n}{b}\left[ a+\sqrt{n\left(b-a^2\right)}\right]
\end{aligned}

and, therefore, $\left[ a+\sqrt{n\left(b-a^2\right)}\right]$ is a multiple of $b$. Choosing $m$ so that $m-1$ is a multiple of $b$ makes $\left[ am+\sqrt{n\left(b-a^2\right)}\right]$ also divisible by $b$.

If $p\in\Bbb R-\Bbb Q$. I'm not very clear as to what needs to happen for $np+\sqrt{npq}\in \Bbb N$. My guess is that this cannot happen except for vey specific $p$ and $n$, but I couldn't figure out an example.


答:
It will be possible if $\dim(V)$  is even because $\dim(T)+\mathrm{null}(T)=\dim(V)$ so $\dim(V)=2\dim(T)$

And then you will have, for example, $2n$ basis for $V$ as $\{\alpha_1,\ldots,\alpha_n,\alpha_{n+1},\ldots,\alpha_{2n}\}$.

For every $T$ that maps $n$ arbitrary basis elements from $\{\alpha_1,\ldots,\alpha_n,\alpha_{n+1},\ldots,\alpha_{2n}\}$ to zero and maps the others (remains basis ) to the first $n$ basis elements, it will be our answer.

Title: Is it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?

Body:
Help would be very appreciated.

It it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?


答: The other direction holds by definition. Suppose $A=B$. Then clearly $\forall x (x\in A \iff x\in B)$.

Title: How to establish (formally) equality of sets?

Body:
The axiom of extensionality establishes that "if two sets have the same elements, then they are identical", but not the second implication (the one I will use to prove that two sets are equal).

If we try to solve this by defining "A and B are equal sets iff they have the same elements", then why do we need to establish the axiom of extensionality?


答:
The second implication is true without the axiom of extensionality.

To see why, given two sets $A,B$ suppose that $A=B$. It follows that the truth value of a sentence involving $A$ is unchanged when $A$ is substituted by $B$. Apply this to the evidently true statement

$\forall x, x \in A \iff x \in A$

by substituting $B$ in place of the second $A$, and we obtain the true statement

$\forall x, x \in A \iff x \in B$

This sentence says that $A$ and $B$ have the same elements.


答:
An alternative approach: it suffices to show that this limit $L = \lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$ is equal to the limit
$$
M = \lim_{n \to \infty} \frac{1+2^2+\dots +n^2}{n^2(n+1)}.
$$
Note that $M$ is the limit of the left-side Riemann sum
$$
\frac{1+2^2+\dots +n^2}{n^2(n+1)} = \sum_{i=0}^{n+1} f(i/n) \cdot \frac 1{n+1},
$$
where $f(x) = x^2$. It follows that $M = \int_0^1 x^2 \,dx = \frac 13$.

Note that $\sqrt[n]{n} \to 1$ as $n \to \infty$ with $\sqrt[n]{n} \geq 1$. It is clear that $L \geq M$, but it remains to be shown that $L \leq M$. To that end, let $\epsilon > 0$. There exists an $N$ such that $n \geq N \implies \sqrt[n]
{n}\leq 1 + \epsilon$. For $n > N$, we have
$$
\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)} =
\frac 1{n+1}\sum_{i=1}^n \frac{i^2 \sqrt[i]{i}}{n^2}\\
=
\frac 1{n+1} \left[ \sum_{i=1}^N \frac{i^2 \sqrt[i]{i}}{n^2}
+ \sum_{i=N+1}^n \frac{i^2 \sqrt[i]{i}}{n^2}\right]
\\ =
\frac{C(\epsilon)}{n+1} +
\frac 1{n+1} \sum_{i=N+1}^n \frac{i^2 \sqrt[i]{i}}{n^2}
\\ \leq
\frac{C(\epsilon)}{n+1} +
\frac 1{n+1} \sum_{i=N+1}^n \frac{i^2 (1 + \epsilon)}{n^2}
\\ =
\frac{C(\epsilon)}{n+1} +
\frac {1 + \epsilon}{n+1} \sum_{i=N+1}^n \frac{i^2 }{n^2}
\\ \leq
\frac{C(\epsilon)}{n+1} +
\frac {1 + \epsilon}{n+1} \sum_{i=1}^n \frac{i^2 }{n^2}.
$$
It follows that
$$
L \leq \lim_{n \to \infty} \frac{C(\epsilon)}{n+1} +
\frac {1 + \epsilon}{n+1} \sum_{i=1}^n \frac{i^2 }{n^2}
\\ = 0 + (1 + \epsilon) M = (1 + \epsilon )M.
$$
That is, we have $L \leq (1 + \epsilon)M$ for all $\epsilon > 0$, which means that we must have $L \leq M$, as desired.

Body: $\lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$ so i've thought of simplifying into smaller fractions as such $$\lim_{n\to \infty}(\frac{1}{n^2(n+1)}+\frac{2^2\sqrt{2}}{n^2(n+1)}+\dots +\frac{n^2\sqrt[n]{n}}{n^2(n+1)})$$ and as all of the terms we  get k/something that approaches infinity, most of the terms approach 0 andon the last one the n^2 cancel out, and we still get all terms to be 0, so the answer is 0? is this logic right? i doubt it's that easy


问:
Linear transformation, norm and matrix
Let $T \in \mathcal{L}(\mathbb{R^{n}},\mathbb{R^{m}})$ and $A=[a_{ij}] \in \mathcal{M}_{m\times n}$ the matrix associated to $T$. Let:

$$\mu := \max_{i=\overline{1,m} \\ j=\overline{1,n}}|a_{ij}| $$

Prove the following relationship:

$$\mu \leq \| T \| \leq \mu \sqrt{mn}$$

答:
I suppose that $||T||$ is supremum of the set $\{|Tx| \hspace{0.3cm} | \hspace{0.3cm} \text{where} \hspace{0.3cm} |x|=1\}$ (say $K$). There is $a_{ij}$ such that $|a_{ij}|=\mu$. So if we choose $v=(0,0,...0,1,0,...,0)$ where only $j^{th}$ term is $1$ others are zero, then $|Tv|$ is in $K$. So $||T||\geq |Tv| \geq |a_{ij}| = \mu$.

Let $B \in \mathcal{M}_{m\times n}$ such that all entries are $\mu$. So $||T||\leq ||B||$. $||B||=\mu||C||$ where $C$ is the matrix whose all entries are only $1$ in $\mathcal{M}_{m\times n}$. $||C||=|C . (1/\sqrt{n},1/\sqrt{n},...,1/\sqrt{n})|=|(\sqrt{n},\sqrt{n},...,\sqrt{n})|=\sqrt{mn}$. Thus $||T||\leq ||B|| =\mu ||C|| \leq \mu \sqrt{mn}$

Hence we have $\mu \leq ||T|| \leq \mu \sqrt{mn}$

Title: Linear transformation, norm and matrix


问:
Constrained extrema and geometric interpretation of constrained stationary points
I'm studying constrained extrema for functions of two variables and with a single constraint(of equality).
So I have a function $f$ and I have to find its constrained extrema in the restriction $E_0=\{(x,y)|g(x,y)=0\}$ where $f,g:A\subseteq \mathbb{R}\to \mathbb{R} $ and $f,g \in C^1(A)$ (where $A$ is open).
If $g(x_0,y_0)=0$ and $\nabla g(x_0,y_0)\neq 0$(in this case the point $(x_0,y_0)$ is called regular), by implicit function theorem I can parametrize $E_0$ in a neighborhood of $(x_0,y_0)$, so let's call this curve $\gamma:\mathbf{r}(t)$ (Without loss of generality $\mathbf{r}(0)=(x_0,y_0)$) .  A constrained stationary point is regular point  such that $(f \circ  \mathbf{r})'(t)=0$. Clearly this is equivalent to $\langle \nabla f(x_0,y_0),\mathbf{r}'(0) \rangle=0$. Then my book says that this is equivalent to the fact that "the derivative of $f$ along the direction that is tangent to $E_0$ in $(x_0,y_0)$ is $0$". I think that the book used the fact that:
$$ \langle \nabla f(x_0,y_0),\mathbf{r}'(0) \rangle=D_{\mathbf{r}'(0)} f(x_0,y_0)=0 $$
But this is valid iff $\mathbf{r}'(0)\neq \mathbf{0}$. How do we know that?
If $\mathbf{r}'(0)= \mathbf{0}$, then it doesn't correspond to a direction.
This is giving me many difficulties. Thank you in advance.

Edit:
Now that I think about it the geometric intepretation of $\mathbf {r}'(t)$ is the tangent vector in $\mathbf{r}(t)$ to the curve $\gamma$, what would it mean if $\mathbf{r}'(t)=0$, geometrically.

答:
You can of course assume ${\bf r}'(0)\ne{\bf 0}$. This comes from the implicit function theorem:

When $\nabla g(x_0,y_0)\ne{\bf 0}$ you have, e.g., $g_y(x_0,y_0)\ne0$. This implies that there is a   window $$W:=[x_0-h,x_0+h]\times[y_0-h,y_0+h]\subset{\mathbb R}^2$$ and a $C^1$ function
$$\psi:\quad[x_0-h,x_0+h]\to[y_0-h,y_0+h],\qquad x\mapsto y=\psi(x);\qquad\psi(x_0)=y_0,$$
such that the set $\bigl\{(x,y)\in W\bigm| g(x,y)=0\bigr\}$ coincides with the graph of $\psi$. This implies that you can take
$$\gamma:\quad[x_0-h,x_0+h]\to W, \qquad x\mapsto {\bf r}(x):=\bigl(x,\psi(x)\bigr)\qquad(x_0-h<x<x_0+h)$$
in your proof. Here ${\bf r}'(x)=(1,\psi'(x))\ne{\bf 0}$.


问:
Need help in finding counterexample
I need to find example that this isn't correct:

Let $R_1,R_2,R_3$ be binary relations on set $A$. Prove that this is not correct: $(R_1\cup R_2)\circ R_3 \supseteq(R_1\circ R_3)\cup(R_2\circ R_3)$

答:
According to one convention for writing the composition of two relations, $\langle a,b\rangle\in(R_1\cup R_2)\circ R_3$ if and only if

$$\exists x\in A\Big(\langle a,x\rangle\in R_3\land\langle x,b\rangle\in R_1\cup R_2\Big)\;.$$

But this is true if and only if

$$\exists x\in A\left(\langle a,x\rangle\in R_3\land\Big(\langle x,b\rangle\in R_1\lor\langle x,b\rangle\in R_2\Big)\right)\;,$$

which in turn is equivalent to

$$\exists x\in A\Big(\langle a,x\rangle\in R_3\land\langle x,b\rangle\in R_1\Big)\lor\exists x\in A\Big(\langle a,x\rangle\in R_3\land\langle x,b\rangle\in R_2\Big)$$

and hence to $\langle a,b\rangle\in(R_1\circ R_3)\cup(R_2\circ R_3)$.

According to the other convention, $\langle a,b\rangle\in(R_1\cup R_2)\circ R_3$ if and only if

$$\exists x\in A\Big(\langle a,x\rangle\in R_1\cup R_2\land\langle x,b\rangle\in R_3\Big)\;,$$

i.e., if and only if

$$\exists x\in A\left(\Big(\langle a,x\rangle\in R_1\lor\langle a,x\rangle\in R_2\Big)\land\langle x,b\rangle\in R_3\right)\;,$$

which in turn is equivalent to

$$\exists x\in A\Big(\langle a,x\rangle\in R_1\land\langle x,b\rangle\in R_3\Big)\lor\exists x\in A\Big(\langle a,x\rangle\in R_2\land\langle x,b\rangle\in R_3\Big)\;,$$

i.e., to $\langle a,b\rangle\in(R_1\circ R_3)\cup(R_2\circ R_3)$.

By either convention, therefore,

$$(R_1\cup R_2)\circ R_3=(R_1\circ R_3)\cup(R_2\circ R_3)\;,$$

and no such counterexample exists.

Title: Need help in finding counterexample


问:
Mistake in Solving Linear Recurrence Relation
I am trying to find a closed for solution to the following:

$$a_n = -4a_{n-1}-3a_{n-2}+4a_{n-3}+4a_{n-4}$$

With $a_0=3, a_1=2, a_2 = 6, a_3 = 8$.

My Attempt:

I want to first find the roots:

$$r^4+4r^3+3r^2-4r-4=0$$

The roots are: $r=-1,r=1,r=-2$.

Then we can set up the following: $a_n = \alpha_1\cdot -1^n+\alpha_2\cdot1^n+ \alpha_3 \cdot-2^n$.

Now I can solve for the constants $\alpha$:

$$a_0 = 3= -\alpha_1+\alpha_2-\alpha_3 \\ a_1 =2= -\alpha_1+\alpha_2-2\alpha_3 \\ a_2 =6= -\alpha_1+\alpha_2-4\alpha_3 \\ a_3=8 = -\alpha_1+\alpha_2-8\alpha_3 \\$$

But when try to solve for the constants I get that there is no solution for the system. I was wondering if someone can point out my mistakes on this recurrence problem.

Body:
Two mistakes are that $-1^n$ and $-2^n$ should instead be $(-1)^n$ and $(-2)^n$.  A third mistake is that you ignored the fact that $-2$ is a double root.

An alternative approach is to use generating functions.  Let $A(z)=\sum_{n=0}^\infty a_n z^n$.  Then the recurrence and initial conditions imply that
\begin{align}
A(z)
&= \sum_{n=0}^3 a_n z^n + \sum_{n=4}^\infty a_n z^n \\
&= \sum_{n=0}^3 a_n z^n + \sum_{n=4}^\infty (-4a_{n-1}-3a_{n-2}+4a_{n-3}+4a_{n-4}) z^n \\
&= \sum_{n=0}^3 a_n z^n - 4 z \sum_{n=4}^\infty a_{n-1}z^{n-1}-3 z^2 \sum_{n=4}^\infty a_{n-2}z^{n-2} +4z^3\sum_{n=4}^\infty a_{n-3}z^{n-3}+4z^4\sum_{n=4}^\infty a_{n-4} z^{n-4} \\
&= \sum_{n=0}^3 a_n z^n - 4 z (A(z)-a_0-a_1z-a_2z^2)-3 z^2 (A(z)-a_0-a_1z) +4z^3 (A(z)-a_0)+4z^4 A(z) \\
&= (-4z-3z^2+4z^3+4z^4)A(z) + a_0 + (4a_0+a_1 )z + (3a_0+4a_1+a_2 )z^2 + (-4a_0+3a_1+4a_2+a_3 )z^3 \\
&= (-4z-3z^2+4z^3+4z^4)A(z) + 3 + 14z + 23z^2 + 26z^3,
\end{align}
so
\begin{align}
A(z)&=\frac{3 + 14z + 23z^2 + 26z^3}{1+4z+3z^2-4z^3-4z^4} \\
&=\frac{11/3}{1 - z} - \frac{7}{1 + z} + \frac{25/3}{1 + 2 z} - \frac{2}{(1 + 2 z)^2} \\
&=\frac{11}{3}\sum_{n \ge 0} z^n - 7\sum_{n \ge 0} (-z)^n + \frac{25}{3}\sum_{n \ge 0} (-2z)^n - 2\sum_{n \ge 0} \binom{n+1}{1}(-2 z)^n \\
&=\sum_{n \ge 0} \left(\frac{11}{3} - 7(-1)^n + \frac{25}{3}(-2)^n - 2(n+1)(-2)^n\right)z^n \\
&=\sum_{n \ge 0} \left(\frac{11}{3} - 7(-1)^n + \frac{19}{3}(-2)^n + n(-2)^{n+1}\right)z^n,
\end{align}
which immediately yields explicit formula
$$a_n = \frac{11}{3} - 7(-1)^n + \frac{19}{3}(-2)^n + n(-2)^{n+1}.$$


问:
Decide whether $\left(\frac{100}{101}\right)^n$ is big O,Ω, or Θ of $\log(n)$
So I have to find if $\left(\frac{100}{101}\right)^n$ is either big O, Ω or Θ of $log(n)$.My thinking went a bit like this: First of all, we can easily see that $\left(\frac{100}{101}\right)^n$ is strictly decreasing $\forall x \in \mathbb{R}$, so $\forall x > 0$. $\log(x)$ is strictly increasing $\forall x > 0$. So eventually there will be a $k$ such that $\log(x) \geq \left(\frac{100}{101}\right)^n$ so $\left(\frac{100}{101}\right)^n = O(\log(x))$.

To be exact, if a proof was needed, I would let $h(x) = (\frac{100}{101})^n - log(x), \forall x > 0$.

Firstly, $h$ is stricly decreasing as $\forall x_1, x_2, x_1 < x_2 \rightarrow (\frac{100}{101})^{x_1} > (\frac{100}{101})^{x_2} \quad (1)$

$x_1 < x_2 \rightarrow log(x_1) < log(x_2) \rightarrow -log(x_1) > - log(x_2) \quad (2)$

From (1) and (2), $h$ is decreasing, and $$lim_{x \rightarrow \infty}h(x) = - \infty, lim_{x \rightarrow 0^+}h(x) = +\infty$$

Now let $a<b$, with $a$ infinitely close to $0^+$ and $b$ infinitely big, so close to $\infty$. Then $h(a) > 0, h(b) < 0$, and since $h$ is continuous  and $h(a)h(b)<0$, there will be a $x_0$ such that h(x_0) = 0 (intermediate value theorem). So, for $x_0 < x \rightarrow h(x_0) > h(x) \rightarrow h(x) < 0 \rightarrow (\frac{100}{101})^x < log(x)$. So there is a point $k > x_0$, with $C=1$, for which $(\frac{100}{101})^x = O(log(x))$

If that was my responce, would that be correct or not? Since I am not finding a exact value for $k$, but instead prove there exists one so big $O$ is satisfied. Also is there a easier way to prove such thing? Thanks a lot!

答: There is a much easier approach.  You can simply notice that $h(n) \to 0$ as $n \to \infty$ and $\log(n) \to \infty$ as $n \to \infty$.  Therefore $\frac{h(n)}{\log(n)} \to 0$ as $n \to \infty$, which implies in particular that $h(n) \leq \log(n)$ for sufficiently large $n$, so $h(n) = O(\log(n))$.


问:
Approximating a double sum by a double integral
Related to this question, I'm interested in bounding from above the following sum
$$
S:=\sum_{x=0}^\infty \sum_{y=0}^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}},
$$
which I hope to do by relating it to the integral
$$
I:=\int_0^\infty \int_0^\infty (x+y)^m e^{-\frac{x^2}{2i} - \frac{y^2}{2j}} dx\,dy.
$$

Answers to the previous questions confirmed my expectation that $I = O\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right)$, the intuition for which is probably that the function behaves approximately like a gaussian around its maximum at $(x_0,y_0) = \left(i \sqrt{\frac{m}{i+j}},j \sqrt{\frac{m}{i+j}} \right)$, where the function takes the value $\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)$.

However I've been unable to show that the difference $|I-S|$ is significantly smaller than this bound. For simple one dimensional integrals, for example with a unique maximum, it's not too hard to bound this difference in terms of the maximum by considering appropriate telescoping sums. However, a naive analogue of this argument doesn't seem to work in two dimensions, and trying to apply this argument to each `slice' of the integral led to some pretty horrendous calculations. I also looked into using the Euler-Maclaurin formula but it's a bit out of my area of expertise.

I suspect that there should be a relatively standard way to approximate $|I-S|$, and I also wouldn't be surprised if someone more proficient in computing can get a CAS to provide a proof. The former would be more useful, just so that I have a tool for approaching similar questions.

So, very explicitly, I would like to know if
$$
|I-S| = o\left(\exp\left(m\log\sqrt{(i+j)(m)}-\frac{m}{2}\right)\sqrt{ij}\right),
$$
where even big-O would be sufficient for the application I have in mind, and I wouldn't be surprised if the difference is even bounded by a multiple of the maximum of the function. I'm interested in the asymptotics for $i$ and $j$ tending to infinity, $m$ can be fixed or also a function of $i$ and $j$. For the application I have in mind it would probably be sufficient to have such a result for $i = (1+o(1))j$ and $m = o(i)$.

答:
I cannot provide an actual answer , but only some considerations and hint
which might hopefully be helpful.

The function
$$
f(x,y) = \left( {x + y} \right)^{\,m} e^{ - \,{{x^{\,2} } \over {2\,i}}\, - {{y^{\,2} } \over {2\,j}}}
$$
having a (cutted) bell-shape in the first quadrant, means that it is concave around the maximum and convex
further from it.

This makes quite difficult to relate the integral to the  Riemann sum with a $>, <$, because the sign
of the inequality changes in the two areas.

Furthermore, at increasing $i, \, j$, while the position of the max move $\approx \sqrt{i}$, and so approximately does its spread
the bell peak increases $\approx i^{m/2}$.

Since the $\Delta x , \, \Delta y$ of the sum are fixed at $1$, I doubt that the sum might converge to the integral.

Concerning the integral I would try the following approach
$$
\eqalign{
  & I = \int_{y\, = \,0}^{\,\infty } {\int_{x\, = \,0}^{\,\infty }
 {\left( {x + y} \right)^{\,m} e^{ - \,{{x^{\,2} } \over {2\,i}}\, - {{y^{\,2} } \over {2\,j}}} dxdy} }  =   \cr
  &  \Rightarrow \left\{ \matrix{
  s = x + y \hfill \cr
  t = x - y \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  x = \left( {s + t} \right)/2 \hfill \cr
  y = \left( {s - t} \right)/2 \hfill \cr}  \right.\quad  \Rightarrow   \cr
  &  = \int_{y\, = \,0}^{\,\infty } {\int_{x\, = \,0}^{\,\infty }
 {s^{\,m} e^{ - \,{{\left( {s + t} \right)^{\,2} } \over {2\,i}}\,
 - {{\left( {s - t} \right)^{\,2} } \over {2\,j}}} {1 \over 2}dsdt} }  =   \cr
  &  = \int_{s\, = \,0}^{\,\infty } {\int_{t\, = \, - s}^{\,s}
 {s^{\,m} e^{ - \,{{\left( {s + t} \right)^{\,2} } \over {2\,i}}\,
 - {{\left( {s - t} \right)^{\,2} } \over {2\,j}}} {1 \over 2}dsdt} }  \cr}
$$
then also consider that
$$
\eqalign{
  &  - \,\left( {{{s^{\,2}  + t^{\,2}  + 2st} \over {2\,i}}\,
 + {{s^{\,2}  + t^{\,2}  - 2st} \over {2\,j}}} \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2\,i\,j}}
\left( {\left( {{t \over s}} \right)^{\,2}  - 2{{i - j} \over {i + j}}
\left( {{t \over s}} \right) + 1} \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2i\,j}}\left( {\left( {{t \over s}} \right)^{\,2}
  - 2{{i - j} \over {i + j}}\left( {{t \over s}} \right) + \left( {{{i - j} \over {i + j}}} \right)^{\,2}  + 1 - \left( {{{i - j} \over {i + j}}} \right)^{\,2} } \right) =   \cr
  &  =  - \,{{\left( {i + j} \right)s^{\,2} } \over {2\,i\,j}}
\left( {\left( {\left( {{t \over s}} \right) - {{i - j} \over {i + j}}} \right)^{\,2}
  + 1 - \left( {{{i - j} \over {i + j}}} \right)^{\,2} } \right) \cr}
$$
we can change the variables again
$$
\left\{ \matrix{
  s = s \hfill \cr
  r = t/s \hfill \cr}  \right.\quad J = \left| {\left( {\matrix{
   1 & 0  \cr
   { - t/s^{\,2} } & {1/s}  \cr
 } } \right)} \right| = {1 \over s}
$$
and then proceed with approximation or series expansion of the Error Function.


问:
Age is fraction of year man dies
My friend sent me a question from an olympiad and im not sure that we have followed the right method, we both did the same thing:

The age of a man was 2/61 of the year in which he died. How old would he have been if he
lived until 1992?

Surly then he dies in 1992 and then his age is 2/61 times 1992 rounded to the nearest integer? I am unsure though, this seems too simple.

答:
If you let $a$ be the age of the man in the year in which he died and $d$ be the year of his death, then $a=\frac {2d}{61}$.

Since $61$ is prime and $a$ is an integer we have to have $d=61k$, in which case $a=2k$.

So if, for example, $k=32$ then $d=1952$ and $a=64$, which would give an age of [ ] in $1992$.

If $k=31$ then $d=1891$ and $a=62$ implying age [ ] in $1992$. To exclude this possibility (and others) you have to use a nonmathematical assumption, for example that human beings do not live beyond age $120$.

Title: Age is fraction of year man dies


问:
SVD with non invertible matrices
My problem:

Assuming A is a $n \times n$ matrix and can be expressed as $A=U\Sigma V^T = \sum_{i=1}^r \sigma_i u_i v_i^T$ and it is not invertible.

Given $B=\sum_{i=1}^r \frac{1}{\sigma_i} v_i u_i^T$, show that $BAx = x$ for all $x = \sum_{i=1}^r \alpha_iv_i$.

What I have done so far:

I rewrote $B$ as $B = V\Sigma^+U^T$ where $\Sigma^+$ is a $n \times n$ diagonal matrix with: $\sigma^+_i =
    \begin{cases}
        \sigma^{-1}_i & \textit{if } i \leq r \\
        0 & \textit{otherwise}
    \end{cases}$

I rewrote $x$ as $x = \sum_{i=1}^r \alpha_i v_i = \bar{\alpha} V$ where $\bar{\alpha} =
\begin{pmatrix}
\alpha_1, & \alpha_2, & \dots, & \alpha_r, & 0, & \dots, & 0
\end{pmatrix}^T =
\begin{pmatrix}
    \underset{(rx1)}{\bar{\alpha}}\\
    0
\end{pmatrix} \in \mathcal{R}^n$

Question 1:

I used
$\underset{n \times n}{V} = \begin{pmatrix}
    \underset{(r\times r)}{V} & 0\\
    0 & 0
\end{pmatrix}$,
$\underset{n \times n}{\Sigma} = \begin{pmatrix}
    \underset{(r\times r)}{\Sigma} & 0\\
    0 & 0
\end{pmatrix}$,
$\underset{n \times n}{\Sigma^+} = \begin{pmatrix}
    \underset{(r\times r)}{\Sigma^+} & 0\\
    0 & 0
\end{pmatrix}$ because I assumed that besides the index $r$, the other values of $\Sigma$ would be all $0$ (in the definition the summatory was until $r$ and $r \leq n$).

Is it correct? If not why?

I substituted the new terms in the expression:

$BAx = x \rightarrow (V\Sigma^+ U^T) (U \Sigma V^T) x = x \rightarrow (V \Sigma^+) (\Sigma V^T) \bar{\alpha} V = \bar{\alpha} V \rightarrow$

$\rightarrow V \underbrace{
\begin{pmatrix}
    \underset{(r\times r)}{\Sigma^{-1}} & 0\\
    0 & 0
\end{pmatrix}
\begin{pmatrix}
    \underset{(r\times r)}{\Sigma} & 0\\
    0 & 0
\end{pmatrix}}_{\mathcal{I}_{r\times r}}
V^T \bar{\alpha} V = \bar{\alpha} V \rightarrow$

$\rightarrow
\underbrace{
\begin{pmatrix}
    \underset{(r\times r)}{V} & 0\\
    0 & 0
\end{pmatrix}
\begin{pmatrix}
    \underset{(r\times r)}{\mathcal{I}} & 0\\
    0 & 0
\end{pmatrix}
\begin{pmatrix}
    \underset{(r\times r)}{V^T} & 0\\
    0 & 0
\end{pmatrix}}_{\mathcal{I}_{r\times r}} \bar{\alpha} V = \bar{\alpha} V \rightarrow \bar{\alpha} V = \bar{\alpha} V$

Question 2

Is it correct? If not, why? How can I prove the assignment?

Thanks for your help.

答:
Regarding question 1: you should write $x = V \bar \alpha$ rather than $x = \bar \alpha V$; note that $x$ must be a column vector. Also, we have no reason to assume that $U$ and $V$ have zero-rows and zero-columns.

Regarding the problem itself, here is a much simpler proof. If $x = \sum_{i=1}^r \alpha_i v_i$, then we can write
$$
\begin{align}
Ax & =
\left(\sum_{i=1}^r \sigma_i u_iv_i^T \right)\left(\sum_{j=1}^r \alpha_j v_j \right)
\\ & = \sum_{i=1}^r \sum_{j=1}^r \alpha_j\sigma_i \cdot u_i(v_i^Tv_j)
\\ & = \sum_{i=1}^r  \alpha_i\sigma_i \cdot u_i,
\end{align}
$$
and similarly
$$
\begin{align}
BAx & = B(Ax) = \left(\sum_{i=1}^r \frac 1{\sigma_i} v_iu_i^T \right)\left(\sum_{j=1}^r \alpha_j \sigma_j u_j \right)
\\ & =
\sum_{i=1}^r \sum_{j=1}^r \frac{\alpha_j \sigma_j}{\sigma_i} v_i(u_i^Tu_j)
\\ & = \sum_{i=1}^r  \frac{\alpha_i \sigma_i}{\sigma_i} v_i = x,
\end{align}
$$
which was what we wanted.

If you prefer to work with matrices, you could also have used $x = V \bar \alpha$ and $B = V \Sigma^+ U^T$ to find that
$$
\begin{align}
BAx & = (V \Sigma^+ U^T)(U\Sigma V^T)(V\bar \alpha)
\\ & = V \Sigma^+ (U^TU)\Sigma (V^TV)\bar \alpha
\\ & = V (\Sigma^+ \Sigma) \bar \alpha
\\ & = V \bar \alpha = x.
\end{align}
$$
However, the fact that $U^TU, V^TV$ are identity matrices and that $(\Sigma^+ \Sigma) \bar \alpha = \bar \alpha$ should be justified.

Title: SVD with non invertible matrices


问:
Find $a_0$ and $a_1$ when given a generating function.
The recursive sequence $a_n$ has the following generating function

$$\begin{align}
&f(x)={\frac{x}{1-2x}+{\frac{4}{1+3x}}}
\end{align}$$

I am to find $a_0$ and $a_1$:

Rule: ${\frac{1}{1-a}}=1+a+a^2+a^3+...+a^n$

So we have

${\frac{1}{1-2x}}=1+2x+4x^2+8x^3+...+2^{n-1}x^{n-1}+2^nx^n$

${\frac{1}{1+3x}}=1-3x+9x^2-27x^3+...+(-3)^{n-1}x^{n-1}+(-3)^nx^n$

Thus,

$$\begin{align}
&f(x)={\frac{x}{1-2x}+{\frac{4}{1+3x}}}\\
&f(x)= x(1+2x+4x^2+8x^3+...) + 4(1-3x+9x^2-27x^3+...)\\
&f(x)= x+2x^2+4x^3+8x^4+... + 4-12x+36x^2-108x^3+...
\end{align}$$

For $a_0$ we have: $x+4=0+4=4$ , but when I try to calculate $a_1$ I get the wrong answer. The right answer should be $a_1=-11$, but I get

$$\begin{align}
&a_1=x+2x^2+4-12x=1+2+4-12=-5
\end{align}$$

What am I doing wrong? What is the correct way to find $a_1$?

答:
\begin{align}f(x) &= x+2x^2 + 4x^3+ \dots + 4 - 12 x + 36 x^2 - 108x^3 + \dots \\&= 4-11x+38x^2-104x^3+\dots\\&=a_0+a_1x+a_2x^2+a_3x^3+\dots\end{align}

where the last line is precisely the definition of a generating function. There is no need to substitute any values of $x$.

Alternatively we have $a_n = \dfrac {f^{(n)}(0)}{n!}$, where $f^{(n)}(x)$ is the $n$th derivative. This might be useful for some other $f$.


问:
notation question (bilinear form)
So I have to proof the following:
for a given Isomorphism $\phi : V\rightarrow V^*$ where $V^*$ is the dual space of $V$ show that $s_{\phi}(v,w)=\phi(v)(w)$ defines a non degenerate bilinear form.

My question : Does $\phi(v)(w)$ denote the map from $v$ to a linear function $w$? (in this case i had serious trubles in showing linearity in the second argument,really confusing.

Or

maybe it just means $\phi(v)$ times $w$ where $w$ is the scalar value ( we get $w$ by applying  $v$ in the linear function it is mapped to)

I just started today with dual spaces and try my best with the notation , but i couldn't figure it out , please if you have any idea please help me with the notation , i will solve the problem on my own.

答:
Note that $\phi$ is a map from $V$ to $V^\ast$. So for each $v \in V$, we get an element $\phi(v) \in V^\ast$. Now $V^\ast$ is the space of linear functionals on $V$, i.e.
$$V^\ast = \{\alpha: V \longrightarrow \Bbb R \mid \alpha \text{ is linear}\}.$$
So each element of $V^\ast$ is a function from $V$ to $\Bbb R$. Then for $v, w \in V$, the notation
$$\phi(v)(w)$$
means
$$(\phi(v))(w),$$
i.e. the function $\phi(v): V \longrightarrow \Bbb R$ takes $w \in V$ as its argument and we get an element of $\Bbb R$.

So $s_\phi$ is really a map of the form
$$s_\phi: V \times V \longrightarrow \Bbb R,$$
$$(v, w) \mapsto (\phi(v))(w).$$

Title: notation question (bilinear form)


问:
Setting up exponential equation
I want to devise a function $f(x)$ such that $f(4) = 160, f(16)=80, f(28)=40$ and $f(40)=20$, as well as all the other values in between $4-40$. As in, the numbers between $4-12$ have a proportional value between $160-80$. Numbers between $28-40$ have a proportional value between $40-20$ etc.

The values on the left go up linearly and the values on the right go down exponentially.

I hope this makes sense. But would appreciate any help. Thanks

答:
Take the logarithm of the values on the right column and plot them vs. the values on the left column.

You should obtain quite a straight line.

in any case draw the line that best approximate the graph and you have the parameters of the exponential function you are looking for.

Title: Setting up exponential equation