问:
Proposition 4.2 in Richard Bass' ''Real Analysis for Graduate Students''
The following is from R. "Bass Real Analysis for Graduate Students" :

Proposition 4.2 Suppose $\mathcal{C}$ is a collection of subsets of $X$ such that $\varnothing$ and $X$ are both in $\mathcal{C}$. Suppose $\ell:\mathcal{C}\longrightarrow[0,\infty]$ with $\ell(\varnothing)=0$. Define
$$\mu^*(E)=\bigl\{\sum_{i=1}^\infty\ell(A_i):\ A_i\in\mathcal{C}\ \text{for each}\ i\ \text{and}\ E\subset\bigcup_{i=1}^\infty A_i\Bigr\}.\tag{4.1}$$

Then $\mu^*$ is an outer measure.

Proof. (1) and (2) of the definition of outer measure are obvious. To prove (3), let $A_1,A_2,\ldots$ be subsets of $X$ and let $\varepsilon>0$. For each $i$ there exist $C_{i1},C_{i2}\ldots\in\mathcal{C}$ such that $A_i\subset\bigcup_{j=1}^\infty C_{ij}$ and $\sum_j\ell(C_{ij})\leq\mu^*(A_i)+\varepsilon/2^i$. Then $\bigcup_{i=1}^\infty A_i\subset\bigcup_i\bigcup_jC_{ij}$ and
\begin{align*}
\mu^*\Bigl(\bigcup_{i=1}^\infty A_i\Bigr) & \color{red}{\leq}\sum_{i,j}\ell(C_{ij})=\sum_i\Bigl(\sum_j\ell(C_{ij})\Bigr)\\
& \leq\sum_{i=1}^\infty\mu^*(A_i)+\sum_{i=1}^\infty\varepsilon/2^i\\
& =\sum_{i=1}^\infty\mu^*(A_i)+\varepsilon.
\end{align*}
Since $\varepsilon$ is arbitrary, $\mu^*(\bigcup_{i=1}^\infty A_i)\leq\sum_{i=1}^\infty\mu^*(A_i)$. $\qquad\square$

Where the red inequality comes from? There are two summations are on RHS of the red inequity how those two sums appears?

PS the book is available online.

答:
Book I looked:  Richard F. Bass - Real analysis for graduate students_ measure and integration theory-CreateSpace (2011).

We get circled inequality, because family of sets  $C_{ij}$ satisfy condition in definition of $\mu^*$, where $E=\bigcup\limits_{i=1}^{\infty} A_i$.

Addition.

This answer to second, added later question, about two sums in rhs: as $C_{ij}$ is countable cover of $E=\bigcup\limits_{i=1}^{\infty} A_i$, then by definition of inf we have $\mu^*(E) \leqslant \sum_{i,j}l(C_{ij})$. Right side is absolute converged so we can write it as $\sum_{i,j}l(C_{ij})=\sum_{i}\sum_{j}C_{ij}$.


问:
How to calculate $\mathbb{E}[X]\mathbb{E}[Y]$ given $\mathbb{P}(X=k, Y=m)={\frac{1}{36}}\cdot\left( {\frac{5}{6}} \right)^{k-2}$
Let $X,Y$ be random variables, s.t for $1\leq m<k$ $$\mathbb{P}(X=k, Y=m)={\frac{1}{36}}\cdot\left( {\frac{5}{6}} \right)^{k-2}$$
Calculate $\mathbb{E}[X]\mathbb{E}[Y]$

I am not sure even how to approach it, so far I wrote the definition of $\mathbb{E}$ but I am not sure how to write the indexes of $k$ and $m$

答:
First I suggest you to tell us any detail that you hidden

the random variables are discrete and integer

the support is

$$1\leq y<x<\infty$$

Now you can start in calculating the marginals. I start in calcuating

$$P(Y=y)=\frac{1}{36}\sum_{x=y+1}^{\infty}\left(\frac{5}{6}\right)^{x-2}=\frac{1}{6}\left(\frac{5}{6}\right)^{y-1}$$

where $y=1,2,3,...$

Thus $Y\sim Geo(1/6)$  with expectation $E(Y)=6$

(in this case the Geometric distribution counts the trials before the first success. It is always important to explain if the geometric distribution counts the trials or the failures)

Now you can proceed by yourself

ContentLicense: CC BY-SA 4.0


问:
Is it analytic ? $f(z)=z^2\bar{z}$
Please find the question and its solution.

My answer gone wrong dont know why, I think the answer marked in the question is wrong.

My try:

I solved for CR equations and replaced x=y=0 and found CR equations satisfied. But not sure why in the solution it says CR equations doesn't satisfy.

Please let me know what should be the correct answer ?

答:
The problem is poorly phrased. It awkwardly mixes real and complex differentiability as well as pointwise and local notions. Let me try to clear things up.

As a map $\mathbb R^2\to\mathbb R^2$ the map $f(z)=z^2\bar z$ has polynomials as component functions. In particular, it is real differentiable everywhere and even real analytic (meaning that all components have convergent real power series expansions near all points in the domain).

However, the Cauchy-Riemann equations are only satisfied at $z=0$. So as a map $\mathbb C\to\mathbb C$,  $f$ is complex differentiable at $z=0$ but not at any $z\ne0$. This means that $f$ is not complex analytic at zero. Otherwise it would have a convergent complex power series expansion in a neighborhood of zero, and would thus be complex differentiable in that neighborhood.

Title: Is it analytic ? $f(z)=z^2\bar{z}$


问:
What is the distribution of the outcome of this game?
We were playing a game yesterday and I am curious about some of the potential outcomes. In this game, there are n players who write words on cards. There is one unique word on each card. Each individual writes on k cards. Then the individuals play this game k rounds where in each round an individual picks a card without replacement from the box. I am wondering how I can calculate the distribution of the number of cards picked by the individual who wrote the word him/herself.

E.g.
Let's say Alice and Hans wrote 3 cards. Then the total number of cards picked by the individual who wrote it ranges from 0 to 6. I think he probability of 0 such cards is $2 \times \frac{3}{6}\frac{3}{5}\frac{2}{4}\frac{2}{3}\frac{1}{2}$. I am not sure how to generalize this to n individuals, and k rounds. Thank you!

答:
There isn't a simple formula for this.  The way to solve the problem involves rook polynomials.

Let me consider an equivalent game.  A deck of cards has $k$ suits of $n$ cards each.  There are $k$ players, each of whom is assigned a different suit.  The deck is shuffled, and dealt out to the players, so that each player receives $n$ cards.  What is the probability that a total of $j$ cards are dealt to to the player assigned that card's suit?

It doesn't matter how we deal the cards, so we can assume the first player gets the first $n$ cards, the second player gets the next $n$ cards, and so on.  In looking at the chessboard, we see $k$ $n\times n$ dark squares representing the forbidden positions.  A very useful fact about rook polynomials is that if the board breaks up into two parts that have no square in the same row or the same column, then the rook polynomial of the board is equal to the product of the rook polynomial of the two parts.  Thus, if $R_n$ is the rook polynomial of an $n\times n$ square, then then rook polynomial of our dark board is $R_n^k$.

It is easy to compute the rook polynomial of an $n\times n$ square, as explained here.  For example, for a $4\times4$ board, there is $1$ way to place $0$ rooks and $16$ ways to place one rook.  For two rooks, after we place the first in $16$ ways, we eliminate a row and column for the next rook, leaving $9$ possibilities.  The order in which we place the two rooks doesn't matter, so there are $\frac{16\cdot9}{2}=72$ possibilities.  Similarly, for $3$ rooks, we have $\frac{4^2\cdot3^2\cdot2^2}{3!}=96$ possibilities, and of course, there are $4!=24$ ways to place $4$ non-attacking rooks.  The rook polynomial is therefore $$24x^4+96x^3+72x^2+16x+1$$.

Once you've computed the rook polynomial, the standard results allow you to compute the number of ways to place $nk$ rooks on the board with exactly $j$ rooks in forbidden positions, for any $j$.  Then just divide by $(nk)!$ to get the probability.

One thing I remember about this problem is that if $n$ remains fixed and $k$ increases, the the limit, as $k\to\infty$ of the probability that no player receives a card of his own suit is $e^{-n}.$


问:
Hardness of a special case of maximum matching
Input:

A set of N Users $\{u_1, ..., u_N\}$. A set of M products $\{i_1, ..., i_M\}$.

Every pair $(u,i)$ is associated with the probability of u purchasing the product i, $p_{u,i}$.

Task: Assign each user with exactly $K$ products (notation: $(u,i) \in A$) so that the following objective function is maximized
$$
\sum_{i: \exists u \ni (u,i) \in A} \left(1-\prod_{u: (u,i) \in A}{(1-p_{u,i})}\right)
$$
Question: Is this problem NP-Hard?

答:
This looks like a version of the "weapon target assignment problem", see e.g.
http://web.mit.edu/sloan-msa/Papers/1.5.pdf
. It seems that no exact algorithms are known.</p>

ContentLicense: CC BY-SA 3.0


问:
A problem with the Boolos/Burges/Jeffrey proof of incompleteness of second order logic
My question concerns the proof of the fact that the set of valid sentences of 2nd order logic (SOL) is not recursively enumerable, as presented in "Computability and Logic" 4th edition in Proposition 22.8

Let $\ulcorner\varphi\urcorner$ be the Gödel number of $\varphi$. Let $\mathrm{PA}_2$ be the conjunction of the axioms for 2nd order Peano arithmetic. The function $f$ that takes $\ulcorner\varphi\urcorner$ to $\ulcorner\mathrm{PA}_2\rightarrow\varphi\urcorner$ is recursive. Let $\mathfrak{N}$ be the standard model of arithmetic. Consider it to be the full model when the SOL is taken into account. Now, consider $\mathrm{Th}(\mathfrak{N})$ - the first-order theory of $\mathfrak{N}$, about which we know that is not definable, the more so is not r.e. We have:
$$\varphi\in\mathrm{Th}(\mathfrak{N})\qquad\text{iff}\qquad\vDash\mathrm{PA}_2\rightarrow\varphi\,.$$
The conclusion that is drawn, is that if the set $\mathrm{Val}_2$ of Gödel numbers of 2nd order logical validities were r.e., $\mathrm{Th}(\mathfrak{N})$ would have to be r.e., but we know it is not. However, does r.e. of $\mathrm{Val}_2$ entail r.e. of $A=\{\ulcorner\mathrm{PA}_2\rightarrow\varphi\urcorner\mid\varphi\in\mathrm{Th}(\mathfrak{N})\}$? For what I can see, the proof demonstrates that the recursive function $f$ sends $\mathrm{Th}(\mathfrak{N})$ onto $A$, which is a proper subset of $\mathrm{Val}_2$ (which contains, e.g., numbers of all formulas $\mathrm{PA}_2\rightarrow\psi$, where $\psi$ is a 2nd order formula true in the stadnard full model):
$$\mathrm{Th}(\mathfrak{N})=f^{-1}[A]\qquad\text{and}\qquad A\subsetneq\mathrm{Val}_2\,.$$
In general, subsets of r.e. sets are not r.e., so I cannot reason like "If $\mathrm{Val}_2$ were r.e., then $A$ would be too, and so $\mathrm{Th}(\mathfrak{N})$ either".

What am I missing to see that the set of valid formulas of SOL cannot be r.e.?

答:
Note first that
$$
A=\{\ulcorner\mathrm{PA}_2\rightarrow\varphi\urcorner\mid\varphi\in\mathrm{Th}(\mathfrak{N})\}=\{\ulcorner\mathrm{PA}_2\rightarrow\varphi\urcorner\mid(\mathrm{PA}_2\rightarrow\varphi)\in\mathrm{Val}_2,\varphi\text{ is a $\mathrm{PA}_1$-formula}\}.
$$
Now assume you had an algorithm $f$ which outputs $\mathrm{Val}_2$. Consider the function $f'$ which works like $f$, but discards all formulas in the output which are not of the form $\mathrm{PA}_2\rightarrow\varphi$ where $\varphi$ is a $\mathrm{PA}_1$-formula. Since this is only a simple syntactic check, $f'$ is computable as well and so $A$ is r.e. From this you get the desired contradiction.


问:
The area between two curves
Hi there's one problem on my study guide that my teacher didn't go over and I don't know how to approach/solve it. Here's the problem:

Find the area between the curves on the given interval. Draw a graph of the functions and the region.

$$y=x^4 , y=x-1, -2 \le x \le 0.$$

答:
Hint: determine if there are any intersection points of these two curves on the specified intervals. If there are none, great! Determine which one of the two is greater on the interval (WLOG call the greater of the two $y_{1}$, and the smaller $y_{2}$; graphing should help you determine this!), and then find $$\int_{-2}^{0}y_{1}-y_{2} dx$$ If there are intersection points, break up the integral into several subintervals of $[0, 2]$ and again determine which curve is greater on each portion of the integral. Then integrate each piece exactly as you did for the case that there are no intersection points.

Edit: at intersection points, the function which is greater might change, and so you would have to break up the integral accordingly to get the right subtraction of functions. In this case, $x^{4}$ and $x-1$ don't intersect each other on the interval $[-2, 0]$, and so you don't have to worry about it; $x^{4} > x - 1$ for all $-2 \leq x \leq 0$.

Title: The area between two curves


问:
Dominant and Injective Morphisms
As mentioned before, I am self-studying some commutative algebra out of "A Course in Commutative Algebra" by Kemper. In the text, Kemper has the following problem:

Let $X$ and $Y$ be affine varieties over a field $ K$, and let $f: X → Y$ be a morphism with induced
homomorphism $ϕ: K[Y ] → K[X].$ We say that $f$ is dominant if the image
$f(X)$ is dense in $Y$ , i.e., $f(X) = Y$ .

Show that $f$ is dominant if and only if ϕ is injective.

He provides a proof of the forward implication namely,

Assume that $f$ is dominant, and let $ϕ(g) = 0$ with $g ∈ K[Y]$. Viewing $g$
as a regular function $Y → K,$ we have $g\circ f = 0,$ so the restriction of $g $ to
$f(X)$ is zero. Since $g$ is continuous and $f(X)$ is dense, this implies $g = 0$.

I am trying to understand the overall structure of this proof however I had 3 questions regarding some details.

In viewing $g: Y \rightarrow K$ as a regular function, how comes he can conclude that  $g\circ f = 0,$?

Having  $g \circ f = 0,$, how does this imply that restriction of $g $ to
$f(X)$ is zero?

I see why $g$ is continuous ( I believe here he means continuous in the Zariski Topology), but how does the continuity of $g$ and the fact by the hypothesis $f(X)$ is dense imply that $g = 0$.

Once we have the result that $g = 0$, the injectivity follows immediately. Any help in coming to terms with my 3 questions would be highly appreciated.

答:
Let me try to answer your questions:

This is just the definition of the morphism $\phi$. Under the assumption that $g\in\ker(\phi)$ this means $g\circ f=0$.

$g\circ f=0$ means that for any point $x\in X$, $g(f(x))=0$, so the restriction of $g$ to $f(X)$ is zero.

There is the following theorem: Let $T$ be a reduced scheme and $S$ a separated scheme, $U\subset T$ dense open set. Then two morphisms $f,g \colon T\rightarrow S$ that agree on $U$ must be equal. As you are working with affine varieties all of these properties are satisfied, so $g=0$.

Hope that helps!

Title: Dominant and Injective Morphisms


问:
Can the equation of a vertical line parallel to the y-axis be $x+y+1=0$?
I found a conics problem in my book which has given me the equation of the directrix and the coordinates of the focus of a rightward opening parabola (open end towards the positive side of the $x$-axis) and has asked me to draw the conic section (parabola).

In the problem, the equation of the directrix is given as $x+y+1=0$, but isn't the directrix of a rightward opening parabola a vertical line parallel to the $y$-axis, and shouldn't it be in the form $x=c$?

答: You are correct, the directrix should be of the form $x=c$. There might be something else to the question. The parabola might have been rotated.


问:
What is the contragredient representation?
Let $V=M_2(\Bbb C)$ be the set of all $2$x$2$-matrices. Let $G=B$x$B$ where $B$ is the group of $2$x$2$ lower triangular matrices with non-zero diagonal entries. Then G acts on $V$ by $\rho (g,h)x=gx^th$ for $x \in V$ and $(g,h) \in G$. What is the representation $\rho^*$ of $G$ contragredient to $\rho$? </p>

答:
First, I'll rewrite the formula for $\rho$ so that I find it more understandable (I know it is also somewhat accepted notation to write the "t" to the left, but in this case I first read it as "$x$ transposed"):
$$\rho(g,h)x=g x h^t.$$ (That is, $h$ is getting transposed).

The space $V$ has a non-degenerate symmetric bilinear form defined by
$$(x,y)=\mathrm{trace}(xy).$$ This form identifies $V$ with its dual. In terms of this identification, the dual ("contragredient") representation $\chi(g,h)$ is given by

$$\chi(g,h)(y)(x)=y(g^{-1} x (h^{-1})^t)=\mathrm{trace}(y g^{-1} x (h^{-1})^t)=\mathrm{trace}((h^{-1})^t y g^{-1} x)=((h^{-1})^t y g^{-1})(x).$$ In other words,
$$\chi(g,h) y =(h^{-1})^t y g^{-1}.$$


问:
Family of RV is uniformly integrable iff there is $\phi$ with $\frac{\phi(x)}{x}\rightarrow\infty$ and $\sup_n E(\phi(\vert X_n\vert))<\infty$
I want to show that a family $(X_n)_{n\in I}$ of real-valued RVs is uniformly integrable if and only if there exists a measurable function $\varphi:[0,\infty)\rightarrow[0,\infty)$ with $\frac{\varphi(x)}{x}\xrightarrow{x\rightarrow\infty}\infty$ and $\sup_{n\in I}E(\varphi(\lvert X_n\rvert))<\infty$.

There is a hint, which says to consider for "suitable" $(C_k)_k\subseteq\mathbb R$ the function
$$\varphi[0,\infty)\rightarrow[0,\infty], x\mapsto\sum_{k=1}^\infty (x-C_k)1_{\{x>C_k\}}.$$
This hint raises may questions for me than it answers.

We have checked two conditions for uniform integrability of a family of RVs, which are that they are (1) bounded in $L^p$ for some $p>1$, or alternatively that they (2) are dominated by an integrable RV.

On a high level, I see the relationship between the first condition and the fact that $\varphi$ grows superlinearly (in the theorem). But I can't begin to formulate a proof. Is this a good place to start? If yes what does the start look like?

Also, I can't find any use for the hint. The function doesn't even match the requirements put on $\varphi$, and I don't see what $(C_k)$ would be "suitable". Does someone see the relationship between the hint and the problem?

答:
First of all, the function $\varphi$ given in the hint should be well-defined: if $x\geq C_k+1$ for all $k$ then $\varphi(x)$ would be infinite hence we should have that $C_k\to\infty$.

Moreover, observe that
$$
\varphi\left(\lvert X_n\rvert\right)=\sum_{k=1}^\infty \left(\lvert X_n\rvert-C_k\right)\mathbf{1}_{\{\lvert X_n\rvert>C_k\}}\leqslant \sum_{k=1}^\infty \lvert X_n\rvert \mathbf{1}_{\{\lvert X_n\rvert>C_k\}}
$$
hence it would be sufficient to find $C_k$ such that $C_k\to\infty$ and for each $n$,
$$
\mathbb E\left[ \lvert X_n\rvert \mathbf{1}_{\{\lvert X_n\rvert>C_k\}}\right]\leqslant 2^{-k}.
$$


问:
Order and degree of a differential equation
Here is a question in my book

Find the order and degree of the differential equation $$y=1+\frac{dy}{dx}+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+\cdots$$

At first sight we can conclude that the order is $1$ and the degree is undefined as  as the power of $\frac{dy}{dx}$ continues to increase and has no limit.However  my book gives the following solution

Rewrite the DE as $$y=\exp\left({\frac{dy}{dx}}\right)$$ $$\implies \frac{dy}{dx}=\ln y$$ whose order and degree is 1 .

Now ,I completely agree with this solution however I find it rather counterintuitive to my first line of thought .If the book is correct how can it be justified to prove my intuition was wrong?

答:
Your intuition about the order of given ODE is correct as writing the RHS into closed form function never changes it. But the degree of an ODE is a parameter which is something vague to talk about as one can always disturb it just by squaring both sides.

Though if you constraint yourself to the topic linear ODE (which seems to be true as per your syllabus of JEE) then $y=1+y'+\frac{y'^2}{2!}+....$ is a non-linear ODE which on linearization gives you the linear version : $y'=\log_e  y$ has degree $1$ (if you still want to define this parameter).


问:
A conditional probability problem where the next day depends on the last 3 days
For many years, Meteorologists have spent long visits (5 days) at the Bigtown.
They have observed that, for three consecutive days, if there are EXACTLY two sunny days, the next day is a sunny day*, while half of all the cloudy days are followed by another cloudy day. Assuming days are either cloudy or sunny, estimate how many days have been cloudy in their last ten visits. Visits are not necessarily following each other.

''half of all the cloudy days are followed by another cloudy day'' so $P(C\mid C)=50\%$

Is this possible? If not why not?

*i.e.
SSC -> S

SCS -> S

CSS -> S

where -> means followed by

答:
Note: edited since the question was clarified.

First, we cannot ever have two or more consecutive sunny days. If we do, then the next cloudy day (which may not be for some time, but will certainly be preceded by two sunny days) must be followed by two sunny days (since SSC->S, SCS->S), and the same for the next cloudy day after that, and so on. In particular, every future cloudy day is followed by a sunny day, contradicting the assumption that $50\%$ of cloudy days are followed by sunny ones.

Second, we cannot ever have two sunny days separated by a single cloudy day, since then the next day will be sunny, contradicting the previous point.

Now we know that in between every two sunny days there are at least two cloudy days. These cloudy days must include one which is followed by a sunny day, and at least one which is followed by another cloudy day. Since these two things are equally common, we must have exactly two cloudy days between every two sunny days.

Thus the unique sequence satisfying these properties is ...SCCSCCSCC... in which exactly $2/3$ are cloudy.


答:
The book is indeed incorrect, when you do such modifications on the equation, the order and degree changes. You must read the order and degree of the equation you are given without any simplification. To illustrate what I mean, suppose I give you:

$$ y' =y$$

I could square the above and get:

$$(y')^2 = y^2$$

Now, look at the two equations you may see first equation was first degree and second is second degree.

However, in case of JEE and board exams (Indian syllabus stuff) , usually they want you to simplify before saying degree or order (this is actually wrong though)

Related

Title: Order and degree of a differential equation

Body:
Here is a question in my book

Find the order and degree of the differential equation $$y=1+\frac{dy}{dx}+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+\cdots$$

At first sight we can conclude that the order is $1$ and the degree is undefined as  as the power of $\frac{dy}{dx}$ continues to increase and has no limit.However  my book gives the following solution

Rewrite the DE as $$y=\exp\left({\frac{dy}{dx}}\right)$$ $$\implies \frac{dy}{dx}=\ln y$$ whose order and degree is 1 .

Now ,I completely agree with this solution however I find it rather counterintuitive to my first line of thought .If the book is correct how can it be justified to prove my intuition was wrong?


问:
Determinant Expression for Grover operator
I want find the characteristic polynomial for the grover quantum operator $U^{N\times N}$

$$\begin{align*}
U=(2|D\rangle\langle D| -I_N)(2|M^{\perp}\rangle\langle M^{\perp}| -I)
\end{align*}$$

where
$|D\rangle = \dfrac{1}{\sqrt{N}}\sum_{i=1}^{N} |i\rangle$; $|M\rangle = \dfrac{1}{\sqrt{N-m}}\sum_{i\notin M} |i\rangle$
; $N=2^n$

答:
I will assume $|M^{\perp}\rangle$ is just a normalized vector. Notice that the subspace $\mathcal{S}$ generated by $|D\rangle$ and $|M^{\perp}\rangle$ is an invariant subspace for $U$. Let me define $|e_1\rangle =|D\rangle$ and $|e_2\rangle = |M^{\perp}\rangle$. Expanding the product you will notice that

$$ U |e_i\rangle = \sum_{j=1}^{2} U_{j,i} |e_j\rangle $$

and the coefficients $U_{i,j}$ are essentially functions of $\langle D|M^{\perp}\rangle$. In this way you obtained a matrix representation of your operator $U$ in $\mathcal{S}$. This is a $2\times2$ matrix and you should be able to find the eigenvalues easily. On the remaining subspace $\mathcal{S}^\perp$ It should be clear that

$$U |\xi\rangle = |\xi\rangle, \quad \mathrm{ for } \quad |\xi\rangle \in \mathcal{S}^\perp$$

So in this subspace you find eigenvalue 1 with multiplicity $N-2$ (its dimension). Now you have all the eigenvalues.

Note that your vectors $|D\rangle$ and $|M^{\perp}\rangle$ don't need to (and in general won't) be orthogonal. But the determinant is defined irrespective of the scalar product and is hence invariant under any change of basis (read similarity transformation). However to realize that the above procedure works it may be helpful to imagine that you could in principle find an orthonormal basis whose first two vector (say) span $\mathcal{S}$ and the remaining span  $\mathcal{S}^\perp$.

Hint

In the subspace $\mathcal{S}$, in the basis $|D\rangle, |M^{\perp}\rangle$, $U$ has the form (if I'm not wrong)

$$
\left. U \right|_{\mathcal{S}}= \left(\begin{array}{cc}
4\left|x\right|^{2}-1 & 2\overline{x}\\
-2x & -1
\end{array}\right)
$$

with $x=\langle D|M^{\perp}\rangle $. Correctly its eigenvalues have modulus one. In fact if $P$ is an orthogonal projector, $2P-I$ is unitary so that your $U$ is unitary being a product of two unitary.


问:
Sharing out money
Say I have $M$ dollars that need to be shared between 4 people. Each person will obviously get $\frac{M}{4}$.

However, each of the 4 people has already taken some so that there is none left.

Person A took $a$ dollars, person B took $b$ dollars, person C took $c$ dollars and person D took $d$ dollars.

If $a=b=c$ then how much do they give $d$ so that $a=b=c=d$?

I tell person A to give $w$ amount to everyone else, if they don't have equal money, then I tell person B to give $x$ to everyone, if they don't have equal money, then I tell person C to give $y$ to everyone else, if they don't have equal money, then I tell person D to give $z$ to everyone else.

What is a formula for $w,x,y$ and $z$ such that once person D gives away $z$ everyone has equal money?

答:
1.If $a > d$ (which seems to be implied by your question), then A, B and C each need to give D the following amount :

$$\frac{\frac{M}{4} - d}{3}$$

(D is missing $\frac{M}{4} - d$ dollars to have a fair share, and since A, B and C each have the same amount, each needs to contribute a third of the missing money to D)

However, it's entirely possible that $d > a$, in which case D would be the one having to give money back to A, B and C. The amount that D would need to give back to each of them is :

$$\frac{M}{4} - a$$

2.If you can devise a system such that, after C has given away $y$ dollars, (i) D has more money than A and (ii) A, B and C have the same amount of money, then $z$ can be calculated using the answer to the previous question.

In a similar fashion, a formula for $y$ that allows one to reach this desired state (where both (i) and (ii) are true) can be calculated assuming that by the time C is giving away money, (i') A and B already have the same amount of cash and (ii') both C and D have more money than A.

Then, to completely solve the puzzle, you only need to figure out values for $w$ and $x$ such that (i') and (ii') are true before C has to share any money.

Such a solution exists and can be found with a bit of trial and error ;)

Title: Sharing out money


问:
Hyperbolic Set as an intersection of Halfspaces
The hyperbolic set in my problem is defined as:

$H = \{x \ \epsilon \ R_+^2 \ | \ x_1x_2 \ \geq \ 1\}$

I need to show that this can be written as an intersection of halfspaces. In the previous part I showed that it is convex using first principles so there must exist such an intersection of halfspaces that gives us this set but I'm not sure where to start. What exactly constitutes as an intersection of halfspaces?

答:
Let me use $x,y$ coordinates, and so your set is
$$H = \{(x,y) \in \mathbb R^2_+ \mid xy \ge 1\}
$$
Consider the bounding hyperbola $xy=1$ (where $x,y > 0$). For each point $(a,b)$ on this hyperbola, first find the tangent line to the hyperbola at that point, and then find the half-space of that tangent line which contains the hyperbola. Your set $H$ is the intersection of those half-spaces, one for each such point $(a,b)$.

In detail, using calculus we find that the tangent line to $xy=1$ at $(a,b)$ has slope equal to $-\frac{1}{a^2}$, and therefore its equation is given in point slope by the formula
$$\frac{y-b}{x-a} = - \frac{1}{a^2}
$$
which can be rewritten
$$x + a^2 y = a + a^2 b
$$
We can eliminate $b=\frac{1}{a}$ to get
$$x + a^2 y = 2a
$$
The corresponding half-space which contains the hyperbola is
$$H_a = \{(x,y) \in \mathbb R^2 \mid x + a^2 y \ge 2a\} \quad\text{for all $a > 0$}
$$
Your set $H$ can then be written as the intersection of these half-spaces:
$$H = \bigcap_{a > 0} H_a
$$
Now there's something to prove: you must show that for all $(x,y) \in \mathbb R^2_+$ the following is true:

$xy \ge 1$ if and only if for all $a > 0$, $x + a^2 y \ge 2a$

Even though this solution was discovered by calculus, I am guessing that this "if and only if" can be proved with straightforward algebra and inequalities, but maybe I'll stop here.


问:
Proving F to be equicontinuous on compact sets
The following question is from Ponnusamy and Silvermann Complex Variables with applications (Pg 395) and I need help in solving it.

Suppose that for each point in domain D there corresponds a neighbourhood in which a family F is equicontinuos. Show that F is equicontinuos on compact subsets of D.

I proved that family is locally  uniformly bounded on D, which implies that F is uniformly bounded on each compact subsets of D. But  F is not given to be a family such that each of it's member is analytic. If such has been given then I would have used: " If F is Locally uniformly bounded
of analytic functions in a domain D, then F is equicontinuous  on compact subsets of D".

Also, for part ii , I thought quite a lot but wasn't able to make any progress.

So, Please help me!

答:
This is actually a general metric space result.

Claim: There exists $r\colon D\to\mathbb{R}$ continuous such that $B(z,r(z))\subseteq D$ for all $z$ and $\mathscr{F}\vert B(z,r(z))$ is equicontinuous.

Proof: Let $R\colon D\to\mathbb{R}\cup\{\infty\}$ be defined by $R(z)=\sup\{r\in\mathbb{R}:\mathscr{F}\vert B(z,R)\text{ is equicontinuous}\}$.  Then the triangle inequality gives $R(z)\geq R(w)-\operatorname{dist}(z,w)$ for all $z,w\in D$ (with obvious meaning when $\infty$ are around) so $R$ is 1-Lipschitz.  Now let $r=\min\{\frac12 R, 1\}$.  $\checkmark$

For any compact $K\subseteq D$, we have $\inf_K r=r_K>0$.  Cover $K$ by finitely many open balls of radius $\frac12r_K$, say centred at $z_1,\dots,z_n$.  Given $\epsilon>0$, we find for each $i=1,\dots,n$ a $\delta_i>0$ by equicontinuity of $\mathscr{F}\vert B(z_i,r_K)$:
$$
(\forall f\in\mathscr{F})(\forall z,w\in B(z_i,r_K))(\operatorname{dist}(z,w)<\delta_i\implies \operatorname{dist}(f(z),f(w))<\epsilon)
$$
and we may assume $\delta_i\leq r_K$.  Let $\delta=\frac12\min\{\delta_1,\dots,\delta_n\}$. Then for every $z\in K$, we have $B(z,\delta)\subseteq B(z_i,r_K)$ for some $i$.  Hence for all $f\in\mathcal{F}$ and all $w\in B(z,\delta)$, we have $\operatorname{dist}(f(z),f(w))<\epsilon$ by the equicontinuity condition on $\mathscr{F}\vert B(z_i,r_K)$.


问:
$X^5-6X+3$ has two complex roots
Show that the polynomial $X^5-6X+3$ has exactly two non-real roots.

I am struggling to solve this question for some time now and I wonder whether there is a theorem that helps to answer it.

Any help is appreciated.

答:
We can apply Descartes' Rule of Signs

$f(x) = x^5 - 6x + 3\qquad$// 2 sign changes, 0 or 2 positive roots.

$f(-x) = -x^5 + 6x + 3\;\,$// 1 sign change, thus 1 negative root.

$f(0) = 3 \\ f(1)=-2$

Since f is continuous, there must have a root between 0 and 1.

→ there are 2 positive roots, 1 negative root.

→ Complex roots = 5 - 2 - 1 = 2

Title: $X^5-6X+3$ has two complex roots


问:
A Question from Mathematical circles Book
A chessboard has the form of a cross, obtained from 4x4 chessboard by deleting the corner squares (see figure https://i.stack.imgur.com/MO0eO.jpg). Can a knight travel around this board, pass through each square exactly once, and end on the same square he starts on?

I tried to solve the problem by making a graph (see figure https://i.stack.imgur.com/Ok5DP.jpg) and then transformed this graph for better visualization (see figure https://i.stack.imgur.com/9MvVI.jpg) but still the answer I am getting is such a path is not possible, while in solutions it is given such a path is possible and can be easily constructed. (see figure https://i.stack.imgur.com/aq79e.jpg)

答:
I don't know if I understood the problem correctly, but this could be a solution:
https://i.stack.imgur.com/K3Fze.jpg</p>


答:
Let $E$ the region given by two curves $f$, $g$, which are functions with domain $[a,b]$ in the $x$ axis and $f>g$. Then:

$$A(E)=\iint_E 1\, dA=\int_a^b\int_{g(x)}^{f(x)} 1\, dydx=\int_a^b (f(x)-g(x))\,dx=\int_a^b f-\int_a^b g $$

Separe in cases when $f>g$ and $f\leq g$.

Title: The area between two curves

Body:
Hi there's one problem on my study guide that my teacher didn't go over and I don't know how to approach/solve it. Here's the problem:

Find the area between the curves on the given interval. Draw a graph of the functions and the region.

$$y=x^4 , y=x-1, -2 \le x \le 0.$$


问:
How to Invert a Polar Lemniscata into a hyperbola
Hello i want invert a lemniscate but i find few information on internet, i know how to find the inverse of a function in rectangular coordinates functions but i dont know how to do it in polar coordiantes and how inverting a lemniscate become a hyperbola?.

i am using python for plot because is an assigment.
this is the equation:

r = (a**2)(np.cos(2rad))

```<code>import numpy as np
import matplotlib.pyplot as plt
import math

plt.axes(projection = 'polar')
a=3
rads = np.arange(0, (2 * np.pi), 0.01)

for rad in rads:
    r = (a**2)*(np.cos(2*rad))
    plt.polar(rad,r,'g.')
plt.show()
</code>```

grafica python

答:
The set of MacLaurin polar curves obtained by inversion $ r\to \dfrac{a^2}{r}$

$$ r^n = a^n \cos n \theta, r^n = a^n \sec n \theta $$

In the particular case $n=2$ Lemniscate of Bernoulli and rectangular hyperbola are mutually invertible about radius of inversion circle $r=a=1$

```<code> PolarPlot[ {1,Sqrt[Sec[2 t]],Sqrt[Cos[2 t]]},{t,0,2 Pi},PolarGridLines->Automatic]
</code>```

An inversion  transformation between Cartesian coordinates can be written

$$ (x,y)\to \dfrac{a^2\;(x,y)}{x^2+y^2}$$

For $n=1,3$ and inversions:

Body:
The set of MacLaurin polar curves obtained by inversion $ r\to \dfrac{a^2}{r}$

$$ r^n = a^n \cos n \theta, r^n = a^n \sec n \theta $$

In the particular case $n=2$ Lemniscate of Bernoulli and rectangular hyperbola are mutually invertible about radius of inversion circle $r=a=1$

 PolarPlot[ {1,Sqrt[Sec[2 t]],Sqrt[Cos[2 t]]},{t,0,2 Pi},PolarGridLines->Automatic]

An inversion  transformation between Cartesian coordinates can be written

$$ (x,y)\to \dfrac{a^2\;(x,y)}{x^2+y^2}$$

For $n=1,3$ and inversions:


问:
The natural sequence {$a_n$} is given. Need to prove that from some point all values of $a_n$ will be the same.
Natural number $a_0$ is given. The natural sequence $\{a_n\}$ is given so that $a_n$ is the smallest possible number that satisfies this:   $$\sqrt[n]{a_0\times a_1\times a_2 \times\cdots\times a_n} = \text{natural number}.$$

Prove that after some point all the values of $a_n$ will be the same.

I've done this:

Variable $a_1$ will be $1$ because $a_0\times a_1 \in \mathbb{N}$.
If $a_0$ is a prime  then $a_0\times a_1\times a_2 = x^2$, that means $a_2 = a_0$, and in the same way $a_3 = a_0,\;\cdots,\; a_n=a_0$. That means for prime $a_0$'s it's true.

Then I tried a non-prime example for $a_0$. It was $a_0 = 2^3\times 3$.
And it really gave me that from some point $a_n$-s are the same.
I don't know how to prove this for $a_0 = p_1^{b_1}\times p_2^{b_2}\times\cdots\times p_k^{b_k}$.

答:
Hint: you can try to prove the inverse, this is, if they aren't the same at from a point forwards, then it's not going to be the smallest $a_n$, but this obvious:

We know that each prime factor $p_k$ for some $k$ needs to appear at least $n$ times, then it's clear that $p_k^n \geq 2^n$ if $p_k > 2$ for example, and for each $p_k$ we take different from all others such that we have a non-decreasing monotone sequence.


问:
Finding equilibrium points of system of equations
I'm asked to solve system of ODE for finding equilibrium points of this

$$\frac{dx}{dt} = a_1x(1-x) + a_2x\frac{z}{1+z}+a_3xy$$
$$\frac{dy}{dt} = b_1y(1-y) + b_2y\frac{z}{1+z}+b_3xy$$
$$\frac{dz}{dt} = c_1z(1-z) + c_2z\frac{x}{1+x}+c_3z\frac{y}{1+y}$$

I checked some related links to solve this problem and i know that i need to extract variables from this system to matrix form: link:https://www.math24.net/linear-autonomous-systems-equilibrium-points/
but problem how to extract them

My question is how correctly do change of variable for further extraction to matrix form

答:
Hint

An equilibrium point is a triple $\left(x , y , z\right)$ satisfying the non-linear system

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{a}_{1} x \left(1-x\right)+{a}_{2} x \displaystyle  \frac{z}{1+z}+{a}_{3} x y&=&0\\
{b}_{1} y \left(1-y\right)+{b}_{2} y \displaystyle  \frac{z}{1+z}+{b}_{3} x y&=&0\\
{c}_{1} z \left(1-z\right)+{c}_{2} z \displaystyle  \frac{x}{1+x}+{c}_{3} z \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}

The most interesting case is when $x \neq  0 , y \neq  0 , z \neq  0$. Then the system reduces to

\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{a}_{1} \left(1-x\right)+{a}_{2} \displaystyle  \frac{z}{1+z}+{a}_{3} y&=&0\\
{b}_{1} \left(1-y\right)+{b}_{2} \displaystyle  \frac{z}{1+z}+{b}_{3} x&=&0\\
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle  \frac{x}{1+x}+{c}_{3} \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}

From the two first equations, one deduces in general a relation of the form
$x = {\alpha} y+{\beta}$. Substituting again in these two first
equations leads to

\begin{equation}
x = \frac{P \left(z\right)}{1+z} \qquad  y = \frac{Q \left(z\right)}{1+z}
\end{equation}

where $P$ and $Q$ are polynomials of degree $1$ which coefficients
depend on ${a}_{1} , {a}_{2}, a_3 , {b}_{1} , {b}_{2}, b_3$.

Injecting these values of $x$ and $y$ in the third equation leads to

\begin{equation}
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle  \frac{P \left(z\right)}{1+z+P \left(z\right)}+{c}_{3} \frac{Q \left(z\right)}{1+z+Q \left(z\right)} = 0
\end{equation}

This gives a cubic equation in $z$ with potentially three complex
roots, producing three equilibrium points.

With the help of a computer algebra system all the equilibrium points of this system can be computed.


问:
Proof of CDF of nakagami-m distribution?
I want to derive the CDF of Nakagami-m distribution. Here is what i had tried:

My approach: We know that to find CDF, we have to integrate the PDF. Hence, first writing the PDF of nakagami random variable (X) as $f_X(x)=\frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)^mx^{(2m-1)}e^{-\left(\frac{m}{\Omega}x^2\right)}$-------(1).

Next, i assume $\alpha = m$ and $\beta = \frac{m}{\Omega}$.

Thus (1) becomes: $f_X(x)=\frac{2}{\Gamma(\alpha)}\left(\beta\right)^{\alpha}x^{(2\alpha-1)}e^{-\beta x^2}$-------(2).

Integrating (2) we get CDF as $F_{X}(x) = $ $\int_0^u\frac{2}{\Gamma(\alpha)}\left(\beta\right)^{\alpha}x^{(2\alpha-1)}e^{-\beta x^2}\text{dx}$

Thus, $F_{X}(x) = \frac{2\beta^{\alpha}}{\Gamma(\alpha)}\int_0^ux^{(2\alpha-1)}e^{-\beta x^2}\text{dx} $------(3)

I am not getting how to further solve Eq. (3).

Any help in this regard is highly appreciated.

答:
$$F_X(u)=\int_{0}^{u} f_X(x)dx=\int_{0}^{u} \frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)^mx^{(2m-1)}e^{-\left(\frac{m}{\Omega}x^2\right)}dx$$

Now, substitute $t=\frac{m x^2}{\Omega}$, $dt=\frac{2 m x}{\Omega}$ to derive

$$F_X(u)
=\int_{0}^{u} \frac{2}{\Gamma(m)}\left(\frac{m}{\Omega}\right)x \left(\frac{m x^2}{\Omega}\right)^{(m-1)}e^{-\left(\frac{m x^2}{\Omega}\right)}dx
=\frac{1}{\Gamma(m)}\int_{0}^{\frac{2u^2}{\Omega}}  t^{(m-1)}e^{-t}dt=\frac{\gamma(m,\frac{2u^2}{\Omega})}{\Gamma(m)}=P(m,\frac{2u^2}{\Omega})$$

$\gamma$ is the lower incomplete gamma function and $P$ is the regularized Gamma function


问:
Is there a simpler closed form for $\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}$
I have the following infinite sum that can be expressed in terms of the generalized hypergeometric function:
$$\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}=\frac{31}8-4\times{_4F_3}\left(-\frac12,\frac12,1,1;\ 2,2,2;\ 1\right)\\\ \\\approx0.008749644047541935203478962326551903908774780849356243615274...$$
I wonder if it can be expressed in terms of simpler functions and well-known mathematical constants.</p>

答:
I can give a partial answer:

My program based on the TranscendentalRecognize algorithm using a wide set of commonly occuring constants after several hours of work discovered the following inequality:

$$\Bigg|\frac{16\ln2}{3}-\frac{32C}{3\pi}+\frac{560}{27\pi}-\frac{56}{9}-{_4F_3}\left(-\frac{1}{2},\frac{1}{2},1,1;2,2,2;1\right)\Bigg|<10^{-1000},$$
where $C$ is the Catalan constant.

I have no idea if the actual difference is exact $0$, or how to (dis-)prove it.


问:
Show that the statistic $T(x_1,...,x_n)=\sum_{i=1}^nx_i^2$ is complete.
Let $X_1,...,X_n$ be iid random sample form $N(\theta,c\theta)$, where $c$ is a known constant.
Show that the statistic $T(x_1,...,x_n)=\sum_{i=1}^nx_i^2$ is complete.

In other words I have to show that if $$E[g(T)]=0\quad \Longrightarrow \quad g(T)=0 \quad \forall t$$

I'm kind of stuck with the distribution of $T$ because $x_i^2$ isn´t a gaussian distribution. Any suggestions on how to approach this problem would be great!

答: This is Proposition 2.1 on page 110 in [Jun Shao]Mathematical Statistics(2nd edition).</p>


答:
Numbering the squares 1-12, a sequence is
$$1,6,8,2,10,4,12,7,5,11,3,9,1$$
The way I found it was first to notice that each of the four center squares can only be reached by two other squares, so any complete path through 4, say, must contain the sequence $10,4,12$ or $12,4,10$. Similarly, the groupings have to include
$$10,4,12$$
$$7, 5, 11$$
$$2, 8, 6$$ and
$$1,9,3$$
From there it was simply connecting the end points of the four segments to find a path.

Title: A Question from Mathematical circles Book

Body:
A chessboard has the form of a cross, obtained from 4x4 chessboard by deleting the corner squares (see figure https://i.stack.imgur.com/MO0eO.jpg). Can a knight travel around this board, pass through each square exactly once, and end on the same square he starts on?

I tried to solve the problem by making a graph (see figure https://i.stack.imgur.com/Ok5DP.jpg) and then transformed this graph for better visualization (see figure https://i.stack.imgur.com/9MvVI.jpg) but still the answer I am getting is such a path is not possible, while in solutions it is given such a path is possible and can be easily constructed. (see figure https://i.stack.imgur.com/aq79e.jpg)


问:
Image of an intersection is the intersection of the image
I'm studying the proof that path-connectedness implies connectedness. However there is one assertion that I don't understand.

Let $f:[0,1] \rightarrow X$ be a continuous mapping where $X$ is a metric space. Let $U$ and $V$ be two disjoint subsets of X such that $X = U \cup V$.

I don't understand why $f^{-1}(U) \cap f^{-1}(V) = \emptyset $.

Is it because:

$$ \emptyset = f^{-1}(\emptyset) = f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$$

That makes sense but I thought the last equality $f^{-1}(U \cap V)=f^{-1}(U) \cap f^{-1}(V)$ only holds iif $f^{-1}$ is injective. Does that mean $f^{-1}$ is injective or am I missing something?

Or is it because $f$ is continuous?

For a reference to the proof: https://topospaces.subwiki.org/wiki/Path-connected_implies_connected

答: $f^{-1}[U \cap V] = f^{-1}[U] \cap f^{-1}[V]$ always holds. The function $f$ need not be 1-1 or onto etc. You’re confused with $f[U \cap V] = f[U] \cap f[V]$ which does only hold if $f$ is injective. Forward and inverse images behave quite differently.


问:
Jacobi Method with error
Preparing for my Numerical Analysis exam,

If the Jacobi's method is used to solve the linear system, $Ax=b$, where
$$A=\begin{pmatrix}5 & -2 & 3\\-3 & 9 & 1\\2&-1&-7\end{pmatrix}$$

will the method be convergent?

This part I think I can do. Since there is no $b$ vector as in $Ax = b$, we cannot iterate toward a solution. Instead we reason that $A$ will converge using the Jacobi method as it is diagonally dominant.

Then it says, Determine the matrix G in the error equation $e_{n+1} = Ge_n$, and test the convergence criterion using either $||\cdot||_\infty$ or $||\cdot||_1$ norm.

I know that the norms are the max of the column and row sums, but I don't understand the "find the matrix G in the error equation" part of the question.

答:
In methods like Jacobi and Gauss-Seidel, you create a fixed point iteration of the form $f(x_k) = x_{k+1} = Tx_k + c$. For it to work, you require that at your solution $x^{*}$ the iteration returns $x^{*}=f(x^*)=Tx^{*} + c$ . We can express the equation $Ax=b$ as

$$Ax = (L + D + U)x = b$$
$$Dx = -(L+U)x + b$$
$$x = -D^{-1}(L+U)x + D^{-1}b$$

If we turn this into an iteration we already have the form we wanted above and we know that at the solution the iteration will return the solution, since that is how we got it in the first place.

This is actually the Jacobi iteration, where the iteration matrix $T=-D^{-1}(L+U)$.

The question now is whether this iteration will converge and for which $x_0$. To answer this question we define the error at step $k$ and then substitute with the equations we derived above.

$$x_{k} - x^{*} = e_{k} = Tx_{k} + c - Tx^{*} - c = T(x_{k} - x^{*}) = Te_{k} = T^{k}e_{0}$$

So the error at each step is transformed by $T$ and is amplified or increased linearly by it. We want the error to decrease, so we need to derive a condition that guarantees that the error will disappear, which we can do using matrix norms and two associated inequalities $|Ax| \leq ||A||\; |x|$ and $||AB|| \leq ||A||\;||B|| $:

$|T^ke_0| \leq ||T^k||\;|e_0| \leq ||T||^k\;|e_0| \implies e_0 \rightarrow 0$ as $k \rightarrow \infty$ iff $||T||<1$ for some norm of T and any $x_0$.

So the $G$ in your case is the $T$ in my explanation and it is the iteration matrix that ends up remapping the error you make at each step of the iteration we have derived above.

Title: Jacobi Method with error


问:
General term of a sequence with multiples of 6
I am struggling to fin the general term of the following sequence:

$$\begin{aligned}
\text{for} \ n = 1,..,5 & \rightarrow n\\
n=6 & \rightarrow n+1 \\
n=7 & \rightarrow n+3 \\
n=8 & \rightarrow n+5 \\
n=9 & \rightarrow n+7 \\
n=10 & \rightarrow n+9 \\
n=11 & \rightarrow n+11 \\
n=12 & \rightarrow n+13+1 \\
n=13 & \rightarrow n+15+3 \\
n=14 & \rightarrow n+17+5 \\
n=15 & \rightarrow n+19+7 \\
n=16 & \rightarrow n+21+9 \\
n=17 & \rightarrow n+23+11 \\
n=18 & \rightarrow n+25+13+1 \\
\vdots
\end{aligned}$$

Any help? Thanks.

答: Let $I$ be the integer part of $\frac{n}{6}$. Then $$x_n=(2I+1)n+I-6I(I+1).$$


问:
Linear Regression Prediction Errors
Suppose that we perform linear regression on data $\mathbf{X}$ (an $N \times {(D+1)}$ matrix) and predictions $\mathbf{y}$ (an $N \times 1$ vector). Let $\mathbf{w}$ ($(D+1) \times 1$ vector) be the optimal parameters for minimizing the expected loss. Why is it true that for any vector $\mathbf{\tilde{w}}$
$$\mathbb{E}_{\rho(x_0,y_0)}\left[(y_0-\mathbf{w}^{T}{x_0})(\mathbf{w}-\mathbf{\tilde{w}})^{T} x_0\right]=0?$$
This implies that the the prediction error is uncorrelated with linear functions, but why is this true?

So far, I've tried using the fact that $\mathbf{w} = (\mathbf{X}^{T}\mathbf{X})^{-1}\mathbf{X}^{T}\mathbf{y}$, but I'm not sure how useful that would be.

答:
This statement is false.

I suppose you assume that $y$ was generated by a linear model, $y=x^Tv+\varepsilon$, with $x$ and $\varepsilon$ independent, and $E[\varepsilon]=0$.
If so, the statement holds for $w=v$:
$$
E_{\rho(x_0,y_0)}[(y_0−v^Tx_0)(v−\tilde{w})^Tx_0]=E_{\rho(x_0,y_0)}[\varepsilon_0(v−\tilde{w})^Tx_0]
$$
$$
=E_{\rho(x_0,y_0)}[\varepsilon_0](v−\tilde{w})^TE_{\rho(x_0,y_0)}[x_0]=0
$$
However, since your $w$ is an estimator, dependent on random observations, you generally won't have $w=v$.
The estimator you propose takes the form:
$$
w=(X^TX)^{-1}X^Ty=(X^TX)^{-1}X^T(Xv+\varepsilon)=v+(X^TX)^{-1}X^T\varepsilon=:v+L\varepsilon
$$
So, in fact you have:
$$
E_{\rho(x_0,y_0)}[(y_0−w^Tx_0)(w−\tilde{w})^Tx_0]
=E_{\rho(x_0,y_0)}[(y_0−(v+L\varepsilon)^Tx_0)(v+L\varepsilon−\tilde{w})^Tx_0]
$$
Exploiting linearity:
$$
=E_{\rho(x_0,y_0)}[(y_0-v^Tx_0)(v−\tilde{w})^Tx_0]+
E_{\rho(x_0,y_0)}[(y_0-v^Tx_0)(L\varepsilon)^Tx_0]-
E_{\rho(x_0,y_0)}[(L\varepsilon)^Tx_0(v−\tilde{w})^Tx_0]-
E_{\rho(x_0,y_0)}[(L\varepsilon)^Tx_0(L\varepsilon)^Tx_0]
$$
We previously saw that the first term is zero. Using the definition of $\varepsilon_0$,
$$
=0+E_{\rho(x_0,y_0)}[\varepsilon_0(L\varepsilon)^Tx_0]-
E_{\rho(x_0,y_0)}[(L\varepsilon)^Tx_0(v−\tilde{w})^Tx_0]-
E_{\rho(x_0,y_0)}[(L\varepsilon)^Tx_0 x_0^T(L\varepsilon)]
$$
This can be further simplified:
$$
=0-(L\varepsilon)^T E_{\rho(x_0,y_0)}[x_0(v−\tilde{w})^Tx_0]-(L\varepsilon)^TE_{\rho(x_0,y_0)}[x_0 x_0^T](L\varepsilon)
$$
If $\Sigma$ is the covariance matrix of $x_0$, we have:
$$
=-\varepsilon^TL^T E_{\rho(x_0,y_0)}[x_0(v−\tilde{w})^Tx_0]-\varepsilon^T L^T\Sigma L\varepsilon
$$
This term is random (as it depends on $\varepsilon$), and generally won't be zero.
If you want, you can take the expectation with respect to $\varepsilon$ or even $L$.
This will eliminate the frist term.
The second term, however, is a quadratic form, and will not be zero almost surely.

Title: Linear Regression Prediction Errors


问:
A region is simply connected if $\theta < 2\pi$, and not simply connected if $\theta \ge 2\pi$
The question is:
Which of the following regions are simply-connected? (...)

e) in polar coordinates, the region where $r>0,0<\theta<\theta_0$.

The solution says:

[simply connected] if $\theta_0<2\pi$, no if $\theta_0\ge2\pi$, since then R is the plane with $(0,0)$ removed.

What is the logic behind the answer? How come that $(0,0)$ sort of removes itself when $\theta_0\ge2\pi$?

I've checked answers like this one and this one, but (although the first adds intuition) they delve into topology, which is too much for me at this point.

This question comes from Multivariable Calculus course (for reference, exercise 4G-5), not from Topology course, so please don't use into (too much) topology.

答:
$(0,0)$ is excluded by the requirement $r>0$. Though technically since you said $0<\theta<\theta_0$ (both bounds strict) the case $\theta_0=2\pi$ is still simply connected, because $\{ (r,\theta) : r>0,0<\theta<2\pi \}$ corresponds to the plane minus the nonnegative $x$ axis.

The intuitive reason why the plane with a hole in the origin isn't simply connected is because you can draw a circle containing the origin and then you can't continuously contract it to a point; no matter what you do, either at some point you will cross the origin or else the origin will be inside the loop throughout the process.

The intuitive reason why a wedge with the origin cut out of it is simply connected is that you can't draw a loop that encloses the origin at all, so any loop can be contracted to a point other than the origin.


问:
If a policyholder filed a claim, what is the probability that it was not a low claim?
An insurance company classifies its incoming claims as low if they under $10,000$ USD, medium if
they are $10,000$ USD or above but under $20,000$ USD, and high otherwise. During the year, $79.2$%
of its policyholders filed no claims, $15.9$% filed low claims, $3.4$% filed medium claims and $1.5$% filed
high claims.

If a policyholder filed a claim, what is the probability that it was not a low claim?

My attempt:

If it was not a low claim, then it either is a medium claim or a high claim. Thus we want P[either Medium or High] $=$ P[M union H] $=$ P[M] + P[H] $= 0.034 + 0.015 = 0.049$

Is it correct?

答: We cannot ignore the probability of a policyholder not filing a claim. The question expects you to calculate $\mathbb P(M\cup H|N^c)$, where $M,H,N$ refers to the events where a policyholder files a medium claim, a high claim and no claims respectively, and $c$ denotes the complementary event. We have $$\mathbb P(M\cup H|N^c)=\frac{\mathbb P(M\cup H)}{\mathbb P(N^c)}=\frac{0.034+0.015}{1-0.792}\approx 0.2356$$


答:
The point $(0,0)$ belongs to none of those regions (and therefore $(0,0)$ doesn't get suddenly removed). However, if $\theta_0\geqslant2\pi$, the region is $\Bbb R^2\setminus\bigl\{(0,0)\bigr\}$, which is not simply connected.

On the other hand, if $\theta_0<2\pi$, then, given any point $p$ of the regien, the line segment that goes from $p$ to $\left(\cos\left(\frac{\theta_0}2\right),\sin\left(\frac{\theta_0}2\right)\right)$ is contained in the region. Therefore, the region is simply connected.

Title: A region is simply connected if $\theta < 2\pi$, and not simply connected if $\theta \ge 2\pi$

Body:
The question is:
Which of the following regions are simply-connected? (...)

e) in polar coordinates, the region where $r>0,0<\theta<\theta_0$.

The solution says:

[simply connected] if $\theta_0<2\pi$, no if $\theta_0\ge2\pi$, since then R is the plane with $(0,0)$ removed.

What is the logic behind the answer? How come that $(0,0)$ sort of removes itself when $\theta_0\ge2\pi$?

I've checked answers like this one and this one, but (although the first adds intuition) they delve into topology, which is too much for me at this point.

This question comes from Multivariable Calculus course (for reference, exercise 4G-5), not from Topology course, so please don't use into (too much) topology.


问:
Is it analytic ? $f(z)=z^2\bar{z}$
Please find the question and its solution.

My answer gone wrong dont know why, I think the answer marked in the question is wrong.

My try:

I solved for CR equations and replaced x=y=0 and found CR equations satisfied. But not sure why in the solution it says CR equations doesn't satisfy.

Please let me know what should be the correct answer ?

答:
Apart from the continuity of all partial derivatives $u_x, v_x, u_y, v_y$ along with $u$ and $v$ everywhere, note that $u_x=v_y \forall x,y$

But $u_y=-v_x$ holds for $xy=0\implies x=0$ or $y=0$ i.e. on coordinate axes. So

CR-equations hold at $z=0$ and the test says that $f(z)$ is differentiable at $z=0$ (not analytic though).

I would say that the question is poorly formatted as all the options stand incorrect.

Title: Is it analytic ? $f(z)=z^2\bar{z}$


问:
In topology what is meant by a square and a circle?
When you have terms such as square or circle in topology does it mean only the boundary of the object or the interior space also?</p>

答: The boundaries are topologically the same, and can be called “circle” or $\Bbb S^1$. The two-dimensional area (so with inside area too) is called a “disk” (or $\Bbb D^2$, topologically the same as a filled square). Product sets like $[0,1] \times [2,3]$ are called “rectangle” in the product topology, ofttimes. That the “lengths” are equal in this case is irrelevant in topology.


问:
Mathematical induction for proving boundedness
So consider a sequence {$x_n$}, where $x_1 = 1$, $x_2 = 2$ and $x_n = \frac{1}{2} (x_{n-2} + x_{n-1})$. Now I would like to show that $1 \le x_n \le 2$ by induction. I first show that $x_3 = \frac{3}{2}$ which is indeed inside the range. Now for $k \ge 3, k \in \mathbb{N}$, assume $1 \le x_k \le 2$, then consider $x_{k+1} = \frac{1}{2} (x_{k-1} + x_{k})$. Since I have already assumed $x_k$ is inside the range, does that mean $x_{k-1}$ is automatically included in the range too?

答:
No it doesn't imply that.

I think that the natural way to go is to use strong induction here. You have $x_1,x_2\in[1,2]$. Now, take $k\in\Bbb N$ and assume that $j\leqslant k\implies a_j\in[1,2]$. Then $a_{k+1}=\frac{a_k+a_{k-1}}2$ and, since both $a_k$ and $a_{k-1}$ belong to $[1,2]$, then so does $a_{k+1}$.


问:
What parameters can be used with the hyperbolic tangent function to enable optimised curve fitting?
If I want to fit a sigmoid curve to data using the logistic function I use something like

$$y = \frac{L}{1 + \exp[-k(x-x_{0})]} + b$$

where $L$, $k$, $x_{0}$ and $b$ are functional parameters that can be varied to achieve the optimal curve fit. In particular, $L$ and $b$ here represent the upper and lower limits respectively of the range of possible values for $y$ (and so may be fixed parameters rather than variable). The other parameters, $k$ and $x_{0}$ affect the gradient of the curve and its mid-point relative to the $x$-axis respectively. These could be varied to optimise a fit of the curve to data points.

Where should I insert parameters into the hyperbolic function to enable similarly arbitrary curve fitting in the same way, if I would like similar behaviour (ie definition of $y_{\rm{min}}$ and $y_{\rm{max}}$, slope and mid-point, or similarly useful parameters)?

$$y = \tanh(x) = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$$

答:
They are the same family, since
$$
\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{-2}{1+\exp(2x)}+1=\frac{2}{1+\exp(-2x)}-1.
$$

So, whereas $y=\dfrac{L}{1 + \exp(-k(x-x_0))} + b$ has range between $b$ and $b+L$ exclusive, $x_0$ is the central value and $k$ is a slope control,
$$
y=A\tanh(k(x-x_0))+B
$$
has range between $B\pm A$ exclusive, same $x_0$ but here $k$ differs by a factor of $2$.


问:
How do I solve $\min_x \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$ for $\lVert x \rVert_2 = 1$.
Let $f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$.
where $x, c_1, c_2, \dots, c_k \in \mathbb R^n$.
What fast iterative methods are available for finding the (approximate) min of $f$ with the constraint $\lVert x \rVert_2 = 1$?

Notes:

$f$ is convex and and non negative, that is, $c_i$ positively span $\mathbb R^n$

For my use case $k \gg n$ and (rougly) $3 \le n \le 50$ and $100 \le k \le 10000$.
I tried projected subgradients (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.

I've also tried approximating $f$ with a smooth maximum LogSumExp and doing gradient descent with projection. I also tried unconstrained gradient descent with a quadratic penalty. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in SQP because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.

Edit:
I believe this problem is equivalent to a linear programming problem with quadratic equality constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a SDP method but I'm not sure.

答:
We need to solve the following optimization problem (P$0$):
\begin{align}
(\mathrm{P0})\qquad &\min_{x} \max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}\\
&\mathrm{s.t.}\quad \|x\|_2 = 1.
\end{align}

Assume that $\{c_1, c_2, \cdots, c_k\}$ satisfies
$\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\} > 0$ on $\mathbb{R}^n\backslash \{0\}$.

(P$0$) is equivalent to $\min_{x} \frac{\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}}{\|x\|_2}$
and $\max_{x} \frac{\|x\|_2}{\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}}$ and
\begin{align}
&\max_{x} \|x\|_2^2\\
&\mathrm{s.t.}\quad \max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\} = 1
\end{align}
and
\begin{align}
(\mathrm{P1})\qquad &\max_{x} \|x\|_2^2\\
&\mathrm{s.t.}\quad c_i^\mathsf{T}x \le 1, \ i=1, 2, \cdots, k.
\end{align}
(P$1$) is a Quadratic Programming (QP) problem (https://en.wikipedia.org/wiki/Quadratic_programming).

Some online and software Resources for QP are given in [1]. Some quadratic programming codes are given in [2].

References

[1] https://neos-guide.org/content/quadratic-programming

[2] http://www.numerical.rl.ac.uk/people/nimg/qp/qp.html


问:
Infinitesimal generator of the standard Brownian motion
As explained in this Wikipedia page, the infinitesimal generator of the standard Brownian motion is $\frac{1}{2}\Delta$ and for the Brownian motion it has an extra $\partial_t f$ term. Can anybody please explain how one can reach to $\frac{1}{2}\Delta$ from $Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t}$ in the standard Brownian motion, for a person who is not that much familiar with the stochastic math?

In the first answer here, the Brownian motion was studied. I would appreciate if i can have the proof (not deep stochastic math) for the standard Brownian motion.

And two points which might be covered in the answer of the question above:

Unlike the Brownian case, in the standard case we have $\mathbb{E}^x(B_t-x)=-x$.

In the equation (1) in that answer, how "one can show $\frac{d}{dt} P_t f(x) = A P_tf(x)$                 "  ?

Thanks in advance.

答:
I'm in the process of learning this myself, and I did the following.

So in the 1D case, we want to show that $Af(x) := \lim_{t\rightarrow 0} \frac{\mathbb{E}^x[f(B_t)] - f(x)}{t} = \frac{1}{2}f''(x)$.

A Taylor expansion about $f(x)$ gives
\begin{align}
\mathbb{E}^x[f(B_t)] &\approx \mathbb{E}^x[ f(x) + (B_t-x) f'(x) + \frac{1}{2}(B_t-x)^2 f''(x)]\\
&= f(x) + \mathbb{E}^x[B_t-x] f'(x) + \frac{1}{2} \mathbb{E}^x[(B_t-x)^2] f''(x)\\
&= f(x) + \frac{t}{2} f''(x)
\end{align}

where the last equality comes from the properties of a Brownian motion with $B_0=x$, $\mathbb{E}^x[B_t-x] = 0$ and $\mathbb{E}^x[(B_t-x)^2] = t$.

Plugging back into $Af(x)$, the $f(x)$'s cancel and we get
$$ Af(x) = \lim_{t\rightarrow 0} \frac{\frac{t}{2}f''(x)}{t} = \frac{1}{2}f''(x)$$

(The t's trivially go away: $\lim_{t\rightarrow 0} \frac{t}{t} = \lim_{t\rightarrow 0} 1 = 1$)

You can extend this reasoning to the multidimensional case $x= (x_1, \dots, x_n)$, where the Taylor expansion becomes
$$\mathbb{E}^x[f(B_t)] = f(x) + \frac{t}{2} \sum_{i=1}^n \frac{\partial}{\partial x_i} f(x).$$


问:
Integral involving two hypergeometric functions of multiple variables.
Let us define an auxiliary functions:
$$f_i(x)=x^{c_i-1}\Phi^{(2)}_2(b_i,b'_i;c_i;v_i x,w_i x)$$
with $\Phi^{(2)}_2(\cdot)$ - standard Humbert function of two variables and $i=1,2$.

The question is: Is it possible to compute the following integral
$$\int_{\alpha}^{\infty}f_1(x)f_2(x-\alpha){\rm d}x.$$
At a first glance it looks like a convolution integral. I can find the Laplace transform of the functions $f_i(x)$ but will this approach be correct since it only "looks like" but not a convolution.
The series representation of Humbert functions does not give any solution since the integral blows up.

I know that the integral is finite, since it represents some combination of probability distributions and the numeric solution can be obtained without any problem. What I am seeking is the closed-form solution.

Addendum: the parameter $x\geq 0$, but $v_i, w_i$ can be negative.

答:
Not an answer, but if you want your cross-correlation integral to look like a convolution, define:

$$g_2(x) = f_2(-x)H(-x) $$

where $H(x)$ is the Heaviside unit step function.

Then

$$\begin{align*} I(\alpha) &= \int_\alpha^\infty f_1(x)f_2(x-\alpha) dx\\
\\
&= \int_{-\infty}^\infty f_1(x)f_2(x-\alpha)H(x-\alpha) dx\\
\\
&= \int_{-\infty}^\infty f_1(x)g_2(\alpha-x) dx\\
\\
&= (f_1*g_2)(\alpha) \\
\end{align*}$$

is your integral expressed as a convolution.


问:
Calculating the limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\tan \theta}$
I'm calculating this limit and would kindly appreciate feedback on my solution

$\lim\limits_{\theta \to 0}\dfrac{\sin \theta}{\tan \theta}$

What I've tried:

given that $\tan \theta = \dfrac{\sin \theta}{\cos \theta}\;,$

then I rearrange the equation like so:

$$\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta \cos \theta }{\sin \theta} = \cos \theta$$

As $\theta$ approaches $0$, then is it true that $\dfrac{\sin \theta}{\tan \theta}=\cos\theta\to1\;?$

答: Your approach is perfect. The reason why you can cancel the $\sin\theta$s is that theyre not exactly zeroes, even though theyre tending to $0$


问:
Finding dim null T for nilpotent operators
I was working through some problems when I came across this one that I was having some trouble with.

Assume that $\dim V = n$. Assume that $n > 20130520$. Let $T$ be an operator on $V$.

(1) If $T^{n-1} \neq  0$, $T^n = 0$. Find $\dim \operatorname{null} T$.

(2) If $T^{n-2} \neq  0$, $T^{n-1} = 0$. Find $\dim \operatorname{null} T$.

I wasn't too sure how I should go about this. One thing I was thinking of was that for (1), the assumption implies that $\operatorname{null} T^{n-1} \neq \operatorname{null} T^n$ which would imply that the left side would be a subset of the right side and you could continue until you get to $\operatorname{null} T$.

Any tips or advice would be appreciated. Thanks in advance.

答:
I am not sure why $n$ is given such a large lower bound. Anyway, consider the chain of subspaces:
$$
T^n(V) \subseteq T^{n-1}(V) \subseteq \ldots \subseteq T(V) \subseteq V.
$$
If we have $T^{N+1}(V) = T^N(V)$ for some $i$, then $T^n(V) = T^N(V)$ for all $n \ge N$. That means if $T^n(V) = 0$ and $T^{n-1}(V) \ne 0$, we must have
$$
0 = \dim T^n(V) < \dim T^{n-1}(V) < \ldots < \dim T(V) < \dim(V) = n.
$$
By simple counting, we must have $\dim T^i(V) = n - i$, so $\dim \ker T = 1$.

For the second case, this argument is not good enough, as it only tells us that $\dim \ker T \le 2$.

By the counting argument, we know that somewhere in the chain, we must have
$$\dim\ker T|_{T^I(V)} = \dim T^I(V) - \dim T^{I+1}(V) = 2$$
with $I \ge 1$. But since $T^I(V) \subseteq V$, we know
$$
\dim\ker T|_{T^I(V)} \le \dim\ker T
$$
and so
$$
2 \le \dim \ker T.
$$
(The general statement is that $\dim \ker T^{i+1}|_{T^i(V)}$ is a non-increasing sequence, and $\sum_{i=0}^{n-1} \dim\ker T^{i+1}|_{T^i(V)} = \dim V = n$ if $T$ is nilpotent.)


问:
Upward Löwenheim-Skolem Theorem for logics without equality
The usual proof of the upward Löwenheim-Skolem theorem rests on the use of the equality symbol as a primitive symbol, interpreted in every relevant structure as true equality. My question is on how the upward Löwenheim-Skolem theorem has to be modified if equality is not a primitive symbol anymore and, say, just introduced as a binary relation with the relevant axioms expressing that it is a congruence with respect to the other non-logical symbols.

EDIT: Based on the answer of t09l, the above paragraph is best augmented by the following: Is there a meaningful reformulation in contexts without equality or does the literature just not consider upward-type theorems without something like a strong equality?

My question in particular comes from the point of view of many-valued predicate logics where the inclusion of a (crisp) equality primitive creates further expressive strength.

Also I wonder how an exclusion of equality affects the Hanf numbers of extensions of classical first-order logic, i.e. of (e.g.) infinitary first-order logic (without equality).

答: Concerning your first question, the upward Löwenheim-Skolem theorem in FOL without equality is much simpler (and less interesting). Let $M$ be an infinite model, and pick an element $m\in M$. Then you can always adjoin an arbitrary collection $\{m_i|i\in I\}$ of copies of $m$ to the model $M$ by stipulating that the $m_i$'s satisfy the same relations as $m$, and are mapped to the same elements as $m$ by any function. The resulting model $M\cup\{m_i|i\in I\}$ satisfies the same sentences (without $=$) as $M$.


问:
Good (simple) approximation to the arithmetic mean in integer arithmetic
I appreciate this is a bit of an open ended question that might veer off towards opinion and not rigour but I hope it's interesting enough to be thought suitable.

Recently I had to write some code to calculate the arithmetic mean of a long run of observed values. Super-accuracy was not too important - being able to see a reasonable approximation of the current value and the longer-term trends were important.

A previous implementation had worked on the basis of:

 NewMean = (OldMean x (Observations - 1) + NewObservation)/Observations

Which is obviously correct but is in danger of overflowing as the product of means and observations grows.

I looked for a canonical replacement in Knuth's The Art of Computer Programming but he doesn't discuss this in terms of integer maths at all, but does suggest a simple recurrence relation using floating point - and I adopted a version of this:

NewMean = OldMean + (NewObservation - OldMean)/Factor

I set Factor as 20 and generate the first 20 means in the old way before switching to this.

As the observations are about 3 orders of magnitude greater than Factor it works reasonably well (though obviously I lose precision) - but is there a better way that is neither very computationally intensive nor likely to overflow?

答:
The second formula is simply a rearrangement of the first, and what you're calling "factor" should simply be the number of observations. Setting the factor equal to $20$ forever instead is just a way of weighting new observations more heavily than old ones, which I take it is not what you want.

Let the old mean be $\mu$, the new observation be $x$, and the total number of observations $n$. Then
$$
   \frac{\mu \cdot (n-1) + x}{n} = \mu \cdot \frac{n-1}{n} + \frac xn = \mu - \frac \mu n + \frac xn = \mu + \frac{x-\mu}{n}.
$$
If we were doing the calculations with real numbers, both formulas would be exact. The only way to improve over rounding at each step would be to keep track of $(x-\mu) \bmod n$, and try to save it in some running total of fractional parts, which is difficult and not worth it in my opinion.

One improvement I have to suggest is language-dependent. If you're using a programming language where 7/5 gives 1 but (-7)/5 gives 2, then integer division always rounds down, introducing a bias which is slight but easy to correct for: just replace $\frac{x-\mu}{n}$ by $\frac{x-\mu + n/2}{n}$ to round to the nearest integer. If you're using a language where 7/5 gives 1 and (-7)/5 gives -1, this is not necessary. But this is a very minor correction either way.


问:
Inverting n-dimensional vector function
As part of a differential geometry problem, I need to inverse this function to get $\vec{u}$ as a function of $\vec{x}$:

$$
\vec{x} = \left(\frac{2\vec{u}}{1 + |\vec{u}|^2}, \frac{1 - |\vec{u}|^2}{1 + |\vec{u}|^2} \right)
$$

where $x\in\mathbb{R}^{n+1}$ and $u\in\mathbb{R}^n$.

I tried writing $\vec{x}$ as $(\vec{x_0}, x_n)$ so I have the relations $\vec{x_0} = \frac{2\vec{u}}{1 + |\vec{u}|^2}$ and $x_n = \frac{1 - |\vec{u}|^2}{1 + |\vec{u}|^2}$ but I'm not sure how to combine these in the right way, and also not quite sure how to handle the inverse when the magnitude is involved.

答:
First of all, notice that you are not going to get a global inverse, otherwise $\mathbb{R}^{n+1}$ and $\mathbb{R}^{n}$ would be diffeomorphic. So your function maps $\mathbb{R}^{n}$ to a subset of $\mathbb{R}^{n+1}$. In fact, it is easy to see that $|\vec{x}|=1$.

Assume $x_n\neq -1$ (otherwise you need to rename your coordinate $x_n$). Then, from the definition of $x_n$ it is straightforward to obtain:
$$
|\vec{u}|^2=\frac{1-x_n}{1+x_n}
$$
whereas for $i=0\dots n-1$,
$$
u_i=\frac{1-|\vec{u}|^2}{2}\frac{x_i}{x_n}=\frac{x_i}{1+x_n}
$$

Title: Inverting n-dimensional vector function


问:
Can you expand induction proofs to the real numbers?
Everyone knows the principle of induction, where you first prove a base case for some $n_0\in \mathbb{N}$, and than show that by assuming the case for an $n \in \mathbb{N}$ the $n+1$ case follows. This is true because of the application of some sort of domino principle.
However, you could also try to combine this principle with an epsilon proof.
For any statement $S_x$, show that $S_{x_{0}}$ holds and than show that from $S_x$, $S_{x+\epsilon}$ follows, for $|\epsilon |>0$, where $\epsilon,x_0,x \in \mathbb{R}$.
If I am not mistaken this should at least work for proving statements for all rational numbers, please correct me if I am wrong(I am still new to this). But does this also work for proofs of real numbers and if not, is there some sort of "induction for the reals"?

I hope that I could make my question clear enough. I am quite sorry for any sort of mistakes and hope nonetheless, that you could help/correct me.

答:
You have to be quite careful where you put the quantifiers in your principle, to make it valid. If you have $S_{x_0}$ and, for some $\varepsilon > 0$, you have that for all $x, y \in \Bbb{Q}$ (or $\Bbb{R}$) $S_x$ implies $S_y$ whenever $|x - y| < \varepsilon$, then you have that $S_x$ for all $x \in \Bbb{Q}$ (or $\Bbb{R}$). This is really just induction over the natural numbers for the proposition for $n \in \Bbb{N}$ that $S_x$ holds for all $x$ with $|x - x_0| < n\epsilon$.

The general replacement for induction in $\Bbb{R}$ is the principal of completeness: every non-empty set that is bounded above has a least upper bound. One application of completeness is to prove the Archimedean property: if $\epsilon > 0$, then for any $x$ there is $n \in \Bbb{N}$ such that $x < n\epsilon$ (the proof begins by assuming that $x$ is an upper bound for $\{n\epsilon \mid n \in \Bbb{N}\}$ and then derives a contradiction using completeness). The Archimedean property justifies the line of argument in the previous paragraph.


问:
Show $ \begin{bmatrix} A & BC(A+BC) \\ I_n & 0 \end{bmatrix} $ and $\begin{bmatrix} A+BC & 0 \\ 0 & -CB \end{bmatrix}$ have same nonzero eigenvalues
Let $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$, and $C\in\mathbb{R}^{m\times n}$.  Define
\begin{equation}
D_1=\begin{bmatrix}
A & BC(A+BC) \\
 I_n & 0
\end{bmatrix},\qquad D_2=\begin{bmatrix} A+BC & 0 \\ 0 & -CB \end{bmatrix}
\end{equation}
Show that $D_1$ and $D_2$ have the same nonzero eigenvalues.

My attemp: If we check the claim numerically, we see that the nonzero eigenvalues of $D_1$ and $D_2$ are the same. Probably, there is a factorization of $D_1$, for example $D_1=EF$ such that $D_2=FE$.

答:
In general, $BC$ and $CB$ share the same (multi)set of nonzero eigenvalues. The conclusion now follows from the observation that your $D_1$ is similar to
\begin{aligned}
&\pmatrix{I_n&-(A+BC)\\ 0&I_n}\pmatrix{A&BC(A+BC)\\ I_n&0}\pmatrix{I_n&A+BC\\ 0&I_n}\\
&=\pmatrix{-BC&BC(A+BC)\\ I_n&0}\pmatrix{I_n&A+BC\\ 0&I_n}\\
&=\pmatrix{-BC&0\\ I_n&A+BC}.
\end{aligned}


问:
Find the probability that exactly $2$ urns remain empty, given that not every box receives a ball.

$10$ indistinguishable balls are placed in $5$ distinguishable urns.
Find the probability that exactly $2$ urns remain empty, given that not every box receives a ball.

So, this is a conditional probability that can be written as $\text{P(exactly $2$ urns remain empty/at least $1$ box remains empty)} = P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

To choose the $2$ urns that remain empty we have $\displaystyle \binom{5}2$ ways and we have $\displaystyle \binom{7+3-1}{3-1}$ ways to put a ball in every other urn.

But I don't know how to calculate the above probability. Any help please? Thank you

答:
Number of ways to distribute all balls in the urns is clearly given by $5^{10}$.

Now suppose if $a_1$ balls go in the first urn, $a_2$ balls go in the second urn and so on and $a_5$ balls go in the fifth urn , we have $$a_1+a_2+a_3+a_4+a_5=10$$

whose number of solutions by stars and bars is given by $\displaystyle \binom{10+5-1}{5-1}=\binom{14}{4}$.

Now suppose no urns remain empty, so we force the condition $a_i \geq 1 $ for all $i$ where $1\le i \le 5$.

So if we assume $a_1=b_1+1, a_2=b_2+1$ and so on so that we have $b_i \geq 0$ for all $i$.

So we have our new equation as $$b_1+b_2+b_3+b_4+b_5=5$$

whose number of solutions by stars and bars is given by $\displaystyle \binom{5+5-1}{5-1}=\binom{9}{4}$.

So this was the number of ways of distributing the balls so that no urn remains empty, so that means the number of ways in which atleast one urn remains empty is $\displaystyle \binom{14}{4}-\binom{9}{4}$.

Therefore the probability that atleast one urn remains empty is $$\displaystyle \dfrac{\binom{14}{4}-\binom{9}{4}}{5^{10}}$$

For computing the probability that exactly $2$ urns remain empty, let $a_4=a_5=0$ and force $a_1,a_2,a_3 \geq 1$, so by similar reasoning we have the equation $$b_1+b_2+b_3=7$$

whose solution by stars and bars is given by $\displaystyle \binom{7+3-1}{3-1}= \binom{9}{2}$. But we could let any two $a_i's$ to be $0$, so we multiply by $\displaystyle \binom{5}{2}=10$, so it's probability becomes $$\dfrac{\binom{5}{2}\binom{9}{2}}{5^{10}}$$

So by dividing the above two probabilities, we get out final answer as $\displaystyle \dfrac{\binom{5}{2}\binom{9}{2}}{\binom{14}{4}-\binom{9}{4}}=\dfrac{72}{175}$.


答: Look up Kahan's method for computing running mean and variance. It computes it quite accurately at the expense of some additional computation.</p>

Title: Good (simple) approximation to the arithmetic mean in integer arithmetic

Body:
I appreciate this is a bit of an open ended question that might veer off towards opinion and not rigour but I hope it's interesting enough to be thought suitable.

Recently I had to write some code to calculate the arithmetic mean of a long run of observed values. Super-accuracy was not too important - being able to see a reasonable approximation of the current value and the longer-term trends were important.

A previous implementation had worked on the basis of:

 NewMean = (OldMean x (Observations - 1) + NewObservation)/Observations

Which is obviously correct but is in danger of overflowing as the product of means and observations grows.

I looked for a canonical replacement in Knuth's The Art of Computer Programming but he doesn't discuss this in terms of integer maths at all, but does suggest a simple recurrence relation using floating point - and I adopted a version of this:

NewMean = OldMean + (NewObservation - OldMean)/Factor

I set Factor as 20 and generate the first 20 means in the old way before switching to this.

As the observations are about 3 orders of magnitude greater than Factor it works reasonably well (though obviously I lose precision) - but is there a better way that is neither very computationally intensive nor likely to overflow?


问:
Justifying implicit limits
Consider the hyperbola given as:
$$\frac{x^2}{4}- \frac{y^2}{12} =1$$

Divide through by $x^2$

$$ \frac14 - \frac{1}{12} \left(\frac{y}{x}\right)^2= \frac{1}{x^2}$$

Now, here is the tricky step, I take limit as $ x \to \infty$ on both sides, this kills the RHS, and doing some algebra in the back, I get:

$$ 3= \lim_{x \to \infty} \left(\frac{y}{x}\right)^2$$

Now, I want to say that the slope of tangent line infinity/ asymptote to parabola has slope of $\sqrt{3}$ i.e: $ \lim_{x \to \infty} y' = 3$ and say that $ \lim_{x \to \infty} \left(y'\right)^2 = \left( \frac{y}{x}\right)^2$ is that possible?

I considered using Stolz Casero theorem for this, but for it I'd need a sequence, now how do I get a sequence such that I can say the above equivalence?

答:
$\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=3$ is a result you derived from the original equation. It contains relationship between x and y. We know that x is a set of constant values in R. In order for the equation to be true, y has to be the same order with x.
For example, if $y=x^2$ then $\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=\lim_{x\rightarrow \infty}\frac{x^4}{x^2}=\lim_{x\rightarrow \infty}x^2=\infty$.
So it actually depends on your definition for y. If you use $y=f(x)=ax, a\in R$, you can solve the limit directly.\

If a function is continuous on $R$, for any point in it's co-domain, there must exist a sequence convergent to it. Since $f(x)$ is continuous, there must exist a sequence convergent to it.

Title: Justifying implicit limits


问:
A Basic Limit From Exponentials
Reading the proof of exponential derivatives I understand this:

To show that $(2^x)'=\ln 2 \cdot 2^x$ in the proof is used the limit:

$$\lim_{x \to 0} \frac{2^x-1}{x}$$

My question is: ¿How do I prove that this limit exist?

I don't care about its value. If I were going to prove that this limit is equal to
$\ln 2$, I would need the number $e$, and again, this number is defined as the number $a$ such that:

$$\lim_{x \to 0} \frac{a^x-1}{x} = 1$$

In another words if I know that for some constant value $a$ the limit:

$$\lim_{x \to 0} \frac{a^x-1}{x} $$

exists, let's say it's a number $L$ then I could find all the limits like these in function of $L$,
But what makes obvious that this limit exist?

Title: A Basic Limit From Exponentials

Body:
$${{2^x-1}\over x}={{e^{xln(2)}-1\over x}}$$

$$\ln(2){{e^{x\ln(2)}-1\over {\ln(2)x}}}$$

Use the variable change $u=\ln(2)x$.


答:
Let $a>0$ and $a\ne1$. We want to prove the existence of  $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.

Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then

$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\
=0 &\text{if }x=1\\
>0 &\text{if }x>1 \end{cases}$$

Therefore, $f$ attains its absolute minimum at $x=1$. So  for all $x>0$, we have

$$f(x)\ge f(1)=0$$

$$x^r\ge rx+1-r$$

So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0
\end{align}

When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence

\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0
\end{align}

Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.

When $h<0$,

$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$

As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.

Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.

Source: How does one prove that $e$ exists?

Title: A Basic Limit From Exponentials

Body:
Reading the proof of exponential derivatives I understand this:

To show that $(2^x)'=\ln 2 \cdot 2^x$ in the proof is used the limit:

$$\lim_{x \to 0} \frac{2^x-1}{x}$$

My question is: ¿How do I prove that this limit exist?

I don't care about its value. If I were going to prove that this limit is equal to
$\ln 2$, I would need the number $e$, and again, this number is defined as the number $a$ such that:

$$\lim_{x \to 0} \frac{a^x-1}{x} = 1$$

In another words if I know that for some constant value $a$ the limit:

$$\lim_{x \to 0} \frac{a^x-1}{x} $$

exists, let's say it's a number $L$ then I could find all the limits like these in function of $L$,
But what makes obvious that this limit exist?


答: To add on to Rob Arthan's answer: the induction principle does not work if you put the quantifiers differently, that is: for all $x$ there is some $\epsilon > 0$ such that if $S_x$ holds, then also $S_y$ for all $y\in \mathbb{R}$ with $\lvert x-y\rvert < \epsilon$. The problem is that now the $\epsilon$ you get might get smaller and smaller when you try to go away from $x_0$ and you don't get very far. A counterexample could be the statement $$S_x = x \in (-1,1),$$ which is certainly true for $x = 0$ and around each $x\in (-1,1)$ you can find a small interval that is still completely contained in $(-1,1)$.

Title: Can you expand induction proofs to the real numbers?

Body:
Everyone knows the principle of induction, where you first prove a base case for some $n_0\in \mathbb{N}$, and than show that by assuming the case for an $n \in \mathbb{N}$ the $n+1$ case follows. This is true because of the application of some sort of domino principle.
However, you could also try to combine this principle with an epsilon proof.
For any statement $S_x$, show that $S_{x_{0}}$ holds and than show that from $S_x$, $S_{x+\epsilon}$ follows, for $|\epsilon |>0$, where $\epsilon,x_0,x \in \mathbb{R}$.
If I am not mistaken this should at least work for proving statements for all rational numbers, please correct me if I am wrong(I am still new to this). But does this also work for proofs of real numbers and if not, is there some sort of "induction for the reals"?

I hope that I could make my question clear enough. I am quite sorry for any sort of mistakes and hope nonetheless, that you could help/correct me.


问:
How to find $A$ such that $A^2$ is the zero matrix?

Let $\space t:\mathbb{R^3} \to \mathbb{R^3}$ such that $\space t(v)=0 \space \forall \space v \in \mathbb{R^3}$. The matrix form would be a $3$ by $3$, zero matrix.

If $\space t=f \circ f$, where $f:\mathbb{R^3} \to \mathbb{R^3}$ and $f\neq t$, what is the matrix form of $f$?

I know that $\space t=f \circ f= f\cdot f$ and if one let $A$ to be the matrix of $f$ and $\space T$ the matrix of $t$, then $T=A^2$. And so $A^2$ would be a zero matrix.

Is there a systematic way to solve this problem? Thanks for the hints.

答: If you multiply two $3 \times 3$ matrices together, the entries are the dot product of the first's rows with the second's columns. For the dot product to be zero, the rows/columns must be perpendicular. So, find a matrix where every column is perpendicular to every row (use zero rows and columns to make things easier).</p>


问:
Knots with infinite crossing
For me knots are embedding of $S^1$ in $\mathbb{R}^3$.

I have following questions:

Will knots have infinite crossing?

If so, Why are we considering only knots with finite crossing ?

Can someone elaborate aes answer on the question :Is there any knot showing infinitely many crossings?

答: As answered in the topic you linked, there are knots with infinite many crossings in a diagram. For example wild knots, and some people consider them. The reason the finite case is much more popular is that any smooth knot has a diagram with finite number of crossings, and any diagram has all information about a knot.</p>

Title: Knots with infinite crossing


问:
calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$
I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:

$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$

These are my steps in hopefully deriving the correct result:

$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$

Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?

EDIT:
Given that:

$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$

because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.

hence, $\frac{\sin \theta}{\theta}$ approaches 1.

So the answer is $\frac{1}{2}$

答:
You are almost done!
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=$$
$$=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta} \to 1\cdot \frac{1}{1+1} = \frac{1}{2}$$
Note (and prove) that indeed
$$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$$


问:
Prove by induction $2^n \le (n+1)!$

Prove by induction that:  $2^n \le (n+1)! \ \ \ \ \ \ \ \ \forall n\in\mathbb{N}$

I tried to solve in this way:
$$P(1): 2 \le 2! \quad\text{(true)}$$
$$P(n) \implies p(n+1), \;\text{ where }\;\; p(n+1): 2^{n+1} \le (n+2)!$$
so
$$2^n \le (n+1)! \\ 2^n \cdot 2\le 2 \cdot (n+1)! \\ (n+1)! \cdot 2 \le 2 \cdot (n+1)! \quad\text{(true)}$$
where tapped that $2^n \le (n+1)!$

Are my steps right?

答:
It's a bit confusing: the base case is correct but I don't think you have proved the inductive step, the inequality that you seek to prove in that case is never reached in the steps you have written.

A more streamlined approach for the inductive step would be the following:

$$(n+2)!=(n+2)(n+1)!\geq (n+2) 2^n\geq 2\cdot 2^n=2^{n+1}$$

where in the first equality we used the definition of factorial, in the first inequality the inductive hypothesis and in the second inequality the fact that $n+2\geq 2\ (n\geq 0) $

Title: Prove by induction $2^n \le (n+1)!$


问:
How to find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$?
so I'm totally lost how to find this question

find the maximum value of $f(x,y) = 39 - x^2 - y^2$ on the line $x+6y=37$

I know how to find local max and min of multivariable equations, but now I don't know what to do given that line equation.

If someone could help me it would be very much appreciated

Thanks in advance

答:
HINT

One obvious approach: on that line, $x=37-6y$, which you can plug into $f(x,y)$, reducing it to $f(x(y),y)$, a function of one variable, and use ordinary Calculus.

Second approach would be to use something like Lagrange multipliers.


问:
Any counter-example of pasting lemma?
This is from Munkres:

Let $X = A \cup B$, where $A$ and $B$ are closed (open) in $X$. Let $f\colon A \rightarrow Y$, $g\colon B \rightarrow Y$ be continuous. If $f(x) = g(x)$ for every $x \in A \cap B$, then $f$ and $g$ combined to give a continuous $h\colon X \rightarrow Y$ by setting $h(x) = f(x)$ if $x \in A$, and $h(x) = g(x)$ if $x \in B$.

The proof is fairly straightforward, using the fact that inverse image of closed (open) map is closed. However, what happens if we drop closed (or open) condition? Are there any examples such that $f|_{A \cap B} = g|_{A \cap B}$, but $h\colon X \rightarrow Y$ is not continuous?

One example I can think of is: $A = [0,1] \cap \mathbb Q$ and $B = [0,1] \setminus \mathbb Q$, and $f = \chi_{A}, g = \chi_B$. Here, $h$ is inevitably discontinuous. However, $A \cap B = \varnothing$ in this case, and it will be wonderful if I can construct a function on $A, B$ such that $A \cap B$ is something non-trivial (by non-trivial, I mean finite). I would really appreciate any other insightful and exciting counterexample that can spark my mind!

Thank you in advance.

答:
If you want $A\cap B\not=\emptyset$, you can just tweak the example you already have in some small way. For example, take $$f:\mathbb{Q}\rightarrow\mathbb{R}: a\mapsto 0,\quad g:(\mathbb{R}\setminus\mathbb{Q})\cup\{0\}\rightarrow\mathbb{R}: a\mapsto a.$$ Here $A=\mathbb{Q}$, $B=(\mathbb{R}\setminus\mathbb{Q})\cup\{0\}$, and so $A\cap B=\{0\}\not=\emptyset$. (In your example you've put everything inside $[0,1]$, but there's no need to do that.)

A similar tweak gets an infinite, indeed uncountable, intersection:

 Just set $A=(-\infty,0]\cup\mathbb{Q}$, $B=(\mathbb{R}\setminus\mathbb{Q})\cup(-\infty,0]$, $f: A\rightarrow\mathbb{R}: a\mapsto 0$, and $$g:B\rightarrow \mathbb{R}: a\mapsto\max\{0,a\}.$$

Title: Any counter-example of pasting lemma?


问:
Differentiate $g(t)= {e^t - e^{-t} \over e^t + e^{-t}}$
I'm having some trouble trying to differentiate the function $g(t)= \dfrac{e^t - e^{-t}}{e^t + e^{-t}}$

Can someone help me? Thanks a lot!

答:
Note that $g(t)=1$ for all $t$ for which it’s defined, i.e., for all $t\ne 0$.

Note: This answer was for the original version, which had

$$g(t)=\frac{e^t-e^{-t}}{e^t-e^{-t}}\;.$$

The corrected version can be differentiated as it stands by application of the quotient rule, or it can be manipulated a bit first to make the differentiation a little simpler:

$$g(t)=\frac{e^t-e^{-t}}{e^t+e^{-t}}\cdot\frac{e^t}{e^t}=\frac{e^{2t}-1}{e^{2t}+1}=1-\frac2{e^{2t}+1}=1-2\left(e^{2t}+1\right)^{-1}\;,$$

so

$$g'(t)=(-2)(-1)\left(e^{2t}+1\right)^{-2}\left(2e^{2t}\right)=\frac{4e^{2t}}{\left(e^{2t}+1\right)^2}\;.$$


答:
Just to show how to solve with Taylor series: as $\theta \to 0$:

$$\cos(\theta) \approx 1 - \dfrac{\theta^2}{2}$$

$$\sin(\theta) \approx \theta$$

Whence

$$\lim_{\theta \to 0} \frac{1 - \left(1 - \dfrac{\theta^2}{2}\right)}{\theta^2} = \lim_{\theta\to 0} \dfrac{1}{2} = \dfrac{1}{2}$$

Title: calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$

Body:
I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:

$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$

These are my steps in hopefully deriving the correct result:

$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$

Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?

EDIT:
Given that:

$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$

because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.

hence, $\frac{\sin \theta}{\theta}$ approaches 1.

So the answer is $\frac{1}{2}$


答:
Given that as $\theta$ approaches $0$ then $\dfrac{\sin \theta}{\theta}\approx 0$

should be

Given that as $\theta$ approaches $0$ then $\dfrac{\sin \theta}{\theta}\approx 1$

More precisely, you have
$$
\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1
$$

Now you can fix your calculation.


答:
There's a few things wrong with this:

Your conclusion should be that $2^{n+1} \le (n + 2)!$, i.e. $P(n + 1)$. This conclusion should at least appear somewhere in the argument (preferably at the end of it!), or else you haven't completed the induction step.

Ending your argument with "... therefore we get this, which is clearly a true statement" is the fallacy of affirming the consequent. You should start with the statements you know (or are assuming) to be true, and ending with the statement you want to show is true. Ending with a true statement tells you nothing.

When you shift from $2^n \cdot 2 \le 2 \cdot (n + 1)!$ to $(n + 1)! \cdot 2 \le 2 \cdot (n + 1)!$, you appear to be substituting $2^n$ with the presumed larger quantity $(n + 1)!$. In the former inequality, $2^n$ is the smaller quantity. You are allowed to replace it with an even smaller quantity, but replacing it with a larger quantity is a crapshoot. For example, $3 \le 7$, and $3 \le 8$, but this doesn't mean that $8 \le 7$.

Title: Prove by induction $2^n \le (n+1)!$

Body:
Prove by induction that:  $2^n \le (n+1)! \ \ \ \ \ \ \ \ \forall n\in\mathbb{N}$

I tried to solve in this way:
$$P(1): 2 \le 2! \quad\text{(true)}$$
$$P(n) \implies p(n+1), \;\text{ where }\;\; p(n+1): 2^{n+1} \le (n+2)!$$
so
$$2^n \le (n+1)! \\ 2^n \cdot 2\le 2 \cdot (n+1)! \\ (n+1)! \cdot 2 \le 2 \cdot (n+1)! \quad\text{(true)}$$
where tapped that $2^n \le (n+1)!$

Are my steps right?


问:
Sum over the primes of $\frac{\ln(p)}{p^2}$
$$\sum^{\infty}_{p\in \mathbb P} \frac{\ln(p)}{p^2}$$
I've been able to prove the sum converges ($P(2)<S<\sum^{\infty}_0 \frac{\ln(n)}{n^2}$  where $P$ is the prime zeta function).

I've been trying to solve it by replacing 2 with $s$ and integrating.
$$\int^\infty_s\sum^{\infty}_{p\in \mathbb P} \frac{\ln(p)}{p^k}dk=\sum^{\infty}_{p\in \mathbb P}\int^\infty_s \frac{\ln(p)}{p^k}dk=\sum^\infty_{p\in \mathbb P} \frac{1}{p^s}$$
but I'm unsure how to continue and I am pretty sure undoing the integral will be rather hard.

Can any of you help?

答:
$\sum \limits_{p \in \mathbb{P}} \frac{\ln p}{p^2}  \leq  \sum \limits_{p \in \mathbb{P}} \frac{\sqrt {p}}{p^2}$ since $ \ln x  \leq \sqrt{x} $ for all $x>0$ and we want to prove that $ \sum \limits_{p \in \mathbb{P}} \frac{\sqrt {p}}{p^2}$ converge !

$\sum \limits_{p \in \mathbb{P}} \frac{\sqrt {p}}{p^2} \leq \sum \limits_{2\leq j \in \mathbb{N}} \frac{1}{j \sqrt{j}}$ since $ \frac{1}{j \sqrt{j}}$ is monotone function then $  \frac{1}{j \sqrt{j}} \leq \int \limits_{j-1}^{j} \frac{dx}{x \sqrt{x}} $

And so $\sum \limits_{2\leq j \in \mathbb{N}} \frac{1}{j \sqrt{j}} \leq \sum \limits_{2\leq j \in \mathbb{N}} \int \limits_{j-1}^{j} \frac{dx}{x \sqrt{x}} = \int \limits_{1}^{\infty} \frac{dx} {x \sqrt{x}}  = 2$

The approach you took is harder because its more accurate and in general used when we want to calculate something like $ \sum \limits_{p > x} \frac{\ln p}{p^2}$ (as function related to $x$) and not just to check convergence though you can do it.


问:
Proving uncountable sum as series using nets.
I saw this question:

The sum of an uncountable number of positive numbers

Asking about a proof of the following:
Let $A = \{a_i\}_{i\in I}$ be a set of positive numbers. If the uncountable sum converges (as a net) to $a$, then only countably many $a_i$ are nonzero.

And was wondering if this proof any good (I have little experience working with nets):

The closure of $A$ is a closed subset of a metric space and $a$ lies in this closure. Thus there is a sequence in $A$ that converges to $a$.
$$ a = \sum_{i=1}^\infty a_i. $$
If $A$ is not countable, we could find a $a_k \in A$ that is not part of the sequence. If $a_k\neq 0$, then
$$ a_k + \sum_{i=1}^\infty a_i > \sum_{a_i\in A}a_i $$
yielding a contradiction.

I did not immediately see this proof being used anywhere. Sorry if I missed something obvious.

Edit: I missed something obvious: $a$ is not in the closure $A$!

It is, however in the closure of the set of finite sums, so we can find a sequence of finite sums converging to $a$, $(s_i)_{i\in\mathbb{N}}$, where each finite sum features each $a_i$ at most once. Each sum has a finite set of summands, so the union of all summands in the entire sequence is a countable union of finite subsets and is countable. Take the sum of all these summands and we have a countable infinite sum
$$ \sum_{i=1}^\infty a_i. $$

The claim is this converges to $a$. Indeed clearly $\sum_{i=1}^\infty a_i \leq \sum_{a_i\in A}a_i = a$. Conversely, for all $s_j$, there is an $n$ such that $\sum_{i=1}^n a_i \geq s_j$, thus $\sum_{i=1}^\infty a_i \geq a$. The rest of the argument then follows.

答: It's not true that $a \in \overline{\{a_i\mid i \in I\}}$. So it breaks down right away.