问:
Derivative of Quadratic Form
For the Quadratic Form $X^TAX; X\in\mathbb{R}^n, A\in\mathbb{R}^{n \times n}$ (which simplifies to $\Sigma_{i=0}^n\Sigma_{j=0}^nA_{ij}x_ix_j$), I tried to take the derivative wrt. X ($\Delta_X X^TAX$) and ended up with the following:

The $k^{th}$ element of the derivative represented as

$\Delta_{X_k}X^TAX=[\Sigma_{i=1}^n(A_{ik}x_k+A_{ki})x_i] + A_{kk}x_k(1-x_k)$

Does this result look right? Is there an alternative form?

I'm trying to get to the $\mu_0$ of Gaussian Discriminant Analysis by maximizing the log likelihood and I need to take the derivative of a Quadratic form. Either the result I mentioned above is wrong (shouldn't be because I went over my arithmetic several times) or the form I arrived at above is not the terribly useful to my problem (because I'm unable to proceed).

I can give more details about the problem or the steps I put down to arrive at the above result, but I didn't want to clutter to start off. Please let me know if more details are necessary.

Any link to related material is also much appreciated.

答:
Yet another approach using the Frobenius product notation.

For a column vector $x \in \mathbb{R}^n$, and a matrix $A \in \mathbb{R}^{n \times n}$  we can write:

$$ x^TAx = Tr(x^TAx) = x:Ax$$

Then we take the differential and derivative as

\begin{align}
    d(x:Ax) & = dx:Ax + x:Adx\\
            & = Ax:dx + A^Tx:dx\\
            & = (Ax + A^Tx):dx\\
         \frac{\partial (x^TAx)}{\partial x} &= (Ax + A^Tx) = (A + A^T)x
\end{align}

Title: Derivative of Quadratic Form


问:
Why doesn't the permutation formula work on questions regarding unique rearrangements of a specific word?
I have not used probabilities for a long time. I have just went through a summary of probabilities, but still have some ambiguities, including when to use the Permutation (or Combination) formula and when not to use them.

For example, below is a question that I think the permutation formula should work on, but it doesn't!

How many unique ways are there to rearrange the word "PRIOR"?

There are 5 slots (letters) in this word to be field, so it would be 5!. However, 2 of the letters being R, makes some combinations the same, so we have to remove those non-unique answers. Therefore, we have to remove those two, so we divide the 5! by 2!.

Now, I think the permutation formula should be able to answer this question, but it produces a different result (5!/3!). What is my mistake here? Which part of the permutations I misunderstood?

Thanks!

答:
In order to count the total number of ways there are to rearrange the letters of the word $PRIOR$, here is what I would do:

Although the letter $R$ is duplicated, count the number of ways there are to rearrange the letters assuming that all the letters are different (I will use $R_0$ and $R_1$ to highlight this). There are $5!$ different ways of doing it. One possible setting could be $$IR_0OPR_1.$$
Now, since $R_0$ and $R_1$ are different letters (but they represent the same symbol $R$), divide by the total number of ways to rearrange the 2-letters-word $R_0R_1$. It can be done in $2!$ different ways. Then the final answer should be $$\dfrac{5!}{2!}.$$
If the original word were $$TOMATOMATO,$$ where the letters $T$ and $O$ appear $3$ times, and the remaining $M$ and $A$ appear twice each, by applying this 'algorithm' we will have that a feasible setting could be $$M_0M_1O_0T_0A_0A_1T_1T_2O_1O_2.$$ This will lead us to a total of $$\dfrac{10!}{2!\cdot 2!\cdot 3!\cdot 3!}$$
different ways of rearranging the word $TOMATOMATO$.

From here, it shouldn´t be hard to obtain a general expression for any word.

ContentLicense: CC BY-SA 4.0


问:
Sum of two Cosine functions is periodic.
I wish to show that the sum of two cosine functions $\cos(a_{1}x+b_{1})$ and $\cos(a_{2}x+b_{2})$ is periodic ($a_{1},a_{2}\neq0$) where both functions are non-zero with their sum also being non-zero.

Well I know this statement itself is incorrect but I wish to specify that periods of both functions are of course rationals and are given by $\displaystyle T_{1}=\frac{2k\pi}{a_{1}}$ for an integer $k$ and $\displaystyle T_{2}=\frac{2n\pi}{a_{2}}$ for an integer $n$. Well, summing their periods would result in :
$$
T_{1}+T_{2}=\frac{2ka_{2}\pi+2na_{1}\pi}{a_{1}+a_{2}}=\frac{2\pi(ka_{2}+na_{1})}{a_{1}+a_{2}}
$$
I believe to complete this proof, I must state something about $ka_{2}+na_{1}$ I would hope if someone could assist me on this proof.

答:
All you need is for the ratio $\ \frac{a_1}{a_2}\ $ to be rational. If $\ \frac{a_1}{a_2}=\frac{k_1}{k_2}\ $, where $\ k_1,k_2\ $ are relatively prime integers, then
$$
T=\frac{2\pi k_1}{a_1}=\frac{2\pi k_2}{a_2}\\
$$
is a period of both $\ \cos\big(a_1x+b_1\big)\ $ and $\ \cos\big(a_2x+b_2\big)\ $, and hence a period of $\ \cos\big(a_1x+b_1\big)+\cos\big(a_2x+b_2\big)\ $:
\begin{align}
\cos\big(a_1(x+nT)&+b_1\big)+\cos\big(a_2(x+nT)+b_2\big)\\
&=\cos\left(a_1\left(x+\frac{2\pi nk_1}{a_1}\right)+b_1\right)+\\
&\hspace{6em}\cos\left(a_2\left(x+\frac{2\pi nk_2}{a_2}\right)+b_2\right)\\
&=\cos\big(a_1x+b_1+2\pi nk_1\big)+\cos\big(a_2x+b_2+2\pi nk_2\big)\\
&=\cos\big(a_1x+b_1\big)+\cos\big(a_2x+b_2\big)
\end{align}


问:
prove the inequality $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$
Positive numbers  $x + y = 2$. Prove that $x^2y^2 (x^2 + y^2)\leq2$

答:
$$y = 2 - x \implies x^2y^2 (x^2 + y^2) = x^2(2-x)^2(x^2 + (2-x)^2)$$ $$= f(x) = 2x^6 - 12x^5 + 28x^4 - 32x^3 + 16x^2.$$

Now if we can find the maximum of $\ f(x)\ $ for $\ 0 < x < 2\ $ then we are done.

$$f'(x) = 12x^5 - 60x^4 + 112x^3 - 96x^2 + 32x.$$

By inspection, this can be factorised fully as:

$$\ f'(x) = 4x(x-1)(x-2)(3x^2-6x+4)$$

$ \implies x = 1\ $ is the only stationary point for $\ 0 < x < 2.\ $ Now note that $f(1) = 2.$

We must also check the end-points: $\displaystyle\lim_{x\to 0^+}f(x) = 0,\ $ and $\ \displaystyle\lim_{x\to 2^-}f(x) = 0.$

Hence, the maximum of $\ f(x)\ $ is $\ 2,\ $ and this occurs when $\ x = 1.$


问:
Conditional expectation of sum of numbers on 2 throws of a dice given their difference
Let$  X$ and $Y$ be the numbers obtained on first and second throws of a fair die .
Calculate $E(X+Y|(X-Y)^2=1)$

I am doing basically $$E(X|(X-Y)^2=1)=E(Y|(X-Y)^2=1)$$
So $E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
$E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
implies $E(X|(X-Y)^2=1)=1/6(1+2+3+4+5+6)$

So I am getting it $7$ please help

答: This can also be seen by symmetry. Conditioning on $(X-Y)^2$ corresponds to the red stripes, and the distribution of the sum $X+Y$ (green stripes), which a priori ys symmetric around $X+Y=7$, is still symmetric around that value after the conditioning.


问:
$D_{2n}$ is isomorphic to a subgroup of $S_n$ (for $n>2$).
All the proofs I've come across of the fact in the title call into play the action of the group on the vertices of the regular $n$-gon, i.e. they rely on the geometrical definition of $D_{n}$. Am I wrong if I say that this result just follows from being $\{1,s\}\cap\{1,r^2s\}=\{1\}$, for $n>2$? In fact, this suffices to get that the action of $D_{n}$ by left multiplication on $X:=\{gH, g\in D_{n}\}$ is faithful, where $H:=\{1,s\}\le D_{n}$ and $|X|=n$.

答:
You can show $D_{2n}$ is a subgroup of $S_n$ directly. Thinking about the elements $S_n$ in cycle notation, $D_{2n} = \langle r,s\rangle$ where
$r = (1\,2\,3\,\dotsc\,n)$ and
$$s = \begin{cases}(1\,n) \dotsb (\lfloor{n \over 2}\rfloor\,\lceil{n \over 2}\rceil)\;\text{if $n$ is odd}\\(2\,n)\dotsb({n \over 2}\,{n\over 2} -1)\text{if $n$ is even}\end{cases}$$
noting that this $s$ and $r$ satisfy your relations.


问:
Showing that stochastic process solves an SDE
I am given

$$
X(t) = (1+W(t)/3)^3
$$

and am asked to show that it solves

$$
dX = \frac{1}{3}X^{1/3}dt + X^{2/3}dW\\
X(0)=1
$$

By using Ito's rule on $X$, it is quite easy to arrive to $dX = \frac{1}{3}X^{1/3}dt + X^{2/3}dW$. So here I go from $X$ to $dX$.

However, this does not make use of the initial condition $X(0)=1$ in any way. In order to use the initial condition, I guess one has to go from $dX$ to $X$. How does one go about that?

答:
In general, the differential expression of SDE's are shorthand for an integral equation. So you actually do have the expression for $X$ already.
$$
\text{d}X_{t} = \frac{1}{3}X_{t}^{\frac{1}{3}}\text{d}t+X^{\frac{2}{3}}\text{d}W_{t} \iff X_{t} - X_{0} = \frac{1}{3}\int_{0}^{t}X_{s}^{\frac{1}{3}}\text{d}s+\int_{0}^{t}X_{s}^{\frac{2}{3}}\text{d}W_{s}
$$
However in your case, it is rather easy to see that $X_{0} = 1$, since you are given
$$
X_{0} = \left( 1 + \frac{W_{0}}{3} \right)^{3} = 1
$$
since $W_{0}=0$ by properties of the Wiener process.


答:
You say you don't understand why the permutation formula wouldn't work, but it would help us answer your question if you clarify why you think it should work. The permutation formula is for a specific type of rearrangement of $k$ distinct letters chosen out of $n$ distinct letters, and doesn't apply when there are repeats unless it's appropriately modified. Or put another way: not every problem involving permutations (colloquial definition) can be directly solved using permutations ($_n P_r$ definition).

Edit responding to OP's comment: The fact that $_5P_3$ (not $_5P_2$) happens to give the same answer as the number of rearrangements of PRIOR is more like a coincidence than it is a generally applicable formula.

It comes from one of them being written as $_5 P_3 = \frac{5!}{2!}$ and the PRIOR rearrangements as $\frac{5!}{2! 1! 1! 1!}$.

If there's exactly one repeated letter that occurs $r > 1$ times in a word of $n$ otherwise distinct letters, then $_n P_{n - r} = \frac{n!}{r!}$ will happen to equal the number of possible rearrangements of that word, but if there's multiple repeats, you need extra factorials in the denominator that $_n P_r$ can't always accomplish.

For instance, the number of ways to rearrange the letters in AURORA, which has 2 A's, 1 U, 2 R's, and 1 O,  is $$\frac{6!}{2!1!2!1!} = 180,$$ and there's no way to write $180$ in the form $_nP_r$ aside from the trivial $n = 180$, $r = 1$.

Title: Why doesn't the permutation formula work on questions regarding unique rearrangements of a specific word?

Body:
I have not used probabilities for a long time. I have just went through a summary of probabilities, but still have some ambiguities, including when to use the Permutation (or Combination) formula and when not to use them.

For example, below is a question that I think the permutation formula should work on, but it doesn't!

How many unique ways are there to rearrange the word "PRIOR"?

There are 5 slots (letters) in this word to be field, so it would be 5!. However, 2 of the letters being R, makes some combinations the same, so we have to remove those non-unique answers. Therefore, we have to remove those two, so we divide the 5! by 2!.

Now, I think the permutation formula should be able to answer this question, but it produces a different result (5!/3!). What is my mistake here? Which part of the permutations I misunderstood?

Thanks!


问:
Why does the integral work for this sum?
Suppose we have the following sum: $$ \sum_{n=1}^{1010^2} \frac{1}{\sqrt{n}} $$ and we want to calculate the very next integer greater than this.

Then one may take the integral, $$ \int_1^{1010^2} \frac{1}{\sqrt{x}}$$ and calculate it to be $2018$, so the very next integer greater than this is $2019$. Why does the integral work so well here?

I know the integral is equal to the sum of $$ f(x) \delta x$$ as $x$ tends to zero, but I still can’t derive how the integral works so well here. I lack intuition, I think.

答:
HINT

$$\frac1{\sqrt{k+1}}<\int_k^{k+1}\frac{\mathrm{d}x}{\sqrt x}<\frac1{\sqrt{k}}$$

Can you do it now?


问:
Compute the integral, $\iint_R x^2+xy^3 dA$ where R is bound $0\leq y \leq 2$

Compute the integral, $\iint_R x^2+xy^3 dA$ where R is bound $0\leq y \leq 2$

I found this question on one of my lecutre tutorials. I just want to know if its possible to calculate this without knowing the range for x? I highly doubt that this is a misprint. I would like to get some ideas?

P.S. I know how to do the double integral if a range for x is given.

答:
The way this is specified, one assumes $R$ is a vertical strip with $x$ anywhere in $\mathbb{R}$, so you get
$$
\int_{-\infty}^\infty \int_0^2 \left(x^2 + xy^3\right)dy\ dx
= \int_{-\infty}^\infty \left(2x^2 + 16x\right) dx,
$$
which diverges. Most likely, this is a misprint, with intent to write $0 \le x,y \le 2$, which would converge...


问:
Calculate third vertex of a triangle from limited information
This should be a simple problem to solve, but I cannot seem to get my head around it. I would like to find the third vertex of the triangle shown below $(x_i, y_i)$ given the following information:

I know the position of two vertices of a triangle: $(x_a, y_a)$ and $(x_b, y_b)$.

I can calculate the angle $\theta=\arctan\left(\frac{y_b-y_a}{x_b-x_a}\right) $

I can also calculate the length of one side of the triangle: $\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}$.

The other two sides are of length $s_b\Delta t$ and $s_a\Delta t$ where $s_a$ and $s_b$ are both known, but $\Delta t$ is not.

I also know the angle $\phi$, as shown.

What isn't clear to me is if I have enough information for a solution. Can this be solved and, if so, can you point me in the right direction of how to do so? (My old math teacher's voice is in my head muttering "Law of cosines!", which I could apply to get angles within my triangle but I'm drawing a blank on how to proceed from there.) Any advice would be greatly appreciated!

答:
Hint

Working with complex numbers. Denote by $a$ the point $(x_a,y_a)$, similarly for $b$ and $m = (x_i,y_i)$ the point to be find. Also denote $\lambda$ the positive real $\Delta t$.

You have $m = b - s_b\lambda e^{i \phi}$ and $\vert m-a \vert = s_a \lambda$. Which leads to the equation

$$\vert b - s_b \lambda e^{i \phi} -a \vert^2 =s_a^2 \lambda^2.$$

This is one quadratic equation in $\lambda$ that you can solve.


问:
A New Proof of Pythagorean theorem?
A right triangle with side lengths $a,b,c$  where the hypotenuse($c$) intercepts the height($h$) such that $c$ is the sum of $2$ side lengths, $c = m + n$.

Image here:

I found an elegant proof of Pythagorean theorem that seems to be related to Einstein's proof:

First, slopes of $a$ and $b$ give:

$$(\frac{h}{m})(\frac{h}{n}) = 1\implies h^2 = mn$$

Than with area formula, we find the intercept $(x,y)$ of the height and hypotenuse:

$$bx = hm\implies x = \frac{am}{c}$$
$$ay = hn\implies y = \frac{bn}{c}$$

Next plugging this into the height's equation:

$$\frac{a}{b}(\frac{am}{c}) =
\frac{bn}{c}\implies a^2m = b^2n$$

Using $c = m  + n$:

$$\frac{c}{\sqrt{mn}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$$
$$\frac{c^2}{ab} = \frac{a}{b} + \frac{b}{a}$$
$$a^2 + b^2 = c^2$$

Q.E.D.

Thoughts?

答: This is really just an overly complex use of similar triangles to show$$m/a=h/b,\,n/b=h/a\implies c=m+n=(a^2+b^2)/c,$$where in the last $=$ I, like you, have used $ab=ch$. As with your other questions I've answered about Pythagorean proofs, the strategy works but is intimately related, once extraneous Cartesian coordinates have been stripped away, to old approaches. In particular, computing $x,\,y$ is equivalent to computing $m,\,n$, so there's no need to worry about how axes are oriented.

Title: A New Proof of Pythagorean theorem?


问:
Help proof by induction
I just got my first induction assignment in a new course. They want me to prove by induction that:
$$\sum_{s=1}^k s*s! = (K+1)!-1.$$

The way I understand induction is that I test for the first value. Then for n and then n+1, to show the given expression is true.
I've done it for 1 and n, however I'm stuck at n+1.
$$\sum_{s=1}^{n+1} s*s!=(n+1)!-1+(n+1)*(n+1)!$$
Using maple I can see the expression is $-1+(n+2)!$ (which is true) however i dont know how to reduce/rewrite $(n+1)!-1+(n+1)(n+1)!$ to $-1+(n+2)!.$$

I've asked my friends and a older student, but to no avail. I'm hoping you guys can help.

答:
First proof the statement for some value, which you did (induction basis).

Then induction step: Let's assume that the statement you want to proof holds for K = n. Then you have to establish that it also holds for n+1.

So we know that $\sum_{s=1}^{n} s \cdot s! = (n+1)! - 1$ (induction hypothesis)

$\sum_{s=1}^{n+1} s \cdot s! = \sum_{s=1}^{n} s \cdot s! + (n+1) \cdot (n+1)! = (n+1)! - 1 + (n+1) \cdot (n+1)! = (n+1)! + (n+1) \cdot (n+1)! - 1 = (1 + (n+1)) \cdot (n+1)! - 1 = (n+2) \cdot (n+1)! - 1 = (n+2)! - 1 = ((n+1)+1)! - 1$

So then we know that the statement also holds for n+1. This concludes the proof.

So in proofs by induction you explicitly use the induction hypothesis for the induction step.

Title: Help proof by induction


问:
2021 AMC 10 A Problem #22

Question: Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$, the second sheet contains pages $3$ and $4$, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$. How many sheets were borrowed?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$

Does anyone have a solution that's easy to understand?

So here's what I've tried so far: I set $x$ as the first page his roommate took and $y$ as the total number of sheets his roommate took. I tried setting up an equation to find the average (19):
$$
\frac{1275-\text{sum of the  pages his roomate took}}{50-2y}=19.
$$
1275 is the sum of all the pages and 50-2y is the number of pages that are left. But I can't figure out how to express sum of the  pages his roommate took in terms of x and y

答:
HINT

I would like you show some work on this problem so it becomes easier to guide you. Here is a hint to get you started. Note that in the taken pages, the first one is odd and the last one must be even. Let's say the pages $[a,b]$ are taken. Can you find the mean of the remaining numbers?

HINT

If $x$ is the first page taken, and total number is $y$, then the taken pages are
$x, x+1, x+2, \ldots, x+y-1$ so sum of total pages taken is the arithmetic series
$$
\sum_{k=x}^{x+y-1} k
 = \sum_{k=0}^{y-1} (k+x)
 = x(y-1) + \sum_{k=1}^{y-1} k
 = x(y-1) + \frac{y(y-1)}{2}
 = \frac{(y-1)(2x + y)}{2}
$$

Title: 2021 AMC 10 A Problem #22


问:
Differential of a function in integrals
In riemann sums,$dx$ is the $\Delta x$ (which are evenly spaced) when goes to $0$. And by $\int_{0}^{x}dx=x$,we mean the sum of those equally spaced infinitesimals is $x$. But when we use $\int g(f(x)) df(x)$ we treat $f(x)=z$ as an independent variable like $x$ and use the properties we did with $x$,for example $dx$ are evenly spaced infinitesimals but $df(x)$ behave irregularly meaning their differences are not same.Why is it then justified to treat $df(x)$ as an independent variable like $x$ although it is a dependent variable and are not equally spaced.

答: Your $dx$ can be interpreted as a (uniformly distributed) weight on the values of $g(f(x))$ and that's the same case for the $df(x)$. If $f$ is differentiable w.r.t $x$ then you know that a value of $g(f(x))$ "participates" $g(f(x))f'(x)$ to the integral value (for every movement in the variable $x$).

Title: Differential of a function in integrals


答: Two little right triangles and the original one are similar. Then proof follows after simply two to three lines arithmetic.</p>

Title: A New Proof of Pythagorean theorem?

Body:
A right triangle with side lengths $a,b,c$  where the hypotenuse($c$) intercepts the height($h$) such that $c$ is the sum of $2$ side lengths, $c = m + n$.

Image here:

I found an elegant proof of Pythagorean theorem that seems to be related to Einstein's proof:

First, slopes of $a$ and $b$ give:

$$(\frac{h}{m})(\frac{h}{n}) = 1\implies h^2 = mn$$

Than with area formula, we find the intercept $(x,y)$ of the height and hypotenuse:

$$bx = hm\implies x = \frac{am}{c}$$
$$ay = hn\implies y = \frac{bn}{c}$$

Next plugging this into the height's equation:

$$\frac{a}{b}(\frac{am}{c}) =
\frac{bn}{c}\implies a^2m = b^2n$$

Using $c = m  + n$:

$$\frac{c}{\sqrt{mn}} = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}$$
$$\frac{c^2}{ab} = \frac{a}{b} + \frac{b}{a}$$
$$a^2 + b^2 = c^2$$

Q.E.D.

Thoughts?


问:
Why the mean value of a Gaussian process is usually set to zero?
In most textbooks (e.g. Rasmussen's book on Gaussian Processes for Machine Learning) the mean value of a gaussian process is set to zero. Of course, this does not mean that all the values are expected to be zero since we are looking for the Maximum a Posteriori estimate of these variables, which do not have any more a zero mean.

Is there a mathematically robust proof or at least an explanation based on mathematics that the assumption of the zero mean value is not too prohibitive for the a posteriori estimate?

To make things a bit more clear, assume that we have the following model where the noise $e$ is uncorrelated with $f(x)$:

$y= f(x) +e $,

$f(x) \sim \mathcal{N} (m, K)$,

$e \sim \mathcal{N} (0, \sigma^2)$.

Then the a posteriori (which is actually the MAP estimate) is given by

$\mathbb{E} (f | y) = m + K (\sigma^2 I + K)^{-1} (y-m)$

The way I see it, by setting the mean equal to zero we usually reduce the number of hyperparameters that have to be estimated, because most of the times the mean is also unknown, so it has to be parametrized by some hyperparameters and then these hyperparameters have to be estimated by a non-convex optimization routine, thus increasing the computational burden. However, this does not explain mathematically if the assumption that the mean is zero does not restrict too much the a posteriori estimate.

Any help would be appreciated!

答:
To cite Chiles and Delfiner in Geostatistics Modeling Spatial Uncertainty (2012) on page 151 for the derivation of Simple Kriging (which essentially is GPR for a Gaussian assumption):

The estimator $Z^*$ becomes

$Z^*=m_0+\sum_\alpha\lambda_\alpha(Z_\alpha-m_\alpha)$

This amounts to estimating the zero-mean variable $Y(x)=Z(x)-m(x)$ by the linear estimator

$Y^*=\sum_\alpha\lambda_\alpha Y_\alpha$

and adding the mean afterwards. Thus we have established that the case of a known mean is equivalent to the case of a zero mean with $\lambda_0=0$, and from now on in this section we will consider that $Z(x)$ has a zero mean.

In other words: it would not make a difference if you use the mean of the data or assume a zero mean with normalized data and add the mean afterwards. Rasmussen and Williams say on page 13 in Gaussian Processes for Machine Learning (2006) that the mean is set to zero only for simplicity, but it does not necesseraily have to be zero. Usually, when you assume a known mean (as in SK or GPR when the mean is not individually modeled), you have two options:

Assume the mean to actually be zero and just perform GPR

Normalize your observed data (so the new mean really is zero), then solve your GPR interpolation on the normalized data and add the mean afterwards (which is the exact same as described in Chiles and Delfiner (2012))

Both options will give a very similar interpolation and variance prediction for interpolation but differ in extrapolation. The reason for this is that the extrapolated value will go towards the assumed mean after a set distance which is proportional to the length scale parameter. For option 1 this will exactly be zero. In contrast to that, option 2 will first go to zero but due to the added mean value the extrapolation will result in the mean before normalization.

ContentLicense: CC BY-SA 3.0


问:
Does a second-countable, locally compact, Hausdorff space admit a countable basis of pre-compact open sets?
Let $X$ be a such space. I know that for a Hausdorff space, being locally compact is equivalent to having a basis of pre-compact open sets. But how do I prove that $X$ can be covered by countably many such sets, by using second-countability of X? Thank you.

答:
Every locally compact Hausdorff space is (completely) regular. Every second-countable regular Hausdorff space is metrisable, and hence separable. Any locally compact metrisable space admits a compatible metric, each of whose (proper) open balls is precompact.

Thus if $X$ is locally compact, Hausdorff and second-countable, then it is Polish. It admits a compatible, but possibly noncomplete, metric $d$ for which there is a countable subset $\{x_n\}_{n\in\mathbb{N}}$ such that $\{B_{\frac{1}{m}}(x_n)\mid m,n\in\mathbb{N}\}$ is a countable base of precompact sets for $X$.


问:
Can you always treat variables as constants in partial derivatives/integrals?
Ok so imagine you are given an expression: $xy : x=y$, & you need to take the partial derivative w.r.t. $x$ but you don't simplify (i.e. substitute x=y) --> $d/dx(xy)=y$.

Now you've gotten a different answer than you would have gotten had you simplified/substituted. But in math you should only be able to get 1 correct answer regardless of simplification. So why does standard partial derivative procedure in this trivial example yield 2 seemingly 'correct' answers?

P.S. Ditto question for partial integrals & cases where $x=y + \epsilon: \epsilon\approx 0$.

答:
The partial derivative with respect to $x$ of $f(x,y) = xy$ is defined to be the result you get by holding $y$ constant and differentiating with respect to $x$:
$$\frac{\partial}{\partial x}f(x,y) = \frac{\partial}{\partial x}(xy) = y.$$  If you are looking for the total rate of change of $f(x,y) = xy$ with respect to $x$, with the additional information that $y = x$, then you are really looking for the total derivative, and will need to use the chain rule:
$$\frac{d}{dx}f(x,y) = \frac{d}{dx}(xy) = y + x\frac{dy}{dx} = y + x = x + x = 2x,$$ or alternatively by substituting $y = x$ in the expression, you have $$\frac{d}{dx}f(x,x) = \frac{d}{dx}x^2 = 2x.$$  It all depends on whether you are in fact looking for the partial derivative or the total derivative.


问:
Definition of diffeomorphism functions
I see the definition of Diffeomorphism in Wikipedia homepage, but I don't understand whether "the differential of f (Dfx : Rn → Rn) should be bijective at each point x in U" or "f itself" should be bijective?</p>

答: I think @JBruce's answer, although elegant and sophisticated, might mislead you. The definition of a diffeomorphism $f\colon M\to N$ is that it is a smooth bijection with a smooth inverse. It now follows that for each $x\in M$, $df_x\colon T_xM\to T_{f(x)}N$ is a vector space isomorphism. Hadamard's Theorem gives some very restrictive conditions under which one can know that a local diffeomorphism is actually a global one.</p>

Title: Definition of diffeomorphism functions


问:
Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid?
Suppose that we have a convex polyhedron $P$, such that the symmetry group of $P$ acts transitively on its vertices, edges, and faces (that is, it is isogonal, isotoxal, and isohedral). It then follows that $P$ is one of the Platonic solids, which have a number of equivalent definitions but usually includes some stipulation that the faces are regular. (Once you have regular faces plus these symmetry conditions, most other properties follow easily.)

This can be shown via a rather gross casework argument on the number of sides at each face. By face-transitivity, all faces are congruent (in fact, we only need that the polyhedron be monohedral for this). By edge-transitivity, said faces are equilateral polygons. Then we break into cases by the number of sides of the polygons:

If the faces are triangles, being equilateral implies being regular.

If the faces are non-regular equilateral quadrilaterals, they are rhombi. Then, at every vertex there must be an equal number of "skinny" and "wide" angles meeting, and there must be at least 3 faces meeting at a vertex. But then there are at least two skinny and two wide angles (since we assume the angles are different), so the angles at each vertex add up to at least 360 degrees, which is not possible in a convex polyhedron.

If the faces are non-regular equilateral pentagons, then they have angles which are not all the same; however, each vertex must have the same angles around it, so each vertex has a proportional fraction of the angles of the pentagons. But for any distribution other than the uniform one, the only way to have a proportional amount of every angle is to have some $n$ copies of every angle with multiplicity, since $5$ is prime. (E.g. if there are $2$ angles of one size and $3$ of another, we can't get a $2:3$ ratio without having at least $5$ angles at every vertex.) But the angles of a pentagon add to more than $360^\circ$, so we can't have them all at once.

If the faces are hexagons or higher, they cannot form a topological sphere by Euler's formula.

However, I don't like this proof very much; I walk away from it with no understanding of why this behavior occurs, and it certainly doesn't suggest any generalizations to other dimensions (where I believe the analogous statement - that transitivity on every degree of cell in a convex polytope implies regularity - is true, though I would appreciate a reference for this fact).

Is there a natural or more elegant proof of this statement?

One thing to note is that any proof must somehow use the fact that we are working on the sphere. In the case of planar tilings, the corresponding implication is false: being transitive on faces, vertices, and edges does not imply that one is looking at a tiling of regular polygons! As a counterexample, consider a rhombic tiling like this one:

As another aside, relaxing any one of these conditions allows for polyhedra whose faces are not congruent regular polygons: the rhombic dodecahedron is face-transitive and edge-transitive, an isosceles tetrahedron is face-transitive and vertex-transitive, and the cuboctahedron is edge-transitive and vertex-transitive.

答:
This is a more elaborate version of my comment below the question.

The first two observations are:

if $P$ is edge-transitive, then all its edges are of the same length.

if $P$ is vertex-transitive, then $P$ is inscribed, that is, all its vertices are one a common sphere.

The next observation is the following: if $P$ has both properties, then also all its faces must have both properties. This is obvious for the edge lengths. For being inscribed consider the following image:

All vertices of $P$ are on a sphere $S$. If $f$ is a face of $P$ and $E$ is the plane that contains $f$, then $S\cap E$ is a circle that contains all the vertices of $f$. In other words, $f$ is inscribed as well.

So all faces of $P$ are inscribed and have all edges of the same length.
The final observation is that the faces must then be regular polygons.
This is easiest seen via an explicit construction:

If the radius of the circle and the edge lengths are fixed, then placing a single edge in the circle inductively determines all other edges as shown in the figure. That is, the inscribed polygon with this edge length is uniquely determined. But a regular polygon has this property, and so the face must be this regular polygon.


问:
Invertible Conditions in Polynomial Rings
Let ${\bf a}(x)\in (\mathbb{Z}/q\mathbb{Z})[x]$, where $q$ is a prime. Suppose that
${\bf a}(1)\equiv 0\pmod{q}$.

Question: How to show that ${\bf a}(x)$ is not invertible in $\frac{\mathbb{Z}/q\mathbb{Z}}{x^N-1}$ where $N$ is prime.

try: Using assumptions we get $(x-1)\mid {\bf a}(x)$ in $(\mathbb{Z}/q\mathbb{Z})[x]$.

But I don't know how to continue the proof.

Thanks for any help

答:
We have $x-1\mid a(x)$.

On the other hand, $x-1\mid x^N-1$ and so the gcd between $a(x)$ and $x^N-1$ is a multiple of $x-1$.

It follows that $a(x)$ is not invertible in the quotient ring $\Bbb Z_q[x]/\langle x^N-1\rangle$, since the invertible elements in this ring have $gcd(a(x),x^N-1)= 1$. All other elements are zero divisors (bar $0$ which is absorbing).

Let $gcd(a(x),x^N-1)=d$.
Then the extended Euclidean algorithm gives $d = c(x)(x^N-1) + b(x)a(x)$.

If $d=1$, then $b(x)$ will be the inverse of $a(x)$ in the quotient ring.

The situation in $\Bbb Z_n$ is similar.


答:
You can also use generalized harmonic numbers $$S_p=\sum_{n=1}^{p^2} \frac{1}{\sqrt{n}}=H_{p^2}^{\left(\frac{1}{2}\right)}$$ Using their asymptotics
$$S_p=2 p+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 p}-\frac{1}{24
   p^3}+O\left(\frac{1}{p^7}\right)$$ Using the "exact" value of $\zeta \left(\frac{1}{2}\right)$ and $p=1010$, the truncated expansion gives
$$S_{1010}\sim \color{blue}{2018.54014054065492242599635}394$$ while the "exact" value is $$S_{1010}\sim \color{blue}{2018.54014054065492242599635638}$$

Title: Why does the integral work for this sum?

Body:
Suppose we have the following sum: $$ \sum_{n=1}^{1010^2} \frac{1}{\sqrt{n}} $$ and we want to calculate the very next integer greater than this.

Then one may take the integral, $$ \int_1^{1010^2} \frac{1}{\sqrt{x}}$$ and calculate it to be $2018$, so the very next integer greater than this is $2019$. Why does the integral work so well here?

I know the integral is equal to the sum of $$ f(x) \delta x$$ as $x$ tends to zero, but I still can’t derive how the integral works so well here. I lack intuition, I think.


问:
Null measure set and integrals
Let $f,g:[a,b] \rightarrow \mathbb{R}$ be integrable functions and $$X=\{x \in [a,b];f(x) \ne g(x)\}$$
be a set with zero measure.

Prove that $$\int_a^b f(x)dx=\int_a^b g(x)dx$$ and show two functions $f \ne g$ such that $X$ is infinite and has zero measure.

How can i do this in a context of Riemman integration? I saw some anserws in other questions but it wasn't very helpfull.

答:
We can prove this in the context of Riemann integration just using one measure-related fact: the measure of a non-empty interval is positive.

If $f$ and $g$ are Riemann integrable, then so is $h := f-g$ and $X = \{x \in [a,b]:h(x) \neq 0\}$ is given to be of measure zero.

For any partition $P: a = x_0 < x_1 < \ldots <x_n = b $, each subinterval $I_j = [x_{j-1},x_j]$ has positive measure and, hence, there is a point $x \in I_j$ where $h(x) = 0$.  This implies that $\inf_{x \in I_j} h(x) \leqslant 0 \leqslant \sup_{x \in I_j} h(x)$ and, consequently for upper and lower Darboux sums we have

$$\tag{1}L(P,h) \leqslant 0 \leqslant U(P,h)$$

Since $h$ is Riemann integrable we have for all $P$,

$$\tag{2}L(P,h) \leqslant \int_a^b h(x) \, dx \leqslant U(P,h)$$

and for any $\epsilon >0$, there exists a partition $P_\epsilon$ where $U(P_\epsilon,h) - L(P_\epsilon,h) < \epsilon$

In view of   (1) and (2) this implies that

$$-\epsilon < -(U(P_\epsilon,h) - L(P_\epsilon,h)) \leqslant \int_a^bh(x) \,dx \leqslant U(P_\epsilon,h) - L(P_\epsilon,h) < \epsilon$$

Since $\epsilon$ can be arbitrarily close to zero, it follows that

$$\int_a^bf(x) \, dx - \int_a^b g(x) \, dx = \int_a^bh(x) \,dx = 0$$

Title: Null measure set and integrals


问:
Is the third order term of a Taylor approximation a covariant tensor of rank 3 restricted to (h,h,h)?
Recently reading Peter Lax's linear algebra text, he had a very concise way of writing the Taylor approximation up to second order: $f(x+h) = f(x) + l(h) + \frac{1}{2}q(h) + \|h\|^2 \epsilon(\|h\|)$, where $l$ is a linear functional, $q$ is a quadratic form, and $\epsilon \to 0$ as $h\to 0$.

How would we describe the third-order term of the approximation as a function of $h$? It couldn't be represented as a matrix, in the way that a bilinear form can; we'd need a 3d array I think. It seems we could express it as a homogeneous polynomial of degree 3 in $h_1, \dotsc, h_n$.  Am I right that it is also a covariant tensor of rank 3, restricted so that it takes arguments $(h,h,h)$, so it is of the form $\left(\sum\limits_{I\in \{1,\dotsc,n\}^3}a_I\varepsilon^*_I \right)(h,h,h)$? Is there a name for such a thing?

答:
Yes, this is the multivariate Taylor expansion. To derive the terms (of whatever order your heart desires) you simply feed a line $a+th$ into a multivariate function $f$ to obtain a single-variable function $g(t)=f(a+th)$ to which the single-variate Taylor expansion applies. After that, systematic application of the multivariate chain-rule reveals the expansion you seek. Just set $t=1$ and you obtain $f(a+h)=f(a)+f'(a) \cdot h + \cdots $. There are a variety of cute ways to express the formula. The one I have in my notes for functions of two variables $x=x_1$ and $y=x_2$ centered at $(a_1,a_2)$
$$ f(x, y) = \sum_{n=0}^{\infty} \sum_{i_1=0}^{n}\sum_{i_2=0}^{n} \cdots \sum_{i_n=0}^{n} \frac{1}{n!}
\frac{\partial^{(n)}f(a_1,a_2)}{\partial x_{i_1}\partial x_{i_2} \cdots \partial x_{i_n}}
(x_{i_1} -a_{i_1})(x_{i_2} -a_{i_2})\cdots (x_{i_n} -a_{i_n}) $$
The formula for functions of three or more variables is similar. I know of no particular name for the third order piece. Intuitively, it becomes important when the expansion is trivial up to second order and yet it is nontrivial. In such a case it is the dominant term (locally). Of course, I have no particular sense of what the graph of such a thing resembles. In contrast, the second order term is nicely captured by the theory of quadratic forms. </p>


问:
Why isn't Bessel's correction used for sample mean absolute deviation
My understanding so far is the (unbiased) sample variance is modified by Bessel's correction to account for the fact that the first observation of $ (X-\overline X)$ in a given sample X, will always equal zero (is not "free to vary"), as i.a. the mean of the first observation and the first observation will always be the same value. So the number of deviations being averaged is n-1 instead of n.

$$
s^2= \frac{∑(X-\overline X)^2}{\rm n-1}
$$

When I look at the formula for the sample mean absolute deviation, however, it is:

$$
Mean\:Absolute\:Deviation = \frac{∑|X-\overline X|}{\rm n}
$$

Is it also possible to have an unbiased estimator here using $n-1$ and why/why not?

Source for equations: Salvatore and Reagle, Statistics and Econometrics Second Edition - Eqs. 2.6b and 2.8b

答:
For the mean absolute deviation, the correction $n-1$ does not give an unbiased estimator in general.
However, it does works in some cases.

Normal distribution For example, take $X$ standard normally distributed.
Then, $E|X-EX|=E|X|=\sqrt{2/\pi}$.
Moreover,
$$
X_1-\bar{X}=\frac{n-1}{n}X_1-\frac{1}{n}\sum_{i=2}^n X_i=:Y
$$
This is random variable $Y$ has a normal distibution with mean zero, and variance:
$$
Var[Y]=\left(\frac{n-1}{n}\right)^2+\frac{1}{n}
$$
So, $Y$ has the same distribution as $\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}X_1$. We calculate:
$$
E\sum_{i=1}^n|X_i-\bar{X}|=nE|Y|=nE\left|\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}X_1\right|
=n\sqrt{\left(\frac{n-1}{n}\right)^2+\frac{1}{n}}\sqrt{\frac{2}{\pi}}
$$
$$
=\sqrt{(n-1)^2+n}\sqrt{\frac{2}{\pi}}
$$
So, an unbiased estimator for $E|X|$ is:
$$
\frac{1}{\sqrt{(n-1)^2+n}}\sum_{i=1}^n|X_i-\bar{X}|
$$

Bernoulli distribution However, it is true for some special cases.
If $X$ is $0.5$-Bernoulli, it works.
We want to estimate $E|X-EX|$.
We have $E|X-EX|=E|X-0.5|=0.5*(0.5+0.5)=0.5$.
So, our estimator should have the same expected value.
Consider $E|X_1-\bar{X}|$.
We can write:
$$
E|X_1-\bar{X}|=E\left[1-\frac{1}{n}\left(1+\sum_{i=2}^nX_i\right)\Bigg|X_1=1\right]\frac{1}{2}+
E\left[\frac{1}{n}\sum_{i=2}^nX_i\Bigg|X_1=0\right]\frac{1}{2}
$$
$$
=\frac{n-1}{2n}-E\left[\frac{1}{n}\sum_{i=2}^nX_i\right]\frac{1}{2}+E\left[\frac{1}{n}\sum_{i=2}^nX_i\right]\frac{1}{2}=\frac{n-1}{2n}
$$
From here, it follows that $\frac{1}{n-1}\sum_{i=1}^n|X_i-\bar{X}|$ has expectation $0.5$, hence it is an unbiased estimator for $E|X-EX|$.


问:
When is $\int_{[0, \infty)}f.d\lambda_1 = \lim_{n\rightarrow \infty} \int_{[0, n]} f.d\lambda_1$
In my Probability course there are some lebesgue Integrals wrt. Lebesgue Measure. In the notes there is a theorem that says:

Let $f : [a, b] \rightarrow \mathbb{R}$ be bounded. If $f$ is Riemann integrable, then it is Lebesgue
integrable and the integrals are the same:
$$\int_{[a, b]}f.d\lambda_1 = \int_{a}^{b} f(x)dx$$

Sometimes we have to evaluate integrals over unbounded intervals, like $[0,\infty)$. My lecturer would rewrite the integral as $$\int_{[0, \infty)}f.d\lambda_1 = \lim_{n\rightarrow \infty} \int_{[0, n]} f.d\lambda_1$$
so that we can use the theorem above. My question is when is this valid? Do we not have to worry about whether the sequence of integrals converge or does it always converge in this form?

答:
In the sense of improper Riemann integration, the RHS is basically the definition of the LHS (not exactly, because in improper Riemann integration you send the limit to infinity continuously, but this distinction usually doesn't matter).

In the sense of Lebesgue integration, you can view the RHS as $\lim_{n \to \infty} \int_{[0,\infty)} f 1_{[0,n]} \, d \lambda_1$ and then apply one of the Lebesgue integral convergence theorems. In this case you can say that $f 1_{[0,n]}$ is dominated by $f$, so by the Lebesgue dominated convergence theorem, the RHS converges to the LHS if the LHS converges absolutely, or in other words if $f$ is Lebesgue integrable on $[0,\infty)$. If it isn't, then the question doesn't really make sense because the LHS is simply undefined, whether the RHS is defined or not.


问:
How to show that (Change of variable) $ rF'(r)=\int_{\partial D(0, r)} \frac{\partial f}{\partial x}dy-\frac{\partial f}{\partial y}dx $
Let $F(r)=\int_0^{2\pi} f(r\cos \theta, r\sin \theta) d\theta$ where $r>0$. Show that

$$
rF'(r)=\int_{\partial D(0, r)} \frac{\partial f}{\partial x}dy-\frac{\partial f}{\partial y}dx
$$

where $\partial D(0, r)$ is the frontier (boundary) of the open ball of radius $r$.

I have no idea how to prove this. I know at some point there is a change of variables, but it is not a standard one with polar coordinates in a double integral, since $r$ is fixed.

答:
Let $x = r \cos \theta$ and $y = r \sin \theta$. For fixed $r$ we have $dx = -r \sin \theta d\theta $ and $dy = r\cos\theta d\theta$ and so
$$ rF'(r) = \int_B r\cos\theta f_x + r\sin\theta f_y d \theta = \int_B f_x dy - f_y dx.  $$


问:
Clarification on the proof of $\pi$ being transcendental.
I was reading a proof on the fact that $\pi$ is transcendental and the author mentioned that

The $\alpha_i$ satisfy a polynomial equation over $\mathbb{Q}$ so their elementary symmetric functions are rational. Hence the elementary symmetric functions of the sums of pairs are symmetric functions of the $\alpha_i$ and are also rational.

I'm not quite sure why the elementary symmetric functions of the sums of pairs are rational given that the elementary symmetric functions of the $\alpha_i$ are rational. For instance, the 2nd elementary symmetric function of the sum of pairs contains $\alpha_i^2$, and how can we deal with these terms?

EDIT: The proof can be found here: https://sixthform.info/maths/files/pitrans.pdf (Theorem 4).

答: The author is using the fundamental theorem of symmetric polynomials, according to which the $A$-subalgebra of $A[X_1,\cdots,X_n]$ generated by the $n$ elementary symmetric polynomials is the ring of symmetric polynomials with coefficients in $A$ (in this case, it's $A=\Bbb Z$ or, at worst, $\Bbb Q$).


问:
Terminology for A line Pattern that Doesnt Cross
I am looking for terminology to the line pattern that we can draw connecting certain points such that the lines doesnt cross. Is there such a name?</p>

答:
As per @player3236's comment who chose not to post as an answer, the term is: Planer Graph

In graph theory, a planar graph is a graph that can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints. In other words, it can be drawn in such a way that no edges cross each other. Such a drawing is called a plane graph or planar embedding of the graph


问:
If a set is "not open", what is it?
Physics background here, no formal pure maths education so please try to reply accordingly.

What are the possibilities for a set which is "not open"? Are they:

1)

closed

both open and closed

neither open nor closed

or

2)

closed

neither open nor closed?

In other words if a set is said to be "not open" does that mean that it could be both open and closed?

Accordingly for "not closed".

Or am I entirely confused?

答:
A set can be open (if it's in the topology, by definition), closed (if its complement, the "outside" of the set in the whole space) is in the topology, both open and closed (if both are the case, for example in $\Bbb N$ with its natural topology inherited from $\Bbb R$), all sets are both open and closed, while in $\Bbb R$ in the usual topology only $\emptyset$ and $\Bbb R$ are both open and closed (as they are in any topology), but in general most sets will be neither open nor closed.

So not open does not imply anything else besides just that: it's not in the topology. It could still be closed, or not closed.

Ditto for non-closed sets: they could be open or not. Open and closed are just the simplest kind of sets topolgosts are interested in, also countable unions of closed sets ($F_\sigma$'s) get a special name, as do countable intersections of open sets ($G_\delta$'s). They play special roles too sometimes, e.g. in studying measures, points of continuity of functions etc.

Don't let language fool you: open is not an opposite of closed, or vice versa.

Title: If a set is "not open", what is it?


答:
Everything can happen!  For instance, in $\mathbb R$, with its usual metric:

$$
\begin{array}{c|cc}
&\text{OPEN} &\text{CLOSED} \\
\hline
(0, 1) & \text{Yes} & \text{No}\\
[0, 1] & \text{No} & \text{Yes}\\
[0, 1) & \text{No} & \text{No}\\
\mathbb R & \text{Yes} & \text{Yes}
\end{array}
$$

Title: If a set is "not open", what is it?

Body:
Physics background here, no formal pure maths education so please try to reply accordingly.

What are the possibilities for a set which is "not open"? Are they:

1)

closed

both open and closed

neither open nor closed

or

2)

closed

neither open nor closed?

In other words if a set is said to be "not open" does that mean that it could be both open and closed?

Accordingly for "not closed".

Or am I entirely confused?


问:
Show that the sum of two conjugate complete sets is conjugate complete
In the proof of the transcendence of $e^a$ for an algebraic nonzero $a$ Burger's Transcendence book gives a very unclear argument that the sum of two conjugate complete sets is a conjugate complete set but I can't understand it at all! For an example each of the two sets ${\{\sqrt{2},-\sqrt{2}}\}$ and ${\{\sqrt{3},-\sqrt{3}}\}$ are conjugate complete since each consists of all zeros of their irreducible polynomial. Consequently sum of the mentioned two sets, which is ${\{\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3},\sqrt{2}+\sqrt{3} }\}$, is all of zeros of a single irreducible polynomial with integer quotient . So how to prove this in general, that is two sets of zeros of two arbitrary irreducible polynomials? I know a little abstract algebra.

答:
Let $\overline {\mathbb Q} / \mathbb Q$ be the field of algebraic numbers. Then the conjugates of an algebraic number $\alpha$ are just the images of $\alpha$ under automorphisms of $\overline {\mathbb Q}$, and a set $S$ of algebraic numbers is conjugate complete if and only if it is stable under the action of this automorphism group.

If $S$ and $T$ are conjugate complete sets, and $\sigma$ is an automorphism of $\overline{\mathbb Q}/ \mathbb Q$, then for any $s \in S$ and $t \in T$, we have
$$
\sigma(s + t) = \sigma(s) + \sigma(t) \in \sigma(S) + \sigma(T) = S + T ,
$$
and so $S+T$ is also conjugate complete.

Adding in response to questions:

In the context of your original question, we can think of all of this happening within a finite extension of $\mathbb Q$. Let $K$ and $L$ be the fields obtained by adjoining to $\mathbb Q$ all the roots of $f, g \in \mathbb Q[x]$. These are Galois extensions. The roots of $f$ are permuted by $G_K = \operatorname{Gal}(K/\mathbb Q)$, and the conjugates of a root $\alpha$ of $f$ are of the form $\sigma(\alpha)$ for $\sigma \in G_K$.
Similarly, the roots of $g$ are permuted by $G_{L} = \operatorname{Gal}(L/\mathbb Q)$.

Let $\alpha$ be a root of $f$ and $\beta$ be a root of $g$.
Then $\alpha + \beta$ lives in the composite field $KL$, and its conjugates are all of the form
$$
  \sigma(\alpha + \beta)
= \sigma(\alpha) + \sigma(\beta)
$$
for some $\sigma$ in $G_{KL} = \operatorname{Gal}(KL/\mathbb Q)$.
Since $\sigma(\alpha) = \sigma|_K(\alpha)$ and $\sigma|_K$ is in $G_K$, $\sigma(\alpha)$ is a conjugate of $\alpha$.
Similarly, $\sigma(\beta)$ is a conjugate of $\beta$.
We conclude that the set of sums $\alpha + \beta$ is conjugate-closed.


问:
Substitution in ordinary differential equation gone wrong
Given

$$
y(t) = \left( \int u(t) \; \text{dt} \right)^2,
$$

the derivative of $y$ over $t$ is

$$
\dot{y} = 2 u(t) \int u(t) \; \text{dt}.
$$

Substituting $\int u$ in the above equation yields

$$
\dot{y} = 2u(t) \sqrt{y}.
$$

But integrating this differential equation with $y(0) = 0$ and $u(t) \equiv 1$ yields $y \equiv 0$ for all $t$, whereas $y(t) = t^2$ in the original equation.

Why is that?

答:
I'm not sure what integration you used to obtain it, but  $y\equiv 0$ is indeed a solution to $y'=2\sqrt y$ with $y(0)=0$. Actually, so is $y(t)=t^2$. It seems then that you have stumbled upon the fact that that going from functions to ODE initial value problems (IVPs) solved by the function does not always allow you to reproduce the function again. This is called non-uniqueness of solutions for the IVP. It is not a simple problem and I'm quite sure there isn't a simple answer.

There is a more elementary example of non-uniqueness of solutions for IVPs: $y(t)=t^2$ also solves
\begin{cases}y''-2=0 \\ y(0)=0\end{cases}
and here there is no unique solution: $y=x^2+kx$ is a solution for all $k\in\mathbb R$. The IVP with the unique solution $y(t)=t^2$ is instead
\begin{cases}y''-2=0 \\ y(0)=0\\y'(0)=0\end{cases}
There is an argument to be made that it is "my fault" for not writing down a good IVP in the above, and it was easily fixed by using the "right" initial data. In the ODE you found however, the problem is more subtle, and it doesn't matter what other information you try to prescribe at $t=0$. In this case there is also an infinity of solutions: they are the two mentioned above, $y=0$, $y(t)=t^2$, and then weird families of solutions that switch from $0$ to $t^2$ at an arbitrary position. That is, for any $a\in(0,\infty)$, the following function is a solution:
$$  y(t)=\begin{cases} (t-a)^2 &\phantom{0\le{}}t\ge a \\ 0 &0\le t
Why this happens is not easy to explain, and the theory of ODEs is much more intricate than it might seem at first (and the theory of partial differential equations is more complex still!)

A basic explanation would be that $y'=2\sqrt y$ does not satisfy the assumptions of the Picard-Lindelöf theorem for ODEs. This says that $y'=f(y,t)$ with an initial condition has unique solutions if $f$ is Lipschitz, and  $\sqrt y$ is indeed not. But there are href="https://encyclopediaofmath.org/wiki/Osgood_criterion" rel="nofollow noreferrer">other ODEs that fail the assumptions of Picard-Lindelöf while still having unique solutions, so this is still far from a perfect explanation.

I'll end with mentioning however that there is a sense in which the "correct" solution to $y'=2\sqrt y$ is in fact $y(t)=t^2$. This is however quite technical; it is the theory of renormalized solutions developed by DiPerna and Lions.

Body:
I'm not sure what integration you used to obtain it, but  $y\equiv 0$ is indeed a solution to $y'=2\sqrt y$ with $y(0)=0$. Actually, so is $y(t)=t^2$. It seems then that you have stumbled upon the fact that that going from functions to ODE initial value problems (IVPs) solved by the function does not always allow you to reproduce the function again. This is called non-uniqueness of solutions for the IVP. It is not a simple problem and I'm quite sure there isn't a simple answer.

There is a more elementary example of non-uniqueness of solutions for IVPs: $y(t)=t^2$ also solves
\begin{cases}y''-2=0 \\ y(0)=0\end{cases}
and here there is no unique solution: $y=x^2+kx$ is a solution for all $k\in\mathbb R$. The IVP with the unique solution $y(t)=t^2$ is instead
\begin{cases}y''-2=0 \\ y(0)=0\\y'(0)=0\end{cases}
There is an argument to be made that it is "my fault" for not writing down a good IVP in the above, and it was easily fixed by using the "right" initial data. In the ODE you found however, the problem is more subtle, and it doesn't matter what other information you try to prescribe at $t=0$. In this case there is also an infinity of solutions: they are the two mentioned above, $y=0$, $y(t)=t^2$, and then weird families of solutions that switch from $0$ to $t^2$ at an arbitrary position. That is, for any $a\in(0,\infty)$, the following function is a solution:
$$  y(t)=\begin{cases} (t-a)^2 &\phantom{0\le{}}t\ge a \\ 0 &0\le t<a \end{cases} $$
Why this happens is not easy to explain, and the theory of ODEs is much more intricate than it might seem at first (and the theory of partial differential equations is more complex still!)

A basic explanation would be that $y'=2\sqrt y$ does not satisfy the assumptions of the Picard-Lindelöf theorem for ODEs. This says that $y'=f(y,t)$ with an initial condition has unique solutions if $f$ is Lipschitz, and  $\sqrt y$ is indeed not. But there are other ODEs that fail the assumptions of Picard-Lindelöf while still having unique solutions, so this is still far from a perfect explanation.

I'll end with mentioning however that there is a sense in which the "correct" solution to $y'=2\sqrt y$ is in fact $y(t)=t^2$. This is however quite technical; it is the theory of renormalized solutions developed by DiPerna and Lions.


答:
A topological space is a pair $(X,T)$ where $X$ is a set and $T$ is a collection of some or all of the subsets of $X$ such that

(1). $X\in T$ and $\emptyset\in T.$

(2). If $S\subseteq T$ then $\bigcup S$ (which means $\cup_{R\in S}R\,$) is a member of $T.$

(3). If $S$ is a $finite$ subset of $T$ then $\cap S$ (the common intersection of the members of $S$) is a member of $T$.

The members of $T$ are called the open sets and their complements in $X$ are called closed sets. $T$ is called a topology on $X.$ Now $T$ could contain every subset of $X$ (Note that $X$ is a subset of $X$), in which case every subset of $X$ is open and closed ("clopen"). But it could also be that $T=\{X,\emptyset\}.$ It is common to say "a space $X$" as if everyone knew which possible $T$ was specified.

Title: If a set is "not open", what is it?


问:
Proving that a proposition is a theorem in $L_I$.
In "Introduction to Mathmatical Logic" by Elliot Mendelson, I was able to prove exercise 1.60 (a) and 1.60(b), but failed to prove 1.60 (c) .It is as follows:

For any m , the wf
$(A_1 \leftrightarrow A_2) \lor...\lor (A_1 \leftrightarrow A_m) \lor (A_2 \leftrightarrow A_3) \lor ...(A_2 \leftrightarrow A_m)\lor...\lor (A_{m-1} \leftrightarrow A_m)$  is not a theorem of $L_I$

I first tried to write it as a summation:
$\bigcup _{j=1}^{m-1} \bigcup _{i=2}^m (A_j \leftrightarrow A_i)$

Then I thought of using induction on this summation.

(i) prove it for $m=2$

(ii) Assume that it is proven for $m=k$

(iii) Prove it for $m = (k+1)$

But still , the problem didn't got anywhere from there. Can someone give me a different idea?

答:
Long comment

$\text {L}_{\text I}$ is Intuitionistic Logic and $\text {IL}$ has the so-called Disjunction property:

The disjunction property is satisfied by a theory if, whenever a sentence $A ∨ B$ is a theorem, then either $A$ is a theorem, or $B$ is a theorem.

The $A_i$s are sentential letters, and thus none of the disjuncts $(A_i \leftrightarrow A_j)$ is a theorem, when $i \ne i$.

Use this fact in a proof by induction on the complexity (number of disjunction signs) of the formula.

Compare with Classical Logic: $(A_1 \leftrightarrow A_2)$ is not a tautology, but  $(A_1 \leftrightarrow A_2) \lor (A_1 \leftrightarrow A_3) \lor (A_2 \leftrightarrow A_3)$ is.


问:
Example of vector space
In almost all the literature that I have seen, one of the examples of vector space is as follows:

Set of all real-valued functions $f(x)$ defined on the real line

what confuses me here is that the word "linear functions" should be used instead of just "functions" because I think we also have non-linear functions and they (non-linear functions)do not follow the rule of addition and multiplication as:

$ (f + g)(x)=f(x) + g(x) $ and $(kf)(x) = k f(x) $

thus, non-linear functions cannot make a vector space.

am I right or wrong?

答:
You're confusing adding the functions with adding their inputs.
A linear function would have the property $f(x+y) = f(x) + f(y)$. However, for a vector space, you add the functions. For example, let $f(x) = x^2$ and $g(x) = x^3$, then $(f + g)(x) = f(x) + g(x) = x^2 + x^3$.</p>

Title: Example of vector space


问:
Prove there are infinitely many squares ending in the digits $54321$
I was training for the math olympiad as I encountered the following, designated simple problem:
Prove there are infintely many squares ending in the digits $54321$.
I tried obvious factorization $3*18107$ and then an approach $\mod 10^5$ but the number is so large, I did not find any example, from which the rest would be trivial.
Factorization of the form $a^2-10^{2n}=54321$ also didn't work, any hints or solutions?

Body:
$11111^2=123454321$

$111111^2=12345654321$

$1111111^2=1234567654321$

$\underbrace{(111\cdot\cdot\cdot\cdot111}_{(n) times})^2=\underbrace{1234567\cdot\cdot\cdot7654321}_{(2n-1) digits}$

This pattern is valid up to $n=11$. From $n=12$ to $n= 18$ we still see a regular pattern in median digits and number of digits safies $2n-1$, but for $n\geq20$ we do not see a regular pattern.

$n=11\rightarrow k^2=12345678987654321$

$n=15\rightarrow: k^2=12345679,012345654320,987654321$

$n=18\rightarrow:k^2=12345679,012345678987654320,987654321$

$n=19\rightarrow:k^2=12345679, 0123456789. 00, 987654320, 987654321$

$n=20\rightarrow: k^2=137174211248285\cdot\cdot\cdot$

This observation shows that the first 5 digits(from right) of n must be $1$, the rest digits can be any number less than $9$. In fact a number in the form $a11111$ when powered by two gives:

$a11111+a11111\times10+a11111\times10^2+a11111\times10^3+a11111\times10^4+a11111\times10^5=(a11111)\times10^5+54321$


问:
Error function etymology: Why the name?
I've recently been introduced to the error function:

$$\operatorname{erf}(z) =\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt$$

Naturally, I wondered about the origin of its name: The error of... what? I'm so used to uncertainty analysis in scientific experimentation that my mind went there first. It seems like the name sprung up in the context of statistical analysis. Here's what I found on Quora:

In the course of the 19th century the function from the theory of errors appeared in several contexts unrelated to probability, e.g. refraction and heat conduction. In 1871 J. W. Glaisher wrote that "Erf(x) may fairly claim at present to rank in importance next to the trigonometrical and logarithmic functions." Glaisher introduced the symbol Erf and the name error function for a particular form of the law as follows:

As it is necessary that the function should have a name, and as I do not know that any has been suggested, I propose to call it the Error-function  ...

("On a Class of Definite Integrals, Philosophical Magazine, 42, 1871, p. 296)

But this quote sounds like Glaisher arbitrarily picked the name. He must have had some motivation in dubbing it error.

Anyway, what does error represent? Does this measure some deviation from the related Gaussian integral from $\int_0^\infty$?

EDIT The link provided by Dilip is useful, yet still did not explain the depth of this name to me. I am wholly unfamiliar with probability and error theory - where $erf(z)$ seems to have arisen - so excuse me if the answer should have been more obvious. (In fact, error theory sounds vaguely alien to me.) Is the function just commonplace in error theory, and for that reason named $erf$? This reasoning just seems a little, well, lackluster to me, but maybe it's true. Does $erf$ have a specific role that reflects its name?

答:
The naming of normal distribution seems to be from Karl E. Pearson (p.112)
see
https://www.maa.org/sites/default/files/images/upload_library/22/Allendoerfer/stahl96.pdf

In addition, the paper hinted the origin of so called the "Error" function. It's indeed the (accumulated) error from the measurement against the "mean" (CDF).

My personal wild guess (just kidding but could be true) is that these gents studied the distance of the star Ceres and they were using their eyes (assisted by their crude apparatus).

Human eyes suffer from various Refractive Errors, hence, the "Error" function.

Title: Error function etymology: Why the name?


问:
Why does the integral work for this sum?
Suppose we have the following sum: $$ \sum_{n=1}^{1010^2} \frac{1}{\sqrt{n}} $$ and we want to calculate the very next integer greater than this.

Then one may take the integral, $$ \int_1^{1010^2} \frac{1}{\sqrt{x}}$$ and calculate it to be $2018$, so the very next integer greater than this is $2019$. Why does the integral work so well here?

I know the integral is equal to the sum of $$ f(x) \delta x$$ as $x$ tends to zero, but I still can’t derive how the integral works so well here. I lack intuition, I think.

答:
The terms of the sum are

$$\frac1{\sqrt n}$$ vs. those of the integral $$\int_n^{n+1}\frac{dx}{\sqrt x}=2{\sqrt{n+1}}-2{\sqrt{n}}\approx2\sqrt n\left(1+\frac1{2n}-\frac1{8n^2}+\frac1{16n^3}-\cdots\right)-2\sqrt n
\\=\frac1{\sqrt n}-\frac1{4n\sqrt n}+\frac1{8n^2\sqrt n}-\cdots$$

When you sum from $1$ to $1010^2$, the first error term is close to $$\frac14\zeta\left(\frac32\right)=0.653094\cdots,$$ which is less than $1$. As the error terms alternate in sign, the global error is even smaller.


问:
Show that algebraic direct sum is $\sigma$-weakly dense.
Consider the abstract von Neumann algebra
$$M:= \ell^\infty-\bigoplus_{i \in I} B(H_i)$$
which consists of elements $(x_i)_i$ with $\sup_i \|x_i\| < \infty$ and $x_i \in B(H_i)$.

Let $N$ be the algebraic direct sum $\bigoplus_i B(H_i)$, thus it consists of elements $(x_i)_i$ such that only finitely many $x_i$ are non-zero. I want to show that $N$ is $\sigma$-weakly dense in $M$.

I guess my main problem is that I don't understand the $\sigma$-weak topology on $M$. By a result of Sakai, it is the unique topology on $M$ coming from a weak$^*$-topology when we realise $M$ as the dual of some other Banach space.

Maybe, we can do the following
$$M \cong \ell^\infty-\bigoplus_{i \in I} T(H_i)^* \cong \left(\ell^1-\bigoplus_{i \in I} T(H_i)\right)^*$$
But this does not seem very practical! Any insight in the matter will be appreciated!

答:
I am posting one more answer, relevant to the double commutant theorem. I review the essentials first and then try to apply it, reaching a conclusion that does not have to do with DCT. You can read the last three paragraphs independently of the rest of the answer.

Theorem: (Von Neumann's DCT) Let $A\subset\mathcal{B(H)}$ be a $*$-algebra that contains the identity operator. Then $\overline{A}^{SOT}=A''$

Note that convex subsets of $\mathcal{B(H)}$ have the same strong and weak closure, so the DCT also says that $\overline{A}^{WOT}=A''$.

Also, if $A$ is a $*$-subalgebra, then the ultraweak closure of $A$ is equal to the strong closure, which is equal to the weak closure.

This follows from Kaplansky's density theorem: since the ultraweak topology is stronger than the weak topology, it is immediate that $\overline{A}^{w^*}\subset\overline{A}^{WOT}$. On the other hand, let $x\in \overline{A}^{WOT}$. By Kaplansky's density theorem we can find a norm-bounded net $(a_i)$ in $A$ so that $a_i\to x$ in the weak topology. But the weak topology coincides with the ultraweak topology on norm-bounded subsets, so $a_i\to x$ ultraweakly, thus $x\in\overline{A}^{w^*}$.

Note that we can also improve the DCT a little bit and require only that the identity operator belongs to the strong closure of $A$, because then
$$\overline{A}^{SOT}=\overline{A+\mathbb{C}1_\mathcal{H}}^{SOT}=(A+\mathbb{C}1_{\mathcal{H}})''=A''.$$

Corollary: If $A\subset\mathcal{B(H)}$ is a $*$-subalgebra and $1_\mathcal{H}$ belongs to the strong/ weak/ ultraweak closure of $A$, then $\overline{A}^{SOT}=\overline{A}^{WOT}=\overline{A}^{w^*}=A''$.

Now we adopt the notation that OP uses in the post. We can take a concrete representation of $M$, so we can find a Hilbert space $\mathcal{K}$ so that $M\subset\mathcal{B(K)}$ and $M=\overline{M}^{SOT}$ and $1_\mathcal{K}\in M$. Note that $1_\mathcal{K}$ will be the same as the unit of $M$, which is the element $(1_{\mathcal{H}_i})_{i\in I}$.

The thing now is this: Obviously $N$ is a $*$-subalgebra of $\mathcal{B(K)}$; if we can show that $1_\mathcal{K}$ belongs to the strong/weak/ultraweak closure of $N$, then we can conclude that $\overline{N}^{w^*}=N''$ and then all we must do is verify that $N''=M$, which I guess is going to be easy.

BUT! Why bother with all this? Observe that $N$ is an ideal in $M$. The ultraweak closure of $N$ is thus an ultraweakly closed ideal in $M$. Therefore, if we manage to show that $1_M$ belongs to the ultraweak closure of $N$, then we immediately have that $\overline{N}^{w^*}=M$.

Conclusion: The essence of what we are trying to prove lies in showing that the unit of $M$ can be approximated ultraweakly by elements of $N$. This is essential in every approach (DCT, the one in the other answer and the one with the ideal observation). I think the technology required to show this is what I present in my other answer. Sure, one can take a concrete representation of $M$ and try to approximate the unit of $M$ through $N$ in the weak topology which might be simpler, but I doubt that it is going to be essentially simpler.

PS: obviously, the net that is going to approximate $(1_{\mathcal{H}_i})$ (the unit of $M$) in any case is the net $(x_\lambda)\subset N$ where $\Lambda$ is the set of finite subsets of $I$ and $x_\lambda$ is the sequence having $0$ everywhere except the slots $i\in\lambda$ where it has $1_{\mathcal{H}_i}$


问:
Finding the range of a matrix $A$
Consider the linear transformation define by

$$u(x) = \begin{bmatrix}
x_1 - \bar{x} \\
x_2 - \bar{x} \\
\vdots \\
x_m - \bar{x}
\end{bmatrix},
$$
for all $x \in R^m$, where $\bar{x} = (1/m) \sum x_i$.

Find the matrix $A$ for which $u(x) = Ax$, and then determine the dimension of the range and null spaces of $A$.

I found $A$ to be:

$$
\begin{bmatrix}
(1-\frac{1}{m}) - \frac{1}{m} \cdots -\frac{1}{m}\\
-\frac{1}{m} + (1-\frac{1}{m}) \cdots - \frac{1}{m}\\
-\frac{1}{m} - \frac{1}{m} + (1-\frac{1}{m})
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\vdots\\
x_m
\end{bmatrix}
=
I_m -\frac{1}{m}1_m 1'_m
$$

I am stuck at how to find the dimension of the range because the book has the answer has $\dim(R) = m-1$ but I'm not seeing where they get that from matrix $A$.

答:
$\newcommand{\tr}{\mathrm{tr}}$
If you are just looking for the rank of $A$, here is a (less direct) way. Since it is easy to verify that $A$ is idempotent, by applying the property of idempotent matrix (its rank is identical to its trace), we claim that:
\begin{align*}
\mathrm{rank}(A) = \tr(A) = \tr(I_{(m)} - m^{-1}1_m1_m') =
\tr(I_{(m)}) - m^{-1}\tr(1_m1_m') = m - m^{-1}\tr(1_m'1_m) = m - 1,
\end{align*}
which is the dimension of the linear operator $A$.

Title: Finding the range of a matrix $A$


答: The functions do not need to be linear. You are confusing the fact that $f(x+y)=f(x)+f(y)$ only holds for linear functions, but this is not required here. What is meant is that the sum of two functions (not the sum of the arguments!) $f$ and $g$ is the function $(f+g)(x)$, which is given by $f(x)+g(x)$.</p>

Title: Example of vector space

Body:
In almost all the literature that I have seen, one of the examples of vector space is as follows:

Set of all real-valued functions $f(x)$ defined on the real line

what confuses me here is that the word "linear functions" should be used instead of just "functions" because I think we also have non-linear functions and they (non-linear functions)do not follow the rule of addition and multiplication as:

$ (f + g)(x)=f(x) + g(x) $ and $(kf)(x) = k f(x) $

thus, non-linear functions cannot make a vector space.

am I right or wrong?


问:
switch finite and infinite rows
Asume to have a function $f(k,l):\mathbb{N} \times \mathbb{N}\rightarrow \mathbb{R}$ s.t.
$$\sum_{l=1}^\infty f(k,l)=c_k<C \,\,\,\forall k\in\mathbb{N}.$$

Does the following hold for any finite $n$?:
$$
\sum_{k=1}^n\sum_{l=1}^\infty f(k,l)=\sum_{l=1}^\infty\sum_{k=1}^n f(k,l)
$$

答: We clearly have that the sum on the LHS is just equal to $\sum_{1 \leq k \leq n} c_k.$ To show that your formula holds, we just need to show that the sequence of the partial sums of the series on the RHS converges to this quantity. Let $\varepsilon > 0.$ Since $\sum_{l \geq 1} f(k, l) = c_k,$ there is $n^\varepsilon_k > 0$ such that $\vert \sum_{1 \leq l \leq m} f(k, l) - c_k \vert < \frac\varepsilon{n}$ for all $m \geq n^\varepsilon_k.$ It follows that for all $m \geq \max_{1 \leq k \leq n} n^\varepsilon_k =: n^\varepsilon,$ we have that $$\vert \sum_{1 \leq l \leq m} \sum_{1 \leq k \leq n} f(k, l) - \sum_{1 \leq k \leq n} c_k \vert \leq \sum_{1 \leq k \leq n} \vert \sum_{1 \leq l \leq m} f(k, l) - c_k \vert < \sum_{1 \leq k \leq n} \frac\varepsilon{n} = \varepsilon,$$ so that $\sum_{l \geq 1} \sum_{1 \leq k \leq n} f(k, l) = \sum_{1 \leq k \leq n} c_k.$ I hope this helps. :)

Title: switch finite and infinite rows


问:
Notation of Hyperplane
I have a notation I don't quite understand. A hyperplane defined by
$$ H:=\{x\in \mathbb{R}^n \mid x_i = c \}(i=1,\dots,n)$$
I guess it is $x = ( x_1, ..., x_n)$. But I am confused whether it means at least one $x_i = c$ of $x$ or every $x_i = c$.
The second option doesnt make any sense to me, because there would be only $x = (c,...,c) \in H$. This isnt even a plane.

I cant see how the first option is a hyperplane. I only know the definition as
$$ H:= \{ x\in \mathbb{R}^n \mid x=p+\sum\limits_{i=1}^{n-1} s_i u_i \text{ and } s_1,\dots,s_{n-1} \in \mathbb{R}\}$$
with $p\in\mathbb{R}^n$ and $u_i\in \mathbb{R}^n$ linear independent.

答:
The definition works, though it doesn't capture all possible hyperplanes.
We can take a low-dimensional example to see this: Say we look at $\{ (x, y) \in \mathbb R^2 \mid y = 1\}$ -- this defines the hyperplane (or line, in this case) parallel to the $x$-axis at $y = 1$.

More formally, a hyperplane $H_{i, c} = \{x \in \mathbb R^n \mid x_i = c\}$ for some $i \in [1, n]$ is the hyperplane spanned e.g. by the unit vectors perpendicular to the $i$-th one where the $i$-th coordinate is fixed to $c$, i.e. we have
$$H_{i, c} = \left\{ x \in \mathbb R^n \mid x = c \cdot e_i + \sum_{j \neq i} a_j e_j \text{ with } a_j \in \mathbb R \text{ for all } j \neq i\right\}$$
where $e_k$ denotes the $k$-th unit vector.

Title: Notation of Hyperplane


问:
Confused about something that has to do with uniform continuity

I am confused about something that has to do with uniform continuity.

The rectangles in the image above represent the regions formed by the following intervals, by the definition of uniform continuity

$$c-\delta<x<c+\delta \Longrightarrow f(c)-\epsilon<f(x)<f(c)+\epsilon$$

The intervals above are illustrated in the graph - makes sense.

However, I am confused about the extremes of the interval in the $y$-axis. Namely, $f(c)-\epsilon \text{ and } f(c)+\epsilon$.

They do not seem like actual outputs that are mapped to the line of the function. Consider the first rectangle. If you start at the top edge of the rectangle ($f(c)+\epsilon$) and try to move horizontally to the point that it is mapped to on the line, there isn't one. There is no point on the line for ($x+\delta$, $f(c)+\epsilon$).

This is what I am confused about. How does this work? I know there is something I am either overthinking or just confused about. I would appreciate an explanation, elaboration, or clarification.

Thank you.

Image from WIKIMEDIA COMMONS by Claudia4 " Gleichmäßig stetige Funktion, visualisiert durch epsilon-delta-Umgebungen gleicher Größe" 22 Sept. 2016

答:
Those are the boxes in which the values of the function must lie if that $\epsilon$ will work for all input values. In the picture the graph of the function does in fact remain in those boxes.

The "uniform" here is reflected by the fact that the boxes all  have the same width. Just plain continuity would call for a box at each point whose width (the value of $\delta$) might vary.


问:
Adding water with constraints
There are $n$ cups of capacity $1$ each. A game proceeds in $n$ rounds. In each round, you can add water to some cups subject to the following rules.

For each $2\le i\le n$, you cannot add more water into cup $i$ than the water that was already in cup $i-1$ in the previous round.

You cannot add more water than the water you added into the first round.

Your score for adding to cup $i$ is the amount of water you added into this cup times $1/i$, and your score for the round is the sum over all cups you added to. Your final score is then the minimum among the scores in the $n$ rounds.

Is it possible for you to score $2/n$ or more for some $n$? Or asymptotically larger than $O(1/n)$?

Scoring $1/n$ is easy, by adding $1/n$ to the first cup $n$ times, or by adding $1$ to cup $i$ in round $i$. You may be able to score above $1/n$, for instance with $n=3$, you can add $1/2$ to the first cup (round 1), $1/4$ to each of the first and second cup (round 2), and again $1/4$ to each of the first and second cup (round 3), for a score of $3/8$.

Title: Adding water with constraints

Body:
Yes, it is possible to score more than $\frac2n$ and the least $n$ for this is $10$.

The optimal configuration (according to Mathematica) is the following (added amount of water is shown):
$$
\begin{array}{cccccccccc}
\frac{669}{940}\\
0 & \frac{669}{1645} \\
0 & \frac{223}{1645} & \frac{669}{1645}\\
0 & 0 & \frac{2007}{6580} & \frac{669}{1645}\\
0 & \frac{2227}{29610} & \frac{271}{940} & 0 & \frac{1031}{2961}\\
\frac{311387}{3849300} & \frac{65113}{2474550} & 0 & 0 & \frac{887923}{3464370} & \frac{1031}{2961} \\
0 & \frac{218021}{1539720} & 0 & \frac{86672}{192465} & 0 & \frac{12295}{102648} & 0 \\
\frac{155117}{1539720} & 0 & 0 & \frac{5504}{38493} & 0 & 0 & \frac{20587}{43992} & 0\\
0 & \frac{14871071}{69287400} & 0 & 0 & \frac{1370177}{3464370} & \frac{781931}{7698600} & 0 & 0 & 0 \\ \frac{821131}{7698600} & 0 & 0 & 0 & 0 & \frac{46027}{106925} & \frac{38401}{219960} & 0 & 0 & 0 \\\end{array}
$$
with the score
$$\frac{669}{3290}>\frac2{10}.
$$

The table of the highest possible scores for $n=1\dots9$ is given below.
$$\begin{array}{ccccccccc}
1&2&3&4&5&6&7&8&9\\
\hline
1&\frac12&\frac5{12}&\frac5{14}&\frac{13}{42}&\frac{77}{276}&\frac{77}{304}&\frac{261}{1120}&\frac{669}{3080}
\end{array}$$


问:
Clarification in Hartshorne II Theorem 8.24 c)
Let $X$ be a nonsingular $k$-variety, $Y \subset X$ a nonsingular closed subvariety cut out by ideal sheaf $\mathcal I$. Let $\pi: \widetilde{X} \to X$ be the blowup of $X$ along $\mathcal I$, and $Y' \subset \widetilde{X}$ the subscheme defined by the ideal sheaf

$$\mathcal I' := \pi^{-1}(\mathcal I) \cdot \mathcal O_{\widetilde{X}}$$

From (7.13) on page 183, we know $\mathcal I' \cong \mathcal O_{\widetilde{X}}(1)$. Hartshorne then claims that this implies

$$\mathcal I'/\mathcal I'^2 \cong\mathcal O_{Y'}(1)$$

Why?

答: This is just the result of pulling the isomorphism $\mathcal I' \cong \mathcal O_{\widetilde{X}}(1)$ back to $Y'$ and then pushing it forward to $\widetilde{X}$ again.  Since $Y'$ is the subscheme defined by $\mathcal{I}'$, pulling back to $Y'$ and then pushing forward to $\widetilde{X}$ is the same as tensoring with $\mathcal{O}_{\widetilde{X}}/\mathcal{I}'$.  So this sends $\mathcal{I}'$ to $\mathcal{I}'/\mathcal{I}'^2$, and by definition it sends $O_{\widetilde{X}}(1)$ to $O_{Y'}(1)$.


问:
simple concave two-variable function
I'm looking for a simple continuous concave two-variable function $f(x,y)$, where $0\leq x,y \leq 1$, that satisfies all of the following:
$$\frac{\partial f(x,y)}{\partial x}>0$$ $$\frac{\partial f(x,y)}{\partial y}<0$$ $$\frac{\partial^2 f(x,y)}{\partial x^2}<0$$ $$\frac{\partial^2 f(x,y)}{\partial y^2}<0$$

I was able to find one: $1-e^{-(x\cdot(1-y))}$. However, I would like to find another one without an exponential function. Could anyone give other suggestions or hints please? Thank you!

EDIT: Forgot to mention one another condition. I need $0\leq f(x,y)\leq 1, \forall 0\leq x\leq 1, 0\leq y\leq 1$

Title: simple concave two-variable function

Body:
What about
$$
f(x,y)=-\frac{(x-1)^2}{2}-\frac{(y+\frac{1}{2})^2}{3}+1,
$$
which satisfies $f_x(x,y)=1-x>0$, $f_y(x,y)=-\frac{2}{3}(y+\frac{1}{2})<0$, $f_{xx}(x,y)=-1$ and $f_{yy}(x,y)=-\frac{2}{3}$ for all $y,x\in[0,1]$ and also $f(x,y)\in[0,1]$ for $x,y\in[0,1]$.


问:
MatLab: Lagrange interpolation function graph deviates when nodes are greater than 28
I don't know if this is the right place to ask this question. But I have the following code that interpolates a function $y$ base on the node $x$, which based on the termNumber.  The function is $exp(3x)*cos(3x)$ and the nodes are base on $x=(2j/n)-1$ where $n=$termNumber and $j=0,...,n$

```<code>termNumber = 28;
x = (2.*(0:termNumber)./termNumber)-1;
y = exp(3.*x).*cos(3.*x);

sum=0;
for i=1:length(x)
    p=1;
    for j=1:length(x)
        if j~=i
            c = poly(x(j))/(x(i)-x(j));
            p = conv(p,c);
        end
    end
    term = p*y(i);
    sum= sum + term;
end
disp(sum);
K=flip(sum);

for  g=length(x):-1:2 %display the coefficient on polynomial form
    fprintf('+(%.2fx^%d)',K(g),g-1);
end

fprintf('+(%.2f)',K(1));

v=linspace(-1,1,100);
b = polyval(sum,v);
b1 = exp(3.*v).*cos(3.*v);
plot(v,b,'o') %display lagrange polynomial
hold on
plot(v,b1) %display actual function

</code>```

The graph is displayed correctly using the above method when the termNumber is less than 28: see below:graph when termNumber = 28  where the red curve is the actual function(blue is the Lagrange interpolating polynomial). But when termNumber >= 29, the two curves started to separate: termNumber = 29 . And when termNumber gets large, both curves rea not recognizable anymore termNumber = 40 . Since I coded the method according to my textbook so I can't seem to figure out what I did wrong. Any inputs are appreciated.

答:
There is no theoretical reason for the error, as that should fall like $(6\sqrt2/n)^{n+1}/(4(n+1))$. The observed error is thus due to cancellation errors and other floating point problems in the construction of the interpolation polynomial.

With a different, more compact computation, the deviation only starts to become visible at $32-35$, the error at $40+1$ nodes is $1/10$ of the error reported in the question.

```<code>  q = poly(x);
  dq = polyder(q)
  sum=0;
  for i=1:length(x)
      sum = sum + y(i)*deconv(q,poly(x(i)))/polyval(dq,x(i))
  end
</code>```

One can get some nodes better by computing the interpolation polynomial of the error and add that to the polynomial, and then iterate that. This helps up to $41$. One might perhaps get to higher node densities using the Newton interpolating scheme, starting outside on both interval ends and moving alternatingly inwards. Or use some dedicated divide-and-conquer approach.

Body:
There is no theoretical reason for the error, as that should fall like $(6\sqrt2/n)^{n+1}/(4(n+1))$. The observed error is thus due to cancellation errors and other floating point problems in the construction of the interpolation polynomial.

With a different, more compact computation, the deviation only starts to become visible at $32-35$, the error at $40+1$ nodes is $1/10$ of the error reported in the question.

  q = poly(x);
  dq = polyder(q)
  sum=0;
  for i=1:length(x)
      sum = sum + y(i)*deconv(q,poly(x(i)))/polyval(dq,x(i))
  end

One can get some nodes better by computing the interpolation polynomial of the error and add that to the polynomial, and then iterate that. This helps up to $41$. One might perhaps get to higher node densities using the Newton interpolating scheme, starting outside on both interval ends and moving alternatingly inwards. Or use some dedicated divide-and-conquer approach.


答:
You can have non-linear functions in the vector space. Why do you say that $(f+g)(x) = f(x) + g(x)$ and $(kf)(x) = kf(x)$ are not true for non-linear functions? This is how one defines addition of the vectors and multiplication with scalars. One can for instance have the vector space of continuous functions on some space.

Maybe you are confusing it with $f(kx) = kf(x)$ and $f(x+y) = f(x) + f(y) $?

Title: Example of vector space


问:
how to find the solution of a double differential equation
I have the following second order equation:

$y'[x] + xy''[x] = 0$

From Mathematica I know the solution is $y[x] = A+ B \log[x]$ and I know
that through substitution this is indeed the correct answer.

However, I can't find a way of doing this without guessing the solution itself.

I have tried substitution i.e. letting $y=x^k$ and this gives a trivial solution of
$y=0$. I also tried letting $x=\exp[t]$ and this doesn't work either.

Is there an elegant way of doing this? I have looked up some advanced calculus textbook
and couldn't find an example similar to mine.

答:
$$y'+xy''=(xy')'=0$$ integrates as

$$xy'=b$$ or $$y'=\frac bx.$$

After a second integration,

$$y=a+b\log x.$$

If this is still "too magical", notice that the equation is of the first order in $y'$. Use $z:=y'$ and solve

$$z+xz'=0,$$ which is separable.


答: A linear function is a function $f$ such that $f(ax+by)=af(x)+bf(y)$, the definition of addition and scalar multiplication in the vector space you describe doesn't say anything about this expression. We simply define addition of two real valued functions, and the scalar multiplication of a real valued function to be as you describe.</p>

Title: Example of vector space


问:
Prove $x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $ for some $x>1$
Prove that $$x=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t $$ for some $x>1.$

The only idea I have is to calculate the derivative of $\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \ \mathrm{d}t $. First, I let $y=\int_{1}^{x}\cos{\frac{1}{t}}\cdot\ln\left(3-\frac{1}{t}\right) \mathrm{d}t -x$ , then I calculate $y'(x)$ , but $y(x)$ isn't a monotonous function, so I got stuck.

Can anyone help or give me a hint on how to approach this problem?

Thank you very much guys

答:
Let's define the function:
$$g(x)=\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}-x \, , \,\, x \in (1, +\infty)$$

Step 1

Notice that
$$
1 \lt t \lt x
$$

since all members of the inequality are positive, it follows
$$
1 \gt \dfrac{1}{t} \gt \dfrac{1}{x} \iff -1 \lt -\dfrac{1}{t} \lt -\dfrac{1}{x} \iff  2 \lt 3-\dfrac{1}{t} \lt 3-\dfrac{1}{x}
$$

Also, since $1 \gt \frac{1}{t}\gt \frac{1}{x} \gt 0$ , then $\frac{1}{t} \in \left(0, 1\right) \subset \left(0, \frac{\pi}{2}\right)$ where $\cos{\frac{1}{t}}$
is monotonically decreasing:
$$
\cos{1}\lt\cos{\dfrac{1}{t}}\lt\cos\dfrac{1}{x}
$$

Since, all members of the above two inequalities are positive, multiplying them leads to:
$$
\cos{1}\cdot\ln{2} \lt \cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)} \lt \cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)}
$$

Now by integrating over $\, t$ , from $1$ to $x \gt 1$ :
$$
\int_1^x{\cos{1}\cdot\ln{2}\ \mathcal{d}t} \lt \int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \int_1^x{\cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)\ \mathcal{d}t}}
$$

Since $x$ is independent of $t$:
$$
\cos{1}\cdot\ln{2}\cdot\int_1^x{1\ \mathcal{d}t} \lt \int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)\cdot\int_1^x{1\ \mathcal{d}t}}
$$

Calculating the integrals:
$$
\cos{1}\cdot\ln{2}\cdot(x-1) \lt \int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)}\cdot(x-1)
$$

Step 2

Now suppose there is some real number $a$ very close to $1^+$ and some finite real number $b$ approaching $+\infty$, such that $x \in [a, b]\subset(1,+\infty)$.

Taking the limit as $x\to +\infty$
$$
\cos{1}\cdot\ln{2}\cdot\lim_{x\to +\infty}(x-1) \lt \lim_{x\to +\infty}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \lim_{x\to +\infty}\left(\cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)}\cdot(x-1)\right)
$$
Since $\cos{1}\cdot\ln{2}\gt 0$ and $\lim_{x\to +\infty}\cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)}= \cos{0}\cdot\ln{3}=\ln{3}\gt 0$, this leads to:

$$
\cos{1}\cdot\ln{2}\cdot\lim_{x\to +\infty}(x-1) \lt \lim_{x\to +\infty}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \ln{3}\cdot\lim_{x\to +\infty}\left(x-1\right)
$$

But $\lim_{x\to +\infty}\left(x-1\right)=+\infty$ , so:
$$
+\infty\lt \lim_{x\to +\infty}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt +\infty
$$

The indefinite integral is continuous in $(1,+\infty)$

By the Squeeze Theorem it follows that:
$$
\lim_{x\to +\infty}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} = +\infty
$$

Because of the above limit that goes to $+\infty$ as $x\to +\infty$ and the fact that $b$ was defined as approaching to $+\infty$ but is a finite real number, this leads to the fact that the integral's limit as $x\to b$ tends to a very large positive real number $\epsilon_{\infty}$ that approaches $+\infty$. So:
$$
\lim_{x\to a}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} = +\infty
$$

and $g(b)=\epsilon_\infty$

Taking the limit as $x\to 1^+$
$$
\cos{1}\cdot\ln{2}\cdot\lim_{x\to 1^+}(x-1) \lt \lim_{x\to 1^+}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \lim_{x\to 1^+}\left(\cos\dfrac{1}{x}\cdot\ln{\left(3-\dfrac{1}{x}\right)}\cdot(x-1)\right)
$$

then:
$$
\cos{1}\cdot\ln{2}\cdot 0 \lt \lim_{x\to 1^+}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt \cos{1}\cdot\ln{2}\cdot 0
$$

finally:
$$
0 \lt \lim_{x\to 1^+}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} \lt 0
$$

The indefinite integral is continuous in $(1,+\infty)$

By the Squeeze Theorem it follows that:
$$
\lim_{x\to 1^+}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} = 0
$$

Because of the above limit that goes to $1^+$ as $x\to 1^+$ and the fact that $a$ was defined as a real number close to $1^+$, this leads to the fact that the integral's limit as $x\to b$ tends to a very small positive real number $\epsilon_0$ that approaches $0$. So:
$$
\lim_{x\to a}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} = 0
$$

and $g(a)=\epsilon_0$

Step 3

Now since $g(x)=\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}-x \, , \,\, x \in (1, +\infty)$ is continuous in $[a,b]\subset(1,+\infty)$
and the following hold:
$$
g(a)=\lim_{x\to 1^+}\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} - 1 = -1
$$

and

$$
\lim_{x\to b}g(x)=\lim_{x\to +\infty}\left(\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t} - x\right) = \lim_{x\to +\infty}\left(\dfrac{x\cdot\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}}{x} - 1\right)
$$

The following limit is indeterminate $\dfrac{\infty}{\infty}$
$$
\lim_{x\to +\infty}\dfrac{x\cdot\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}}{x}
$$

But the numerator and denominator are both continuous and differentiable functions, so applying $\mathcal{De\, L'H\hat{o}spital's}$ rule for indeterminate limits:
$$
\begin{aligned}
\lim_{x\to +\infty}\dfrac{x\cdot\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}}{x}
&=\lim_{x\to +\infty}\dfrac{\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}+x\cdot \cos{\dfrac{1}{x}}\cdot\ln{\left(3-\dfrac{1}{x}\right)}}{1}
\\&=\lim_{x\to +\infty}\left(\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}+x\cdot \cos{\dfrac{1}{x}}\cdot\ln{\left(3-\dfrac{1}{x}\right)}\right)
\\&=\lim_{x\to +\infty}\int_1^x{\cos{\frac{1}{t}}\cdot\ln{\left(3-\frac{1}{t}\right)}\ \mathcal{d}t}+\lim_{x\to +\infty}x\cos{\frac{1}{x}}\ln{\left(3-\frac{1}{x}\right)}
\\&=\lim_{x\to +\infty}\int_1^x{\cos{\frac{1}{t}}\cdot\ln{\left(3-\frac{1}{t}\right)}\ \mathcal{d}t}+\ln{3}\cdot\lim_{x\to +\infty}x \\&= +\infty
\end{aligned}
$$

Finally:
$$
g(b) = \lim_{x\to +\infty}\left(\dfrac{x\cdot\int_1^x{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}}{x}\right) - 1= +\infty -1 = +\infty
$$

This means that:
$$
g(a) \lt 0 \lt g(b)
$$
in addition to $g(x)$ being continuous in $[a,b]\subset(1,+\infty)$, then by $\mathcal{Bolzano's}$ theorem there is at least one $\xi \in (a,b)$ such that $g(\xi)=0$.

Which completes the proof as:

$$
\xi = \int_1^{\xi}{\cos{\dfrac{1}{t}}\cdot\ln{\left(3-\dfrac{1}{t}\right)}\ \mathcal{d}t}
$$

where $\xi > b > 1$


问:
$T(O)$ is not open where $T$ is an inclusion map
I want to show that $T(O)$ is not open in $Y$ where $T:X\to Y$ is a linear function given by $Tx = x$, and $$
X = Y = \{x = (x_1,x_2,\dots), \sum_i |x_i|<\infty\}$$
is normed linear spaces with norms $\|x\|_X = \sum_i|x_i|$ and $\|x\|_Y = \sup_{n\in\mathbb{N}}|x_n|$ (they are not Banach).

I want to show that there exists a point $y\in T(O)$ such that $B_{Y}(y; r)\not\subset T(O)$ where $O = \{x\in X : \|x\|_X < 1\}$

Here I cannot proceed anymore. Any helps will be more than welcome.

答:
So we want to show that the set $O=\{x\in X: x=(x_n), \sum_n|x_n|<1\}$ is not open with respect to the supremum norm. Consider the point $x=(\frac{1}{3},\frac{1}{3},0,0,0,\dots)$. Then $x\in O$. We will show that there exists a sequence $(x_n)$ such that $x_n\to x$ in the supremum norm but $x_n\not\in O$ for all $n$. We set

$$x_n=(\frac{1}{3},\frac{1}{3},\frac{1}{n},\frac{1}{n},\dots,\frac{1}{n},0,0,0,\dots) $$

where we have $1/n$ appearing $n$ times. Note that $1/3+1/3+1/n+\dots+1/n+0+0+0+\dots=1/3+1/3+1>1$, so $x_n\not\in O$ for any $n$. On the other hand,
$$\|x_n-x\|_\infty=\frac{1}{n}\to0.$$

What we have shown is that the open ball $B(x,\varepsilon)$ with respect to the supremum norm is not a subset of $O$ for any $\varepsilon>0$, so $O$ is not open with respect to the supremum norm.


答:
Yes this is true:
Define the sequences $(a_m)_m , (b_m)_m \in \mathbb{R}^{\mathbb{N}}$ (so just real valued sequences) by
$$
a_m := \sum_{i=1}^{n} \sum_{j = 1}^{m} f(i,j) \quad \text{ and } \quad b_m := \sum_{j=1}^{m} \sum_{i = 1}^{n} f(i,j).
$$
What you have given me in the question is that $\lim_{m\rightarrow \infty} a_m$ converges, but we do not now this yet for $\lim_{m\rightarrow \infty} b_m$.

For finite sums you can switch summation (if you would like to prove this rigorously then you can use induction), which implies that $a_m = b_m$ for all $m \in \mathbb{N}$.
In other words the sequence $c_m \subset \mathbb{R}$ (or equivalently $c_m \in \mathbb{R}^{\mathbb{N}}$) defined by $c_m = a_m - b_m$ is constant zero.
This implies that $\lim_{m\rightarrow \infty} c_m = 0$.

From this fact we find
$$
0 = \lim_{m\rightarrow \infty} c_m = \lim_{m\rightarrow \infty} (a_m  - b_m) = \lim_{m\rightarrow \infty} a_m -\lim_{m\rightarrow \infty} b_m,
$$
where in the last equality I used the fact that $\lim_{m\rightarrow \infty} a_m$ exists (which implies that $\lim_{m\rightarrow \infty} b_m$ exists) so we can use the sum rule for limits.

Therefore $\lim_{m\rightarrow \infty} b_m  = \lim_{m\rightarrow \infty}\sum_{i=1}^{n} \sum_{j = 1}^{m} f(i,j) = \sum_{i=1}^{n} \sum_{j = 1}^{\infty} f(i,j)$ exists and equals $\lim_{m\rightarrow \infty} a_m $ (convince yourself that this is what you were asking for).

Some general advice: A priori, it is not clear that the summation on the right hand side (i.e. the sequence $b_m$) converges. So before/after asking yourself if equality occurs you could also ask if this infinite sum even exists. What i'm saying is, do not skip the question of existence too quickly. I tried to emphasize this by explicitly stating that the sequence $b_m$ converges and that I did not assume that the sequence $b_m$ converges.

Title: switch finite and infinite rows

Body:
Asume to have a function $f(k,l):\mathbb{N} \times \mathbb{N}\rightarrow \mathbb{R}$ s.t.
$$\sum_{l=1}^\infty f(k,l)=c_k<C \,\,\,\forall k\in\mathbb{N}.$$

Does the following hold for any finite $n$?:
$$
\sum_{k=1}^n\sum_{l=1}^\infty f(k,l)=\sum_{l=1}^\infty\sum_{k=1}^n f(k,l)
$$


问:
How to show that finite holomorphic maps are proper?
I would like to solve Exercise 1.1.5 of Introduction to Singularities and Deformations, which states

Show that finite maps between complex spaces are proper.

A map is called finite if it is closed, and quasi-finite, i.e. every point has only finitely many preimages, and it is called proper if the preimage of every compact subspace is compact.

So suppose $f: X \to Y$ is finite, and $K \subset Y$ is compact. We would like to show that $f^{-1}(K)$ is compact, and I had some hopes that this goes similar to proving that a finite covering of a compact space is compact. In particular since Lemma 1.54 states

Let $f: X \to Y$ be a finite map of topological spaces where $X$ is hausdorff, let $y \in Y$ and let $f^{-1}(y) = \{x_1, \dotsc, x_s\}$. Further let $U_i' \subset X$ be pairwise disjoint open neigbourhoods of $x_i$. Then, for each open neighbourhood $V'$ of $y$, there exists an open neighbourhood $V \subset V'$ of $y$ such that with $U_i = U_i' \cap f^{-1}(V)$

$f^{-1}(V) = U_1 \cup \dotsb \cup U_s$

the restrictions $f: U_i \to V$ are finite

I had the hope that those $U_i$ somehow replace the evenly covered neighourhoods in the case of a covering, but the proofs I found all use that a covering is locally homeomorphic.

答:
If $f^{-1}(K)$ is not compact, there is a sequence $(x_n)$ in $f^{-1}(K)$ with no accumulation point in $f^{-1}(K)$ (and hence no accumulation point in $X$ since $f^{-1}(K)$ is closed in $X$).  Since $K$ is compact, the sequence $(f(x_n))$ must accumulate at some point $y\in Y$.  Now since $(x_n)$ has no accumulation point, there is no value that is repeated infinitely many times among the $x_n$'s.  Since $f$ is quasifinite, there is also no value that is repeated infinitely many times among the $f(x_n)$'s.  In particular, only finitely many of the $f(x_n)$'s can be equal to $y$; omitting those finitely many values from our sequence, we may assume $f(x_n)\neq y$ for all $n$.

Now consider the set $A=\{x_n:k\in\mathbb{N}\}$.  It is closed in $X$ since our sequence $(x_n)$ has no accumulation point.  Thus since $f$ is closed, $f(A)$ is also closed.  But this is a contradiction, since $(f(x_n))$ accumulates at the point $y$ and $y\not\in f(A)$.

This argument uses only the assumption that $f:X\to Y$ is a continuous closed map from a metric space to a Hausdorff space whose fibers or finite (in fact, with minor modifications, the assumption that $Y$ is Hausdorff can be dropped and it suffices to assume the fibers are compact rather than finite).  More generally, in fact, it can be shown that if $f:X\to Y$ is a continuous closed map of arbitrary topological spaces whose fibers are compact, then $f$ is proper.  The proof is a bit trickier though since you can't just use sequences.

Here's how the argument goes.  Suppose $K\subseteq Y$ is compact and $f^{-1}(K)$ is not compact.  Then there is a family $(C_i)$ of closed subsets of $X$, closed under finite intersections, such that $C_i\cap f^{-1}(K)$ is nonempty for all $i$ but $\bigcap_i C_i\cap f^{-1}(K)$ is empty.  Now since $f$ is closed, the sets $f(C_i)$ are all closed, and they also all have nonempty intersection with $K$.  Since $K$ is compact, this means there is some point $y\in \bigcap_i f(C_i)\cap K$.  But now since $f^{-1}(\{y\})$ is compact and $\bigcap C_i\cap f^{-1}(\{y\})\subseteq\bigcap_i C_i\cap f^{-1}(K)=\emptyset$, there is some $i$ such that $C_i\cap f^{-1}(\{y\})$ is empty.  This is a contradiction, since by our choice of $y$, $y\in f(C_i)$.


问:
$M\subset\Bbb R^n$ contains an open ball iff its Hausdorff measure is $n$
$\Rightarrow$ is proved here.

I suspect that if a subset $M\subset\Bbb R^n$ has Hausdorff dimension $n$, then it must contain an open ball. Is that true?

答:
This is false - in fact, there are compact subsets of the real line containing no open intervals but which have positive Lebesgue measure - in particular fat Cantor sets have this property.

The usual Cantor set construction starts with an interval and removes the middle third of that interval, then the middle thirds of the two resulting intervals and so on - but this construction can be modified so that instead of removing a third each time we remove, say, a proportion of $\frac{1}{4^n}$ from the middle of each line at the $n^{th}$ step of the construction, which leaves an area of
$$\prod_{n=1}^{\infty}\left(1 - 4^{-n}\right) > 0$$
in the set. However, after the $n^{th}$ step, the remaining set is the union of $2^n$ intervals of length at most $2^{-n}$ - hence the final result, which is the intersection of all of these steps, does not contain any open interval.


问:
Example of vector space
In almost all the literature that I have seen, one of the examples of vector space is as follows:

Set of all real-valued functions $f(x)$ defined on the real line

what confuses me here is that the word "linear functions" should be used instead of just "functions" because I think we also have non-linear functions and they (non-linear functions)do not follow the rule of addition and multiplication as:

$ (f + g)(x)=f(x) + g(x) $ and $(kf)(x) = k f(x) $

thus, non-linear functions cannot make a vector space.

am I right or wrong?

答:
I think you're confusing stuff: one thing is the vectors in the linear (vector) space, which in this case are real functions, and way another is the kind of function you can define on this vector space that will keep the vector space structure defined, and these are linear functions.

Thus, $\,f(x)=|x|\;,\;g(x)=-1.5\;,\;h(x):=x-\cos(x+1)+e^x\,\ldots\;$ are examples of vectors (real functions), and an example of linear functions (better known as linear transformations) on this vector space could be

$$T(f(x)):=f(0)\;,\;\;T:V\to\Bbb R$$

Title: Example of vector space