问:
Greater precision with regression?
I used my TI-83 to find the quadratic regression of two data columns. The accuracy wasn't close at all. So I tried cubic and then finally quartic regression. The accuracy still isn't close enough. Is there any way I can improve the accuracy? I can't add any more data, unfortunetly.

Edit 1:

To address Henry's comment, my plotted data looks similar to y=x^(1/3)

答: This may mean your data is poorly fit by a polynomial.  Three ways it can fail is either to be exponential, to have poles, or be like a sine wave.  Plotting the data with the fits you have overlayed may give you some ideas.  If it is exponential, taking the logarithm of one will render it linear (or close to polynomial).  If it has poles, using a rational function will help.  Procedures are given in chapter 3.2 of Numerical Recipes.  The obsolete versions are free.  If it is a sine wave, an FFT can help-chapter 12 of NR.</p>

Title: Greater precision with regression?


问:
Uniform integrability of RV's
$\mathbf{Theorem}$: Let $Y \in \mathbb{L}_1$, then the RV $(\mathbb{E}[Y \mid \mathcal{F}], \mathcal{F} \subset \mathcal{A} \space \sigma\text{-algebra})$ are uniformly integrable.

$\mathbf{Proof}$: Choose $K \in (0,\infty)$:

\begin{align}
\mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{|\mathbb{E}[Y \mid \mathcal{F}]| \ge K}] &\le \mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{\mathbb{E}[|Y| \mid \mathcal{F}] \ge K}] \\
&=\mathbb{E}[Y  \mathbf{1}_{\mathbb{E}[|Y| \mid \mathcal{F}] \ge K}]
\end{align}

Next, we know, that
\begin{align}
\mathbb{P}(\mathbb{E}[Y \mid \mathcal{F}] \ge K) &=\int_\Omega \mathbf{1}_{\mathbb{E}[Y \mid \mathcal{F}] \ge K}] \, d \mathbb{P}\\
&\le \frac{\mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}]]}{K} = \frac{\mathbb{E}[|Y|]}{K}
\end{align}

($\mathbf{Q}$: How do we know that? Is it an application of the Martingale inequality - generalized Kolmogorov for $\psi(K) =K$ and $U = Y$ ?) My guess is no - then by what property do we obtain the above inequality?)

Hence,

$$\sup\left\{ \mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{|\mathbb{E}[Y \mid \mathcal{F}]| \ge K}] \right\} \le \sup \left\{\mathbb{E}[|Y| \mathbf{1}_A ]: \mathbb{P}(A) \le \frac{\mathbb{E}[|Y|]}{K}\right\}$$

And uniform integrability follows from uniform continuity of expectation.

答:
I think you're missing some absolute values throughout your post. The inequality is just Markov's inequality applied to the non-negative variable ${\rm E}[|Y|\mid\mathcal{F}]$. For completeness: If $X$ is any non-negative random variable and $K>0$, then $$X\geq K \iff \frac{X}{K}\geq 1$$ and hence
$$
1_{\{X\geq K\}}\leq \frac{X}{K}1_{\{X\geq K\}}\leq \frac{X}{K}.
$$
Apply this to $X={\rm E}[|Y|\mid\mathcal{F}]$ and integrate both sides.</p>

ContentLicense: CC BY-SA 3.0

Title: Uniform integrability of RV's


问:
Soluions over ring of integers from an non archimedean field
This is Proposition 1.9.  I am confused with the forward direction:

If $K$ is alg. closed narc field with $x$ a nonzero element with posiive norm. Then:

$K$ is not discrete.

Every polynomial over $O_K/(x)$ has a root in $O_K/(x)$.

where $O_K$ is ring of inegers of $K$.

My problems/ thoughts. Suppose I have a polynomial over $O_K/(x)$ of form
$\sum a_i t^i$ when lifted to $K$. It has root $y \in K$. But I don't see how we can scale $y$ so as to obtain new root in $O_K/(x)$.

any hints would be appreciated.

答:
The valuation/norm on $K$ is not discrete because $v(x^{n/m}) = v(x) n/m$.

Every monic polynomial $\in O_K[t]$ has a root in $O_K$.

$pt-1\in O_{\overline{Q}_p}/(p^2)[t]$ doesn't have a root in $O_{\overline{Q}_p}/(p^2)$.

ContentLicense: CC BY-SA 4.0


问:
Finding $\tan(x+90^\circ)$ when $\sin x = k$ and $x$ is obtuse. I get the reciprocal of the correct answer.

If $\sin x = k$ and $x$ is obtuse, find an expression for $\tan(x+90^circ)$.

So $$\tan(x+90^\circ)=\frac{\sin(x+90^\circ)}{\cos(x+90^\circ)}$$
and after drawing a triangle, I obtain
$${\sin(x+90^\circ)} = -k \qquad\text{and}\qquad\cos(x+90^\circ)= -\sqrt{1-k^2}$$ because $x+90^\circ$ is in Quadrant 3 where both $\sin$ and $\cos$ are negative.

So $$\tan(x+90) = \frac{k}{\sqrt{1-k^2}}$$

However, the answer is the reciprocal of this. Why is that so?

Body:
Given $\tan(x+90) = \frac{\sin(90+x)}{\\cos(90+x)}$,

$sin(+90)= sin(x)cos(90)+cos(x)sin(90)$=cos(x)
$cos(+90) =cos(x)cos(90)-sin(90)sin(x)$=-sin(x)

$\tan(x+90) = -\frac{\cos(x)}{\sin(x)}$,

x in Quad II, cos(x) <0, $\sin(x)=k, \cos(x) = - \sqrt{1- k^2}$

$\tan(90+x) = \frac{\sqrt{1-k^2}}{k}$


问:
$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}$ has any three consecutive terms coprime
Let $ a \in \mathbb N $. For $ n \in \mathbb N $, let $$ x_n = 1 + 2a + 3a ^ 2 + \cdots + na ^ {n-1}. $$

$ x_1 = 1, x_2 = 1 + 2a, \ldots $.

For $ n \in \mathbb N $, show that $ \gcd (x_n, x_ {n + 1}, x_ {n + 2}) = 1 $.

Any help, suggestion on how to start solving that exercise.

Body:
Hint: put $\,c_k = k\,$ below, which uses the Euclidean algorithm / Euclid's Lemma / gcd laws.

If $\ x_n = c_1 + c_2 a + \cdots + c_n a^{n-1}\ $ and $\ \color{#0a0}{(c_1,a)=1}\,$ and $\ \color{#c00}{(c_j,c_{j+1})=1\,\ {\rm for\ all}\,\ j}\:$ then

$$\begin{align} (x_n,x_{n+1},x_{n+2})
&= (x_n,\,x_{n+1}-x_n,\,x_{n+2}-x_{n+1})\\[.1em]
&= (x_n,\ \ \  c_{n+1} a^n,\ \ \ \  c_{n+2}a^{n+1})\\[.1em]
&= (x_n,\ \ \ c_{n+1},\ \ \ \ \ \ \ \ c_{n+2})\ \ {\rm by}\ \ (x_n,a) = \color{#0a0}{(c_1,a) = 1}\\[.1em]
&=\, 1,\ \  {\rm by}\ \ \color{#c00}{(c_{n+1}, c_{n+2}) = 1}
\end{align}\qquad$$


问:
$95\%$ confidence interval with unknown standard deviation

The Question

We have the results for the heart beats per minute of a sample of $15$ people: $$54, 63, 58, 72, 49, 92, 70, 73, 69, 104, 48, 66, 80, 64, 77$$ Give a $95\%$ confidence interval for the mean number of heart beat per minute

i. if the standard deviation is known and equal to $\sigma=15$

ii. if the standard deviation is not known.

(Assume normal distribution.)

My Understanding

The mean value is $\overline{x}\approx 69.27$.

In case of known standard deviation the interval is $$\left (\overline{x}-z\cdot \frac{\sigma}{\sqrt{n}}, \overline{x}+z\cdot \frac{\sigma}{\sqrt{n}}\right )=\left (69.27-1.960\cdot \sqrt{15}, 69.27+1.960\cdot \sqrt{15}\right )$$

In case of unknown standard deviation we calculate the estimated from the given data, $s\approx 15.17$. Is the formula again tha same? Or is the $z$ value different?

答:
If the standard deviation is not known, the pivotal quantity to be used is the following:

$$t=\frac{\overline{X}_n-\mu}{S}\sqrt{n -1}$$

which follows a Student T distribution with $n-1$ degrees of freedom.

Thus, the confidence interval is the same but the quantiles are the two ones taken from a Student $T$ with 14 dof: $\pm2.14$ instead of $\pm 1.96$


问:
Can we multiply a matrix by an imaginary or complex number not purely real?
I am currently reading the topic of scalar multiplication of matrices. It always says that if k is a scalar then k times matrix = matrix with all the elements multiplied by k.
My question is what is this scalar. Like we know complex numbers are very similar to the vectors.
so is it only the purely real ks we are talking about or we can multiply a matrix by 2+3i also for example.</p>

答:
The short answer is yes, you can multiply a real matrix by a complex scalar. You can even multiply it by a vector whose components are complex numbers. But this simple answer gives little explanation.

The thorough answer is that when we do linear algebra, we really need to specify over which field we are constructing our matrices (and vectors). This is important because it affects the question of the dimension of the relevant vector space. So to answer your question, in which space the possible scalars (and entries) reside must be declared when defining the relevant vector space. If you are working with $\mathbb{R}^{m \times n}$, i.e. the set of all $m \times n$ matrices with real entries, viewed as a real vector space, it is understand that the scalars reside in $\mathbb{R}$. However, you could also work with $\mathbb{R}^{m \times n}$ viewed as a subspace of $\mathbb{C}^{m \times n}$, in which case it would be natural to allow multiplication by complex scalars.


问:
Why is weak convergence of bounded measures defined for $C_b(\mathbb{R}^d)$?

Definition: For any sequence of measures $(\mu_n)_n \subset M_b(\mathbb{R}^d)$ where $M_b(\mathbb{R}^d)$ is the space of bounded measures we say $(\mu_n)$ converges weakly to $\mu$ if for all $f \in C_b(\mathbb{R}^d)$ it holds that

$\lim_{n \rightarrow \infty}\int_{\mathbb{R}^d} f d\mu_n = \int_{\mathbb{R}^d} f d\mu$

My question or issue for discussion is that I wonder why we use $f \in C_b(\mathbb{R}^d)$ instead of say $f \in L_1$ since for bounded measures $L_1 (\mathbb{R}^d)$ contains almost all useful functions (including $C_b(\mathbb{R}^d)$).

An idea I also had was to e.g. have the definition with $C_c(\mathbb{R}^d)$ since it lies dense in $L_1$.

What do you think of these ideas and why was this definition used as a standard instead?

答:
You should note that what is  weak convergence for Probabilists is actually weak* convergence for Functional Analysists.

Consider the simple example $\mu_k=\delta_{1/k}$ (the delta measure at $\frac 1 k$) and $\mu=\delta_0$. Any reasonable definition of weak convergence should give $\mu_n \to \mu$. Now $f(\frac  1 k) =\int fd\mu_n \to f(0)= \int fd\mu $ if $f$ is continuous at $0$ but this fails if $f$ is just integrable.

Now consider $\mu_n =\delta_n$ and $\mu=0$. Then $\int fd\mu_n \to \int fd\mu$ for all $f \in C_c(\mathbb R)$ but not for all bounded continuous functions $f$. ($f\equiv 1$ for example).  In this case we don't want $\mu_n$ to converge to the zero measure so we don't want to use $C_c(\mathbb R)$  instead of $C_b(\mathbb R)$


问:
Solve ${x}^{\lfloor x \rfloor} = a$
I have found this weird equation in a math book. Could you give me any hints?
$${x}^{\lfloor x \rfloor}  = a$$ for a given a. I have dealt with the trivial cases where $x$ is an integer but cannot solve it in $\mathbb{R}$.

Graph of $x\mapsto {x}^{\lfloor x \rfloor}$

答:
I'll show how to find solutions $x>0$.

Note that for $0<x<1$ we have $\lfloor x\rfloor = 0$ so $f(x) = x^0 = 1$, and for $x\ge1$ the funcion $f(x) =x^{\lfloor x\rfloor}$ is increasing - that is because if $1\le x<y$ then $1<\lfloor x\rfloor \le \lfloor y\rfloor$ and
$$ x^{\lfloor x\rfloor} < y^{\lfloor x\rfloor} \le y^{\lfloor y\rfloor}$$
The function is also piecewise continuous; specifically, it is continuous on every interval $[n,n+1)$, $n\in\mathbb{N}$ and has discontinuiities at $n\in\mathbb{N}$, $n\ge 2$ that is because
$$\lim_{x\to n_-} f(x) = \lim_{x\to n_-} x^{n-1} = n^{n-1} $$
$$\lim_{x\to n_+} f(x) = \lim_{x\to n_+} x^{n} = n^{n} $$
If $n > 1$ it means that there's a discontinuity at $n$. We can therefore conclude that

If $a = 1$ then equation $f(x) = a$ is satisfied by all $x\in(0,1]$.

If $a>1$ and $a \in [n^n,(n+1)^n)$ for some $n\in\mathbb N$ then equation $f(x) = a$ has one solution $x>0$, and this solution satisfies $x\in[n,n+1)$ , therefore we have $\lfloor x \rfloor =n$ and we have $x^n=a$, that is $x=\sqrt[n]{a}$.

If $a<1$ or $a \in [(n+1)^n,(n+1)^{n+1})$ for some $n\in\mathbb N$ then equation $f(x) = a$ has no solutions $x>0$.

I'll leave the searching for solutions $x<0$ to you, it can be done in a similar fashion.

Title: Solve ${x}^{\lfloor x \rfloor} = a$


问:
What is the similarity between an actual riffle shuffle and a riffle shuffle permutation?
A riffle shuffle permutation is sometimes cited as the mathematical equivalent of a riffle shuffle in real life, but I'm having a hard time understanding the similarity between the two.

In real life, a riffle shuffle cuts the ordered deck in two, then interleaves the two decks by adding a card from the bottom of each deck to the top of the (soon-to-be) shuffled deck.

For instance, starting with $1, 2, 3, 4, 5, 6, 7, 8$, cut into $1, 2, 3, 4, 5$ and $6, 7, 8$, we might get $1, 6, 2, 3, 7, 4, 8, 5$.

In mathematics, the riffle shuffle permutation is a permutation that turns an ordered sequence into two consecutive ordered sequences.

For instance, starting with $1, 2, 3, 4, 5, 6, 7, 8$, we might get $1, 4, 7, 2, 3, 5, 6, 8$.

The real-life riffle shuffle, unless I'm misinterpreting the link between the two, will never yield a mathematical riffle shuffle permutation unless we do not interleave the two cut decks at all. On the other hand, the mathematical riffle shuffle looks more similar to a kind of "reverse" riffle shuffle, where you extract two decks which are interleaved from the initial ordered deck, then stack them on top of eachother.

答:
In your first example, card 1 ended up in position 1 after the shuffle, card 2 ended up in position 3, card 3 in position 4, card 4 in position 6, etc., so the riffle shuffle permutation $\pi$ corresponding to that shuffle is
$$
\begin{aligned}
\pi(1) = 1 \\
\pi(2) = 3 \\
\pi(3) = 4 \\
\pi(4) = 6 \\
\pi(5) = 8 \\
\\
\pi(6) = 2 \\
\pi(7) = 5 \\
\pi(8) = 7
\end{aligned}
$$
with two increasing sequences giving the positions of cards 1 to 5 and of cards 6 to 8, respectively.


问:
Null and alternative hypotheses : Car model fuel usage
Some people assume that a specific car model does at least $\mu_0=120$ km with $1$ Lt of gasoline (petrol).

$10$ independent tests give the following results: $$104, \ 96, \ 80, \ 100, \ 108, \ 100, \ 112, \ 120, \ 130, \ 132$$

(a) Give the Null Hypothesis $H_0$ and the alternative Hypothesis $H_1$, for the test of that assumption.

(b) Give the statistic function of that test.

(c) late the p-value of the test.

(d) In what confidence level can the assumption be rejected?

$$$$

For (a) is the null hypothesis $H_0: \ m_0=120$ and the alternative $H_1: \ \mu_1\neq 120$ ?

答:
The ten observations 104, 96, 80, 100, 108, 100, 112, 120, 130, 132
have mean $\bar X = 108.2,$ which is less than 120. The question
is whether 108.2 is significantly smaller than 120 in a statistical sense. That is, the question whether to reject
$H_0: \mu \ge 120$ in favor of $H_1: \mu < 120.$

It seems safe to assume the population from which these ten
observations were taken is normal. There are not enough observations for a convincing goodness-of-fit test, but a boxplot shows no outliers and a normal probability
plot is not far from linear. [Plot and test use R.]

```<code>par(mfrow=c(1,2))
 boxplot(x, col="skyblue2")
 qqnorm(x);  qqline(x, col="blue")
par(mfrow=c(1,2))
</code>```

It seems safe to do a t.test, which gives P-value $0.02206 < 0.05 = 5\%,$ rejecting $H_0$ at the 5% level of significance. [@tommik gives the formula for the t statistic.]

```<code>t.test(x, mu=120, alt="less")

        One Sample t-test

data:  x
t = -2.3385, df = 9, p-value = 0.02206
alternative hypothesis: true mean is less than 120
95 percent confidence interval:
   -Inf 117.4499
sample estimates:
mean of x
    108.2
</code>```

The 5% critical value for this one-sided test is $-1.833 > -2.338;$ another indication to reject at the 5% level.

The exact P-value cannot be found from printed tables, but it can be bracketed between two values (e.g., in my table between 0.01 and 0.025). A direct computation in R is shown below:

```<code>qt(.05, 9)
[1] -1.833113   # 5% left-sided critical value

pt(-2.3385, 9)
[1] 0.02206428  # consistent with output above
</code>```

If you are expected to show how to compute the t statistic, determine the degrees of freedom, or find the 5% critical
value of the test, I will leave that to you.

Notes on alternative methods: (1) If one questions normality, the data seem
sufficiently symmetrical that one might do a one-sample
Wilcoxon signed-rank test for the population median.
The P-value $0.025,$ shown below indicates rejection at the 5% level, but there are warning messages
that ties and a data value exactly at 120 prevent the P-value from being exact.

```<code>wilcox.test(x, mu = 120, alt="less")$p.val
[1] 0.02510388
</code>```

(2) One pivot-quantile style of 95% nonparametric bootstrap one-sided CI is $(-\infty, 116.2),$ which does not include the hypothetical value $120,$ suggesting rejection of $H_0$ at the 5% level. (I say 'suggest' because ten observations may not be enough for a reliable bootstrap CI. An advantage is that the bootstrap does not assume normal data.)

```<code>a.obs = mean(x)
set.seed(225)
d.re = replicate(5000, mean(sample(x,10,rep=T)-a.obs))
L = quantile(d.re, .05)
a.obs - L
   5%
116.2
</code>```

(3) A one-sided sign test for median 120 looks at 2 of 9
observations (different from 120) greater than 120 and
gives P-value $0.0898,$ not significant at the 5% level.

```<code>pbinom(2, 9, .5)
[1] 0.08984375
</code>```

Because the sign test is easy to do as above, the core
of R lacks a formal sign test. Here is output from Minitab:

```<code>Sign test of median =  120.0 versus < 120.0

    N  Below  Equal  Above       P  Median
x  10      7      1      2  0.0898   106.0
</code>```

Note on testing normality in small samples: With only ten observations it is difficult
to judge normality using a formal test. The Shapiro-Wilk
test is one of the best, but its power against (often skewed) exponential samples is only about 44% and
against (often symmetrical) uniform samples only
about 8%. Consequently, many statisticians
prefer graphical methods (boxplots, normal probability
plots) for judging normality of small samples.

```<code>set.seed(1234)
pv = replicate(10^5, shapiro.test(rexp(10))<span class="math-container">$p.val)
mean(pv <= .05)
[1] 0.4424
pv = replicate(10^5, shapiro.test(runif(10))$</span>p.val)
mean(pv <= .05)
[1] 0.0833
</code>```

Body:
The ten observations 104, 96, 80, 100, 108, 100, 112, 120, 130, 132
have mean $\bar X = 108.2,$ which is less than 120. The question
is whether 108.2 is significantly smaller than 120 in a statistical sense. That is, the question whether to reject
$H_0: \mu \ge 120$ in favor of $H_1: \mu < 120.$

It seems safe to assume the population from which these ten
observations were taken is normal. There are not enough observations for a convincing goodness-of-fit test, but a boxplot shows no outliers and a normal probability
plot is not far from linear. [Plot and test use R.]

par(mfrow=c(1,2))
 boxplot(x, col="skyblue2")
 qqnorm(x);  qqline(x, col="blue")
par(mfrow=c(1,2))

It seems safe to do a t.test, which gives P-value $0.02206 < 0.05 = 5\%,$ rejecting $H_0$ at the 5% level of significance. [@tommik gives the formula for the t statistic.]

t.test(x, mu=120, alt="less")

        One Sample t-test

data:  x
t = -2.3385, df = 9, p-value = 0.02206
alternative hypothesis: true mean is less than 120
95 percent confidence interval:
   -Inf 117.4499
sample estimates:
mean of x
    108.2

The 5% critical value for this one-sided test is $-1.833 > -2.338;$ another indication to reject at the 5% level.

The exact P-value cannot be found from printed tables, but it can be bracketed between two values (e.g., in my table between 0.01 and 0.025). A direct computation in R is shown below:

qt(.05, 9)
[1] -1.833113   # 5% left-sided critical value

pt(-2.3385, 9)
[1] 0.02206428  # consistent with output above

If you are expected to show how to compute the t statistic, determine the degrees of freedom, or find the 5% critical
value of the test, I will leave that to you.

Notes on alternative methods: (1) If one questions normality, the data seem
sufficiently symmetrical that one might do a one-sample
Wilcoxon signed-rank test for the population median.
The P-value $0.025,$ shown below indicates rejection at the 5% level, but there are warning messages
that ties and a data value exactly at 120 prevent the P-value from being exact.

wilcox.test(x, mu = 120, alt="less")$p.val
[1] 0.02510388

(2) One pivot-quantile style of 95% nonparametric bootstrap one-sided CI is $(-\infty, 116.2),$ which does not include the hypothetical value $120,$ suggesting rejection of $H_0$ at the 5% level. (I say 'suggest' because ten observations may not be enough for a reliable bootstrap CI. An advantage is that the bootstrap does not assume normal data.)

a.obs = mean(x)
set.seed(225)
d.re = replicate(5000, mean(sample(x,10,rep=T)-a.obs))
L = quantile(d.re, .05)
a.obs - L
   5%
116.2

(3) A one-sided sign test for median 120 looks at 2 of 9
observations (different from 120) greater than 120 and
gives P-value $0.0898,$ not significant at the 5% level.

pbinom(2, 9, .5)
[1] 0.08984375

Because the sign test is easy to do as above, the core
of R lacks a formal sign test. Here is output from Minitab:

Sign test of median =  120.0 versus < 120.0

    N  Below  Equal  Above       P  Median
x  10      7      1      2  0.0898   106.0

Note on testing normality in small samples: With only ten observations it is difficult
to judge normality using a formal test. The Shapiro-Wilk
test is one of the best, but its power against (often skewed) exponential samples is only about 44% and
against (often symmetrical) uniform samples only
about 8%. Consequently, many statisticians
prefer graphical methods (boxplots, normal probability
plots) for judging normality of small samples.

set.seed(1234)
pv = replicate(10^5, shapiro.test(rexp(10))$p.val)
mean(pv <= .05)
[1] 0.4424
pv = replicate(10^5, shapiro.test(runif(10))$p.val)
mean(pv <= .05)
[1] 0.0833


问:
Restriction of homotopy equivalence which is surjective is homotopy equivalence?
I have a question related to my previous post. The question is

Let $f:X\to Y$ be a homotopy equivalence and $A\subset X$ be a subspace such that $f|_A:A\to Y$ is surjective. Then $f|_A$ is a homotopy equivalence between $A$ and $Y$?

The original problem I concerned in my previous post is: restriction of $X\to X_1\to X_1/q(U)$ to $V$ is surjective so $X_1/q(U)\simeq V/W$ (3) in my post). Could you answer this question?

答:
Consider the following counterexample : let $X$ be a contractible space; then $X\to *$ is by definition a homotopy equivalence.

Take a subspace $Y$; as long as $Y\neq X$, the restriction to $X\setminus Y$ will be surjective.

However it is rarely the case that all $X\setminus Y$ are also contractible.

Consider, e.g. $X= \mathbb R, Y=\{0\}$, $X= \mathbb R^2, Y= \{(0,0)\}$. In fact, continuing this way with $\mathbb R^{n+1}$ shows that you could have a space that is still $n$-connected, but not contractible anymore.

Here's also a slightly less trivial example : consider the projection $S^1\times [0,1]\to S^1$. This is a homotopy equivalence, however if you remove an interior point on the left hand side you will get something very different, which is in fact homotopy equivalent to a wedge of two circles $S^1\vee S^1$.


问:
Find all functions $f:\mathbb N\to\mathbb N$ such that $f\left(f(m)+f(n)\right)=m+n$, for all $m,n \in\mathbb N$.

Find all functions $f:\mathbb N\to\mathbb N$ such that $f\left(f(m)+f(n)\right)=m+n$, for all $m,n \in\mathbb N$.

I did not get any idea. I was just plugging out values for $m=n=1$ and so on
no idea.

答:
This is not an answer, mostly just results I found interesting, feel free to edit this post to complete it.

I'm going to assume there's no $0$ in $\mathbb{N}$ (if there was then the function would not be well-defined unless $f(0)=0$, if $0 \mapsto e \not=0$ then $2e \mapsto 0$ and then $0\color{red}{\mapsto} 4e \not= e$):

$f$ by definition must be injective, injective basically means that $f$ doesn't map different elements in the domain $(\mathbb{N})$ to the same element in the range/codomain $(\mathbb{N})$. Say you had a function $g$ s.t. $g(1)=1$ and $g(2)=1$, that cannot happen if $g$ is injective because $g$ doesn't map different elements $(1\not=2)$ to the same element $(\text{in this case } 1)$.

As mentioned by hellofriends in the comments:

Suppose $f(m)=f(n)$ then $f(m)+f(m)=f(n)+f(m)$ and applying $f$ to both sides $$f(f(m)+f(m))=f(f(n)+f(m))$$
$$m+m=n+m$$
$$m=n$$
after subtracting $m$ from both sides, thus if $f(m)=f(n)$ then $m=n$ and $f$ must be injective.

By definition of a function every input must have an output, thus $1$ must be mapped to something, call it $s\in \mathbb{N}$. We can assume $s \not= 1$ because then $f(n)=n\:$ $\forall n \in \mathbb{N}$ which clearly satisfies $f(f(m)+f(n))=f(m+n)=m+n$, and we can assume that $s\not= 2$ because then $f$ would not be injective or well-defined, $1\mapsto 2$ then $4 \mapsto 2$ and then $2+2 = 4 \mapsto 1+4 =5$. Let $\alpha \geq 0$.

Given $\color{blue}{(1)}1 \mapsto s$, then $\color{blue}{(2)}2s \mapsto 2$ and $4 \mapsto 4s$, from then on we can use
$\color{blue}{(1)},\color{blue}{(2)}$ consecutively to get $5s\mapsto 5, 8s \mapsto 8,..., (3\alpha + 2)s \mapsto (3\alpha + 2),...$ thus we have $f$ must be surjective onto $[2]_3 \cap \mathbb{N}$.

Also $2$ must be mapped to something, $s' \in \mathbb{N}$ where $s' \not= s$ because $f$ is injective. $s’ \not= 1$ because then $2\mapsto 1$ implies $2 \mapsto 4$ and $f$ would not be well-defined.

Then $2 \mapsto s'$ and using $\color{blue}{(1)}$ we get $s+s'\mapsto 3$, then
$\color{blue}{(1)},\color{blue}{(2)}$ consecutively $4s + s' \mapsto 6, 7s + s' \mapsto 9, 10s+s'\mapsto 12,...,(3\alpha +1)s+s'\mapsto 3\alpha +3,...$ consequently $f$ must be surjective onto $[0]_3 \cap \mathbb{N}$.

Also since $2 \mapsto s'$, $2s' \mapsto 4$ and using $\color{blue}{(1)},\color{blue}{(2)}$ consecutively $3s + 2s' \mapsto 7, 6s+2s'\mapsto 10,..., 3\alpha s+2s'\mapsto 3(\alpha +1)+1$ and thus $f$ must be surjective onto $[1]_3\cap \mathbb{N} \setminus \{1\}$.

$f$ is pretty close to surjective, all that's missing is the existence of $\gamma \in \mathbb{N}$ s.t. $f(\gamma)=1$.

edit:

Thanks to Carl’s proof it looks like my post was half-baked, I started by letting $1\mapsto 2$ and derived a contradiction, I should of kept going and tried other values because it turns out that as you try other values for $f(1)>1$ they all somehow lead to a contradiction and the above arguments can be used to demonstrate this:

Suppose $1\mapsto (3\alpha +2) \in [2]_3 \cap \mathbb{N}$ then $(3\alpha+2)^2 \mapsto 3\alpha +2$ and $(3\alpha+2)^2 \geq 4 >1$. $f$ would not be injective, contradiction.

Suppose $1\mapsto 3\alpha +3 \in [0]_3 \cap \mathbb{N}$ then $(3\alpha +1)(3\alpha +3)+s’ \mapsto 3\alpha +3$ and $(3\alpha+1)(3\alpha+3)+s’\geq 3+s’>3>1$ because $s’>0$. $f$ would not be injective, contradiction.

Suppose $1\mapsto 3(\alpha +1)+1 \in [1]_3 \cap \mathbb{N}$ then $3\alpha(3(\alpha +1)+1)+2s’ \mapsto 3(\alpha +1)+1$ and $3\alpha(3(\alpha +1)+1)+2s’\geq 0+2s’ \geq 2\cdot 1 >1$. $f$ would not be injective, contradiction.

Since we ran through all possiblities $1\mapsto s \in \mathbb{N}\setminus\{1\}$ and they all led to a contradiction, one way or another, we are left only with the original case we knew that worked $1 \mapsto 1$, i.e. $f$ is the identity.


答: Of course. Changing the integrand on a set of measure zero does nothing to the integral, as you've essentially already noted. You could write the second integral as using the integrand $g(z) := f(z) 1_{z \neq 1}$ where $1_{z \neq 1}$ is $1$ if $z \neq 1$ and is $0$ if $z = 1$.

Title: Prove that $\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1) }f(z)dz $

Body:
Define $C(0,1)= \{z: \mid z\mid=1 \}$
$D(0,1)= \{z: \mid z\mid=1, z\neq 1+0i \}$

Prove that,
$$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1)
  }f(z)dz  $$
My try

$z=1+0i$ is of measure zero. So, $$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1)
  }f(z)dz  $$


问:
Probability that the circuit will not fail
When connected to the network circuit each element fails with a probability of 0.8. Find the probability that the circuit will not open at the time of inclusion.

Using basic formulas I got answer 0,83: P(X)=$0,8+(1-0,8^2)^2-0,8(1-0,8^2)^2$

答:
The upper branch works when both the subsystems work, thus with probability

$(1-0.8^2)\cdot(1-0.8^2)=0.1296$

the lower branch fails with probability

$0.8$

The system is in parallel thus it work always but both branches not working, thus

$$\mathbb{P}[\text{The System Works}]=1-(1-0.1296)\cdot 0.8=30.37\%$$


问:
Why does the definition of topological space require that the empty set be open?
I'm taking my first course in general topology, and so obviously I came across the definition of topological space $(T, \tau)$:

$\tau \in \mathcal P \mathcal P (T)$

$\emptyset, T \in \tau$

$(A_i)_{i \in I} \in \tau \implies \bigcup_{i \in I} A_i \in \tau$

$A_1, \dots, A_n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau$

This is very abstract, so I tried to find a few examples, and these examples I found caught my eye:

$T = \mathbb R^n$, $\tau = \{ \text{every set whose complement is compact} \}$

$T = [0,1]$ $\tau = \{ \text{every set whose complement is a finite set}  \}$

Now these satisfy every part of the definition, except that one has to add the empty set to $\tau$ explicitly as well. This gives the natural question: why is this needed?

答:
First of all, in your examples $\emptyset$ is not a set with a compact complement, as $\Bbb R^n$ is not compact, and it does not have a finite complement $[0,1]$ in the second case.. So they should be explicily added to $\tau$ as one of the axioms states it should be in there.

Also, in many cases, like topologies from metric spaces or ordered spaces, there will be many cases where the topology contains disjoint sets and we want open sets to be closed under finite intersections, so for that reason alone we have to add $\emptyset$ to $\tau$ or accept that a topology reduces to a weakened sort of filter on $X$ (another type of families of subsets, also interesting, but not topology; your second example as stated (without the empty set) is indeed a filter BTW and the first a so-called filter-base ). Also formally, in the arbitrary union axiom we could take $I=\emptyset$ and then we're also "forced" to add $\emptyset$ to $\tau$ in order to fulfill it.

In short, we have the empty set in $\tau$ for practical and internal logic/consistency reasons. We consider mostly non-empty open sets in practice (as these are the neighbourhoods used in continuity/convergence etc.), but we do need it in there.

This question has many more reasons in its answers.


问:
Showing that the derivative function is not continuous w.r.t. the sup norm
Let $\mathcal{C}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f \text{ is continuous}\}$, $\mathcal{C}^{1}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f' \text{ is continuous}\}$, where both sets are equipped with the supremum norm $||\cdot||_\infty$. Let $h: \mathcal{C}^{1}([-1, 1]): \to \mathcal{C}([-1, 1]), [h(f)](t) = f'(t), t \in [-1, 1]$ be the derivative mapping from the set of continuously differentiable functions to continuous functions. I want to show that $h$ is not a continuous mapping w.r.t. the supremum norm, but I have not yet managed to "hack" the intuition for functional analysis. Specifically, what we want to demonstrate is that $\exists \epsilon > 0: \forall \delta > 0: \forall f, g \in \mathcal{C}^{1}([-1 ,1]): d_{\mathcal{C}^{1}}(f, g) < \delta\land d_{\mathcal{C}}(h(f), h(g)) \geq \epsilon$.

Where I am currently stuck is that before inspecting the derivative mapping, let $f(t) = \cos(t)$ and $h(t) = 0$. Both are evidently continuous, but $||f - h||_\infty = \sup_{t \in [-1, 1]}\left|\cos(t) - 0\right| = 1$. So to my understanding, even a redefined $[h(f)](t) = f(t)$ should not be continous, as for no $\delta$ is $1 < 0.5$. Or am I mistaken? In the case that I am correct, does this same claim show that the original (derivative mapping) $h$ is not continuous?

答:
Let $f_n(t): \frac{1}{n} \sin (nt).$ Then $||f_n||_{\infty} \le  \frac{1}{n}$ for all $n$. Hence

$$ ||f_n -0||_{\infty} \to 0.$$

But we have $(h(f_n))(t)= \cos (nt)$ and therefore

$(h(f_n)))$ does not converge to $(h(0))$ in the norm $||\cdot||_\infty$


问:
Expectation of X choose k
Let $X$ be a random variable, assume that we know $\mathbb{E}\left[ X \choose k \right] $.
Can I assume
$$ \mathbb{E}\left[ X \choose k \right] =  {\mathbb{E}\left[  X \right]\choose k}$$ as k is not a random variable?

答:
No, this is not true. If, for instance $X \sim \text{Binomial}(n,p)$ with $n = 10$ and $p=0.55$ then $\mathbb E[X] = 5.5$ and ${\mathbb{E}\left[  X \right]\choose k}$ is not defined.

Applying the LOTUS, we have that
$$ \mathbb{E}\left[ X \choose k \right] = \sum_{i = 0}^{\infty} {i \choose k}\mathbb P(X=i) $$
And there is no reason for that sum to be equal to ${\mathbb{E}\left[  X \right]\choose k}$ in general.

Title: Expectation of X choose k


答: Taking a categorical perspective we should ask why are we even defining topological spaces. In a category we are primarily interested in the morphisms and the objects are merely there to serve as domains and codomains. So, the reason we define topological spaces is in order to define continuous functions between them. A function $f\colon X\to Y$ is continuous if it pulls back every open set in $Y$ to an open set in $Y$. Namely once topologies have been chosen, for $f$ to be continuous it must be the case that $f^{-1}(U)$, the inverse image of $U\subseteq Y$, is open in $X$. At first site this may seem like an odd definition but it just justified in many ways, chiefly, at the elementary level, by the desire to mimic the usual $\varepsilon - \delta $ notion of continuity which also goes in reverse. In any case, one certainly wants all constant functions to be continuous. If we did not demand the empty set to always be open, then constant functions would typically fail to be continuous since you can find, typically, plenty of open sets in the codomain whose inver image is empty.

Title: Why does the definition of topological space require that the empty set be open?

Body:
I'm taking my first course in general topology, and so obviously I came across the definition of topological space $(T, \tau)$:

$\tau \in \mathcal P \mathcal P (T)$

$\emptyset, T \in \tau$

$(A_i)_{i \in I} \in \tau \implies \bigcup_{i \in I} A_i \in \tau$

$A_1, \dots, A_n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau$

This is very abstract, so I tried to find a few examples, and these examples I found caught my eye:

$T = \mathbb R^n$, $\tau = \{ \text{every set whose complement is compact} \}$

$T = [0,1]$ $\tau = \{ \text{every set whose complement is a finite set}  \}$

Now these satisfy every part of the definition, except that one has to add the empty set to $\tau$ explicitly as well. This gives the natural question: why is this needed?


答:
Technically we could define topology without the empty set. But then everywhere an empty set appears we would have to deal with it in a special way. For example a continuous function $f:X\to Y$ is a function such that preimage of any open set is open. Without the "$\emptyset\in\tau$" assumption we would have to modify this definition: preimage of every open set is open if nonempty. And this special "if nonempty" now has to appear everywhere.

This is an unnecessary noise. Makes proofs and definitions longer, less readable and for no good reason.

Also note that similar reasoning applies to "$T\in\tau$". Since we often deal with closed sets which by definition are complements of open sets. We now have to deal with two special cases: "if nonempty" and "if not entire $T$". Which can be avoided by two simple axioms.


问:
how to show convergence and evaluate $n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2}$
I tried this :
$$\lim\limits_{n \rightarrow +\infty}n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2} = \int_{0}^{1} \frac{e^{-1/x}}{x^2} dx$$ to evaluate, and to show convergence, I think $$n\sum_{k=1}^{2n} \frac{e^{-n/k}}{k^2} < \sum_{k>1}{} \frac{1}{k^2}$$ which converges.

Body:
$$
n\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{n}{k}} }}{{k^2 }}}  = \frac{1}{2}\frac{1}{{2n}}\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{1}{2}\frac{{2n}}{k}} }}{{(k/2n)^2 }}}  \to \frac{1}{2}\int_0^1 {\frac{{e^{ - 1/(2x)} }}{{x^2 }}dx} \mathop  = \limits^{t = 1/(2x)} \int_{1/2}^{ + \infty } {e^{ - t} dt}  = e^{ - 1/2} .
$$


答:
Continuity of $h$ means given $f$ and $\epsilon >0$ there exist $\delta >0$ such that $\|g-f\|_{\infty} <\delta$ implies $\|h(f)-h(g)\|_{\infty} <\epsilon$. In your example $\|h-f\|_{\infty} <\delta$ is not satisfied for any $\delta \in (0,1)$ so the example says nothing about continuity of $h$. [BTW, you should avoid using $h$ with two diferent meanings].

To prove that $h$ is not continuous  consider the funsions $f_n(x)=\frac {x^{n}} n$. These converge in the norm to the $0$ function but their derivatives do not converge in the norm to the $0$ function since $\sup\{|x^{n-1}-0|: 0 \leq x \leq 1\}=1$ for all $n$.

Title: Showing that the derivative function is not continuous w.r.t. the sup norm

Body:
Let $\mathcal{C}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f \text{ is continuous}\}$, $\mathcal{C}^{1}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f' \text{ is continuous}\}$, where both sets are equipped with the supremum norm $||\cdot||_\infty$. Let $h: \mathcal{C}^{1}([-1, 1]): \to \mathcal{C}([-1, 1]), [h(f)](t) = f'(t), t \in [-1, 1]$ be the derivative mapping from the set of continuously differentiable functions to continuous functions. I want to show that $h$ is not a continuous mapping w.r.t. the supremum norm, but I have not yet managed to "hack" the intuition for functional analysis. Specifically, what we want to demonstrate is that $\exists \epsilon > 0: \forall \delta > 0: \forall f, g \in \mathcal{C}^{1}([-1 ,1]): d_{\mathcal{C}^{1}}(f, g) < \delta\land d_{\mathcal{C}}(h(f), h(g)) \geq \epsilon$.

Where I am currently stuck is that before inspecting the derivative mapping, let $f(t) = \cos(t)$ and $h(t) = 0$. Both are evidently continuous, but $||f - h||_\infty = \sup_{t \in [-1, 1]}\left|\cos(t) - 0\right| = 1$. So to my understanding, even a redefined $[h(f)](t) = f(t)$ should not be continous, as for no $\delta$ is $1 < 0.5$. Or am I mistaken? In the case that I am correct, does this same claim show that the original (derivative mapping) $h$ is not continuous?


问:
How to prove that the line outside a convex polygon, having the minimum sum of distances is one of the edges?
As part of a computational geometry question, I need a result for an intermediate step.

Suppose there is a convex polygon S, with finitely many points inside the polygon. Now, we want to find the line outside the polygon (i.e, the line can pass through a vertex or edge but cannot intersect) which minimizes the sum of perpendicular distances between the line and (vertices + interior points).

I have a hunch that we can prove that this line must be one of the edges, but I am unable to find a proof.

The simplest approach I tried was checking if rotating a line passing through vertex (but outside polygon) always decreases the sum till it reaches the edge. But this is clearly not true.

How to prove this?

Note 1: I think that this theorem must be true, but I have not seen it anywhere. It might be possible that it is actually false and the original question can be solved by a different method. If so, I would like to see a counterexample.

Note 2: The original question, the only answer proposes the theorem but does not provide a proof.

答:
A distance from a point $(x_i,y_i) $ to a line given by the equation $ax+by+c=0 $ is
$$
\frac{|ax_i+by_i+c|}{\sqrt {a^2+b^2}}.\tag1
$$

Now because the line is outside of the hull polygon the expression $ax_i+by_i+c$ has the same sign for all points, which we without loss of generality will assume to be positive. Hence we may drop taking the absolute value and write for the sum of distances:
$$
D (a,b,c)=\sum_{i=1}^N\frac{ax_i+by_i+c} {\sqrt {a^2+b^2}}=N\frac{aX+bY+c} {\sqrt {a^2+b^2}},\tag2
$$
where $(X,Y) $ are coordinates of the centroid of the points.

Now observe that the right hand side in (2) is nothing else but the multiplied by $N$ distance from the centroid to the line which is indeed minimized by one of the polygon edges, as claimed.


问:
How to find range of values for where line cuts circle?
I am doing a question on circles. The question is:

The circle  has equation $x^2 + y^2 - 6x + 10y + 9 = 0
$
The line with equation $y= kx$ , where  is a constant, cuts  at two distinct points.
Find the range of values for k

My working so far:

Found centre: (3, -5) radius = 5

Now what do I do after this?

答: Hint: Plug the equation of line into circle, you will get a quadratic in 'x'... now you need two distinct points.. what's the condition on discriminant for that?</p>


问:
What is the most general unitary that commutes with a one dimensional projector in a finite dimensional Vector Space
Given a vector space of finite dimension $N$ with an orthonormal basis $\mathcal{B}=\{|0\rangle,\ldots,|N-1\rangle \}$ what is the most general unitary operation that commutes with the projector onto one of the basis elements (say $|0\rangle$), i.e., what is the most general $\mathcal{U}$ such that $\mathcal{U}^\dagger \mathcal{U}=\mathcal{U} \mathcal{U}^\dagger=\mathbb{I}$ and $\left[\mathcal{U},|0\rangle \langle 0| \right]\equiv \mathcal{U} |0\rangle  \langle 0|- |0\rangle  \langle 0|\mathcal{U}=0$. A partial answer to the question is:
\begin{eqnarray*}
\mathcal{U(\lambda,\theta,\gamma)}&=&\exp\left(i G(\lambda,\theta,\gamma )\right)\\
G(\lambda,\theta,\gamma)&=&\sum_{k=1}^{N-1}\lambda_k |k \rangle \langle k|\\
&&+\sum_{k=1}^{N-1}\sum_{l>k}^{N-1} \theta_{l,k} \left(|k \rangle \langle l|+|l \rangle \langle k|\right)\\
&&+i\sum_{k=1}^{N-1}\sum_{l>k}^{N-1} \gamma_{l,k} \left(|k \rangle \langle l|-|l \rangle \langle k|\right)
\end{eqnarray*}
For $N=2$ it is indeed the answer to the question, but I am not sure for higher dimensions.

答: I am not familiar with the bra-ket notation. However, since $\mathcal{U}$ commutes with the projector into the span of the first basis vector, the matrix of $\mathcal{U}$ with respect to the given orthonormal basis is a unitary matrix in the form of $\pmatrix{z\\ &V}$, where $z$ is a complex number with unit modulus and $V$ is an $(n-1)\times(n-1)$ unitary matrix.


答:
You have a parallel circuit with two paths. If any of the path does not fail, the circuit remains connected.

The upper path fails if at least one of the two loops in the upper path fails. Probability that a loops fails is $0.8^2 = 0.64$ and that a loop does not fail is $0.36$. So the probability that at least one of the loop fails is $1 - 0.36^2 = 0.8704$.

Hence the probability that the upper path fails is $0.8704$.

Probability that the lower path fails is $0.8$.

So the probability that the circuit does not fail is $(1 - 0.8704 \times 0.8) = 0.30368$.

Title: Probability that the circuit will not fail

Body:
When connected to the network circuit each element fails with a probability of 0.8. Find the probability that the circuit will not open at the time of inclusion.

Using basic formulas I got answer 0,83: P(X)=$0,8+(1-0,8^2)^2-0,8(1-0,8^2)^2$


问:
Circle of Apollonius; proving a tangent line
Quick recap, since there are several circles of Apollonius:

Given a triangle with fixed base and the other two sides in known ratio, the circle of Apollonius (of the first type) gives the possible positions of the third point.

Now constructing that circle from the two base points $A$ and $B$ and another point $D$ that divides $AB$ in the appropriate ratio is a question considered here, which can be achieved in various different ways. $D$ is clearly on the Apollonian circle,  so finding one more point and knowing that the center of the circle $O$ is on $AB$ is sufficient to construct it.

The construction I gave is this:

Construct the circle $\color{#0B2}{\Gamma}$ centred on $D$ through the nearer of $A$ and $B$, which I will assume is point $B$. Also construct a line perpendicular to $AB$ at $B$. Then form the tangent to circle $\color{#0B2}{\Gamma}$ from $A$. This intersects the perpendicular at $E$, which is on the Apollonius circle because $\triangle AEB$ is similar to the right triangle $\triangle ADT$ (with $T$ the tangent point of $A$ on $\color{#0B2}{\Gamma}$), so $|AE|:|EB| = |AD|:|DT| = |AD|:|BD|$ as required.

Now the extra claim I make - visible in the diagram - is that $AE$ is a tangent to the Apollonius circle. I initially thought I had proved this via the power of point $A$ to the circle with a bit of Pythagoras, but I don't really think this works. However I do think that $AE$ is actually tangent and certainly this reliably works in interactive geometry tools.

So my question is: can you prove (or disprove) that $AE$ is tangent to the circle of Apollonius shown?

答:
From $|AE|:|EB| = |AD|:|BD|$, $DE$ is angle bisector of $\angle AEB$.

Then $$\angle BEO = \angle DEO - \angle DEB=\angle ODE - \angle AED=\angle EAB$$

This implies $OE$ is tangent to circumcircle of $\triangle ABE$. But $ABE$ is a right triangle with its circumcenter on $AE$. So $OE \perp AE$ which means $AE$ is tangent to the Apollonian circle $(O)$.


问:
Closed formula for LDU decomposition
I know how LDU decomposition works and moreover, I know that these closed formulas are more difficult to compute and hence it is not done in practice, but I want to understand the following cite from Wikipedia:

Closed formula
When an LDU factorization exists and is unique, there is a closed (explicit) formula for the elements of L, D, and U in terms of ratios of determinants of certain submatrices of the original matrix A.In particular, $D_{1}=A_{1,1}$ and for i = 2 , … , n $D_i$ is the ratio of the i -th principal submatrix to the ( i − 1 )-th principal submatrix. Computation of the determinants is computationally expensive, so this explicit formula is not used in practice.

Unfortunately I do not understand how this formula looks. Especially what is the i-th principal submatrix. I am not familiar with this term and it is not explained in more detail. I tried reformulating the recursive formulas that are usually used to explicit ones but this gets very fast very messy and I did not find an easy technique for doing so.

Here is the full wikipedia article: https://en.wikipedia.org/wiki/LU_decomposition

Thank you very much in advance for your help.

EDIT: Hello,

I was thinking about it since yesterday and as soon as I post the question, I realize what part of the answer must be. The definition of principal submatrix that is meant, it also called leading principal submatrix and obtained by keeping the first i rows and column of a matrix and eliminating the others. Hence $D_i = \frac{det A_i}{det A_{i-1}}$. The part that is still missing is the formula for $L_i$ and $U_i$.

答:
Use slice notation $m:n=[m,m+1,..,n]$, so that for example $A_i=A_{1:i,1:i}$. Then we want to achieve the decomposition in the step $i\to i+1$
$$
A_{i+1}=\pmatrix{A_i&A_{1:i,i+1}\\A_{i+1,1:i}&A_{i+1,i+1}}
=
\pmatrix{L_i&0\\L_{i+1,1:i}&1}
\pmatrix{D_i&0\\0&d_{i+1}}
\pmatrix{U_i&U_{1:i,i+1}\\0&1}
$$
Of course, $A_i=L_iD_iU_i$ by iteration assumption. Also from the determinant laws, $\det(A_i)=\det(D_i)$ so that $d_{i+1}=\det(A_{i+1})/\det(A_i)$.

For the other parts multiply out the matrix product
$$
=\pmatrix{L_iD_i&0\\ L_{i+1,1:i}D_i&d_{i+1}}
\pmatrix{U_i&U_{1:i,i+1}\\0&1}
=
\pmatrix{A_i&L_iD_iU_{1:i,i+1}\\L_{i+1,1:i}D_iU_i&L_{i+1,1:i}D_iU_{1:i,i+1}+d_{i+1}}
$$
to then get
$$
L_iD_iU_{1:i,i+1}=A_{:i,i+1}=(L_iD_iU_i)A_i^{-1}A_{1:i,i+1}\implies U_{1:i,i+1}=U_iA_i^{-1}A_{1:i,i+1}
\\
L_{i+1,1:i}D_iU_i=A_{i+1,1:i}=A_{i+1,1:i}A_i^{-1}(L_iD_iU_i)\implies L_{i+1,1:i}=A_{i+1,1:i}A_i^{-1}L_i
\\
d_{i+1}=A_{i+1,i+1}-L_{i+1,1:i}D_iU_{1:i,i+1}
$$
Remember that the entries of $A_i^{-1}=\frac{A_i^\#}{\det(A_i)}$ (using the adjoint/adjunct/adjugate which is the transpose of the co-factor matrix) are fractions of determinants, and for further consideration, that constructs like $A_i^\#A_{1:i,i+1}$ can be interpreted as consisting of Lagrange determinant formulas, giving $-\bigl(A_{i+1}^\#\bigr)_{1:i,i+1}$. Thus in total
$$
U_{1:i,i+1} = -\frac{U_i\bigl(A_{i+1}^\#\bigr)_{1:i,i+1}}{\det(A_i)}\\
L_{i+1,1:i} = -\frac{\bigl(A_{i+1}^\#\bigr)_{i+1,1:i}L_i}{\det(A_i)}
$$

Title: Closed formula for LDU decomposition


问:
$A$-content topology
Let $X$ be a non-empty set and $\emptyset\subset A\subset X$. Define
$${\cal T}=\{Y\ ;\ A\subseteq Y\subseteq X \}\cup\{\emptyset\}$$
we know $\cal T$ is a topology on $X$.
I want know more about structure of this topology. but I can't find this example on any books. Please help me to find more about this. In particular I want to find a book written about this example.

答:
If $A= \{p\}$ this example is known as the included point topology (wrt $p \in X$).
Or the particular point topology in Counterexamples in Topology, mirrored here e.g.

This is a slight generalisation of it.

$X$ is $T_0$ iff $|A|= 1$ in fact. It will rarely fulfill any higher separation axioms (like complete normality or normality).

$X$ is hyperconnected (all non-open sets intersect), so in particular connected

$X$ is not compact if $X\setminus A$ is infinite.

$X$ is separable as any $\{a\}$ with $a \in A$ is dense.

$X$ is second countable iff $X \setminus A$ is countable.

Etc etc. All this without a book (don't rely on books, just check the definitions!). This space is so simple, almost all properties that are commonly used are easy to verify or disprove for it.

Title: $A$-content topology


问:
How many different pairs of advisors can be selected for a seven-story building from $12$ male candidates and $15$ female candidates?
I'm stuck on this question:

"At a certain college, the housing office has decided to appoint, for each floor, one male and one female residential advisor. How many different pairs of advisors can be selected for a seven-story building from $12$ male candidates and $15$ female candidates?"

OK, so would i use permutation?
I cant seem to figure out what's the $n$ and what's the $r$... Or combinations? or something else?

答:
The answer above is wrong for anyone looking here. You are supposed to add them, not multiply them. So:

$12*15+11*14+10*13+9*12+8*11+7*10+6*9=784$.


问:
Rudin Theorem 3.43 - Alternating series test
Suppose $(a): \vert c_1 \vert\geq\vert c_2\vert\geq \vert c_3 \vert\cdots$, $(b):c_{2m-1}\geq0,\enspace c_{2m}\leq0 \enspace(m=1,2,3,...)$, and $(c):=\lim_{n\to\infty}c_n=0$. Then $\sum c_n$ converges.

What is the meaning of $(b)$? If we take a look at the alternating series $\sum \frac{(-1)^n}{n}$, then clearly condition $b$ is left unsatisfied. The odd terms are in fact negative, while the even terms are positive. However, ignoring $b$, we can prove this series converges using $(a)$ and $(c)$. What am I missing?

答:
If you ignore b) you cannot conclude that $\sum c_n$ is convregent, an easy counter-example being $c_n=\frac  1 n$.

The fact that $c_{2m} \geq 0$ and $c_{2m-1} \leq 0$ in your example does not affect applicabilty of the theorem. This is because $\sum c_n$ is convergent iff $\sum (-c_n)$ is convergent. The main point here is that the signs of the terms have to alternate.


问:
Evaluating $\int_{0}^{1} \frac{\ln^{n} x}{(1-x)^{m}} \, \mathrm dx$
On another site, someone asked about proving that

$$ \int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx = (-1)^{n+m-1} \frac{n!}{(m-1)!} \sum_{j=1}^{m-1} (-1)^{j} s (m-1,j) \zeta(n+1-j), \tag{1} $$

where $n, m \in \mathbb{N}$, $n \ge m$, $m \ge 2$, and $s(m-1,j)$ are the Stirling numbers of the first kind.

My attempt at proving $(1)$:

$$ \begin{align}\int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx  &= \frac{1}{(m-1)!} \int_{0}^{1} \ln^{n} x \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2) \  x^{k-m+1} \ dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1)  \cdots (k-m+2)  \int_{0}^{1} x^{k-m+1} \ln^{n} x \, dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \frac{(-1)^{n} n!}{(k-m+2)^{n+1}}\\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} \sum_{j=0}^{m-1} s(m-1,j) \  k^{j}  \ \frac{(-1)^{n} n!}{(k-m+2)^{n+1}} \\ &= (-1)^{n}  \frac{n!}{(m-1)!} \sum_{j=0}^{m-1} s(m-1,j)  \sum_{k=m-1}^{\infty}   \frac{k^{j}}{(k-m+2)^{n+1}} \end{align} $$

But I don't quite see how the last line is equivalent to the right side of $(1)$. (Wolfram Alpha does say they are equivalent for at least a few particular values of $m$ and $n$.)

答:
Substitute $x=e^{-t}$ and get that the integral is equal to

$$(-1)^n \int_0^{\infty} dt \, e^{-t} \frac{t^n}{(1-e^{-t})^m} $$

Now use the expansion

$$(1-y)^{-m} = \sum_{k=0}^{\infty} \binom{m+k-1}{k} y^k$$

and reverse the order of summation and integration to get

$$\sum_{k=0}^{\infty} \binom{m+k-1}{k} \int_0^{\infty} dt \, t^n \, e^{-(k+1) t}$$

I then get as the value of the integral:

$$\int_0^1 dx \, \frac{\ln^n{x}}{(1-x)^m} = (-1)^n\, n!\, \sum_{k=0}^{\infty} \binom{m+k-1}{k} \frac{1}{(k+1)^{n+1}}$$

Note that when $m=0$, the sum reduces to $1$; every term in the sum save that at $k=0$ is zero.

Note also that this sum gives you the ability to express the integral in terms of a Riemann zeta function for various values of $m$, which will provide the Stirling coefficients.


问:
How do I prove that the vectors $w,\,u,\,v$ are positively oriented, if $w$ is the cross product of $u$ and $v$?
Pretty much the question in the title. My textbook claims (without a reference or even an argument) that

The cross product $w$ of vectors $u$ and $v$ is orthogonal towards both vectors.

$||w||=||u||||v||\sin \alpha$

If vectors $u$ and $v$ are not parallel, the vector triple $u,\,v,\,w$ is positively oriented.

I managed to prove the first claim myself, and the second through this wonderful video by khan academy. I have no idea however of how I'm going to prove the third claim, and my web searches have come up short.

Any ideas?

答: Hint: work in a coordinate system where$$u=\Vert u\Vert\vec{i},\,v=a\vec{i}+b\vec{j},\,a^2+b^2=\Vert v\Vert^2$$so $w=\Vert u\Vert b\vec{k}$. Now separately consider the cases $b<0,\,b=0,\,b>0$.


答:
If you want this to hold for all $k$, then the answer is NO, even if $X$ takes value in integers, unless $X$ is constant. For example we take $k=2$, what you suggest will be $\mathbb{E}\frac{X(X-1)}{2}=\frac{(\mathbb{E}X)(\mathbb{E}X-1)}{2}$, which is equivalent to

$$\mathbb{E}X^2-\mathbb{E}X=(\mathbb{E}X)^2-\mathbb{E}X\Longleftrightarrow \mathbb{E}X^2=(\mathbb{E}X)^2\Longleftrightarrow Var(X)=0,$$

thus $X$ should be constant.

Title: Expectation of X choose k

Body:
Let $X$ be a random variable, assume that we know $\mathbb{E}\left[ X \choose k \right] $.
Can I assume
$$ \mathbb{E}\left[ X \choose k \right] =  {\mathbb{E}\left[  X \right]\choose k}$$ as k is not a random variable?


问:
Solution to matrix equation with exponentials
Assuming
$$
A, M \in \mathcal{R}^{n \times n }
$$
with A, M being invertible. Could someone tell me if it is possible to find a closed form solution for the matrix $M$ in the following equation and if so, provide me with some hints on how to obtain it
$$
A = e^M + e^{M^T}.
$$

I have tried a few things but can´t seem to make it work.

Thanks.

答:
If $A=e^M + e^{M^t}$ then $A$ is a real symmetric matrice.

You can use spectral theorem to diagonalize it that is there is $D$ a diagonal matrix and $P$ a invertible matrice, such that $A=PDP^{-1}$.

Then you can rewrite you equation as $D=e^{P^{-1}MP}+e^{P^{-1}M^{t}P}$.

If $D$ have positive coefficients on the diagonal you can search N=P^{-1}MP also as a diagonal matrice whose coefficient will be $n_{i,j}=1/2 log(d_{i,j})$. Then $M=PNP^{-1}$.

I am not there is a solution if one coefficient of $D$ is negative.


问:
Can there be non open affine subset in a scheme?
Now this may sound like a stupid question but I am studying schemes' theory and open affine subset of a scheme arise naturally in many statements. In many cases I wonder if the request on openness of an affine subset of a scheme in the hypothesis of a theorem is really necessary or if any affine subset of a scheme is necessarily open.</p>

答:
There are two kinds of "natural" affine subsets of (affine) schemes:

Localisations: This is the one you are alluding to probably. For $R$ a ring and $f\in R$, one has a natural inclusion $$Spec(R_f)\to Spec(R)$$ and this map is an embedding with image $D(f)=Spec(R)-V(f)$, thus open

Quotients by ideals: Take any ideal $I\subset R$, then there is again a natural inclusion $$Spec(R/I)\to Spec(R).$$ You can check that this is an embedding with image $V(I)\subset Spec(R)$, which by definition is closed.

Thus we need to find some $Spec(R)$ with $V(I)$ not open.

Take for example $R=k[x]$ for some field $k$ and $I=(x)$. Then this corresponds to the inclusion of a point into the affine line $$ *\to \mathbb{A}^1_k.$$


问:
Proving $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$
It is straightforward to prove using a connectedness argument that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^n$, for $n>1$.

How do you prove that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$?

Note: I'm looking for a proof which does not use any algebraic topology,not even Brouwer's fixed point theorem. [I know that there are proofs of Brouwer's theorem using purely analytic methods, but I still do not want to include it - I'm looking for something even simpler]

Me and a few friends have been at this for a couple of weeks, but kept running in circles. Unfortunately, this was a long time ago, and I have completely forgotten what we tried. Any ideas?

Edit: Thanks for all the answers, but I still didn't get what I was looking for. I should have been more explicit : I don't want to use the Jordan curve theorem either. (The simplest proof of that which I've seen involves Brouwer's fixed point theorem). I'm looking for a proof which does not use anything apart from ideas of connectedness and compactness.

Looking at the answers, a second question came to my mind: Given that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$, can you deduce the Jordan curve theorem?

答:
A elementary proof can be found in the following article:

An interesting proof on the nonexistence of a continuous function between $\mathbb{R}^2$ and $\mathbb{R}^n$ for $n \neq 2$, by F. Malek, H. Daneshpajouh, H.R. Daneshpajouh and J. Hahn.


问:
A question about "rearranging" finite unions.
I am working on this simple problem:

Show that
$$\overline{A\cup B}=\overline{A} \cup \overline{B}$$
where $\overline{X}$ denotes the closure of $X$.

Now I have already proved it by showing that LHS is a subset of RHS and vice-versa and using the fact that a finite union or intersection of closed sets is closed. I'm wondering if it can't be solved in an even more direct way as follows:

Let $L_{X}$ denote the set of all limit points of the set $X$. Then $\overline{A\cup B}=(A\cup B)\cup L_{A\cup B}$. $L_{A\cup B}=L_{A}\cup L_{B}$, so $\overline{A\cup B}=(A\cup B)\cup (L_{A}\cup L_{B})$. Now can I write this expression as $(A\cup L_{A})\cup(B\cup L_{B})$ so that it becomes $\overline{A}\cup\overline{B}$, which is what I need?

Apologies if this seems like a trivial question.

答:
Union is indeed a commutative and associative operation on sets (well known from set theory, we use set theory all the time in topology), and denoting the set of limit points of any set $C$ by $C'$, as is more common, you can indeed say

$$\overline{A \cup B} = (A \cup B) \cup (A\cup B)' = (A \cup B) \cup (A'\cup B')= (A \cup A') \cup (B \cup B)' = \overline{A} \cup \overline{B}$$

provided you indeed know that limit points sets are preserved by finite unions.

An alternative proof using not limt points but the definition of a $\overline{C}$ as the smallest (by inclusion) closed superset of $C$:

$\overline{A} \cup \overline{B}$ is closed (as finite sets are closed under finite unions which follows (by de Morgan) from the fact that finite intersections of open sets are open). It contains $A \cup B$ trivially, so by minimality $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$. OTOH: $A \subseteq A \cup B$ so $\overline{A} \subseteq \overline{A \cup B}$ and also $B \subseteq A \cup B$ so $\overline{B} \subseteq \overline{A \cup B}$, and combine these to give $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$ and we have both inclusions and equality.


问:
First order logic reference request
I need a textbook (or any other material) about first order logic, that includes these precise parts: $LK$ sequent calculus, and substitution. I already searched on Shoenfield and Mendelson and I didn't found anything too similar to what we are doing in our course, so I would like to know if there is other valid material.

答:
See:

G.Takeuti, Proof Theory,

A.Troelstra & H.Schwichtenberg, Basic Proof Theory, as well as

S.Negri & J.von Plato, Structural Proof Theory.

You can see also F.Pfenning's Lecture Notes on Sequent Calculus.

Title: First order logic reference request


答:
The other answers give fine examples, but I think it is worth to first point out that "an affine subset" of a scheme does not truly make sense. A subscheme can be affine or not, but a subset of a scheme does not in general have a natural structure of subscheme: it can have none, and it can have several.

It turns out that if the subset is open, then it always has a canonical structure of subscheme, so for open subsets it makes sense to ask if it is affine or not.

Now to highlight how far affine subschemes are from being open: if $X$ is an affine scheme, every closed subscheme is affine. So an affine subscheme not being open is not a weird phenomenon that you can find examples of if you look hard enough: it is completely the norm, and is a fundamental part of the theory.

Title: Can there be non open affine subset in a scheme?

Body: Now this may sound like a stupid question but I am studying schemes' theory and open affine subset of a scheme arise naturally in many statements. In many cases I wonder if the request on openness of an affine subset of a scheme in the hypothesis of a theorem is really necessary or if any affine subset of a scheme is necessarily open.</p>


问:
Why does the definition of topological space require that the empty set be open?
I'm taking my first course in general topology, and so obviously I came across the definition of topological space $(T, \tau)$:

$\tau \in \mathcal P \mathcal P (T)$

$\emptyset, T \in \tau$

$(A_i)_{i \in I} \in \tau \implies \bigcup_{i \in I} A_i \in \tau$

$A_1, \dots, A_n \in \tau \implies \bigcap_{i=1}^n A_i \in \tau$

This is very abstract, so I tried to find a few examples, and these examples I found caught my eye:

$T = \mathbb R^n$, $\tau = \{ \text{every set whose complement is compact} \}$

$T = [0,1]$ $\tau = \{ \text{every set whose complement is a finite set}  \}$

Now these satisfy every part of the definition, except that one has to add the empty set to $\tau$ explicitly as well. This gives the natural question: why is this needed?

答:
You could do it without requiring $\emptyset \in \tau$. But if you do not want to exclude more or less all interesting spaces (like the reals with its standard topology), then the standard intersection axiom $(I)$ must be replaced by

$(I')$ $\phantom{x}$ If $A_1, \dots, A_n \in \tau$, then $\bigcap_{i=1}^n A_i \in \tau$ or $\bigcap_{i=1}^n A_i = \emptyset$.

Anyway, the system $\tau' = \tau \cup \{ \emptyset\}$ is again a topology in your modified sense, and it satisfies the intersection axiom in its original form. This is a much more elegant formulation which eliminates the special status of $\emptyset$ in the intersection axiom. Moreover, as the other answers have shown, the absence of $\emptyset$ as an open set would require to add another unnecessary special case to the definition of a continuous function.

Also note that if we work with $\emptyset \in \tau$, then the union axiom is satisfied also for the empty index set. This is just a very formal reason, but again it allows to avoid to consider an unnecessary special case.

Similarly you could omit the requirement $X \in \tau$. In that case you should replace the standard union axiom $(U)$ by the weaker version

$(U')$ $\phantom{x}$ If $(A_i)_{i \in I} \in \tau$, then $\bigcup_{i \in I} A_i \in \tau$ or $\bigcup_{i \in I} A_i = X$.

Once more there is no benefit in doing so. The system $\tau' = \tau \cup \{ X \}$ is again a topology in the modified sense, and it satisfies the union axiom in its original form without a special treatment of $X$. And the absence of $X$ as an open set would require to add another unnecessary special case to the definition of a continuous function. In fact, $f^{-1}(V) = X$ is possible for a continuous function $f : X \to Y$ and open $V \subsetneqq Y$. Finally, if we work with $X \in \tau$, then the intersection axiom is satisfied also for the empty index set.

Let us call a system $\tau \subset \mathcal P (X)$ a quasi-topology if it satisfies $(I')$ and $(U')$. We do require neither $\emptyset \in \tau$  nor $X\in \tau$. The following facts show that it does not make much difference to work with topologies or with quasi-topologies.

Each topology is a quasi-topology.

For each topology $\tau$, the "reduced set" $r(\tau)  = \tau \setminus \{ \emptyset, X\}$ is a  quasi-topology contained in $\tau$. Other quasi-topologies contained in $\tau$ are $r_U(\tau) = \tau \setminus \{ \emptyset\}$ (which satisfies $(U)$) and $r_I(\tau) = \tau \setminus \{ X \}$ (which satisfies $(I)$).

For each quasi-topology $\tau'$, the "extended  set" $e(\tau')  = \tau' \cup \{ \emptyset, X\}$ is a  topology.

One has $e(r(\tau)) = \tau$ and $r(e(\tau')) \subset \tau'$. It is possible that $r(e(\tau')) \ne \tau'$.


问:
Computational complexity of least square regression operation
In a least square regression algorithm, I have to do the following operations to compute regression coefficients:

Matrix multiplication, complexity: $O(C^2N)$

Matrix inversion, complexity: $O(C^3)$

Matrix multiplication, complexity: $O(C^2N)$

Matrix multiplication, complexity: $O(CN)$

where, N are the training examples and C is total number of features/variables.

How can I determine the overall computational complexity of this algorithm?

EDIT:
I studied least square regression from the book Introduction to Data Mining by Pang Ning Tan. The explanation about linear least square regression is available in the appendix, where a solution by the use of normal equation is provided (something of the form $a=(X^TX)^{-1}X^Ty)$, which involves 3 matrix multiplications and 1 matrix inversion).

My goal is to determine the overall computational complexity of the algorithm.
Above, I have listed the 4 operations needed to compute the regression coefficients with their own complexity. Based on this information, can we determine the overall complexity of the algorithm?

Thanks!

答: In this work https://www.research-collection.ethz.ch/bitstream/handle/20.500.11850/153646/eth-6011-01.pdf?sequence=1&isAllowed=y two implementation possibilities (the Gaussian elimination alternative vs. using the QR decomposition) are discussed in pages 32 and 33 if you are interested in the actual cost DFLOP-wise.</p>


问:
Definitions of well ordered set, maximal element and upper bound
I am currently studying equivalents of the Axiom of choice such as the well ordering theorem and the Zorn's lemma. I understand partially and linearly ordered set is a primitive definition in both definitions.

So this are the definitons i have got so far:

A  partial ordering on a non-empty set $S$ is a binary relation $R$ such that for all  $ a, b , c \in S$ satisfies:

$R$ is reflexive, i.e.,  $aRa$.

$R$ is antisymmetric, i.e.,  if $aRb$ and $bRa$ then $ a = b $.

$R$ is transitive, i.e., if $aRb$ and $bRc$ then $aRc$.

A partially ordered set, $(S,R)$ is a linear ordered set or a  totally ordered set if any two elements in $S $ is comparable

A linearly ordered set $(S,\leq)$ is a well ordered set, if for every $A \subseteq S$, there exists an element $s_0$ where $s_0 < s \forall s \in T$.

From my understanding when we consider a partially ordered set or a linealy ordered set the relation it's defined under can be any binary relation such as addition, subset etc. However, when considering a well ordered set we only consider the partially ordered set (S,<), i.e. the set under the less than ordering.

This is also the case when we look at upper bound and maximal element. The partially ordered set we consider is the set defiend under the less than relation

Is this right?

Thank you in advance.

答:
You appear to believe that $\le$ has some specific meaning - perhaps you're thinking that $a \le b$ if and only if $b - a$ is nonnegative, for example.

On the contrary, $\le$ is used as a symbol to indicate any relation that is intended to be interpreted as some kind of ordering. The meaning of $\le$ will be defined before it is used. It need not be a relation on $\mathbb{N}$ or $\mathbb{R}$; I could define the relation-class $\le$ on the class of all sets by $X \le Y$ if and only if $X$ injects into $Y$, for example. Or I could define $\le$ on the genuine set $\{\text{the ordinals less than $\alpha$}\}$ by $\beta \le \gamma$ iff $\beta$ is isomorphic to an initial segment of $\gamma$.


问:
curves and surfaces. curvature of a regular curve
Let $\gamma(t)$ be a regular curve lies on a sphere $S^2$ with center $(0, 0, 0)$ (origin) and
radius $r$. Show that the curvature of $\gamma$ is non-zero, i.e., $κ \ne 0$.

Furthermore, if the torsion of the curve $\tau \ne 0$ we have:

$\gamma(t)= - p \overrightarrow{n} - p'\alpha \overrightarrow{b} $

where: $p=1/k, \\ \alpha=1/\tau, \\ r^2=p^2+(p'\alpha)^2  $

答: This is a standard sort of problem in differential geometry. Assume $\gamma$ is arclength parametrized and write $\gamma = aT+bN+cB$ where $T,N,B$ is the Frenet frame, and $a,b,c$ are smooth. What does the fact that $\gamma$ lies on $S^2$ tell you?</p>


问:
How to find range of values for where line cuts circle?
I am doing a question on circles. The question is:

The circle  has equation $x^2 + y^2 - 6x + 10y + 9 = 0
$
The line with equation $y= kx$ , where  is a constant, cuts  at two distinct points.
Find the range of values for k

My working so far:

Found centre: (3, -5) radius = 5

Now what do I do after this?

答:
Road map.

0)Draw the problem; By inspection $k=0$ i.e. $y=0$ is a tangent.

Limits of $k$:

1)Tangential distance from line to point (center of circle): $5$

$d=\dfrac{|Ax_{0}+By_{0}+C|}{\sqrt{A^2+B^2}}$ where

$Ax+By+C=0$ is your line, and $(x_0,y_0)$ is the point.

2)This is a quadratic for $k;$ find  $k$. These are the limiting $k's.$

3)Determine the range of $k.$


问:
How to write a Circle Equation?
How can we write a circle equation (in 3D) with the know values of circle radius, center position in 3D, and the normal vector that is perpendicular to the plane that circle is lying on? Thanks in advance.</p>

答:
$$((\vec p-\vec c)^2-r^2)^2+(\vec n(\vec p-\vec c))^2=0$$

or

$$\vec p=\vec a\cos\theta+\vec b\sin\theta+\vec c$$

where

$$\vec a=r\frac{\vec n\times\vec m}{\|\vec n\times\vec m\|},\vec b=r\frac{\vec a\times\vec n}{\|\vec a\times\vec n\|}$$ and $\vec m$ is an arbitrary vector not parallel to $\vec n$.

Title: How to write a Circle Equation?


问:
Sum of two coprime squares
In this post it is stated that

A positive integer $n$ can be written primitively as the sum of two squares, meaning $n = x^2 + y^2$ with $\gcd(x,y)=1,$ precisely when $n$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4.$

I am looking for a proof of this statement.

What I have done: I can show that, if $n=x^2+y^2$ with $\gcd(x,y)=1$, any prime factor $p$ of $n$ is congruent to $1\pmod{4}$. But, I am stuck on how to prove that $n$ is also not divisible by $4$ in this case, in addition to the converse implication. Any help would be appreciated.

答: If both $x,y$ are odd, then $x^2+y^2\equiv 2\pmod4$, so if $4\,|\,x^2+y^2$, then both $x,y$ have to be even.

Title: Sum of two coprime squares


问:
Function f(x) is continuous on the interval [0,2c]. If f(c) > f(0) and f(c) > f(2c), show that there is a ∈ [0,c] such that f(c+a) = f(a).
I need help with this question. Is there something to do with Intermediate Value Theorem? Not sure how to approach this. Appreciate any help, thanks!

Assume $c>0$ is a constant and the function $f(x)$ is continuous on the interval $[0,2c]$. If $f(c) > f(0)$ and $f(c) > f(2c)$, show that there is $a ∈ [0,c]$ such that $$f(c+a) = f(a)$$.

答:
Consider the function $g(x):=f(x+c)-f(x)$ for $x \in [0,c]$

Show that $g(0)>0$ and $g(c)<0.$ Can you proceed ?


答:
Is there something to do with Intermediate Value Theorem?

Yes, very much.

Take a look at

$$f(c+a) = f(a)\\f(c+a) - f(a) = 0$$
Can you make something intermediate-theoremy out of that?

Title: Function f(x) is continuous on the interval [0,2c]. If f(c) > f(0) and f(c) > f(2c), show that there is a ∈ [0,c] such that f(c+a) = f(a).

Body:
I need help with this question. Is there something to do with Intermediate Value Theorem? Not sure how to approach this. Appreciate any help, thanks!

Assume $c>0$ is a constant and the function $f(x)$ is continuous on the interval $[0,2c]$. If $f(c) > f(0)$ and $f(c) > f(2c)$, show that there is $a ∈ [0,c]$ such that $$f(c+a) = f(a)$$.


问:
Estimation in uniform distribution
Is there a case when uniform distribution posses unbiased and umvue estimate of $\theta$?

Suppose
$X_1, X_2, \dots, X_{45}$ ~ uniform on interval $[\theta-1/2\ , \theta+1/2]$

My views: I know $\max(X_1, X_2, \dots, X_{45})$ and $\min(X_1, X_2, \dots, X_{45})$ are mle.

2 sample mean is unbiased in case when lower limit is $0$ for uniform distribution.

What gave me doubt is.. this is a symmetric distribution, so what property does sample mean and median has for this distribution over this limits?

Edit : Adding more details,
My problem is I want to know if sample mean and sample median are unbiased for this particular distribution.? My intuition says yes, but I couldn’t connect it with the theory.

答:
Actually the likelihood is

$$L(\theta)=\mathbb{1}_{[x_{(n)}-0.5;x_{(1)}0.5]}(\theta)$$

Thus any value in the interval is MLE. As an example you can take the midrange

$$T_1=\frac{x_{(1)}+x_{(n)}}{2}$$

as a MLE.

Also $T_2=(X_{(1)},X_{(n)})$ is sufficient, minimal but the problem is that this estimator is NOT complete.

In fact the given density belongs to the Location Family and $(x_{(n)}-x_{(1)})$ is location invariant too, thus it is also ancillary and thus

$$\mathbb{E}[X_{(n)}-X_{(1)}]=C$$

where C does not depend on $\theta$ which implies that, $\forall \theta$

$$\mathbb{E}[X_{(n)}-X_{(1)}-C]=0$$

and clearly, $\forall \theta$

$$\mathbb{P}[X_{(n)}-X_{(1)}-C=0] \neq 1$$

With these preliminary statements, perhaps it is better that you add all the details of your problem.

Title: Estimation in uniform distribution


问:
Is $\sqrt{i \cdot \frac{1+z}{1-z}}$ in the Hardy space $H^1(\mathbb{D})$
I am trying to prove that
$$\sqrt{i \cdot \frac{1+z}{1-z}}$$ is in the Hardy space $H^1(\mathbb{D})$, where we take the principal branch of the square root. I think that this is true but I am stuck at the moment.

Thanks for the help!

Body:
Sure, let $f(z) = \frac{(1+z)^{1/2}}{(1-z)^{1/2}}$ analytic on $|z|<1$ and continuous on $|z|\le 1,z\ne 1$

as $t\to 0$, $f(e^{it}) \sim C t^{-1/2}$

whence $f\in L^1(|z|=1)$.


问:
If the first three columns of matrix A are linearly dependent, then are the first three column of a matrix AB also linearly dependent?
Full Question

If the first three columns of matrix A are linearly dependent, then are the first three column of a matrix AB also linearly dependent for any matrix B (such that AB is defined)?

AB refers to matrix multiplication.

This seems to be true for me intuitively but no matter how I think I can approach the "first three columns" part of the argument.

I also tried the contrapositive: If matrix AB first three columns are linearly independent are the first three rows of A (and presumably also B) linearly independent.

答:
The result is not correct, Consider the following

$A$ = $\begin{bmatrix}1 & 2 & 3 &1 & 0 & 0 \\2 &4 &6 & 0 &1 &0\\ 3 &6 &9 &0 &0 & 1\end{bmatrix}$

The first 3 columns are linearly dependent

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

$AB$ = $I$ which is linearly independent


问:
Prove that certain elements are not in some ideal
I have the following question:

Is there a simple way to prove that

$x+1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$ and

$x-1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$

without using the fact that $\mathbb{Z}[x]/\langle2, x^2+1\rangle$ is an integral domain?

Thanks for the help.

答:
First a comment: it looks like the goal of the problem is to show that $(2,x^2+1)$ isn't a prime ideal (and hence the quotient isn't a domain), because $(x+1)^2=x^2+1+2x$ is in the ideal, and we hope to show that $x+1$ isn't in the ideal.

If $x+1\in (2,x^2+1)$, then we would be able to find two polynomimals in $\Bbb Z[x]$, say $a$ and $b$, such that $x+1=2a+(x^2+1)b$.

Looking at the equation mod 2, you get $x+1=(x^2+1)\overline{b}$, where all the terms are in $\Bbb F_2[x]$. But since $\Bbb F_2[x]$ is a domain, the degrees on both sides have to match. If $\deg(b)>0$, then the degree of the right hand side would be at least 3, and even if the degree of $b$ were 0, the right hand side would have degree 2. It is impossible then, for such an expression to be equal to $x+1$. By this contradiction, we conclude $x+1$ is not in the ideal.

Finally, you can note that $x+1$ is in the ideal iff $x-1$ is, since $x-1+2=x+1$.

Thus we have shown that $(2,x^2+1)$ is not a prime ideal, and it isn't even a semiprime ideal.


问:
Intuitive way to eliminate a parameter
I need to eliminate $\theta$ from the equations $x=\sin\theta+\cos\theta$ and $y=\tan\theta+\cot\theta$. I am actually provide with a hint: consider $x^2y$ , which worked nicely for me. However I am wondering if it was possible to "see" this solution without a hint. I was very nice when $\tan\theta+\cot\theta$ popped back out and I was able to let that simply equal to $y$, but What if the hint wasn't given?

Unless this is simply one of those "know the famous first step" questions

答:
You have $\tan \theta$ and $\cot\theta$ and they're in addition. So first express them in terms of $\sin$ and $\cos$.

$y =  \tan \theta + \cot\theta = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} = \dfrac{1}{\sin\theta \cos\theta} $

Now we need, $\sin\theta \cos\theta$ to be eliminated.

Remember $(a+b)^2 = a^2+2\color{red}{ab} + b^2$ and luckily here, $a^2 + b^2 = \sin^2\theta + \cos^2\theta$ would yield $1$.

So, $\boxed{x^2 = 1+ 2\sin\theta \cos\theta = 1+ \dfrac{2}{y}} \Rightarrow \boxed{x^2y = y+2}$

Title: Intuitive way to eliminate a parameter


答:
You could say $$y=\tan \theta+\cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta+\cos^2 \theta}{\sin \theta \cos \theta} =\frac{1}{\sin \theta \cos \theta}$$

So we want the other equation to give something related to $\sin \theta \cos \theta$, and we know $\sin^2 \theta+\cos^2 \theta=1$ so squaring looks useful.  We get $$x^2 = (\sin \theta +\cos \theta)^2 =  \sin^2 \theta + 2\sin \theta \cos \theta +\cos^2 \theta =1 + 2\sin \theta \cos \theta$$

and thus $$x^2 = 1+\frac{2}{y}$$ which you can tidy up to $x^2y = y +2$

Title: Intuitive way to eliminate a parameter

Body:
I need to eliminate $\theta$ from the equations $x=\sin\theta+\cos\theta$ and $y=\tan\theta+\cot\theta$. I am actually provide with a hint: consider $x^2y$ , which worked nicely for me. However I am wondering if it was possible to "see" this solution without a hint. I was very nice when $\tan\theta+\cot\theta$ popped back out and I was able to let that simply equal to $y$, but What if the hint wasn't given?

Unless this is simply one of those "know the famous first step" questions


答:
There is an important way to view topology in which

an open set is  a property that can be recognized with a finite amount of information.

In the usual topology on $\Bbb R$, membership in the open set $(2,4)$ is such a property. Given any element $x\in(2,4)$, we can recognize this fact by examining only a finite number of digits of $x$.  In contrast. membership in the non-open set $[2,4]$ is not one of these properties, because if we happen to be given  $x=4$, we cannot recognize it.  No matter how many digits of $4.000\ldots$ we examine, we can never be sure that there won't be some nonzero digit later that will prove that $x$ is not in the closed set $[2,4]$.  (A similar problem occurs with $x=2$, if it is presented to us in the form $1.999\ldots$.)

Viewed in this way, the familiar union and intersection properties appear naturally:

Suppose $G_1$ and $G_2$ are open sets.  Given $x\in G_1$, we can recognize this fact using only finite information from $x$, and similarly for $x\in G_2$. Can we also recognize that $x\in G_1\cap G_2$ using only a finite amount of information about $x$?  Clearly we can: just check both properties, using a finite amount of information for each.

Suppose membership in each member of some family  $G_i$ can be recognized with only a finite amount of information.  Can membership in $\bigcup G_i$ be recognized?  Yes, because if $x\in\bigcup G_i$ then it is in $G_k$ for some particular $k$, this fact can be recognized with finite information, and that amount, plus the information about $k$ itself, suffices for the $x\in\bigcup G_i$ question.

Suppose membership in each member of some family  $G_i$ can be recognized with only a finite amount of information.  Can membership in $\bigcap G_i$ be recognized?  Not necessarily.  Even if the amount of information required for each individual  $G_i$ is finite, the total amount of information required to place $x$ in all of the $G_i$ might be infinite.

The other example you invoked (the cofinite topology) can also be understood in this way.

Regardless of what the elements of the topology are or what sorts of questions we can answer about them, there are always two trivial properties that are recognizable. There are the property that is always true and the property that is always false.  These properties correspond to the entire space and to the empty  set.


问:
Regularity of distributions with $L^2$ curl or divergence
Let $\Omega$ be a domain in $\mathbb R^n$. A classical result tells us that if a distribution $T \in \mathcal{D}'(\Omega)$ has partial derivatives (say gradient) in $L^2(\Omega)$ then $T \in L^2_{loc}(\mathbb R^n)$. Is there any known result for distributions satisfying $\operatorname{div} T \in L^2(\Omega)$ or $\operatorname{curl} T \in L^2(\Omega)^3$?

If the answer is affirmative, I expect the conclusion to be be weaker (e.g. $T \in L^1_{loc}$) since here we assume summability only for a linear combination of derivatives.

Thank you in advance for any kind of reference or insight.

EDIT

Recall that if $T \in \mathcal{D}'(\Omega; \mathbb R^3)$, then its distributional curl and divergence are defined as:

$$ \langle \operatorname{div} T, \phi \rangle := - \langle T, \nabla \phi  \rangle, \ \phi \in \mathcal{D}(\Omega), \quad \text{and } \quad \langle \operatorname{curl} T, \psi \rangle := \langle T, \operatorname{curl} \psi \rangle, \ \psi \in \mathcal{D}(\Omega; \mathbb R^3). $$

答:
Since I came up with a solution, I will answer my own question.

As suggested in comments, distributions with integrable curl or divergence can be in fact irregular. To see this (for the divergence), define:

$$ T : \mathcal{D}(\mathbb{R}^3; \mathbb{R}^3) \rightarrow \mathbb{R}, \ \ \phi \mapsto \sum_{i=1}^3 \delta_{\mathbf{0}}(\phi_i); \qquad H := \operatorname{curl} T \in \mathcal{D}'(\mathbb{R}^3; \mathbb{R}^3). $$

Then $\operatorname{div} H = \operatorname{div} \operatorname{curl} T = 0 \in L^2 $, on the other hand $H$ is not represented by any $L^1_{loc}$ since an explicit computation shows that:

$$ \langle H, \psi \rangle = \langle \operatorname{curl} T, \psi \rangle = \sum_{i=1}^3 (\operatorname{curl} \psi)_i (\mathbf{0}) $$

for $\psi \in \mathcal{D}(\mathbb{R}^3; \mathbb{R}^3) $, which is clearly irregular just like a Dirac mass.


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

答: (Intuitive approach with an hidden use of AM-GM)If $n$ is even you can put togheter couples which are equidistant from $n/2$ and then $x_i/x_{n+1-i}+x_{n+1-i}/x_i \geq2$ (and you have $n/2$ of such couples); if n is odd it's almost the same since you'll have a central 1.</p>


问:
Relation between $\mathbb{C}$-automorphisms and Möbius transformations.
Today in my Algebraic Ecuations course, the proffesor mentioned that while there is only one field automorphism (the trivial one) in $\mathbb{R}$, there are infinite field automorphisms in $\mathbb{C}$. I was curious, is this somewhat related to Möbius transformations? (also known as linear transformations, ie $T(z)=\frac{az+b}{cz+d}$ where $a,b,c,d,z\in \mathbb{C}$). Intuitively, I think this as composing in some way with the stereographic projection in the same way one can do it when talking about the fundamental group of the circunference. Thanks in advance!

答:
That is not true. Consider the conjugation map $\phi:\Bbb C \to \Bbb C$ given by $\phi(z) = \bar{z}$.

One can see that this is indeed a field homomorphism and is clearly bijective since $\phi^2$ is identity. However, $\phi$ is not holomorphic.

On the other hand, all the Möbius transformations are holomorphic and thus, composing them with the stereographic projection also won't give you conjugation.


问:
If $W$,$V$ are independent, geometric random variables, what is the distribution of $Z \sim W + V$?
Question: If $W$,$V$ are independent geometric random variables, what is the distribution of  $Z \sim W + V$?

Please note, $W \sim Geometric(3/5)$ and $V \sim Geometric(7/12)$.

Also,

$P(W=w) = (1-\frac{3}{5})^{w}\frac{3}{5}$ for $w = 0,1,2, \dots$

$P(V=v) = (1-\frac{7}{12})^{v}\frac{7}{12}$ for $v = 0,1,2, \dots$

What I have so far: I have looked around already on  mathexchange and usually the geometric pdf is different to the one I am defining (with $1-p$ to the power of $w-1$ instead of just $w$). I know to use convolution however I am unsure of how to simplify the sum, I am also unsure of if my limits are wrong.

My working:

$P(W+V=z)= \sum_{w=0}^z P(W=w)P(V=z-w) $

$= \frac{3}{5} \frac{7}{12}  \sum_{w=0}^z (\frac{2}{5})^{w}(\frac{5}{12})^{z-w}$

$= \frac{3}{5} \frac{7}{12}  \sum_{w=0}^z (\frac{24}{25})^{w}(\frac{5}{12})^{z}$

$= \frac{3}{5} \frac{7}{12} (\frac{5}{12})^{z} \sum_{w=0}^z (\frac{24}{25})^{w}$

My biggest question is if my limits on the sum are correct and where do I go from here?

答:
Hint:

The sum of a geometric series is $$\sum_{i=0}^{n} a^i = \frac{a^{n+1}-1}{a-1}$$.

You can use this on the step where you're stuck.


问:
Prove that $P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$
Let $(Ω,S,P)$ be a probability space. Let $A, B, C ∈ S$ with $P(B)$ and $P(C) > 0$. If $B$ and
$C$ are independent show that
$$P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$

$\textbf{My Solution:}$

[Note: All the arrows used in the solution represent the direction of flow of inference in a graphical model.]

[Note: I have used standard notation which are generally in the literature of probabilistic graphical models. Take a look. https://ermongroup.github.io/cs228-notes/representation/directed/]

It is given that A depends on B. So we can start with the fact that $B \rightarrow A$.

Take RHS
$$P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B,C)}P(C) + \frac{P(A,B,\overline C)}{P(B,\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)P(C)}P(C) + \frac{P(A,B,\overline C)}{P(B)P(\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}\ \ \ \text{(eq. 1)}$$

$\textbf{Case I}: A \perp C$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(A,B)P(C)}{P(B)} + \frac{P(A,B)P(\overline C)}{P(B)}$$
$$= \frac{P(A,B)}{P(B)}$$
$$= P(A|B) = LHS$$

$\textbf{Case II}: C\leftarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(A|B)P(C|A)}{P(B)} + \frac{P(B)P(A|B)P(\overline C|A)}{P(B)}$$
$$= P(A|B)P(C|A) + P(A|B)P(\overline C |A)$$
$$= P(A|B) = LHS$$

$\textbf{Case III}: C\rightarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(C)P(A|B,C)}{P(B)} + \frac{P(B)P(\overline C)P(A|B,\overline C)}{P(B)}$$
$$= P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
How to proceed from here?

答:
I send this question to my instructor, he answered. The answer is as follow:

$\textbf{Case III}:\ C \rightarrow A$

from (eq. 1)

$$={P(A,B,C) \over P(B)} + {P(A,B,\overline C) \over P(B)}$$
$$={P(A,B,C)P(C) \over P(B)P(C)} + {P(A,B,\overline C)P(\overline C) \over P(B)P(\overline C)}$$
$$={P(A,B,C) P(C) \over P(B,C)} + {P(A,B,\overline C) P(\overline C)\over P(B, \overline C)}$$
$$=P(A| B,C) P(C) + P(A|B,\overline C) P(\overline C)$$
$$=P(A| B) P(C) + P(A|B) P(\overline C)$$
$$=P(A| B) = LHS$$


问:
Solve equation $2^{x} - 2^{-x} = 2*\cos(x/5)$
I'm trying to solve it by log two parts:

$$
\log_2(2^{x}/2^{-x}) = \log_2(2*\cos(x/5))
$$
Simplifying:
$$
2x - 1 = \log_2(\cos(x/5))
$$

And I'm stuck here, I think we could solve it by investigating functions(linear and logarithmic) and trying to find border value. Could you give the right direction?

答:
We have the equation
$$2^x+2^{-x}=2\cos\frac{x}{5}$$
and we want to either find the sum of the solutions or the solutions themselves. If we are only concerned with real solutions, then due to your excellent use of inequalities we have the only solution: $x=0$, and we are answering the second problem.

Alternatively, if the question also wants you to consider complex roots, then the second problem is not applicable, as there are multiple solutions, and in fact $0$ is the answer to the first problem. I will demonstrate how to find all these complex solutions below, and also very simply why $0$ is the answer to the first problem.

To find the solutions to this equation:

First observe that $2^x=e^{x\ln2}$ and similarly, we have $2^{-x}=e^{-x\ln2}$. Hence our equation is equivalent to
$$2\left(\frac{e^{x\ln2}+e^{-x\ln2}}{2}\right)=2\cos\frac{x}{5}$$
But if you recall the definition of $\cosh A$ then you should see that actually our equation is equivalent to
$$2\cosh(x\ln2)=2\cos\frac{x}{5}\iff \cosh(x\ln2)=\cos\frac{x}{5}$$
However, we also have a useful identity $$\cosh(A)\equiv \cos(iA)$$
Thus, we actually have
$$\cos{ix\ln2}=\cos\frac{x}{5}\implies ix\ln2=\pm\frac{x}{5}+2n\pi\implies x=\frac{2n\pi}{i\ln2\mp 0.2}=\frac{10n\pi}{5i\ln 2\mp1}$$
for integer $n$.

To answer the question:

Note that if $\alpha$ is a solution of the equation, then so is $-\alpha$. Therefore there must be an even number of roots (in fact there are infinitely many complex roots, but since every roots is paired up with another one which is equal to the negative of that root, loosely speaking we have an even number of roots). So if there are $2n$ roots,
$$\alpha_1,-\alpha_1,\ldots,\alpha_n,-\alpha_n$$ then the sum of the roots is
$$\alpha_1+(-\alpha_1)+\cdots+\alpha_n+(-\alpha_n)=0$$
as required.


问:
How to apply pigeonhole principle to this problem
Let $S$ be any set, How large does $S$ need to be to guarantee that any function $f:S\to [100]=\{1,2,\cdots,100\}$ has the property that there is some $i\in [100]$ that gets associated with $i$ elements of $S$?

答:
For the case where there must be exactly $i$ elements that map to $i$, there is no maximum, because you can simply have a million elements (or an arbitrarily large number) map to each.

For the case where there must be at least $I$ elements that map to $I$, the simplest way to think of these problems is to imagine avoiding the outcome for as long as possible. You can't have 1 element map to 1, so that's zero.  You can't have 2 elements map to 2, so you have at most 1 of those.  You can't have 3 elements map to 3, so that's 2 elements, and so on.  So you can have $$\sum_{0}^{99} i = 99(100)/2 = 4950 $$ without satisfying the condition, and your answer is $4951$.


问:
Is the space of matrices with rank less than 2 a subspace?
Let $W = \{A ∈ M_{2\times2}(\mathbb{R}) : \mathrm{rank}(A) < 2\}$.

Is $W$ a subspace of $M_{2\times 2}(\mathbb{R})$?

答:
No. Consider $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

Then $A,B\in W$ but $A+B=I\not\in W$ since $\mathrm{rank}(I)=2$.


问:
How to find a power series representation for a divergent product?
Euler used the identity $$ \frac{ \sin(x) }{x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2 \pi^2 } \right) = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n + 1)! } x^{2n} $$ to solve the Basel problem. The product is obtained by noting that the sine function is 'just' an infinite polynomial, which can be rewritten as the product of its zeroes. The sum is found by writing down the taylor series expansion of the sine function and dividing by $x$.

Now, I am interested in finding the sum representation of the following product: $$ \prod_{n=1}^{\infty} \left(1 - \frac{x}{n \pi} \right) ,$$ which is divergent (see
this article).

The infinite sum representation of this product is not as easily found (at least not by me) because it does not have an obvious formal representation like $\frac{\sin(x)}{x}$ above.

Questions: what is the infinite sum  representation of the second product I mentioned? How does one obtain this sum? And is there any 'formal' represenation for these formulae (like  $\frac{\sin(x)}{x}$ above).

答:
I never gave the full answer :)

$$\prod_{n=1}^{\infty} (1-x/n)$$
When analysing a product it's often easiest to consider the form $\prod_{n=1}^{\infty} (1+f(n))$ given that $\sum_{n=1}^{\infty}f(n)^m=G(m)$; $f(n)=-x/n$ Then $G(m)=(-x)^m\zeta(m)$.

$$\prod_{n=1}^{\infty} (1-x/n)=e^{\sum_{m=1}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}$$
Because $\zeta(1)$ is the only part of the product that goes to 0 (e.g. e^-infty), the regularized product will tend to 0 just as $\zeta(1)$ goes to infinity. However this is easy fixable:

$$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=e^{\sum_{m=2}^{\infty}\frac{(-1)^{m+1}x^m\sum_{n=1}^{\infty}f(n)^m}{m}}=e^{\sum_{m=1}^{\infty}\frac{- x^{m+1}\zeta(m+1)}{m+1}}$$ which does converge. To show the above more well known representation,

$$\sum_{n=1}^{\infty}-\frac{x^{n+1}\zeta(n+1)}{n+1}=\int_{0}^{-x}\sum_{n=1}^{\infty}(-1)^nz^n\zeta(n+1)dz=\int_{0}^{-x}-H_zdz=-\ln((-x)!)+x\gamma$$
$$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^
{x\gamma}}{(-x)!}$$
And now with the answer I posted 6 years ago, and using the refined stirling numbers:
$$\prod_{n=1}^{\infty} (1-x/n)e^{x/n}=\frac{e^
{x\gamma}}{(-x)!}=\bigg(1-x^2\zeta(2)/2-x^3 2\zeta(3)/6 +x^4 \big(3\zeta(2)^2-6\zeta(4)\big)/24+...  \bigg)$$

So we can also say with p goes to inf:
$$p^x\prod_{n=1}^{p} (1-x/n)=\frac{1}{(-x)!}=$$

We can rewrite our G(m) with $G(1)=-x\gamma$ and use this in the formula given above to write a polynomal representation, which is easier then multiply our previous polynome by $e^{-x\gamma}$.

$$p^x\prod_{n=1}^{p} (1-x/n)= \bigg(1+-x\gamma+x^2 \frac{\gamma^2-\zeta(2)}{2}+x^3 \frac{-\gamma^3+3\gamma\zeta(2)-2\zeta(3)}{3!}+...\big)=\frac{1}{(-x)!}$$

I find it hard to clearly explain these refined stirling numbers but if you are throwing in a bit of time, consider the following ways to represent the found product representation. It's going to be vague but i clarify it if you want.

The "nice" form is in knowing g(m) for all m. There are a lot of way to represent the refined stirling numbers, especially within this context. Lots of it is related to partitions and ways to "write" a number. e.g. for the x^4 term, $4=1+1+1+1$ so we get $g(1)^4*4!/4!/1^4$ and the next term is $1+1+2; G(1)^2 G(2) * 4!/1^2/2^1/2$ as coefficient. So we divide by the a*s^a if you have a  G(s)^a. another more intunitive way is to see it as the "unique" combination of all outcomes. Another way is to write it as these unique combination, but use sum of sums (particular cool! And very easy to image things). You can also achieve it algebraric with writing the e powers out, but that's a real hassle. And if you want another way to represent these refined stirling numbers, you can construct them by using previous found stirling numbers and binominals, which is the most efficient.

I always wondered why there was no wikipedia page about them.


问:
Does convexity of all projections imply convexity in higher dimensions?
If I have $n$ 2-dimensional convex "regions" that are the projections along $n$ (independent) dimensions of a n-dimensional compact subspace:

Does that imply that the latter is convex? Is the orthogonality of the dimensions needed?

What about the reciprocal? (I'm convinced it's true, but if you have a conter-example, shoot :)

答:
Consider the closed unit ball, and remove from it all points a distance at most $1/10$ from the open line segment between $(0,0,0)$ and $(1,1,1)$. This set is compact and simply-connected, but all its projections are convex, despite the set not being convex.

If a set is convex, then all its projections are convex. This follows from the definition of convexity. If $x,y$ are two points in the set, then all points on the line segment between $x$ and $y$ are also in the set. When we project the set onto some plane, then the line segment is projected onto a new straight line segment from the projection of $x$ to the projection of $y$. Thus the projection of the set is convex.


问:
Understanding vital part of the proof that a closed interval is connected.
Part of the proof that a closed interval in $ℝ$ is connected usually relies on the below fact.

Given that $W \neq \varnothing$ is closed and open in a closed interval.

There exists such $x_0$ and $\epsilon > 0$ that $(x_0-\epsilon, x_0 + \epsilon) \subset W$. We then define $r$ as:
$$
r = \sup \{ p: [x_0, x_0 + p) \subset W \}
$$

Then it is stated that since $W$ is closed, then it follows that $x_0 + r \in W$.

I can't understand why the latter holds. I know it has to do with the fact that there is a sequence from $W$ and since it's closed then the limit of that sequence, which is $sup$ will belong to it.

I am confused, since the sequence is supposed to consist of $p$'s from above definition but then it is stated that $x_0 + r$ belongs to $W$ and not just $r$.

Would appreciate any help on that. Thanks.

答: Let $O=(a,b)$ (or maybe $O=(a,1]$) be a basic neighbourhood of $x_0 + r$ in $[0,1]$, so at least we know $a < x_0 + r$ so $a- x_0 < r$. From the latter we know that $a-x_0$ cannot be an upper bound for $\{p: [x_0, x_0+p) \subseteq W\}$ (as sup is the least upper bound, and $a-x_0$ is smaller) so there is some $p' (\le r)$ such that $p' > a-x_0$ and $[x_0, x_0 + p') \subseteq W$, and so $[x_0, x_0 + p'] \subseteq \overline{W} = W$. So $a < x_0 + p'$ and so $x_0 + p' \in O$. So $O$ intersects $W$ in $x_0+p’$ and so as $O$ was an arbitrary basic neighbourhood of $x_0+ r$, $x_0 + r \in \overline{W}=W$ again using that $W$ is closed.