问:
Formal Definition of Summation over a Set
Formal definition of summation over a sequence is very clear. Suppose we have a sequence $\left\{a_{i}\right\}_{0}^{\infty}$. Then the summation of this sequence can be defined as a new sequence $\left\{ b_i \right\}_0^\infty$ in a recursive way:

$$b_0 = a_0,\ b_{i+1} = b_{i} + a_{i+1},\ i \geq 0.$$

Then we may write
$$\sum_{i=0}^n a_i = b_n,$$

and

$$\sum_{i=0}^\infty a_i = \lim_{i\to\infty} b_i.$$

However, sometimes we see summation over a set. For example, in defining the determinant, we use the following notation:

$$\det(A) = \sum_{\sigma\in S_n} \left(\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma_i}\right),$$

where $S_n$ is the set of all permutations of the sequence $\left\{1,2,\dots,n\right\}$. How are summation symbols over sets like $\sum_{\sigma\in S_n}$ formally defined?

答:
The set in your example is finite.

Suppose you have an infinite set $S$ and you want to understand $\sum_{i\in S} a_i.$

If $a_i\ge0$ for all $i$ then one can define this sum as $$\sup \left\{ \sum_{i\in S_0} a_i : S_0\subseteq S \text{ and } S_0 \text{ is finite} \right\}.$$

One can do a similar thing if every term is negative. Then if some are positive and some are negative, you have a well defined sum unless both sums are infinite.


问:
Let $X$ and $Y$ be subspaces of a vector space $V$. Show that $X\cup Y$ is a subspace if and only if $X\subset Y$ or $Y \subset X$
I'm currently learning linear algebra using textbook Linear Algebra Done Wrong, and I encountered this problem.

Let $X$ and $Y$ be subspaces of a vector space $V$. Show that $X\cup Y$ is a subspace if and only if $X\subset Y$ or $Y\subset X$

I don't really have any idea. I only figured out that 0 is in the set, so it is not empty. What else should I do? Thanks!!

答:
Elaborating on @Dave 's answer with slightly different approach:

Suppose $X\cup Y$ is a subspace ,Let $x ,y \in X \cup Y$ such that $x \in X$ and $y \in Y$. Then $x + y \in X \cup Y$ since it is a subspace. Now if $x + y \in X$ and we have $-x \in X$ we must have $ y = x + y - x \in X$ and by which we can say that $ Y \subseteq X$. Otherwise if $x + y \in Y$ similarly we can conclude that $x \in Y$ by which we can say $X \subseteq Y$

ContentLicense: CC BY-SA 4.0


问:
Locally convex implies convex?
There is a question about convex set.

Let $\Omega\subset\mathbb R^n$ be an open, connected set. If for any $x\in\overline{\Omega}$, there is a neighborhood $U_x$ such that $\Omega\cap U_x$ is convex, then $\Omega$ is convex.

Intuitively, if $\Omega$ is not convex, there must be a point on $\partial\Omega$ such that $\Omega$ is not convex locally. But I have no idea how to write it clearly. Thanks to your help.

答:
I will follow the proof of the theorem from the answer in this post. The accepted answer in the aforementioned post is much cleaner but relies on some ideas from abstract functional analysis to generalize the result to arbitrary topological vector spaces, which personally I find thoroughly depressing. The proof I present is more on the concrete, hard analysis side.

It suffices to show the result for bounded $\Omega$, since convex sets are closed under taking ascending chains (just intersect $\Omega$ with open balls of increasing radii). By the Lebesgue covering lemma, there exists $\epsilon > 0$ such that for every $x \in \overline\Omega$, there exists $y \in \overline\Omega$ such that $B_\epsilon (x) \subseteq U_y$, so in particular $B_\epsilon (x) \cap \Omega$ is convex. In other words, $\Omega$ is uniformly locally convex.

Recall open connected implies path connected and $\overline{\Omega \cap U_x}$ is convex, so $\overline \Omega$ is path connected (this is not true in general!). The path length function, given by
$$\text{length}(\gamma) = \sup_{\Pi} \sum_{i} |\gamma(t_i) - \gamma(t_{i + 1})|$$
where the supremum is taken over partitions $\Pi$ of whatever the domain of a continuous path $\gamma: [a, b] \to \overline \Omega$, is lower semicontinuous as the supremum of continuous functions.

Fix $x, y \in \overline\Omega$, then arguing by Arzela-Ascoli and arclength parametrization (cf. for example this post), there exists a path $\gamma : [0, 1] \to \overline\Omega$ of minimum length between $x$ and $y$. We claim that $\gamma$ is a straight line, which would prove convexity of $\overline\Omega$ and therefore its interior $\Omega$.

By uniform continuity, we can find $n \in \mathbb N$ sufficiently large such that
$$ |\gamma(i/2^n) - \gamma((i + 1)/2^n)| < \epsilon $$
for all $i = 0, \dots, 2^n - 1$. For brevity, denote $a_i = \gamma(i/2^n)$. We have $a_i \in \overline\Omega$ and $a_{i - 1} , a_{i + 1} \in B_\epsilon (a_i)$. Since $\overline \Omega \cap B_\epsilon (a_i)$ is convex (follow the usual proof of showing convex sets are closed under closure), there exists a line between $a_{i - 1}, a_{i + 1}$ in $\overline \Omega \cap B_\epsilon (a_i)$, call it $\Gamma_i$. But then we can parametrize
$$ \gamma([0, (i - 1)/2^n]) \cup \Gamma_i \cup \gamma([(i + 1)/2^n, 1]) $$
as a path from $x$ to $y$ in $\overline\Omega$ which has length
$$ \text{length} \ \gamma([0, (i - 1)/2^n]) + |a_{i + 1} - a_{i - 1}| + \text{length} \ \gamma([(i + 1)/2^n, 1]) \geq \text{length} (\gamma). $$
Rearranging,
$$ |a_{i + 1} - a_{i - 1}| \geq \text{length} \ \gamma([(i - 1)/2^n, (i + 1)/2^n]). $$
But of course, straight lines are the unique shortest paths in Euclidean space, so this implies the section of the path $\gamma([(i - 1)/2^n, (i + 1)/2^n])$ is straight. Arguing for each $i$ tells us $\gamma$ is in fact a straight line, completing the proof.

Edit (minor technicality): it is not entirely obvious that a rectifiable (i.e. finite length) path exists between any two points, but this follows from the uniform continuity + local convexity argument from the final part of the proof, which for arbitrary paths allows us to construct piecewise linear (often called "polygonal") paths in $\overline\Omega$ between two points, which clearly has finite length. The existence of this construction is often times convenient to use in complex analysis, and of course in this scenario.

Title: Locally convex implies convex?


问:
Where is the flaw in my logic in this collatz conjecture idea?
Using this article from math stack exchange ( open problem - What does proving the Collatz Conjecture entail? - Mathematics Stack Exchange fourth answer down that starts "I think, for the question"), we can see that the problem to figure out if there are no non-trivial cycles in the Collatz conjecture is solving $\frac{3^N-2^N}{2^S-3^N} = a$ for $ N,S,a \in Z$.

Since a must be an integer, we can add one to our fraction, and still have integers. We add $\frac{2^S-3^N}{2^S-3^N}$ and get $\frac{3^N-2^N+2^S-3^N}{2^S-3^N}$ which reduces to $\frac{2^S-2^N}{2^S-3^N}$. Since the numerator has to be even no matter what and the denominator has to be odd no matter what, the denominator must be 1 to get an integer for a. The only possible solution (using Catalan's conjecture - Catalan's conjecture - Wikipedia) is $S=2,N=1,a=2$. Since we added 1 to a earlier, this shows that the only possible cycle ending for the Collatz conjecture is 1.

I'm guessing there is a flaw here somewhere, but I figured I'd put it out there to find out what it is.

答:
Well, as @MiloBrandt has already mentioned, that formula was about the so-called 1-cycles, a special simple form of general cycles.

But your operation gives a nice aspect, which helps to make the disproof of 1-cycles a bit simpler.
We start with a supposed solution for an 1-cycle of length $N$ odd steps and $S$ as sum of the involved exponents of $2$:
$$ a = { 3^N - 2^N \over 2^S - 3^N } \tag {1.1}
$$
Now you add $1$ to each side
$$ a + 1 = { 3^N - 2^N \over 2^S - 3^N } +1 ={ 3^N - 2^N + 2^S-3^N \over 2^S - 3^N } ={ 2^S - 2^N  \over 2^S - 3^N } \tag {1.2}
$$
Now we can extract the factor $2^N$ on the rhs and divide:
$$ {a + 1 \over 2^N} = { 2^{A-1} - 1 \over 2^S - 3^N } \qquad \Tiny \text{ where } A-1=S-N \tag {1.3}
$$
Because the denominators have no common factor ,$a+1$ must be a $k$'th multiple of $2^N$ , so the initial (and smallest by definition) element of a 1-cycle must have the form $a=2^N\cdot k -1$ (with $k \gt 0$). So, for a 1-cycle of length $N=10$ the smallest element must be equal or larger than $1023$ (actually of the form $ 1023 + k \cdot 2^{10}$).

From another formula, i.e. that for the general cycle on elements $a_1,a_2,...a_N$ we must have
$$ 2^S = (3+1/a_1)(3+1/a_2) \cdots (3+1/a_N) \tag {2.1}$$
we can for each given $N$ determine some (roughly) mean value $\alpha$ of all $a_k$ by
$$ 2^{S/N} = (3+ 1/\alpha) \\
   \alpha = {1 \over 2^{S/N}-3  } \tag {2.2}$$
Here (by (2.1)) $2^S$must always be larger than $3^N$, and we take the smallest solution for $S$:
$$ S= \lceil N \cdot \log_2(3)  \rceil = 16 \tag {2.3}$$
This gives for the mean-value $\alpha = 1/(2^{1.6}-3) \approx 31.81 $ (!).

But by the definition of the cycle (be it general or 1-cycle) $a_1$ is the smallest element and thus it must also be $a_1 \lt \alpha$.

Since for the 1-cycle of length $10$ the smallest element $a_1\ge 1023$ is required, we cannot have a 1-cycle of length $N=10$.

This style of consideration can be generalized, and it can be shown that the discrepancy from the requirements for $a_1$ in a 1-cycle of length $N$ ($a_1=2^N k -1$) and the upper bound $\alpha$ for $a_1$ (by 2.3) for a general cycle of length $N$ even increases (heuristically) and this can then lead to the proof that no (nontrivial) 1-cycle can exist at all in the positive numbers.

(A bit more explanation I have in a small treatize 1cycledisproof )


问:
Quantization to minimize multiplicative error
Consider a random variable $X$ taking values in $[0, 1]$ with unknown distribution.

Consider the set $S_n = \left \{\frac{1}{n}, \frac{2}{n}, ..., \frac{n-1}{n}  \right \}$ to be used to quantize $X$. The following is the standard quantization function $f_n : [0, 1] \to S_n$

$$f_n(X) = \min_{s \in S} |X - s|$$

So the worst case quantization error we get decays as $O\left(\frac{1}{n} \right)$.

Suppose we measure multiplicative error as

$ e(x) = \frac{x}{f_n(x)} $

I think the worst possible value of $e(x)$ can be if $x = \frac{3}{2n}$ $\implies $ $f_n(x) = \frac{1}{n}$ so that

$$e\left( \frac{3}{2}n \right) \text{ is always }  = \frac{3}{2} \text{ no matter how large $n$ is } $$

But one usually expects that large $n$ leads to finer quantization leads to smaller error. Additive error does go down but multiplicative error remains the same.

Question: Is there a different quantization set $S_n$ or mapping $f_n : [0, 1] \to S_n$ which results in decaying multiplicative error as $n \to \infty$ ?

答:
As you have defined $e(x)$ it can range up to $\frac 32$ if $x$ is just smaller than $\frac 3{2n}$ because then $f_n(x)=\frac 1n$.  But it can range down to $0$ if $x$ is $0$ which is multiplicatively infinitely far away from $\frac 1n$.

You can't solve the problem with $0$ with your definition of error, but you can move more points of $S_n$ downward.  A natural approach is to make the ratio between successive elements of $S_n$ constant.  You still need to define what the smallest member is and there doesn't seem to be any natural way to choose it because $0$ is infinitely far away from the smallest element.  Take $n=2$ so $S_2=\{a^2,a\}$.  Should $a$ be $\frac 12, \frac 1{10}, \frac 1{10000}$ or what?


答:
Nice result. Note that this problem has many applications in the solution of problems in linear algebra.

Theorem: Let $V$ a vector space and let  $X$ and $Y$ vectorial subspaces of $V$, so $$X\cup Y\quad \text{vector subspace of V} \iff X \subseteq Y \quad \text{or} \quad Y\subseteq X$$

Proof:

$(\Rightarrow)$ Suppose that $X\cup Y$ is a vector subspace of $V$. Now, we can prove this by contradiction: Suppose that $X\not\subseteq Y$ and that $Y\not\subseteq X$. So, there exists $x \in X\setminus Y$ and $y\in Y\setminus X$. Then, $x+y\in X\cup Y$ because $X\cup Y$ is vector subspace of $V$. But, if $x+y\in X\cup Y$ so, $x+y \in X$ or $x+y\in Y$ by definition of union between sets. Now, suppose that $x+y\in Y$ so $x=(x+y)-y\in Y$ and this contradicts the choice of $X\in X\setminus Y$ and similarly if $x+y\in X$. So, in either case, we have reached a contradiction. Therefore, we can see that $X\subseteq Y$ or $Y\subseteq X$.

$(\Leftarrow )$ Suppose that $X\subseteq Y$ or $Y\subseteq X$, so if $X\subseteq Y=Y$ and by hypothesis $Y$ is a vector subspace. Similarly, if $Y\subseteq X=X$ and by hypothesis we have that $X$ is a vector subspace of $V$.

An aplication of the theorem:

Example: Let $V=\mathbb{R}^{2}$ a vector space and let $$H=\left\{\begin{pmatrix} x \\ y \end{pmatrix}\in \mathbb{R}^{2}: x+y=0\right\}$$
and $$W=\left\{\begin{pmatrix} x \\ y \end{pmatrix}\in \mathbb{R}^{2}: x-y=0\right\}$$
it's clear that $H$ and $W$ is a vector space. But, what about $H\cup W$?

So,  that statement gives us an operational way of studying whether the union between two vector subspaces is or is not a vector subspace. For this, the theorem tells us we must know:

A basis for $W$: $\beta_{W}$.

A basis for $H$: $\beta_{H}$.

The space $W$ with its respective conditional equations: $W=\left\{\begin{pmatrix} x \\ y \end{pmatrix}\in \mathbb{R}^{2}: x-y=0\right\}$.

The space $H$ with its respective conditional equations: $H=\left\{\begin{pmatrix} x \\ y \end{pmatrix}\in \mathbb{R}^{2}: x+y=0\right\}$.

By calculating a basis for $W$ and a basis for $H$, we can see that $$\beta_{H}=\left\{ \begin{pmatrix} -1 \\ 1 \end{pmatrix}\right\} \quad \text{and} \quad \beta_{W}=\left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}\right\} $$
basis for $H$ and $W$ respectively.

Now, using the theorem we have $$H\cup W \quad \text{s.v.e of V} \iff H\subseteq W \quad \text{or} \quad W \subseteq H$$

$W\subseteq H \iff x\in W  \implies x\in H$. But, since that $$x=\begin{pmatrix}1 \\ 1 \end{pmatrix} \implies 1+(1)=2\not=0 \implies x\notin H$$

$H\subseteq W \iff x\in H  \implies x\in W$. But, since that $$x=\begin{pmatrix}-1 \\ 1 \end{pmatrix} \implies -1-(1)=-2\not=0 \implies x\notin W$$

So, by the theorem we can conclude that $H\cup W$ is not vectorial subspace of $V=\mathbb{R}^{2}$.

Title: Let $X$ and $Y$ be subspaces of a vector space $V$. Show that $X\cup Y$ is a subspace if and only if $X\subset Y$ or $Y \subset X$

Body:
I'm currently learning linear algebra using textbook Linear Algebra Done Wrong, and I encountered this problem.

Let $X$ and $Y$ be subspaces of a vector space $V$. Show that $X\cup Y$ is a subspace if and only if $X\subset Y$ or $Y\subset X$

I don't really have any idea. I only figured out that 0 is in the set, so it is not empty. What else should I do? Thanks!!


问:
Farkle Probability of rolling 1 or 5 on at least one die with x number of dice
So I was playing this game with my family, its called farkle. I don't really need to get into the rules of the game but the gist is you need to roll a 1 or a 5 with a die to score. I was thinking about this and wondering what the chances are of rolling a 1 or a 5 with x amount of dice. I thought, well the probability of rolling a 1 or 5 on one die is 2/6, but then with 2 dice it is not 2/6. I don't know exactly why it isn't 2/6, it just feels like something isn't right because of the fact that you can roll a 1 or 5 on one die and not the other, and I wanted to know how this works. Can anyone please help me understand the mathematical relation between these probabilities?</p>

答: Each die has a $\frac 23$ chance not to show a $1$ or $5$.  If you have $x$ dice, the chance that none of them show $1$ or $5$ is $\left(\frac 23\right)^x$ by the multiplication principle.  The chance that you get at least one die showing $1$ or $5$ is then $1-\left(\frac 23\right)^x$


问:
Existence or non existence of uniformly convergent subsequence
Consider the sequence of functions defined from $[0,1]$ to $\mathbb{R}$, given by $$f_n(x) = \frac{2x^2}{x^2+(1-2nx)^2}~,\quad \text{where} ~~n=1,2,\dots$$
I could conclude that this sequence of functions converges pointwise to zero. But supremum of $|f_n - f|$, (here $f=0$) does not converge to zero, hence convergence is not uniform.

But I'm not able to conclude if there exist a subsequence that converges uniformly or not.  Any help would be appreciated.

Thanks in advance.

答: I think not, because for every $n$ other than $0$, function $f_n(x)$ has a maximum at some point $x_{max}$ such that $$f_n(x_{max})=2$$


问:
formula for relating number of successes to number of tries
Imagine we have a jar of marbles, where there are 10 different colors of marbles in the jar.
Let N be the number of marbles drawn in a sample and
K be the number of distinct colors in the sample.
Let C be the total number of colors in the jar (which is 10 in this example).

We know that:
K = F(C,N) + chance.

Intuitively, the more marbles we draw at a time, the more colors we are likely to obtain, up to the maximum.

This strikes me as a binomial. I am looking for a formula to express in more detail how
K = F(C,N).
An obvious feature is that C is the maximum value. So I figure this is a model where K is an increasing proportion of C as N increases.

I already know how to simulate the data, but wish to provide an expression of K depending on C and N.

Any suggestions will be appreciated.

Simulated data for the situation I intend will be as follows. I assume sampling with replacement. For this example, we have 10 colors in the jar (C=10). Note that we start by drawing one marble from the jar, which gives us a sample where N=1 and K=1. The number of colors found in a sample depends on the size of the sample. In this case it is seq(1:20). In this particular simulation, we have an equal representation for each color. The snip of R code I use to simulate the data is:

###snip of modified code for stack exchange exchange

T <- c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10)

for (i in 1:N){

for (s in 1:n.sims) {

plot.sample <- sample(T, size=N[i])

K[s,i] <- length(unique(plot.sample))

}

}

Note that the line is just a loess plot at this point and the proper function would level off at C, the maximum number of colors.

If I use the suggested formula
K=N*C/(N+C-1)
I get the figure on the right, which has the right shape, but finds colors very slowly.

If I estimate the function with a scaling coefficient, I get the following graph. Excellent fit: explained variance = 97%, though the visual suggests there could be a better function?

答:
I'll assume the jar contains a sufficiently large number of balls equally distributed among the colors. If we don't assume equi-distribution of colors or if we take into account the fact we are drawing without replacement the problem becomes significantly more complicated. Supposing we draw $n$ balls from a jar containing $c$ colors, what is the distribution of the random variable $K$, the number of colors in the sample?  (assuming $\boldsymbol{n\geq c}$)

This is a stars-and-bars calculation. Given $c$ different colors possible and our sample being of size $n$, the number of different samples we can get is
$$\mathrm{C}(n+c-1,c-1)$$
With $\mathrm{C}$ being of course the binomial coefficient. Of these possible samples, how many of them contain exactly $k$ different colors? Well, we basically don't get to pick the first $k$ balls - they have to be the $k$ distinct colors. But, the remaining $n-k$ balls can be any mixture of the $k$ colors we already have. The number of ways of selecting $n-k$ balls when we have $k$ color choices is
$$\mathrm{C}(n-1,k-1)$$
Finally, the number of ways of selecting which $k$ colors out of a pool of $c$ our sample will consist of is precisely
$$\mathrm{C}(c,k)$$
Meaning that the number of ways of picking a sample of size $n$ that has exactly $k$ different colors out of a possible $c$ is exactly
$$\mathrm{C}(c,k)\mathrm{C}(n-1,k-1)$$
Meaning that
$$\Pr(K=k)=\frac{\mathrm{C}(c,k)\mathrm{C}(n-1,k-1)}{\mathrm{C}(n+c-1,c-1)}$$
As a sanity check, it can be verified that
$$\sum_{k=1}^c \frac{\mathrm{C}(c,k)\mathrm{C}(n-1,k-1)}{\mathrm{C}(n+c-1,c-1)}=1$$
As long as $n\geq c$. Here is a plot for the $n=50,c=20$ case:

It can also be computed directly that
$$\mathrm{E}[K]=\frac{nc}{n+c-1}$$
And
$$\operatorname{Var}[K]=\frac{nc(nc-1)}{(n+c-1)(n+c-2)}-\left(\frac{nc}{n+c-1}\right)^2$$

EDIT: Numerical evidence seems to suggest my probability mass function gives correct results even for $n<c$.


问:
Semirings of small orders
A semiring is a structure $(R, +, 0, *)$ such that $(R, +, 0)$ is a commutative monoid, $(R, *, 0)$ is a semigroup with zero, and the distributive laws hold.

I know that there were attempts at computing semigroups of small order, and all semigroups up to order $8$ has been computed. Was there a similar attempt at computing semirings of small orders? If yes, where can I find results of that?

答:
Andrej Bauer's program alg exists to calculate these sorts of numbers, it will also output the addition and multiplication tables for finite algberaic structures too. If you look at the file theories/semiring.th there you'll see the axioms it assumes for a semiring

```<code>Theory semiring.

Constant 0.
Binary + *.

Axiom: 0 + x = x.
Axiom: x + 0 = x.
Axiom: x + (y + z) = (x + y) + z.
Axiom: x + y = y + x.
Axiom: x * (y * z) = (x * y) * z.
Axiom: (x + y) * z = x * z + y * z.
Axiom: x * (y + z) = x * y + x * z.
Axiom: 0 * x = 0.
Axiom: x * 0 = 0.
</code>```

I just ran this on my computer up to size six, up to five is fairly quick if you consider the number of structures it is processing, but as you can see the numbers are getting large, and if you consider the number of possible pairs of binary operations on a set of size 7, there are a lot to consider, even if some are easy to see that they fail. Edit: I left it longer overnight and the output is now:

```<code># Statistics

    size | count
    -----|------
       1 | 1
       2 | 4
       3 | 22
       4 | 283
       5 | 4717
       6 | 109010
</code>```

But this sequence is not in the oeis so I have no idea how to find more terms, except by computing further, seven might take a while!

Title: Semirings of small orders

Body:
Andrej Bauer's program alg exists to calculate these sorts of numbers, it will also output the addition and multiplication tables for finite algberaic structures too. If you look at the file theories/semiring.th there you'll see the axioms it assumes for a semiring

Theory semiring.

Constant 0.
Binary + *.

Axiom: 0 + x = x.
Axiom: x + 0 = x.
Axiom: x + (y + z) = (x + y) + z.
Axiom: x + y = y + x.
Axiom: x * (y * z) = (x * y) * z.
Axiom: (x + y) * z = x * z + y * z.
Axiom: x * (y + z) = x * y + x * z.
Axiom: 0 * x = 0.
Axiom: x * 0 = 0.

I just ran this on my computer up to size six, up to five is fairly quick if you consider the number of structures it is processing, but as you can see the numbers are getting large, and if you consider the number of possible pairs of binary operations on a set of size 7, there are a lot to consider, even if some are easy to see that they fail. Edit: I left it longer overnight and the output is now:

# Statistics

    size | count
    -----|------
       1 | 1
       2 | 4
       3 | 22
       4 | 283
       5 | 4717
       6 | 109010

But this sequence is not in the oeis so I have no idea how to find more terms, except by computing further, seven might take a while!


问:
Need help understanding formula for probability that something good happens before something bad happens.
My book says this.

Consider a sequence of independent trials, each of which can
be classified as good, bad, or neutral, which happen (on any given trial) with
probabilities $p, q,$ and $1 − p − q$, respectively. (We do not necessarily have
$q = 1 − p$ here, although that case is allowed.) Then the probability that
something good happens before something bad happens is $p/(p + q)$.

My question is, how did they derive this formula $p/(p+q)$ for probability that something good happens before something bad happens?

can someone show me a derivation/explain to me this?

答: Let $B$ denote the first trial where something bad happens, and let $X$ denote the number of times something good happens within the first $B-1$ trials. Then $B\sim \text{Geometric}(q)$ and $X|B\sim \text{Binomial}\Big(B-1,\frac{p}{1-q}\Big)$ and you're looking to compute $$P(X\geq 1)=1-P(X=0)$$ From the total law of probability, $$P(X=0)=\sum_{b=1}^{\infty}P(X=0|B=b)P(B=b)=\sum_{b=1}^{\infty}\Big(\frac{1-p-q}{1-q}\Big)^{b-1}(1-q)^{b-1}q$$ The RHS evaluates to $\frac{q}{p+q}$ and so $$P(X\geq 1)=1-\frac{q}{p+q}=\frac{p}{p+q}$$


问:
Which set operation generates the $\mathbb{Z}_4$ group?
The Klein four-group is the group generated by the symmetric difference as the binary operation on the subsets of a power set of a set with two elements $\{a,b\}$, that means over its superset with four elements: $\{\{\},\{a\},\{b\},\{a,b\}\}$. $\mathbb Z_4$ is the group generated for which set operation??? I could not find it, any try was meaningless. The two tables:

答: You're right that we can use set theory to ground the foundation of mathematics, but that doesn't mean that every mathematical object can be expressed simply in terms of set operations. For example, we can use axiomatic set theory to show that $1+1 = 2$, but when Whitehead and Russel tried to give a super rigorous set theoretic undergirding to mathematics it took them 379 pages to show that that formula holds(https://en.wikipedia.org/wiki/Principia_Mathematica page count is roughly accurate, depends on the edition). So it set theory the foundation of mathematics? Yes, it is at least a foundation for mathematics, but that doesn't mean that all ideas in mathematics can be simply expressed in terms of foundational principles.


答: The pointwise limit is  $f(0)=2, f(x)=0$ for $x \neq 0$. Every subsequence converges to the same limit. Since this limit is not continuous it follows that the convergence is not uniform.

Title: Existence or non existence of uniformly convergent subsequence

Body:
Consider the sequence of functions defined from $[0,1]$ to $\mathbb{R}$, given by $$f_n(x) = \frac{2x^2}{x^2+(1-2nx)^2}~,\quad \text{where} ~~n=1,2,\dots$$
I could conclude that this sequence of functions converges pointwise to zero. But supremum of $|f_n - f|$, (here $f=0$) does not converge to zero, hence convergence is not uniform.

But I'm not able to conclude if there exist a subsequence that converges uniformly or not.  Any help would be appreciated.

Thanks in advance.


问:
Local maxima at positive-semidefinite Hessian
For a function $f$ having Hessian $H_f$, I have read at a few places that if $H_f$ is positive-semidefinite at a point, then $f$ will either have a local minima or a saddle point. But $f$ cannot have a local maxima at that point.

I just can't understand why is that so conceptually. Why can't $f$ have a local maxima under any circumstances at that point?

答:
Technically if $f$ is constant, then the Hessian is the zero matrix (which is PSD), so every point is a local maximum and a local minimum, although these are not strict maxima/minima. It is also possible to have a strict local maximum when the Hessian is zero: consider $f(x) = -x^4$ at $x=0$. (Thanks Robert Israel.)

Excluding this edge case of a zero Hessian however, you cannot have a local maximum. Suppose the Hessian $H_f$ at $x_0$ has a strictly positive eigenvalue. Let $v$ be the associated eigenvector. Then the univariate function $g(t) = f(x_0 + tv)$ has second derivative equal to $g''(t) = v^\top H_f v > 0$. Since the curvature of $g$ at zero is positive, there exist points $t$ arbitrarily close to $0$ such that $g(t)>g(0)$. This means $f(x_0 + tv) > f(x_0)$ for such $t$ as well.


问:
Regularizing a $L^p(B)$ function in $B$ and in $\partial B$.
Let $B_1=\{x\in\mathbb{R}^N:\ |x|<1\}$. Let $u\in L^p(B_1)$ with $p\in (1,\infty)$ and suppose that $u$ is also defined in the boundary of $B_1$ and satisfies $u_{|\partial B_1}\in L^p(\partial B_1)$, where I'm assuming the surface measure in $\partial B_1$.

I want to find a sequence $u_n\in C^\infty(\overline{B_1})$ such that $u_n\to u$ in $L^p(B_1)$ and $(u_n)_{|\partial B_1}\to u_{|\partial B_1}$ in $L^p(\partial B_1)$

I was trying to do something like this: Let $\epsilon>0$ be a small number and consider the ball $B_{1+\epsilon}$. Extend $u$ in $B_{1+\epsilon}$ by reflecting it along the radial axis and call it $v$. Let $h_\delta$ be a mollifier sequence  and define $$u_\delta(x)=(h_\delta\ast u)(x)=\int_{B_{1+\epsilon}}h_\delta(x-y)v(y)dy$$

The last sequence does converge to $u$ in $L^p(B_1)$ but I dont know how to argue with the boundary. For example, by using Fubini's theorem, I can show that for almost $t\in (0,1+\epsilon)$, there exist a subsequence of $u_\delta$ (not relabaled) such that $(u_n)_{|\partial B_t}\to u_{|\partial B_t}$ in $L^p(\partial B_t)$.

Any to how to go further or even another idea is appreciated.

Thank you

答:
Extend $u$ to  $B_2$ so that $u(x)=u(x/|x|)$ when $1<|x|<2$. Show that the sequence $v_n(x)=u(x/(1-n^{-1}))$ converges to $u$ in  $L^p(B_1)$. (Same proof as for continuity of translations in $L^p$.)  Then convolve $v_n$ with a mollifier $\varphi_n$ whose support contained in $B_{1/n}$. On the boundary $|x|=1$, the value of $v_n*\varphi_n$ is obtained by averaging the boundary values of $u$, and therefore will be close to $u|_{\partial B_1}$.

For example, consider $u=0$ on $(-1,1)$, $u(\pm 1)=1$.

Step 1: Extension

Step 2: Scaling

Step 3: Mollification

ContentLicense: CC BY-SA 3.0


问:
markov chain problem conditioned on future
We know that by Markov property that $P ( X_n = x_n | X_{n-1} = x_{n-1} , . . . , X_0 = x_0 ) = P ( X_n = x_n | X_{n-1} = x_{n-1})$ but can we say that $P ( X_0 = x_0 | X_1 = x_1 , . . . , X_n = x_n ) = P ( X_0 = x_0 | X_1 = x_1)$, any hint on how to prove this?

答:
Hint:
\begin{align}
P(X_0=x_0 \mid X_1 = x_1,  \ldots, X_n = x_n)
&= \frac{P(X_0=x_0, X_1=x\ldots, X_n=x_n)}{P(X_1 = x_1, \ldots, X_n = x_n)}
\\
&= \frac{P(X_0=x_0)\prod_{i=1}^n P(X_i=x_i \mid X_{i-1} = x_{i-1})}{P(X_1=x_1)\prod_{i=2}^n P(X_i=x_i \mid X_{i-1} = x_{i-1})}.
\end{align}
The last step uses the Markov property. Cancelling terms will lead you to $P(X_0 = x_0 \mid X_1=x_1)$.


问:
Proving following ring isomorphism
Suppose  R is an integral domain and $0\neq f\in R$.  Show that
$$R[x]/(xf-1)\cong R[\frac{1}{f}]$$
I am able to get the LHS from the isomorphism theorem by using following homomorphism $\phi(x)=\frac{1}{f}$. The kernel will then be $(xf-1)$ and hence the isomorphism.

This question is from Cutkosky $(1.7)$ the notation on RHS means the smallest subring containing $R$ and $\frac{1}{f}$. Obvioulsy R is an integral domain so it may not contain $f$ inverse but Cutkosky uses an homomorphism from $R[x]$ to a quotient field of $R$ denoted by $K$. Using quotient field will only make sense if $f$ inverse is contained in R otherwise it is not a well defined statement.

How to make sense of RHS of above isomorphism?

答:
"the notation on RHS means the smallest subring containing $R$ and $\frac{1}{f}$"

This is not clear. Perhaps you mean the smallest subring of the field of fractions $K$ of $R$ containing $R$ and $\frac{1}{f}$. The homomorphism $\phi \colon R[x] \to K$ given by $x \mapsto \frac{1}{f}$ is well-defined and has kernel $(fx-1)$. The image is a ring which contains $R$ and $\frac{1}{f}$, so $\phi(R[x]) \supset R[1/f]$. Clearly $\phi(R[x]) = \{a/f^k : a \in R, k \in \mathbb{Z}_{\geq 0}\}$. Just as clearly, $\{a/f^k : a \in R, k \in \mathbb{Z}_{\geq 0}\} \subset R[1/f]$. The result now follows from the first isomorphism theorem.

Title: Proving following ring isomorphism


问:
Derive the condition for a line to be tangent to a circle
Given the
circle
$(x-a)^2+(y-b)^2 = r^2$
and the line,
$ux+vy+w = 0$,
show that the condition
for the line to be
tangent to the circle is
$(w+vb+ua)^2
=r^2(v^2+u^2)
$.

I have a solution
but,
as usual,
it is messy
and I am hoping for
a nicer one.

I'll post mine in a few days
if no one has a better one.

Note:
When I entered this question,
a number of similar questions
were suggested,
but none of them
ask precisely
this question.

答: This is straightforward: the distance of the centre from the line should be $r$, so$$\frac{|ua+vb+w|}{\sqrt{u^2+v^2}}=r$$


问:
Definition of simple spectrum
From the book "Spinning Tops" by Audin, given Lax equation $[A_{\lambda},B_{\lambda}]$ where $\lambda$ is a parameter (so called spectral parameter), she claims that we have spectral curve $P(\lambda,\nu)=0$ where $P(\lambda,\nu)$ is a characteristic polynomial of $A_{\lambda}$. Then, she talks about when $A_{\lambda}$ has a simple spectrum. I tried to look for definition of it, but I couldn't find it. Can you please tell me what does it mean for a spectrum to be simple?

答: I suspect she means that the eigenvalues are all simple, i.e. the characteristic polynomial has no multiple roots.</p>

Title: Definition of simple spectrum


问:
Prove using the limit definition of a derivative an equation relating total and partial derivatives of a two variable function

I've been trying to prove the following equation which relates the total derivative to the partial derivatives of a two variable function $$V(r, h)$$, namely $$\frac{dV}{dr} = \frac{\partial V}{\partial r} + \frac{\partial V}{\partial h}\frac{dh}{dr} $$ using only limits. This isn't a homework question but just something I want to learn how to do.

I was able to do so for the special case of $$V(r, h) = \frac{\pi r^2h}{3}$$ as follows (volume of a cone with height h and radius r):
$$\frac{dV}{dr} = \lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h(r+\Delta r))-V(r,h(r))}{\Delta r}
\\=\lim\limits_{\Delta r \to 0} \left(\frac{\pi(r+\Delta r)^2h(r+\Delta r)}{3\Delta r} - \frac{\pi r^2}{3\Delta r}\right)
\\=\lim\limits_{\Delta r \to 0} \frac{\pi r^2h(r+\Delta r) + 2\pi r\Delta h(r+\Delta r)+\pi(\Delta r)^2h(r+\Delta r) - \pi r^2h(r)}{3\Delta r}
\\=\lim\limits_{\Delta r \to 0} \left(\frac{\pi r^2}{3}\frac{h(r+\Delta r)-h(r)}{\Delta r}\right) + \lim\limits_{\Delta r \to 0} \frac{2\pi rh(r+\Delta r)}{3} + \lim\limits_{\Delta r \to 0} \frac{\pi(\Delta r)h(r+\Delta r)}{3}
\\=\frac{\pi r^2}{3}\frac{dh}{dr}+\frac{2\pi rh(r)}{3} + 0
\\= \frac{\partial V}{\partial h}\frac{dh}{dr} + \frac{\partial V}{\partial r}
$$

I wasn't able to get anywhere with an arbitrary function. The thing that's making it difficult for me is there being two variables h and r in the limit expression. I'm thinking I have to do a change of variables or use the composition limit law but it's just a guess and I haven't been making any progress.

$$\frac{dV}{dr} = \frac{\partial V}{\partial r} + \frac{\partial V}{\partial h}\frac{dh}{dr}$$ which in limit form is:

$$
\lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h(r+\Delta r))-V(r,h(r))}{\Delta r}
\\=\lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h)-V(r,h)}{\Delta r}
+\lim\limits_{\Delta h \to 0} \frac{V(r, h+\Delta h)-V(r,h)}{\Delta h}\lim\limits_{\Delta r \to 0} \frac{h(r+\Delta r)-h(r)}{\Delta r}
$$
which is what I want to prove.

Help would be much appreciated.
-Jay

答:
<s> derivative has a characterization that if $g : A \subseteq \mathbb{R} \to \mathbb{R}$ is differentiable at a point $x \in A$, then
$$g(x + h) = g(x) + g'(x)h + o(h),$$
where $r(h) = o(q(h))$ means $\frac{r(h)}{q(h)} \to 0$ as $h \to 0$
Now assume $f : A \subseteq \mathbb{R^2} \to \mathbb{R}$, and $x : B \subseteq \mathbb{R} \to \mathbb{R}$ and $y : B \subseteq \mathbb{R} \to \mathbb{R}$. Then
\begin{align}
f(x(t + h), y(t + h)) &= f(x + x'h + o(h), y + y'h + o(h)) \\
&= f(x + x'h + o(h), y) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(y'h + o(h)) \\
&= f(x + x'h + o(h), y) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + o(x'h + o(h)) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + o(h) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h)
\end{align}
Now subtract $f(x, y)$ and divide by $h$ and let $h \to 0$ to get
$$\frac{d}{dt}f(x, y) = f_x(x, y)x'(t) + f_y(x, y)y'(t).$$
Whenever I omitted the argument of $x$ and $y$ I mean it to be $t$.
There is a flaw, in that I assumed $f_y$ to be continuous at $(x(t), y(t))$. I'm not sure how to avoid this or if this is necessary for the result to hold. </s>

The answer is wrong.


问:
How to validate a random graph's chromatic number
I developed an algorithm that can calculate the chromatic number of a graph with n vertices in roughly n^2 time. I verified that this works for smaller graphs, but I am looking for a systematic way to verify that this works for larger, more complex graphs. Is there an existing, systematic way to determine chromatic number that I could use to validate my answers?</p>

答:
If this were really true, it would be big news, but it's almost certainly not.
Graph colouring is well-known to be NP-complete, so if you really have a polynomial-time algorithm (and can prove that!) you have proved that P=NP.  You can then apply for your million dollar Millenium Prize from the Clay Mathematics Institute.

But, as I said, it's almost certainly not true.

As for your question, you can find some challenging graph-colouring benchmarks at the Google Graph Coloring Benchmarks site.


问:
Solving $\lim_{x \to 0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $ without L'Hopitals
$$\lim_{x \to  0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $$

This question is very easy by just using L'Hopital's, but I'm trying to do the question without it but I'm stuck on what to do. Any help would be greatly appreciated.

答: $$\lim_{x\to 0}\frac{x+x\cos x}{\sin x\cos x} = \lim_{x\to 0}\frac {x}{\sin x}\cdot\frac{1 +\cos x}{\cos x} = 1\cdot\frac 21$$


问:
What is formal notation for rounding to the nearest hundredth?
For example, if =(X*Y), how would you denote formally that the  is to be rounded to the nearest hundredth in this operation?</p>

答:
For rounding to the nearest hundredth, we have:

$$\mathrm{rd}_{1/100}(x) := \frac{1}{100}\left\lfloor100x + \frac 12\right\rfloor $$

This is how the formula works. First multiply by $100$ so that the hundredth digit of $x$ becomes the unit digit. Then add $\frac{1}{2}$ and round down (that's what the floor function does). If the fractional part of $x$ (call it $u$) satisfies $0 ≤ u < \frac{1}{2}$, then $u + \frac{1}{2} < 1$, and so the floor function leaves only the integer part, rounding it down. Conversely, if $\frac{1}{2} ≤ u < 1$, then $u + \frac{1}{2} > 1$, and so the integer part is increased by $1$, rounding it up.

The last step is to convert the unit digit back into hundredths, which is done by multiplying everything by $\frac{1}{100}$. Let's try this process with $x = 23.456$:

$$100x = 2345.6 \implies 100x + \frac{1}{2} = 2346.6 \implies \lfloor100x + \frac{1}{2}\rfloor = 2346$$
$$\implies \frac{1}{100}\lfloor100x + \frac{1}{2}\rfloor = 23.46$$

and this is correct as $23.456$ rounds up to $23.46$.


问:
Finding $R \circ S$, $S \circ R$, and $R \circ R$ for given relations $R,S$
For the following relations on $\{1,2,3,4,5,6,7,8,9,10\}$.

$R=\{(1,2),(3,6),(4,1),(5,5),(6,4),(7,5)\}$

$S=\{(2,1),(3,6),(9,4)\}$

What I got:

$R∘S = \{(2,2),(3,4),(9,1)\}$

$S∘R = \{(1,1)\}$

$R∘R = \{(3,4),(4,2),(5,5),(6,1),(7,5)\}$

The question stated that "If it is not possible to determine the relation then explain the reason." So I would like to ask is there are any answer not to possible to determine the relation?

答:
$
\newcommand{\CC}{\mathcal{C}}
$
Your work is correct. I would just like to suggest an approach that might help cement understanding of the topic of relation composition where possible, and develop and intuition for it (and in particular see analogies to function composition since, after all, functions are fundamentally relations).

Personally, when possible, I like to approach this sort of problem visually. Write out the elements of the relevant set, in columns, four times over. (For convenience, I will also label the columns $\mathcal C_i$, for $i=1,2,3,4$.) Then, in the gaps between $\CC_1$ and $\CC_2$, draw arrows between the two, where $x$ in $\CC_1$ points to $y$ in $\CC_2$ if and only if $(x,y) \in R$. Replicate this exactly for the gap between $\CC_3$ and $\CC_4$. Finally, draw arrows similarly in the gap for $\CC_2$ and $\CC_3$ representing the relation $S$: $x$ in $\CC_2$ points to $y$ in $\CC_3$ if and only if $(x,y) \in S$.

The diagram you should get is something like this:

You might be guessing what comes next. Essentially, if $x$ in $\CC_1$ points to $y$ in $\CC_2$, and then to $z$ in $\CC_3$, then $(x,z) \in S \circ R$. Or put differently: if there is a path from $x$ in $\CC_1$ to $z$ in $\CC_3$ you can walk along, then $(x,z) \in S \circ R$.

Similarly, the existence of such a path for $x$ in $\CC_2$ to $z$ in $\CC_4$ gives $(x,z) \in R \circ S$.

It will usually be easiest to just start in $\CC_3$, and see if you can trace backwards to $\CC_1$ (and similar for $\CC_4$ and $\CC_2$), but that's up to you. Anyhow, with this in mind, it becomes clear that

$$R \circ S = \{(2,2),(3,4),(9,1)\} \qquad S\circ R = \{(1,1)\}$$

after all.

A similar idea can be used for $R \circ R$. Make three columns with your elements, and in each gap between the columns, draw lines representing the relation $R$. (I drew the bolded lines to indicate the relevant bits of the relation.)

From this we easily see that

$$R \circ R = \left\{ (3,4), (4,2), (5,5), (6,1), (7,5) \right\}$$

as desired. And, of course, even though you didn't ask about that in particular, $S \circ S$ can also be found:

Namely, since no paths can be completed, $S \circ S = \varnothing$.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


答:
$\lim\limits_{x\to0}\dfrac{x+x\cos x}{\sin x\cos x}=\lim\limits_{x\to0}\left(\dfrac x{\sin x}\dfrac1{\cos x}+ \dfrac x{\sin x}\right)=\lim\limits_{x\to0}\dfrac x{\sin x}\lim\limits_{x\to0}\dfrac1{\cos x}+\lim\limits_{x\to0}\dfrac x{\sin x}$

$=1+1=2$

Title: Solving $\lim_{x \to 0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $ without L'Hopitals

Body:
$$\lim_{x \to  0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $$

This question is very easy by just using L'Hopital's, but I'm trying to do the question without it but I'm stuck on what to do. Any help would be greatly appreciated.


问:
Invert Trigonometric System of Equations
In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.

$$\cos \alpha  = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$

I would like to find an inversion in the form:
$$\text{trig}(x) =\; ...$$
$$\text{trig}(y) = \;...$$

How can this be done?

答:
$$\cos(\alpha)  = \cos (\theta) \cos( x) + \sin (\theta) \sin (x) \cos( y)\tag 1$$
$$\tan (\beta) = \frac{\cos (x) +\cos (\theta) \cos (\alpha)}{\sin (\theta) \sin (x) \sin (y)}\tag 2$$

From $(1)$
$$\cos(y)=\csc (\theta ) \csc (x) (\cos (\alpha )-\cos (\theta ) \cos (x))\tag 3$$
From $(2)$
$$\sin(y)=\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
   \cot (x)\tag 4$$
Using $\cos^2(y)+\sin^2(y)=1$, we end with
$$1=(\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
   \cot (x))^2+(\cot (\theta ) \cot (x)-\cos (\alpha ) \csc (\theta ) \csc (x))^2$$

Now, using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, we end with a quadratic in $t^2$
$$a + b t^2+c t^4 =0$$ where
$$a=16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
   \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
   \cot ^2(\theta )-3$$
$$b=-2 \cos (2 (\alpha -\beta ))-2 \cos (2 (\alpha +\beta ))-16 \sin ^2(\alpha ) \csc
   ^2(\theta )-$$ $$4 \cos (2 \alpha )+4 \cos (2 \beta )-12$$
$$c=-16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
   \cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
   \cot ^2(\theta )-3$$

Applied to the case use by @Jean Marie $\alpha=\theta=\frac\pi 4 $ and $\beta=\tan ^{-1}(2))$, this leads to
$$132 t^4-152 t^2+36=0$$ giving the four roots
$$t=\pm  \frac{1}{\sqrt{3}} \qquad \text{and} \qquad t=\pm \frac{3}{\sqrt{11}} $$ that is to say
$$x=\pm \frac \pi 3\approx \pm1.04720 \qquad \text{and} \qquad x=\pm2 \tan ^{-1}\left(\frac{3}{\sqrt{11}}\right)\approx 1.47063$$ and then the corresponding $y$'s using $(3)$ or $(4)$.


问:
On the integrability of $x^{x}$
By the definition of the integral we have:

If $f$ is a function defined for $a \leq x \leq b$, we divide the interval $[a,b]$ into $n$ subintervals of equal width ${\Delta}x = \frac{b - a}{n}$. We let $x_{0} (= a), x_{1}, x_{2}, {\dots}, x_{n} (= b)$ be the endpoints of these subintervals and we let $x_{1}^{*}, x_{2}^{*}, \dots, x_{n}^{*}$ be any sample points in these subintervals, so $x_{i}^{*}$ lies in the $i$th subinterval $[x_{i - 1}, x_{i}]$. Then the definite integral of $f$ from $a$ to $b$ is

\begin{equation}
\int_{a}^{b} f(x) \;\mathrm{d}x = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_{i}^{*}){\Delta}x
\end{equation}

provided that the limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that $f$ is integrable on $[a, b]$.

From this, if $f(x) = x^{x}$, $f$ is defined on $[a, b]$ if $a = 0$ and $b = \infty$ so $f$ should be integrable.

\begin{align}
\int_{0}^{1} x^{x} \;\mathrm{d}x &= \lim_{n \to \infty} \sum_{i = 1}^{n} {x_{i}^{*}}^{x_{i}^{*}}{\Delta}x
\end{align}
with ${\Delta}x = \frac{b - a}{n} = \frac{1 - 0}{n} = \frac{1}{n}$

and $x_{i}^{*} = a + i{\Delta}x = 0 + i \times \frac{1}{n} = \frac{i}{n}$
\begin{align}
\int_{0}^{1} x^{x} \;\mathrm{d}x &= \lim_{n \to \infty} \sum_{i = 1}^{n} \left(\frac{i}{n}\right)^{\left(\frac{i}{n}\right)}\left(\frac{1}{n}\right)\\
                                 &= \lim_{n \to \infty} \left(\frac{1}{n}\right) \sum_{i = 1}^{n} \left(\frac{i}{n}\right)^{\left(\frac{i}{n}\right)}
\end{align}

The issue now is that I'm not sure how to go about evaluating this Riemann Sum to determine whether the limit exists or not. I know that it's difficult (if not impossible) to find the indefinite integral of $x^{x}$ as there is no antiderivative using elementary functions. No function exists which when differentiated yields $x^{x}$. But this shouldn't have any effect on the integrability of the function since by the theorem stated above the function is defined on the interval therefore it should be integrable.

What information could you give me to point me in the right direction?

答:
AFAIK, there is no closed form of this, but I have a uselful result.

This integral is known as sophomore's dream.

We have
$$\int_{0}^{1}x^xdx=\int_{0}^{1}e^{x\log x}dx=\sum_{k=0}^{\infty}\int_{0}^{1}\frac{x^k\log ^k x}{k!}dx$$
I will cheat a bit by using mathematica, which gives

$$\int_{0}^{1}x^k\log^kxdx=(-1)^k(1+k)^{-(1+k)}k!$$
So
$$\int_{0}^{\infty}x^xdx=\sum_{k=0}^{\infty}\frac{(-1)^k}{(1+k)^{1+k}}=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k^k}\label{a}\tag{1}$$
This is a simple and nice sum representation of that integral.

You can also see this, this and this.

Title: On the integrability of $x^{x}$


问:
For any non-singular $A$, $\frac{1}{\kappa(A)}\leq\frac{\|E\|}{\|A\|}$ if $E+A$ is singular

Let $A$ be a non singular, and $A+E$ be singular matrix.
Prove that $\frac{1}{\kappa(A)}\leq\frac{\|E\|}{\|A\|}$.

I thought of assuming $\|A+E\|=0$, but it doesn't seem to be right.

答: I suppose that $\|\cdot\|$ is an operator norm, i.e. $\|M\|=\sup_{\|x\|=1}\|Mx\|$. By the given assumptions, $I+A^{-1}E$ is singular. Therefore $A^{-1}Ex=-x$ for some unit vector $x$. Hence $\|A^{-1}\|\|E\|\ge\|A^{-1}E\|\ge\|A^{-1}Ex\|=\|-x\|=1$ and $\frac{\|E\|}{\|A\|}\ge\frac{1}{\|A\|\|A^{-1}\|}=\frac{1}{\kappa(A)}$.


问:
Least concave majorant of a non-decreasing function
Consider a non-decreasing discrete function $f$ defined over $\{0,1,\ldots,n\}$.
Let $g$ be the least concave majorant of $f$. Assume that for some $i\in \{1,\ldots,n-1\}$, $g(i)\neq f(i)$. How to prove that $g$ is linear over $\{i-1, i ,i+1 \}$, i.e., $g(i)-g(i-1) = g(i+1)-g(i)$?

I tried the following: Suppose that $g(i)-g(i-1) \neq g(i+1)-g(i)$, then we must have that $g(i)$ is above the line that interpolates $g(i-1)$ and $g(i+1)$ by concavity of $g$. Then, I think that one might be able to find a contradiction with the fact that $g$ is the least concave majorant but cannot formalize it.

答:
If $g(i) \ne f(i)$ for some $i \in \{ 1, \ldots, n-1 \}$ then $g(i) > f(i)$ since $g$ is a majorant of $f$. Now assume that $g$ is not linear on $[i-1, i+1]$, i.e. it is strictly concave on that interval:
$$
 g(i) > \frac 12 (g(i-1)+g(i+1)) \, .
$$

The idea is that we can decrease the value of $g(i)$ slightly so that $g$ is still a concave majorant of $f$. More precisely: Define  the function $h: \{ 0, \ldots , n \} \to \Bbb R$ by
$$
 h(j) = \begin{cases}
 g(j) & \text{if } j \ne i \, ,\\
\max\left(\frac 12 (g(i-1)+g(i+1)), f(i) \right) & \text{if } j = i \, .
\end{cases}
$$
$h$ is concave, is a majorant of $f$, and satisfies $h(i) < g(i)$. This is a contradiction to the minimality of $g$.

Proof that $h$ is concave: For $j=i$ we have
$$
 h(i) \ge \frac 12 (g(i-1)+g(i+1)) = \frac 12 (h(i-1)+h(i+1))
$$
and for $j \ne i$ we have
$$
 h(j) = g(j) \ge \frac 12 (g(j-1)+g(j+1)) \ge \frac 12 (h(j-1)+h(j+1)) \, .
$$

Remark: The condition that $f$ is non-decreasing is not needed for this conclusion.


问:
Evaluate $\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$
Suppose $\alpha>1$, and $a_n$ be the largest natrual number $k$ such that $k!\le \alpha^n$.Evaluate $$\lim\limits_{n \to \infty} \frac{\sqrt[n]{a_1a_2\cdots a_n}}{a_n}.$$

According to the assumption, we obtain

$$(a_n)!\le \alpha^n<(a_n+1)!=(a_n+1)(a_n)!.$$
Therefore
$$ (a_1)!(a_2)!\cdots(a_n)!\le \alpha^{1+2+\cdots+n}<(a_1+1)(a_2+1)\cdots(a_n+1)(a_1)!(a_2)!\cdots(a_n)!.$$
But how to go on?

答:
Hint:  When $n = 1$, $\ln \sqrt{a_1/a_1} = 0$.

When $n = 2$,
$$  \ln \sqrt{\frac{a_1}{a_2} \cdot \frac{a_2}{a_2}} = \frac{1}{2}\left( \ln(a_1) - \ln(a_2) + 0 \right) < 0  \text{.}  $$

Now, $\log_\alpha (a_j) \ll \log_\alpha (a_j!) \leq \log_\alpha (\alpha^j) = j$ for all $j$.  Can you get a sharper version of "$\ll$"?


问:
Notation for direct sum of direct sums
Suppose you have a direct sum of two terms $A$ and $B$: $$ A \oplus B.$$
Now suppose that in fact $B = \bigoplus_i B_i$. Then I guess we could write the above sum as
$$ A \oplus \bigoplus_i B_i.$$
However, this looks terrible -- much worse than, for instance,
$$ A \bigoplus_i B_i. $$
Is there a precedent to writing the latter, or some other notation better than the former?

Cheers.

P.S.: what about when B is a direct product instead of a direct sum?

答:
I'm sure it depends a bit person to person as notation tends to, but I think you should definitely avoid the notation $$A \bigoplus_i B_i.$$ To demonstrate why, lets suppose our indexing set is the natural numbers. What I understand you to mean is
$$
A \oplus B_1 \oplus B_2 \oplus ...
$$
But when I see the notation $A \bigoplus_i B_i$ I think that reads as
$$
(A \oplus B_1) \oplus (A \oplus B_2) \oplus (A \oplus B_3 \oplus) ...
$$
And obviously  this ambiguity is not specific to $\mathbb{N}$ as an indexing set, it's just a familiar example. I don't think this is how most people would notate that sum, but I think it's potentially confusing enough to avoid. I think either defining $B = \oplus B_i$ or using parentheses to make it clear, i.e. $A\oplus (\bigoplus B_i)$ seem the most sensible, in line with what @runway44 said in the comments.

Title: Notation for direct sum of direct sums


问:
Isomorphisms define an equivalence relation between modules
I have recently started theory of modules and I have the following doubt:

Is "being isomorphic" an equivalence relation among $R$-modules?

Concretely, I am not sure that if $f$ is an $R$-module isomorphism, then $f^{-1}$ exists and it is also a $R$-module isomorphism.

Thanks.

答:
I am very late but I would like to answer this question.

Definition: Let $R$ be a ring and let $M$ and $N$ be $R-$modules.
A map $\psi:M\to N$ is an $R$-module isomorphism if $\psi$ is bijective and respects the $R$-module structures of $M$ and $N$, i.e.,

$\psi(x+y)=\psi(x)+\psi(y), \forall x,y\in M$

$\psi(rx)=r\psi(x),\forall r\in R,\forall m\in M.$
I'll denote traditionally 'isomorphism' by $'\cong'$.
Here we will show that $R$-module isomorphism is an equivalence relation.
To prove this, we will show that$R$-module isomorphism is reflexive, symmetric and transitive.

Reflexivity: Let $M$ be an $R$-module.
Let $i:M\to M$ be an identity map. $i$ is bijective and
$i(x+y)=x+y=i(x)+i(y), \forall x,y\in M$ and $i(rx)=rx=r(x),\forall r\in R, \forall x\in M$.
Thus $M\cong M.$

Symmetry: Let $M$ and $N$ be $R-$modules and $M$ is $R$-module isomorphic to $N$.
The map $\psi:M\to N $ is bijective and $R$-module homomorphism.
Now we claim that $N$ is $R$-module isomorphic to $M$.
Since $\psi$ is bijective therefore $\psi$ is invertible.
That is, $\psi^{-1}:N\to M$. It remains to show that $\psi^{-1}$ is $R$-module homomorphism.
\begin{equation*}\psi^{-1}(n_1+n_2)=\psi^{-1}(\psi(m_1)+\psi(m_2))=\psi^{-1}\psi(m_1+m_2)=m_1+m_2=\psi^{-1}(n_1)+\psi^{-1}(n_2).
\end{equation*}
Transitivity: Let $M,N$ and $O$ be $R$-modules.
We have $M$ is $R$-module isomorphic to $N$ and $N$ is $R$-module isomorphic to $O$. We'll claim that $M$ is $R$-module isomorphic to $O$.
$\phi:M\to N$, $\psi:N\to O$ and $\psi o\phi:M\to O$.
Let $h=\psi o\phi$. We know that composition of bijective maps is bijective. Therefore $h$ is bijective. Now we'll show that $h$ is $R$-module homomorphism,
\begin{equation}
h(m_1+m_2)=\psi o\phi(m_1+m_2)=\psi(\phi((m_1+m_2)))=\psi(\phi(m_1))+\psi(\psi(m_2))=h(m_1)+h(m_2)
\end{equation}.
I left condition second to prove in both symmetry and transitivity. That is pretty easy to do.


问:
Expectation and Standard deviation of lifetime of a bulb with uniform probability distribution
I came across a problem in the book "Understanding Probability" in the section on uniform continuous variables:

The lifetime of a light bulb has an uniform probability density
on (2, 12). The light bulb will be replaced upon failure or upon reaching the age 10, whichever occurs first. What are the expected value and the standard deviation of the age of the light bulb at the time of replacement?

I saw a possible explanation at A doubt in finding the expected value of lifetime but I still don't get it.

I applied the formula for expectation $E(X) = \frac{a + b}{2}$ to get 7, and for variance to get $\sigma = \sqrt \frac {(b - a)^2}{12}$ to get 2.88, but apparently the correct answers are 6.8 and 2.613 respectively. I just can't  wrap my mind around it. I tried "relocating" the distribution to (0,10) but it still doesn't work!

Can anyone help me here? Thanks!

答: Let $L$ be the lifetime of the bulb. What you are asked to find are the mean and the standard deviation  of the random variable $X=\min \{L, 10\}$. This random variable does not have density. The reason is $X=10$  iff $L >10$ so $P(X=10)=P(L >10)=\frac {12-10} {12-2}=\frac  2 {10}=0.2$. For $2<x <10$ we have $P(X<10)=P(L<10)=\frac {x-2} {12-2}$. Thus the distributon jis  amxituer of a cnsta and  discerte distribution.. You can  find the mean as follows: $EX=\int_2^{10} x \frac d {dx} (\frac {x-2} {12-2})dx+10P(L=10)$ and this works out to $6.8.$ I will let you compute $EX^{2}$ in a similar way and find the standard deviation.


问:
Are 4-chromatic 3-connected unit distance graphs always rigid?
At braced heptagon we determined a graph was rigid. That graph is unit-distance, 3-connected, and 4-chromatic.  Here are some more graphs, all extremely constrained. I'm pretty sure they are all rigid graphs, but I'm not sure.

Are any of these graphs not rigid?
Are all unit-distance, 3-connected, 4-chromatic graphs rigid?

答:
I've programmed the first seven graphs into my new Shibuya repository of unit-distance and integral embeddings of graphs. Strictly speaking I've only programmed six, but {UnitDistance, {15, 1}} contains {UnitDistance, {12, 2}} as a subgraph and the latter is (as will be shown) rigid, so the former is also rigid.

Using the same least-parameters method that I used to prove the rigidity of the braced heptagon, I get the following results. All the rigid graphs are first-order rigid:

Graph
Least parameters required
Rigid?

{UnitDistance, {9, 3}}
1
yes

{UnitDistance, {12, 2}}
2
yes

Exoo–Ismailescu 17
2
yes

Exoo–Ismailescu 19
2+1
yes

Exoo–Ismailescu 21
2+1+1
no (1 DOF)

Hochberg–O'Donnell fish
3+1+1
no (1 DOF)

The notation $a+b+c+\cdots$ means that a subgraph was parametrised using $a$ parameters, then based on this another subgraph was parametrised using $b$ parameters and so on – the result of all these phases is the whole graph.

Exoo–Ismailescu 21 is not rigid because there are two parameters but only one constraint in the first phase; the second and third phases each have one parameter and one constraint, so they don't help with rigidity:

The reason for the "hodfish" graph not being rigid is similar, only the first phase has three parameters and two constraints:

You would have done well to check the number of edges of the two graphs shown to be non-rigid above. They fail the Laman condition (one fewer edge than $2v-3$ where $v$ is the number of vertices), which is necessary for a graph to be rigid.


问:
Injection proof $f(m/n)$
Prove $f(m/n)=2^m3^n$, $m/n$ is in lowest term, is an injection.
I have learnt injection for $x_1=x_2, f(x_1)=f(x_2)$. I would like to ask how to deal with $m/n$. Thanks.

答:
Lets prove that your function is one to one function.
That is:
$f(a/b) = f(c/d) \implies a/b = c/d$
$$f(a/b) = 2^a\cdot 3^b = 2^c\cdot 3^d = f(c/d)$$
Then $2^{a-c} = 3^{d-b}$
Since the left and the right side have no common divisor, the only solution is that both must be 1.

$a-c = 0$ and $d-b=0$ or $a=c$ and $b=d$

Which means $\frac{a}{b} = \frac{c}{d}$

Title: Injection proof $f(m/n)$


问:
Differential equation with the Substitution method
I'm trying to solve this equation below :

$$ y' = \sin\left(\frac{y}{x}\right) + \frac{y}{x}  \tag 1 $$

The first step was to to substitute $\frac{y}{x}$ with $u$

=>   $ u = \frac{y}{x}$     => $y' = u'x + u \tag 2$

Then with
$(1) = (2)$,  I got at the end

$$\dfrac{\ln\left(\cos\left(y\right)+1\right)-\ln\left(1-\cos\left(y\right)\right)}{2} = \ln|x| + C$$

And after more simplification :
$$\frac{\cos(y)}{1-\cos(y)} = 2Cx   $$

How can I get the general function from this equation ?

答:
There is an amazing manner for solving
$$y' = \sin\left(\frac{y}{x}\right) + \frac{y}{x}  $$
First use $y=x\,z$ to make
$$x\,z'=\sin(z)$$ Now, switch variables
$$\frac x {x'}=\sin(z) \implies \frac{x'}x=\csc(z)\implies \log(|x|)+c=\log \left(\tan \left(\frac{z}{2}\right)\right)$$


问:
Mean value inequality in the framework of Banach spaces.

Let $E,F$ be Banach spaces and $U \subseteq F$ be an open set. Let $f : U \longrightarrow F$ be a Gateaux differentiable function then $$\|f(x) - f(y)\|_F \leq \|x - y\|_E \sup\limits_{0 \leq \theta \leq 1} \left \|Df(\theta x + (1 - \theta) y) \right \|$$ whenever $[x,y]: = \{tx + (1-t)y\ |\ t \in [0,1] \},$ the line joining connecting $x$ and $y$ lies in $U.$

At the very beginning of the proof it has been stated that by Hahn-Banach theorem we can find out $\phi \in F^*$ such that $\|\phi\| = 1$ and $$\|f(x) - f(y)\|_F = \phi (f(x) - f(y)).$$

How do I find such a $\phi\ $? Which version of Hahn-Banach theorem is used here? I also don't understand how $\phi$ is linear. Can anybody please help me in this regard?

Thanks for your time.

答: If $u$ is a non-zero vector in a normed linear space $X$. Let $M$ be the one-dimensional space spanned by $u$. Define $\phi (cu)=c\|u\|$. This is a  continuous linear functional on $M$ and its norm is $1$. By Hahn Banach Theorem this extends to a continuous linear functional on the whole of $X$ with norm $1$. Note that $\phi (u)=\|u\|$ by definition. Now take $u=f(x)-f(y)$.


答: The answer is morally yes, but technically no, because equivalence relations technically have to be defined on sets and the collection of modules isn't a set. But nevertheless you can verify that all of the equivalence relation axioms hold, as in Ryszard's answer, if you ignore the distinction between sets and proper classes.

Title: Isomorphisms define an equivalence relation between modules

Body:
I have recently started theory of modules and I have the following doubt:

Is "being isomorphic" an equivalence relation among $R$-modules?

Concretely, I am not sure that if $f$ is an $R$-module isomorphism, then $f^{-1}$ exists and it is also a $R$-module isomorphism.

Thanks.


答:
The error  you did is to confuse the Uniform distribution of the bulb lifetime with its actual lifetime...if the lifetime is $X>10$ they substitute the bulb thus its lifetime remains 10.

Formally, the lifetime duration is represented by the following "mixed" density

$$d(x)=\begin{cases}
\frac{1}{10},  & \text{if $2\leq x<10$} \\
\frac{2}{10},  & \text{if $x=10$}\\
0,  & \text{elsewhere}
\end{cases}$$

Thus, as already written,

$$\mathbb{E}[X]=\int_2^{10}\frac{x}{10}dx+10\frac{2}{10}=6.8$$

$$\mathbb{E}[X^2]=\int_2^{10}\frac{x^2}{10}dx+10^2\frac{2}{10}=\frac{496}{15}+20$$

that is

$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$

Title: Expectation and Standard deviation of lifetime of a bulb with uniform probability distribution

Body:
I came across a problem in the book "Understanding Probability" in the section on uniform continuous variables:

The lifetime of a light bulb has an uniform probability density
on (2, 12). The light bulb will be replaced upon failure or upon reaching the age 10, whichever occurs first. What are the expected value and the standard deviation of the age of the light bulb at the time of replacement?

I saw a possible explanation at A doubt in finding the expected value of lifetime but I still don't get it.

I applied the formula for expectation $E(X) = \frac{a + b}{2}$ to get 7, and for variance to get $\sigma = \sqrt \frac {(b - a)^2}{12}$ to get 2.88, but apparently the correct answers are 6.8 and 2.613 respectively. I just can't  wrap my mind around it. I tried "relocating" the distribution to (0,10) but it still doesn't work!

Can anyone help me here? Thanks!


问:
When is it solvable:$10^a+10^b\equiv -1 \pmod p$
If $p$ is a prime, $(a,p)=1$,denote $ord(a,p)=d,$  where $d$ is the smallest positive integer solution to the equation $a^d\equiv 1 \pmod p$.We can prove that $$10^n\equiv -1 \pmod p\tag1$$ is solvable iff $ord(10,p)$ is even.

Now,consider this equation,$$10^a+10^b\equiv -1 \pmod p.\tag2$$
If $10$ is a primitive root modulo $p$, then there is an integer $a$ for every $b≠\dfrac{p-1}{2}\pmod {p-1}$ so that $a,b$ satisfies $(2)$.

My question is,what's the necessary and sufficient condition that $(2)$ has at least $1$ solution?

If we are given a prime $p$,how to determine whether $(2)$ is solvabe?

This is a way,but not effectively: for every positive integer $b\leq\frac{1}{2}ord(10,p)$,determine whether $(2)$ is solvabe for $a$.By this way,I find $(2)$ is solvable for these primes,which $10^n\equiv -1 \pmod p$ is not solvable:

$3,31,37,43,53,67,71,83,107,151,163,173,191,199,227,277,283,307,311,317,347,359,397,431,439,443,467,479,523,547,563,587,599,613,631,643,683,719,751,757,773,787,797,827,839,853,883,907,911,919,947,991,\cdots$

My original problem is: how many "$1$" is need at least for a decimal number which is consisting of "$0$" and "$1$" and divisible by $p$?This question is to find these primes that three "$1$" is need at least.

Thank you.

答:
The first few primes for which it is not solvable are
$$5, 11, 41, 73, 79, 101, 137, 239, 271, 281, 641, 733, 859, 1321, 1409, 2531, 2791, 3191, 3541, 4013, 4093, 4637, 4649, 5051, 5171, 5237, 6163, 6299, 7253, 7841, 8779, 9091, 9161, 11831, 12517, 12671$$

The sequence does not seem to be in the OEIS.

On the other hand, if you replace $10$ by $2$ there is https://oeis.org/A179113

EDIT: I doubt that there's a simple necessary and sufficient condition.  But here might be the start of a heuristic analysis which might suggest there should be infinitely many.

Heuristically, if the order of $10$ mod $p$ is $m$, we evaluate
$10^a + 10^b \mod p$ for approximately $m^2/2$ unordered pairs $(a,b)$, so the
probability that none of those is congruent to $-1 \mod p$ should be approximately
$\exp(-m^2/(2p))$.  So if you want to find primes $p$ for which your equation is not solvable,
you might look at those where  the order of $10 \mod p$ is less than about $\sqrt{p}$.

The order of $10 \mod p$ is $m$ or one of its divisors if $p$ divides $10^m - 1$.
So we might look for  primes $ p > m^2$ dividing $10^m - 1$.  Nearly all positive integers $x$ will have at least one prime factor greater than $\log_{10}(x)^2$ (see Dickman's theorem).


答:
I think $A \oplus \left( \bigoplus_i B_i \right)$ is the least bad-looking and also least ambiguous option. I think $A \bigoplus_i B_i$ is going to be confusing to parse. The two uses of $\oplus$ in $A \oplus \left( \bigoplus_i B_i \right)$ are referring to two different uses of the direct sum operation and the notation should reflect that. Similarly for ordinary sums I would write

$$a + \sum_i b_i$$

and not attempt to somehow collapse the $+$ and the $\sum$ into a single symbol. In the case of a direct product you can write $A \oplus \left( \prod_i B_i \right)$, of course.

Title: Notation for direct sum of direct sums

Body:
Suppose you have a direct sum of two terms $A$ and $B$: $$ A \oplus B.$$
Now suppose that in fact $B = \bigoplus_i B_i$. Then I guess we could write the above sum as
$$ A \oplus \bigoplus_i B_i.$$
However, this looks terrible -- much worse than, for instance,
$$ A \bigoplus_i B_i. $$
Is there a precedent to writing the latter, or some other notation better than the former?

Cheers.

P.S.: what about when B is a direct product instead of a direct sum?


问:
Evaluating a Lie Derivative of a Meromorphic Complex Form over a Holomorphic Vector Field.
Given a meromorphic form $\omega : \Omega^{1, 0}$, we can get the form $\; \mathrm{d}\omega = \partial \omega + \bar \partial \omega \;$ through the use of Dolbeault operators.

I'm trying to find the lie derivative $\mathcal{L}_v \; \omega$, where $v$ is a holomorphic vector field. How can I find this for non-holomorphic forms?

For example, suppose we allow the following definitions: $$ \omega = \frac{z+4}{z-3i} \; \mathrm{d}z, \quad v = 2z-1.$$

What is the Lie derivative $\mathcal{L}_v \;\omega$?

答:
Ironically enough, I feel able enough to finally respond  with an answer.

Given complex 1-form $$\omega = \alpha(z) \mathrm \; {d}z, \qquad \alpha \in \mathcal{M}(\mathbb{C}), \qquad \frac{\partial \alpha}{\partial \bar z} = 0 \; $$ and given vector field $v = \beta(z) \; \partial_z, $ where $\beta$ is holomorphic to $\mathbb{C}$, define the Lie derivative like such:

$$\mathcal{L}_v \; \omega = i_v \circ d\omega + d \circ i_v\omega.$$

$$\mathcal{L}_v \; \omega = \; \Big(\frac{\partial \alpha}{\partial z} + \frac{\partial \alpha}{\partial \bar{z}}\Big)\beta \;\mathrm{d}\bar z + \frac{\partial(\alpha \beta)}{\partial z} \, \mathrm{d}z + \frac{\partial(\alpha \beta)}{\partial \bar z} \, \mathrm{d}\bar z.$$

Using the fact that $\frac{\partial \alpha}{\partial \bar z} = \frac{\partial \beta}{\partial \bar z} = 0$, the final result can be determined:

$$\mathcal{L}_v \; \omega = (\alpha \beta)'_z \; \mathrm{d}z + \alpha'_z \beta \; \mathrm{d}\bar{z}.$$

This is a surprisingly beautiful and simple result.


问:
To derive quantity $k=\frac{2-p^2}{p+2}$ for positive rational $p$ such that $p^2>2$ and $(p+k)^2>2$
I am reading Baby Rudin and stuck at one small point.
$A$ is set of all positive rationals $p$ such that $p^2<2$ and $B$ is set of all positive rationals $p$ such that $p^2>2$. We wanted to show $A$ contains no largest element and $B$ contains no smallest element. So for any such $p$ in either set, author associated number $q$ as,

$q=p-\frac{p^2-2}{p+2}$, which works fine.

I was trying to figure out how to obtain the quantity $\frac{p^2-2}{p+2}$.

I did it for set $A$ as follows.

So, we have positive rational $p$ such that $p^2<2$. We want some $k>0$ such that $(p+k)^2 <2$

i.e. $p^2+2pk+k^2<2$

As all three quantities are positive, we can say

$p^2+2pk<2$, which gives

$k<\frac{2-p^2}{2p}$.

But as $p^2<2$, we have $p<2 \Rightarrow p+p=2p<p+2$

Therefore, we can choose $k=\frac{2-p^2}{p+2}$.

But I am stuck while deriving it for set $B$.

I did following.

So, we have positive rational $p$ such that $p^2>2$. We want some $k<0$ such that $(p+k)^2 >2$

i.e. $p^2+2pk+k^2>2$.

From this point, I can say $p^2+k^2>2$, but not sure how this will take me to the desired result.

Any hint or help. Thank you.

答:
We get $$(p+k)^2 >2\implies p+k\gt\sqrt 2\implies k\gt \sqrt 2-p$$

Multiplying $\sqrt 2-p$ by $\dfrac{\sqrt 2+p}{\sqrt 2+p}\ (=1)$ gives $\dfrac{2-p^2}{\sqrt 2+p}$, and so we have
$$k\gt \dfrac{2-p^2}{\sqrt 2+p}$$

It follows from $2\gt \sqrt 2$ and $2-p^2\lt 0$ that $\dfrac{2-p^2}{2+p}\gt\dfrac{2-p^2}{\sqrt 2+p}$.

Therefore, we can choose $k=\dfrac{2-p^2}{2+p}$.


问:
How can I become good at combinatorics and probability?
I apologize in advance if this question is too discussion-based, however, I do ask purely to get an answer/suggestion, and the site did prompt "Ask about... Understanding mathematical concepts and theorems."

So it's not that I struggle immensely with combinatorics but it seems like a branch of maths that either you're good at or you're not good at, and there's no way to improve unless you learn to think in a specific way. There's this concept of "combinatorial thinking" that I'm struggling to grasp. Basically, how can you ensure that you have considered all the cases? And there aren't really any "silly mistakes" you can make because it's just that you are thinking/interpreting the question incorrectly.

I could just do a bunch of problems but the problem with that is I usually attempt around 5 questions and then when I look at the solutions they are all wrong. For other topics, a question I get wrong might be because of careless algebra or reading the question incorrectly. Are there any excellent resources for learning combinatorics to a high level out there? I want to be someone who can solve "all" (most) combinatorics questions, and someone who people come to asking for help with combinatorics/probability questions.

答:
For the problems you do know how to solve, or for textbook examples, one method is to try to find a second way to solve to the problem. You can almost always do this in combinatorics and probability. This way, you know what you are trying to aim for and you are forcing yourself to learn many different techniques for these types of problems.

As you do more exercises you can continue to do this and you should hopefully see your confidence towards this topic building, as long as you also pay attention to your mistakes and trace your steps to see where you went wrong and adjust accordingly.


问:
How to solve a polynomial with fractional exponents that vary with time
I am trying to solve an equation for $R$ where all the parameters in the equation $(f, \gamma, c)$ vary with time. The equation is described as

$R^{(f+2\gamma+f)/\gamma}-R^{(f+2\gamma+f\gamma)/\gamma}=c$

The above equation needs to be solved at every time step. I have tried using a substitution of $z=R^{1/\gamma}$, but since the signals are time varying and of type double/float, the exponents might still be fraction at some time step.

I am trying to implement this equation in Simulink with $R$ as Output while others being the inputs. Is there a way to solve the equation described above (of type $x^a-x^b=c, $where $a,b$ are fractions) for each time step?

Since this is a very small part of the entire Simulink model, I am looking for something that can solve it without any explicit iterations or uses very little iterations.

I would greatly appreciate if any insights can be made about uniqueness of the solution.

答:
Consider that you look for the zero's of function
$$f(x)=x^a-x^b-c$$ Since the exponents are not integers, we must consider $x>0$.

To make it simpler, let $x^b=y$, $k=\frac a b$ and consider the function
$$g(y)=y^k-y-c$$ the solution of which being larger that $c^{\frac 1k}$. This will be the $y_0$ of Newton method.

Just for an example, using $a=\pi$, $b=e$, $c=12345$. The iterates will be
$$\left(
\begin{array}{cc}
n & y_n \\
 0 & 3468.8629 \\
 1 & 4583.1758 \\
 2 & 4551.1385 \\
 3 & 4551.1157
\end{array}
\right)$$ and then $x=22.1690$.

Edit

We could probably do much better, generating the starting point using one single iteration of Halley' method. This would give as a new starting point for Newton method
$$y_0=c^{\frac 1k}+\frac{2 c^{\frac 2k} \left(c k-c^{\frac{1}{k}}\right)}{k(k-5)  c^{\frac{1}{k}+1}+2
   c^{\frac 2k}+2 c^2 k^2}$$ Applied to the working case, this would give $y_0=4547.53$.


问:
Prove or disprove: if $R$ and $S$ are two equivalence relations on a set $A$ then $R\cup S$ is also an equivalence relation.
Does my disproof below suffice to disprove the statement below? Also, does anyone have a simpler way to disprove this?

If $R$ and $S$ are two equivalence relations on a set $A$ then $R\cup S$ is also an equivalence relation.

Disproof. Suppose $A=\{a,b,c\}$ and $R$ and $S$ are equivalence relations on $A$. Let

\begin{align*}
R&=\{(a,a),(b,b),(c,c),(a,b),(b,a)\} \\
S&=\{(a,a),(b,b),(c,c),(b,c),(c,b)\}
\end{align*}

Then

$$R\cup S=\{(a,a),(b,b),(c,c),(a,b),(b,a),(b,c),(c,b)\}$$

But then $a(R\cup S)b \land b(R\cup S)c \nRightarrow a(R\cup S)c$, thus $R\cup S$ is not transitive and therefore not an equivalence relation.

答:
This counterexample is perfectly valid.

To help see this, visualize each relation as a directed graph like below, where $x$ points to $y$ if and only if $x$ is  related to $y$ under the relevant relation. We have, for $R$ and $S$ respectively,

You can see these are equivalence relations because each property is demonstrated:

Reflexivity follows since every node points to itself.

Symmetry follows since, whenever a path extends from one node to another, a path going the opposite direction also exists.

Transitivity follows since, whenever three nodes are connected along a single path, there's also a path from the start point to the end point. (Loosely this looks like, whenever two sides of a triangle exist, a path exists by which you can "cut" from the start of the path to the end as well, filling in the triangle in a sense.)

The graph representing the union $R \cup S$ is given by effectively overlapping the pair:

Now we have two triangles where each is missing a side: the sides $a \to b$ and $b \to c$ are missing $a \to c$; likewise, the sides $c \to b$ and $b \to a$ are missing $c \to a$.

(That is, $(a,b),(b,c) \in R \cup S$ but $(a,c) \not \in R \cup S$. And similarly, $(c,b),(b,a) \in R \cup S$ but $(c,a) \not \in R \cup S$.)

Thus $R \cup S$ is not transitive. And, indeed, in general, unions of equivalence relations are not always themselves equivalences.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Automorphism of $\mathbb Q (\sqrt[5]{2}, \zeta_5)$

Show that if $\sigma \in \mathrm{Aut}(\mathbb Q (\sqrt[5]{2}, \zeta_5)/\mathbb Q)$, then $\sigma (\zeta_5) = \zeta^i_5$ for some $i=1,2,3,4$ and $\sigma (\sqrt[5]{2}) = \zeta^j_5 \sqrt[5]{2}$ for some $j=1,2,3,4,5$.

Help please.

答:
Hints:

You know that $\zeta_5^5=1$. Therefore
$$
1=\sigma(1)=\sigma(\zeta_5^5)=\sigma(\zeta_5)^5,
$$
so $\sigma(\zeta_5)$ has to be a fifth root of unity. It cannot be equal to $1$, for $\sigma$ is injective, and $\sigma(1)=1$. Can you see that this leaves the four listed alternatives for $\sigma(\zeta_5)$.

In the same way $\sigma(\root 5\of 2)$ has to be a fifth root of $2$. Can you see, why this leaves the five listed alternatives for  $\sigma(\root 5\of 2)$.

To get all 20 automorphisms (that you asked about in a comment to Zev's answer) you need to be able to argue in favor of the following claims:

Why is an automorphism $\sigma$ fully specified, when we know $\sigma(\zeta_5)$ and $\sigma(\root 5\of 2)$?

Why are we free to use all the $4\cdot5=20$ pairs of choices for $\sigma(\zeta_5)$ and $\sigma(\root 5\of 2)$?

The first item in the last bullet is straightforward field theory. For its part the second item is most conveniently handled by referring to suitable theorems of Galois theory, and is perhaps a bit trickier (although standard).


问:
Inductive algorithm: linear independent versus spanning lists
I'm trying to understand a proof in Axler's textbook which states that the size of any linearly independent list cannot exceed the size of any spanning list. He proves it algorithmically. The idea is to start with the full spanning s et $w_1, \ldots, w_n$, add the first element, $u_1$ of the linearly independent set, and then remove some element (say, $w_n$, without loss of generality) so that $u_1, w_1, \ldots, w_{n-1}$ still spans.

The "base step" involves removing one vector and adding $u_1$, which is fine. But then Axler assumes that it holds for steps $1$ through $j-1$. This is something I'be never been able to understand. In induction, I assumed that we say that $P(n)$ implies $P(n+1)$. If I started with $n-1$ and $n \in \{1, \ldots,\}$, then I need to take a base case of $n = 0$. But what exactly is the "zeroth" step of the algorithm? Do nothing, and notice that we still span?

If that is the case, and the $0$th step is just vacuously true and doesn't merit mention, I'm fine with this. Otherwise, should the proof technically address the zeroth step and/or add a $2nd$ step in and then assume $j \geq 2$?

答:
There is a conceptual problem here, which has nothing to do with Linear Algebra. It's about induction. The most common way of proving that all statements from a sequence $\{P(n)\mid n\in\Bbb N\}$ of statements are true consists of:

proving that $P(1)$ is true;

proving that, for each $n\in\Bbb N$, if $P(n)$ is true, then $P(n+1)$ is also true.

But we can do induction in a slightly different way. It consists of:

proving that $P(1)$ is true;

proving that, for each $n\in\Bbb N\setminus\{1\}$, if $P(n-1)$ is true, then $P(n)$ is also true.

It's actually the same thing, express in a slightly different way. And it's this approach that Axler uses.


问:
Radon-Nikodym derivative: Lebesgue measure
If we have two functions $F(t)=1/2(1-\exp(-t))$ and $G(t)=(1/2)(1+t)\exp(-t)$, where $2F$ and $G$ are two cumulative distribution functions. We know that, if $F=∫_{0}^{t} k(x)dG(x)$, $k$ becomes the Radon-Nikodym derivative. What is the exact value of $k$ (Lebesgue measure)? Is there a way to calculate this?

答:
You forgot to say that $F$ and $G$ are supported by $(0,\infty)$.

$\frac1  2(1-e^{-t})=\int_0^{t} k(x) \frac 1 2 (e^{-x}-(1+x)e^{-x})
dx$ so you get $k(.)$ by differentiating both sides w.r.t. $t$. This gives $k(x)=\frac 1 x, 0<x<\infty$.


问:
When is $C(X)$ reflexive?
Let $X$ be a compact Hausdorff space. What are sufficient and necessary conditions on $X$ under which $C(X)$ would be a reflexive Banach space.

Is there a non reflexive Banach space $C(X)$ such that $C(X)$ is isomorphic to its bidual?

答:
Theorem. Let $X$ be a compact Hausdorff space. Then $C(X)$ is reflexive iff $X$ is finite.

Proof (elementary, I hope): A finite Hausdorff space $X$ is clearly discrete, so if $X$ is finite, then $C(X)$ is isomorphic to $\mathbb{K}^X$ (where $\mathbb{K}=\mathbb C$ or $\mathbb{K}=\mathbb R$), hence reflexive.

Assume then that $X$ is infinite. By the Banach-Alaoglu theorem [Rudin FA: 3.15], it suffices to show that the weak unit ball $U$ of $C(X)$ is not compact, e.g., that $\mu(U)$ is not compact by some $\mu\in C(X)^*$, as "Continuous images of compact sets are compact" [Google]. (Indeed, then $C(X)$ cannot be a dual space, by Banach-Alaoglu.)

But $C(X)^*=rca(X)$. https://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space

$\mu=\sum_k 2^{-k} (-1)^k 1_{x_k}\in rca(X)$ has supremum $1$ on $U$ but no maximum, so we are done. (Here $1_x$ is the unit mass at $x$.)

Indeed, let $A_n=1,3,...,2n-1$, $B_n=2,4,...,2n$ (both closed, being finite unions of singletons), and let $f_n(A_n)=-1$, $f_n(B_n)=1$, $f_n\in C(X,[-1,1])$, by https://en.wikipedia.org/wiki/Urysohn%27s_lemma

Then $f_n\in U$ and $\mu(f_n)\to 1$, as $n\to\infty$.

But if $g\in U$, $\mu(g)=1$, then $g(x_n)=(-1)^n$. Hence, then $g(x_n)$ is not a Cauchy sequence, violating $g(x_n)\to g(x)$ (i.e., $g$ is discontinuous), a contradiction, so $1\not\in \mu(U)$, QED.

Title: When is $C(X)$ reflexive?


问:
A sum of inversion preserving summands using CS inequality
Suppose that $p_k>0$ for $1\leq k\leq n$ and $\sum_{k=1}^np_k=1$. Show that
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\geq n^3+2n+\frac1n$$

Here's my try using Cauchy-Schwarz inequality:
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\cdot\sum_{k=1}^np_k^2\geq\left(\sum_{k=1}^n\left(p_k^2+1\right)\right)^2=n^2+2n\sum_{k=1}^np_k^2+\left(\sum_{k=1}^np_k^2\right)^2$$
However,
$$\sqrt{\frac{\sum_{k=1}^np_k^2}n}\geq\frac{\sum_{k=1}^np_k}n=\frac1n$$
Or,
$$\sum_{k=1}^np_k^2\geq\frac1n$$
Hence we have
$$\sum_{k=1}^n\left(p_k+\frac1{p_k}\right)^2\cdot\sum_{k=1}^np_k^2\geq n^2+2+\frac1{n^2}$$

I'm stuck here now.

答:
Noticed that $RHS=\dfrac{\left(n^2+1\right)^2}{n}$, We only need to prove
$$
n\sum_{k=1}^n \left(p_k+\frac{1}{p_k}\right)^2 \ge \left( n^2+1 \right) ^2
$$

Since $n = \displaystyle\sum_{k=1}^n 1$,

$$
\sum_{k=1}^n {1} \cdot \sum_{k=1}^n {\left(p_k+\frac{1}{p_k}\right)^2} \ge \left[ \sum_{k=1}^n {\left(p_k+\frac{1}{p_k}\right)} \right]^2 = \left( 1+\sum_{k=1}^n {\frac{1}{p_k}} \right)^2
$$

and
$$
\sum_{k=1}^n {\frac{1}{p_k}} = \sum_{k=1}^n {p_k} \cdot \sum_{k=1}^n {\frac{1}{p_k}} \ge \left( \sum_{k=1}^n \left({p_k\cdot\frac{1}{p_k}}\right) \right)^2 = n^2
$$


问:
Why do we exclude $0$ and units from the definition of irreducible elements?
Why do we exclude $0$ and units from the definition of irreducible elements? More clearly how can it be shown that for the units and $0,$ $c=ab$ doesn't always imply either $a$ or $b$ is a unit (where $c = 0$ or $c$ is a unit)?

For $0$ my guess is that $0=0\cdot c$ for some non-unit $c$.

答:
Zero and units are excluded because it would generally make factorisation meaningless.  For example, we say $6$ has two factors $2$ and $3$.  These are the normal forms of the spaces of 2 by a unit, and 3 by a unit.

Were units to be included, the factorisation could be $2\cdot-3\cdot-1$, which suggests three factors.  However $-3$ is a variety of $3$, in that each divides the other, and the product with $-1$ divides it without $-1$.

So factorisation is looking for numbers whose product divides the number and vice versa, but any lesser product is not divisible by the number.  It is these that function as prime-like things.

The only time one includes units, if if the product of the primes in their normal forms do not give the expressed number, so eg $-6=-1\cdot2\cdot3$.

$0$ is also excluded, simply because it does not divide anything unless it's zero itself.  So any non-zero thing is not a product of elements including $0$


问:
Conservative force: path-independent integration implies force equal to gradient
Following the wikipedia page of a Conservative force, I want to prove that given a force such that for any closed path $C$ we have
$$
\oint_C \mathbf{F}\cdot \mathbf{r}=0
$$
there is a potential $U(\mathbf{r})$ such that
$$
\mathbf{F}=-\nabla U(\mathbf{r}).
$$

Proof.
The webpage start by defining
$$
U(\mathbf{r}):=-\int_{c}\mathbf{F}\cdot d \mathbf{r}
$$
and conclude that the claim follows from the fundamental theorem of calculus.

I want to deepen the last step because the derivation is far from obvious for me.

First I adapt the definition of potential by choosing some zero potential point $\mathbf{x}_0$:
$$
U(\mathbf{r}):=-\int_{\mathbf{x}_0}^\mathbf{x}\mathbf{F}\cdot d \mathbf{r}
$$
then I introduce a parametrisation of the path
$$
\mathbf{s}(u),\quad u\in[0,1]
$$
Thus we get
$$
-\nabla \left( -\int_{c}\mathbf{F}\cdot d \mathbf{r} \right)
=\nabla\int_0^1 \left( F_x\frac{dx}{du} +  F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du
$$
For the $x$ component, intuitively but with a lot of errors, we must have something like this
$$
\frac{d}{dx}\int_0^1 \left( F_x\frac{dx}{du} +  F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du
=\frac{du}{dx}\frac{d}{du}\int_0^1 \left( F_x\frac{dx}{du} +  F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)du\\
=\frac{du}{dx}\left( F_x\frac{dx}{du} +  F_y\frac{dy}{du}+ F_z\frac{dz}{du}\right)_{\mathbf{x}_0}^\mathbf{x}=\cdots=F_x
$$

How does the derivation looks like using formal mathematics, but remaining in calculus?

EDIT My solution going back to the definition of derivative:

答:
Since no one else has written an answer here, I'll write a quick one now and I might edit it later if it isn't clear.  I'll give an intuitive rather than  a formal proof; it seems like this is what you're interested in.

Calculating the gradient

What does $-\nabla U(\textbf{r})$ mean for $U$ as defined in your question?  Taking a derivative means that we are modifying the endpoint of the contour used in the integral and observing how $U$ changes when we do this.

The way we formalize this is to consider two contours, $c$ and $c + \delta c$, which end at slightly different locations $\textbf{r}$ and $\textbf{r} + \delta\textbf{r}$.

For simplicity, I'll assume that $\delta c$ is just a straight line from $\textbf{r}$ to $\textbf{r} + \delta\textbf{r}$.  It actually doesn't matter because we are taking a derivative, so $\delta \textbf{r}$ is arbitrarily small.  (Since $c$ is generally assumed to be piecewise differentiable, then if it is small it is basically a straight line.)

In a more formal treatment we would probably discuss the gradient by considering derivatives along an arbitrary direction, but for now just consider the derivative of $U$ with respect to $x$.  In this case, $y$ and $z$ are not changing, so we can write write $\delta \textbf{r} = (\delta x, 0, 0)$.
$$ \begin{align} \frac{\text{d}U}{\text{d}x} = - \frac{\text{d}}{\text{d}x} \int^\textbf{r} \textbf{F} \cdot \text{d}\textbf{r} &= - \lim_{\delta x \to 0} \frac{1}{\delta x} \left( \int^{\textbf{r} + \delta \textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} - \int^{\textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} \right) \\
&= - \lim_{\delta x \to 0} \frac{1}{\delta x} \int^{\textbf{r} + \delta \textbf{r}}_{\textbf{r}} \textbf{F} \cdot \text{d}\textbf{r} \\
&= - \lim_{\delta x \to 0} \frac{1}{\delta x} \int_0^1 \left( F_x \delta x + F_y \cdot 0 + F_z \cdot 0 \right) \text{d}t \\
&= - \lim_{\delta x \to 0} \frac{1}{\delta x} F_x \delta x \\
&= - F_x \end{align} $$

The first line is just applying the definition of a derivative to this case.  The second line is a basic property of integrals.  The third line is just parameterizing, along with $\delta y = 0$ and $\delta z = 0$.  From there it's trivial.

Now you could do the same thing for $y$ and $z$ and find that
$$ \nabla U = - \textbf{F} $$
which is what you are trying to show.  Basically, the thing does boil down to the fundamental theorem of calculus in one dimension, but you just have to set it up correctly.

Path-independence

Now, it's important to point out that there is one really important implicit assumption that I made above, which is that when I write the integral as $\int^\textbf{r}$, I am implicitly assuming that the result does not depend on the particular path taken.  This is only true if $\textbf{F}$ is a "conservative" force, which is the whole point of your question; I'll briefly explain why this is justified with this assumption.

If we define a conservative force as one where every closed path has a contour integral of zero, as you have above, then it's very easy to show that this is true.  Suppose that I have two paths $c_1$ and $c_2$ that start and end at the same point.  Do they result in the same value of $U$ at the endpoint?  Well, if you go "forwards" along $c_1$ and "backwards" along $c_2$, then you have a closed path and the integral along the total path $c_3$ has to be zero.  This gives us
$$ \int_{c_1} \textbf{F} \cdot \text{d}\textbf{r} - \int_{c_2} \textbf{F} \cdot \text{d}\textbf{r} = \int_{c_3} \textbf{F} \cdot \text{d}\textbf{r} = 0 $$
so the integrals along any two paths $c_1$ and $c_2$ with the same start and end points must be equal, and we can ignore the particular path taken, as I did in the derivation above.


问:
Finding the equation of a plane that passes through two points and is parallel to a line
Find the algebraic equation of the plane that passes through the origin, is parallel to the line $\mathbf X= (-1, 0, 2) +t(2,-1,1)$, and contains the point $P=(1,2,3)$.

I suspect you can do this with a cross-product, but our professor isn't letting us use that as he hasn't taught it to us yet. Is there a way to get another point on the plane from the parallel line, so you can make a system of equations to solve for an orthogonal vector?

答:
Given the points $\{p_0,p_1\}\in \Pi$ and the line $L\to p = p_2+\lambda \vec v$, determine a plane $\Pi$ parallel to $L$.

Characterizing the plane as $\Pi\to(p-p_0)\cdot\vec n=0$ we have the properties

$$
\cases{
(p_1-p_0)\cdot \vec n = 0\\
\vec v\cdot \vec n = 0\\
\|\vec n\|=1
}
$$

The last comes from normalization purposes.

Now as

$$
\cases{
p=(x,y,z)\\
p_0=(0,0,0)\\
p_1 = (1,2,3)\\
\vec v = (2,-1,1)
}
$$

we have to solve

$$
\cases{
(1,2,3)\cdot\vec n = 0\\
(2,-1,1)\cdot\vec n = 0\\
\|\vec n\|=1
}
$$


问:
Application of Induction in the analyze of the convergence a sequence defined recursive.

Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.

How can find $a_{n}$?

Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.

答:
Without induction.

If you follow the steps described here which I used answering this question making the story short
$$a_{n+1}=\frac{1}{4-3 a_{n}} \qquad \text{with} \qquad a_1=c$$
$$a_n=\frac 13\frac{c \left(3^n-9\right)-(3^n-3) } {c(3^n-3)-(3^n-1) }$$

Now, you need to consider the various cases to arrive to the nice results from  @Brian M. Scott's nice analysis.


问:
$f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

if $f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

What I know from Rolle's theorem that between every 2 roots of the function theres atleast 1 root of the derivative, so for example $f'(x)=0$ has atleast 3 roots.
 I'm struggling to get how im supposed to know its Exactly 3 roots and not atleast without differentiating since that's what the question is asking for and I feel like im missing something.
 any help is really appreciated.

答:
$p(x)$ has $4$ real roots.

$ x<2 :$ $ p(x) >0$;

$x \in (2,7);$ $p(x) <0;$

$x \in (7,9);$ $p(x)>0;$

$x \in (9,12):$ $p(x)<0;$

$x>12;$ $p(x)>0;$

$p(x)$ has a rel. minimum $\in (2,7);$  a rel, maximum $\in (7,9);$ a relative minimum $\in (9,12)$.

$p'(x)$ has degree $3$ and has $3$ real roots in between the zeroes of $p(x)$ (Rolle).

$p''(x)$ has degree $2$ and has $2$ real roots in between the zeroes of $p'(x)$(Rolle) .


答: The signs of $f$ on $(-\infty,\,2),\,(2,\,7),\,(7,\,9),\,(9,\,12),\,(12,\,\infty)$ are $+-+-+$, so there are three turning points, i.e. three roots of $f^\prime=0$. Similarly, $f^\prime$'s left-to-right signs progression is $-+-+$, so between its three roots are two roots of $f^{\prime\prime}=0$. This may be easier to visualize if you sketch $f$ against $x$, which will look like this.

Body:
if $f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

What I know from Rolle's theorem that between every 2 roots of the function theres atleast 1 root of the derivative, so for example $f'(x)=0$ has atleast 3 roots.
 I'm struggling to get how im supposed to know its Exactly 3 roots and not atleast without differentiating since that's what the question is asking for and I feel like im missing something.
 any help is really appreciated.


答:
Rudin presented the following equation which represents an approximation of $\sqrt{2}$ :
$$
q = p - \frac{p^{2} - 2}{p + 2} = \frac{2p+2}{p+2}
\tag3$$
The equation presented in the book with no explanation from where it came from will be explained here. There are a number of ways to derive equation I chose to derive it geometrically using what is known as the secant method

This method involves drawing a secant into the curve of the function $f(p)=p^{2}-2$ this will result in formation of three important points, two points formed by the intersection of secant and the curve will be projected at $x-axis$ and will be denoted as $S_{n-1}$ and $S_{n}$ and a third point which is formed from the intersection of the secant line with $x-axis$ will be denoted as $S_{n+1}$. A recursive relation is established and is given by :
\begin{equation*}
S_{n+1}=\frac{S_{n-1}f(S_{n}) - S_{n}f(S_{n-1})}{f(S_{n})-f(S_{n-1})}
\end{equation*}
We will denote $S_{n-1} = t$ , $S_{n} = p$ , $S_{n+1} = q$ then our equation becomes :
\begin{equation*}
q=\frac{t(p^{2}-2)-p(t^{2} - 2)}{p^{2}-2-(t^{2}-2)} = \frac{(p-t)(pt+2)}{p^{2}-t^{2}} = \frac{pt+2}{p+t}
\end{equation*}
from the expression $q= \frac{pt+2}{p+t}$ we want to study when $q$ is more accurate than $p$ and when $p$ is more accurate than $q$ and thus we would conclude to which set they would belong in these two cases. So we would want to study the sign of $q-p$ whether its positive or negative. Thus we have that :
\begin{equation*}
q-p= \frac{pt+2}{p+t} - p = \frac{pt+2-p^{2}-pt}{p+t} = \frac{2-p^{2}}{p+t}
\end{equation*}
\begin{equation*}
\therefore q-p=-\frac{p^{2}-2}{p+t}
\end{equation*}
Now notice how the numerator is independent of $t$ yet the denominator is so whatever positive value of $t$ you choose it will not affect the comparison between $p$ and $q$ because what we really wanna look at is when $p^{2}-2 <0$ which is case of set A and when $p^{2} - 2>0$ which is case of set B. So had the author used the Secant Method to derive his expression, he would have picked the easiest number he could think of  $t=2$ in order to obtain the required equation $(3)$.

Title: To derive quantity $k=\frac{2-p^2}{p+2}$ for positive rational $p$ such that $p^2>2$ and $(p+k)^2>2$

Body:
I am reading Baby Rudin and stuck at one small point.
$A$ is set of all positive rationals $p$ such that $p^2<2$ and $B$ is set of all positive rationals $p$ such that $p^2>2$. We wanted to show $A$ contains no largest element and $B$ contains no smallest element. So for any such $p$ in either set, author associated number $q$ as,

$q=p-\frac{p^2-2}{p+2}$, which works fine.

I was trying to figure out how to obtain the quantity $\frac{p^2-2}{p+2}$.

I did it for set $A$ as follows.

So, we have positive rational $p$ such that $p^2<2$. We want some $k>0$ such that $(p+k)^2 <2$

i.e. $p^2+2pk+k^2<2$

As all three quantities are positive, we can say

$p^2+2pk<2$, which gives

$k<\frac{2-p^2}{2p}$.

But as $p^2<2$, we have $p<2 \Rightarrow p+p=2p<p+2$

Therefore, we can choose $k=\frac{2-p^2}{p+2}$.

But I am stuck while deriving it for set $B$.

I did following.

So, we have positive rational $p$ such that $p^2>2$. We want some $k<0$ such that $(p+k)^2 >2$

i.e. $p^2+2pk+k^2>2$.

From this point, I can say $p^2+k^2>2$, but not sure how this will take me to the desired result.

Any hint or help. Thank you.


问:
Are there real numbers a and b such that $f(x,y,t) = x^a t^b$ satisfies the heat equation?
The question is in the title. The heat equation is as follows:
$$
\frac{\partial f}{\partial t} = k \left( \frac{\partial^2 f}{\partial x^2} +  \frac{\partial^2 f}{\partial y^2} \right),\quad k\in\mathbb{R}
$$

Attempt at solution

Plugging the requested form into the above equation yields:

$$
\frac{\partial f}{\partial t} = b\ t^{b-1} x^a\\
\frac{\partial^2 f}{\partial x^2} = a (a-1)\ t^b x^{a-2}\\
\frac{\partial^2 f}{\partial y^2} = 0
$$

Which leads to showing that:
$$
b\ x^2 = k\ a(a-1)\ t
$$

I'm not sure how to proceed from this point. Is this the correct procedure to solve this problem? Thanks!

答: From your derivations, it seems to me that $b=0$ and $a=1$ would work fine for instance...</p>