问:
How can I determine the cardinality of a set of polymorphic functions?
It seems obvious to me that the set of functions with the signature $\forall A. A \rightarrow A$ is "once-inhabited", i.e. there is only one such polymorphic function which "works" for any set $A$, which is the identity function. (I think perhaps mathematicians would consider this to be actually a whole family of functions; I'm coming from CS where we tend to talk about single polymorphic functions, but feel free to explain how a mathematician would phrase the problem). However I don't know how to prove it except in a very hand-wavy manner.

A slightly more complex example would be to show that $$\forall A B C. (A \rightarrow B) \rightarrow (B \rightarrow C) \rightarrow (A \rightarrow C)$$ is also once-inhabited, and that the only such function with this signature is the function composition operator.

答:
I'll try my best to answer your question, but we're on some shaky ground, with regards to foundations of what you call "sets".

However, I've found similar questions are framed better in terms of category theory, which has a much better notion of what you mean by polymorphisms and a "once-inhabited" function space. In the terminology of categories, polymorphic functions are called "natural transformations". There's a bit more going on, but the idea is essentially the same. Similarly, instead of saying "once-inhabited", you'd say that such a natural transformation is "unique up to isomorphism".

Regardless, the way I usually see things like that proved is to let $A$ have only on element. In such a case, there is only one function defined on $A \to A$. So that special case gives you the universal case. A similar technique will probably work for your composition example.

UPDATE:

Here's an example of how I'd prove this using the language of category theory.

Take the category $\textbf{Set}$ of all sets and functions between them. It is well known that the terminal object in $\textbf{Set}$ is the singleton set. In other words, for any set $|A| = 1, A = \{ a \}$, there is at most one function $f : X \to \{ a \}$, for any other set $X$. In other words, $f$ must be the constant function.

Some authors take the empty function ($\emptyset \to \emptyset$) to be a well defined concept (although I personally don't) to make sure that $\textbf{Set}$ is, in fact, a category.

Personally, I find the notion to be a bit silly. If you're coming from a CS background, it's probably closer to the intended semantics to work in $\textbf{Set}_{\bot}$, the category of pointed sets and partially defined functions, since most programs people write are not total functions. And $\bot$ is everywhere you'd want $\emptyset$ to be.

Now consider the function space $\forall A, B, C, (A \to B) \to (B \to C) \to (A \to C)$ where $A, B, C$ are objects in $\textbf{Set}$. To determine the cardinality of this space, as $A, B$ and $C$ vary on $\textbf{Set}$, consider the extreme case where $A, B$ and $C$ are all terminal objects. Then the individual function spaces $f : A \to B$, $g : B \to C$ and $h : A \to C$ all contain, at most, one element. Hence $\forall f, g, h, f \to g \to h$ can, at most, range over one element each: $f, g$, and $h$. Thus, this function space also only has one element.

Side-note

If you find questions like this interesting, and find yourself asking similar questions, I would take some time to read up on categories. They are a lot of fun and can make precise a lot of vague ideas that are not well defined in computer science, making many hard-to-define questions worth asking.


问:
Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$ is unconditionally convergent.
The problem I have is I don't see at which point I'm making a mistake. Since absolute convergence implies unconditional convergence, I want to show that
$$\sum_{n=1}^{\infty} |\frac{(-1)^{n-1}}{n} i^n|$$
converges. But
$$\sum_{n=1}^{\infty} |\frac{(-1)^{n-1}}{n} i^n| = \sum_{n=1}^{\infty} \frac{|(-1)^{n-1}|}{|n|} |i^n| = \sum_{n=1}^{\infty} \frac{1}{n} \to \infty.$$

答:
In $ \mathbb C$, unconditional convergence is equivalent to absolut convergence.

If $a_n=\frac{(-1)^{n-1}}{n} i^n$, then $|a_n|= \frac{1}{n} .$

Hence $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$ is not  unconditionally convergent !

ContentLicense: CC BY-SA 4.0


问:
Inverse limit without category theory
In Hirsch's book Differential Topology, he by and large does not use any category theory, with the exception of one passage on pg. 52 which I am trying to understand.  It is as follows (paraphrased):

Let $X$ be a set and $\mathcal{A}$ a family of subsets, such that $\mathcal{A}$ contains $X$ and is closed under unions.  Suppose we have a contravariant functor from the partially ordered (by inclusion) set $\mathcal{A}$ to the category of sets.  That is, to every $A\in\mathcal{A}$ there is associated a set $\mathcal{F}(A)$, and to every pair of sets $A,B\in\mathcal{A}$ with $A\subset B$ there is assigned a map of sets $\mathcal{F}_{AB}:\mathcal{F}(B)\rightarrow \mathcal{F}(A)$, denoted by $x\mapsto x|A$, where $x\in\mathcal{F}(B)$ such that $\mathcal{F}_{AB}\mathcal{F}_{BC}=\mathcal{F}_{AC}$ whenever $A\subset B\subset C$, and $\mathcal{F}_{AA}=$ identity map of $\mathcal{F}(A)$.

...

A structure functor is continuous if the following holds.  If $\{Y_\alpha\}$ is any simply ordered family of elements of $\mathcal{A}$, and $\cup Y_\alpha=Y$, then the inverse limit of the maps $\mathcal{F}_{Y_{\alpha}Y}:\mathcal{F}(Y)\rightarrow \mathcal{F}(Y_\alpha)$ is a map $\mathcal{F}(Y)\rightarrow \text{inv lim} \mathcal{F}(Y_\alpha)$.

(It seems kind of strange that he explains what a functor is, but assumes the reader knows what an inverse limit is, but whatever.)

So I am trying to understand the last paragraph--in particular, what an inverse limit is, and what a continuous structure functor is--without knowing anything really about category theory beyond the definition of a functor given in the first paragraph.  Is there a way to understand the inverse limit in this specific instance without knowing category theory more generally?

(If the answer is simply that I need to go read up on category theory, that's fine, and I would appreciate if someone would tell me so.)

答:
Since $\mathcal{F}(A)$ is a set for every set $A$, we have a a functor $\mathcal{F}:Sets\to Sets$.

Therefore, we can use the explicit construction of the inverse limits for the category of sets as a subset of a product set. As the wikipedia secton that I have linked says, this construction works for a number of categories such as topological spaces, rings and many others.

The inverse limit satisfies some properties that characterize it up to unique isomorphism, so in particular there might be instances in which a different definition of the inverse limit set is more convinient as long as it satisfies those properties. For example, in the situation of your book, when the chain of maps is a chain of inclusions, the inverse limit is characterized by the union of the sets in the chain. Therefore $Y=\cup Y_\alpha$ is precisely the inverse limit of the $Y_\alpha$'s with inclusions.

Your book defines a continuous functor as one that satisfies $\mathcal{F}(Y)=\lim \mathcal{F}(Y_\alpha)$. More generally, in analogy with continuous functions in topology, a continuous functor is one that preserves limits. This is the case here, since $Y=\lim Y_\alpha$, we have $\mathcal{F}(\lim Y_\alpha)=\lim\mathcal{F}(Y_\alpha)$. I don't think the author really adds any mathematical meaning by saying "structure functor" other than just saying that functors preserve structure (in this case, it sends sets to sets and maps of sets to maps of sets). I might be wrong if he uses this expression before to mean something in particular.

And of course I suggest you to read some category theory, at least the basic definitions of categories and functors, but for this particular question I think you can skip it for now.

Title: Inverse limit without category theory


问:
Is every bounded operator normal?
Let $A: {\scr H} \to {\scr H}$ be a linear bounded endomorphism on a Hilbert space $\scr H$, in brief:
$A \in {\frak B} ({\scr H})$.

If it's true that $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A \in {\frak B} ({\scr H}) $ and $A^*A \geq 0 $, which implies $A^*A$ self-adjoint (every positive bounded operator is self-adjoint), and hence $A^*A=A A^*$ ($^*$ is an antilinear involution)

Then  $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A=AA^*$, which is precisely the definition of normal operator.

So every bounded linear operator is normal. It can't be true... yet I don't see the error.

答:
Note that $A^*A$ is self-adjoint, not $A$ itself. So we can conclude
$$(A^* A)^*(A^* A) = (A^* A)(A A^*)^*$$
for any $A \in \mathscr{B}(\mathscr{H})$, which is an entirely more reasonable proposition.

Title: Is every bounded operator normal?


答:
In general, $ A^*A=AA^*$ is not true !

For example let ${\scr H}= \ell^2$ and $A$ given by $A(x_1,x_2,x_3,...)=(0,x_1,x_2,x_3,...).$

Title: Is every bounded operator normal?

Body:
Let $A: {\scr H} \to {\scr H}$ be a linear bounded endomorphism on a Hilbert space $\scr H$, in brief:
$A \in {\frak B} ({\scr H})$.

If it's true that $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A \in {\frak B} ({\scr H}) $ and $A^*A \geq 0 $, which implies $A^*A$ self-adjoint (every positive bounded operator is self-adjoint), and hence $A^*A=A A^*$ ($^*$ is an antilinear involution)

Then  $\ \forall A \in {\frak B} ({\scr H}) \quad A^*A=AA^*$, which is precisely the definition of normal operator.

So every bounded linear operator is normal. It can't be true... yet I don't see the error.


问:
Limit from both sides in theorem on conditional expectation
In order to prove that

a) $Y = E[X|\mathfrak{F}]$

iff

b) For all $Z \in L_2$: $\langle X-Y,Z \rangle = 0$

our script introduces $t \in \mathbb{R}$ and the proof of a)->b) ends with an argument that is based on taking the limit from both sides of $0$.

I understand everything up to the part where we get:

(*) $t \Vert Z \Vert_2^2 -2\langle X-Y,Z \rangle \geq 0$

But now we say that for $t \searrow 0 $ we get:

$\langle X-Y,Z \rangle \leq 0$

and for $t \nearrow 0$:

$\langle X-Y,Z \rangle \geq 0$

This last step confuses me. I can't figure out how we got from (*) to the last inequality.  My mind is also a bit hazy on how this argument works in the first place. So if you could, please tell me how the last inequality comes to pass and maybe explain shortly how this argument works logically. Thank you

答: $(*)$ does not yield the proof. I think you already divided an inequlity by $t$ before arriving at $(*)$. That was a mistake. You cannot divide an inequality by a negative number (unless you change the inequality sign). So you have to use the fact that $(*)$ holds when $t >0$ and $t\|Z\|_2^{2}-2\langle (X-Y), Z \rangle \leq 0$ when $t <0$.


问:
prove a sequence $(f_n(x))_{n=1}^\infty$ converges
I am new to math stack exchange. I have been introduced to functions $f_{1},f_{2},f_{3}...$, domain is $[0, \infty)$, and just summarized as $$f_{n}(x)=\begin{cases}
      \frac{1}{n} & \text{if } x \in [0, n) \\
      \frac{-1}{n}x+1+\frac{1}{n} & \text{if } x \in [n, n+1) \\
      0 & \text{if } x \in [n+1, \infty) \\
   \end{cases}$$

How do I prove a sequence $(f_n(x))_{n=1}^\infty$ converges to the function $f:[0, \infty)\to \mathbb{R}$ given by $f(x)=0, \forall x\in [0, \infty)$?

This form of the question is very confusing, so I think it is better changing it to:
$$\forall x \in [0, \infty), \lim_{n \to \infty}f_n(x)=f(x).$$
What are the next steps?

答:
First, we would want to start with the pointwise limit. Taking the limit as n tends to infinity we get:
Let $x$ be a real number. Then there exists $n  \in \mathbb{N}$ such that $x<n$. Then
$$\lim f_{n}(x)=\lim{\frac{1}{n}}=0$$
Thus, the only candidate for an uniform limit is the function $f(x)=0$. We have that:
$$||f_n-f||_{x\in\mathbb{R}}>0$$ This holds for all $\mathbb{N}$, since there is a part of the function (in a set of nonzero measure for al $n\in\mathbb{N}$) that is a line different from the constant line $y=0$, so the sequence does not converge uniformly.


问:
Determine whether the given set of image of piecewise function as follow is bounded
Let $f:\Bbb R \to \Bbb R$ be a function with
\begin{equation*}
f(x) =
\begin{cases}
0, &\text{if} \ x \lt -1 \\
x^2+1, &\text{if} \ x \in [-1,1] \\
\frac{1}{2x}, &\text{if} \ x \gt 1
\end{cases}
\end{equation*}

Is it true that the set $A=\{f(x) \mid x \in \Bbb R\}$ bounded? Explain it.

Find the infimum and the supremum of the set $B = \{f(x)+2 \mid x \in \Bbb R\}$.

Attempt:
For $1$, the first think I do is to partition $f$ into three cases as follows.

Case 1. $x \in (-\infty,-1)$.
Let $C = \{f(x) \mid x \in (-\infty,-1)\}$.
By definition, $f$ is a constant (zero) function.
Then, $C = \{0\}$ and so, $C$ is bounded.

Case 2. $x \in [-1,1]$. Let $D = \{f(x) \mid x \in [-1,1]\}$. Then,
$D = [1,2]$ and so, $D$ is bounded.

Case 3. $x \in (1, \infty)$. Let $E = \{f(x) \mid x \in (1,\infty)\}$.
Then, $E = \left(0,\frac{1}{2}\right)$ and so, $E$ is bounded.

Hence, from three cases above, we conclude that $A$ is bounded.

${}$

For $2$, notice that $B = \{2\} \cup [3,4] \cup \left(0,\frac{5}{2}\right)$. Hence, $\inf(B) = 0$ and $\sup(B) = 4$.

Am I true?

答: Your answer to the first question is correct. Bu the other on is not. In fact, $B=\{2\}\cup[3,4]\cup\left(2,\frac52\right)$. Therefore, $\inf(B)=2$ and $\sup(B)=4$.


问:
Relation between entourages and finite open covers of compact Hausdorff spaces
Let $X$ be a compact Hausdorff space and $\mathcal{V}$ be a finite open cover of $X$. Since $X$ is a compact Hausdorff space, hence it has a uniformity $\mathcal{U}$. Is there an entourage $D$ in  uniformity $\mathcal{U}$, such that if $(x, y)\in D$, then there is open set $V$ in finite open cover $\mathcal{V}$, $V\in\mathcal{V}$, such that $\{x, y\}\subseteq V$.

Would you please help me to know it?

答:
For every $x \in X$ we find some $V_x \in \mathcal{V}$ such that $x \in V_x$.

We can find an entourage $U^x_1 \in \mathcal{U}$ such that $U^x_1[x] \subseteq V_x$, because $\mathcal{U}$ is the uniformity for this topology.

By standard axioms we can find a symmetric entourage $U^x_2 \in \mathcal{U}$ such that $U^x_2 \circ U^x_2 \subseteq U^x_1$.

Then $\{U^x_2[x]\mid x \in X\}$ is a cover of $X$ that has a finite subcover, say indexed by $x_1,\ldots, x_n$.

Then $U=\bigcap_{i=1}^n U^{x_i}_2$ is also in $\mathcal{U}$ and  has the required property that $(x,y) \in U$ implies there is some $V \in \mathcal{V}$ that contains $\{x,y\}$.


问:
computation of the inverse of large dimensional Kronecker sum
Suppose that both $A$ and $B$ are $p \times p$ positive definite matrices.

I want to compute
$X=\left[ (A\otimes I_p) + (I_p \otimes A)  \right] \left[ (B \otimes I_p) + (I_p \otimes B) \right]^{-1} (B^2 \otimes B^2)$.

When $p$ is small, it is straightforward to obtain $X$. However, when $p$ is large, since $X$ includes a huge matrix of dimension $p^2 \times p^2$, computation of $X$ is very time consuming or almost impossible.

I'm wondering if there are efficient ways to compute $X$ even for large $p$.

答:
By the spectral theorem we can write
$$
B = U D U^*
$$
for some $D>0$ and unitary $U$. Then
$$
B \otimes I + I \otimes B = (U \otimes U) (D \otimes I + I \otimes D) (U^* \otimes U^*).
$$
Thus To compute the inverse of $B \otimes  I + I \otimes B$ we need only compute
$$
(U \otimes U) (D \otimes I + I \otimes D)^{-1} (U^* \otimes U^*)
$$
and now the inverse is on a diagonal matrix (which we compute from $B$ only) and hence is much faster. We can also absorb the multiplication by $B^2 \otimes B^2$ into this computation as
$$
(B \otimes I + I \otimes B)^{-1}(B^2 \otimes B^2) = (U \otimes U) (D \otimes I + I \otimes D)^{-1}(D^2 \otimes D^2) (U^* \otimes U^*),
$$
which then implies the final multiplication can be reduced to multplication by a diagonal matrix.


答:
Long Comment On Coaxing an Answer From Mathematica 12: (in answer to comment from O.P.)

Mathematica can't symbolically evaluate
$$I(\alpha)=\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$
directly.

Therefore trying integrating by parts with $\int x^{\alpha } \log (1-x) \, dx=\frac{x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)}{\alpha +1}$ and $\frac{d\, \text{Li}_2(x)}{dx}=-\frac{\log (1-x)}{x}$, both of which Mathematica is happy with.

Mathematica can find the limits $x\to 1$ and $x\to 0$ to evaluate

$$\left[\frac{\text{Li}_2(x) \left(x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)\right)}{\alpha +1}\right]^1_0$$

e.g. using

```<code>Limit[((Beta[x, 2 + \[Alpha], 0] + x^(1 + \[Alpha]) Log[1 - x])/(1 + \[Alpha])) PolyLog[2, x], x -> 1]
</code>```

Mathematica can evaluate the integral,

$$-\int_0^1 \frac{x^{\alpha } \log ^2(1-x)}{\alpha +1} \, dx = -\frac{-6 \psi ^{(1)}(\alpha +2)+6 \left(H_{\alpha +1}\right){}^2+\pi ^2}{(\alpha +1) (6 \alpha +6)}$$

but then fails to evaluate:

$$\int_0^1 \frac{(-\log (1-x)) B_x(\alpha +2,0)}{(\alpha +1) x} \, dx.$$

However this is easily solved by substituting in the infinite series for the Incomplete Beta Function in this case

$$B_x(\alpha +2,0)=\sum _{n=0}^{\infty } \frac{x^{\alpha +n+2}}{\alpha +n+2}$$

and then by changing the order of the resulting integral and summation we have

$$\sum _{n=0}^{\infty } \int_0^1 \frac{x^{\alpha +n+1}(-\log (1-x)) }{(\alpha +1) (\alpha +n+2)} \, dx=\sum _{n=0}^{\infty } \frac{H_{n+\alpha +2}}{(\alpha +1) (\alpha +n+2)^2}$$

Then its just a matter of rearranging, changing the summations to start from $n=1$ and simplifying where necessary.

Title: General formula for $\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$

Body:
Consider a following definite integral

$$I(\alpha)  = \int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$

Mathematica is able to provide a result for many $\alpha$ integers. See table here:

$$\begin{array}{c}
 I(0)=-2 \zeta (3)+3-\frac{\pi ^2}{6} \\
 I(1)=-\zeta (3)+\frac{25}{16}-\frac{\pi ^2}{8} \\
 I(2)=-\frac{2 \zeta (3)}{3}+\frac{227}{216}-\frac{11 \pi ^2}{108} \\
 I(3)=-\frac{\zeta (3)}{2}+\frac{5443}{6912}-\frac{25 \pi ^2}{288} \\
 I(4)=-\frac{2 \zeta (3)}{5}+\frac{677309}{1080000}-\frac{137 \pi ^2}{1800} \\
 I(5)=-\frac{\zeta (3)}{3}+\frac{673333}{1296000}-\frac{49 \pi ^2}{720} \\
 I(6)=-\frac{2 \zeta (3)}{7}+\frac{229495199}{518616000}-\frac{121 \pi ^2}{1960} \\
 I(7)=-\frac{\zeta (3)}{4}+\frac{1824315697}{4741632000}-\frac{761 \pi ^2}{13440} \\
 I(8)=-\frac{2 \zeta (3)}{9}+\frac{16317321883}{48009024000}-\frac{7129 \pi ^2}{136080} \\
 I(9)=-\frac{\zeta (3)}{5}+\frac{1946452513}{6401203200}-\frac{7381 \pi ^2}{151200} \\
 I(10)=-\frac{2 \zeta (3)}{11}+\frac{64401732356723}{234300040128000}-\frac{83711 \pi ^2}{1829520} \\
\end{array}$$

Is there a closed form expression for $f,g,h$ in $I(\alpha) = f(\alpha) + g(\alpha)\pi^2 + h(\alpha)\zeta(3)$?

Body:
Long Comment On Coaxing an Answer From Mathematica 12: (in answer to comment from O.P.)

Mathematica can't symbolically evaluate
$$I(\alpha)=\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$$
directly.

Therefore trying integrating by parts with $\int x^{\alpha } \log (1-x) \, dx=\frac{x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)}{\alpha +1}$ and $\frac{d\, \text{Li}_2(x)}{dx}=-\frac{\log (1-x)}{x}$, both of which Mathematica is happy with.

Mathematica can find the limits $x\to 1$ and $x\to 0$ to evaluate

$$\left[\frac{\text{Li}_2(x) \left(x^{\alpha +1} \log (1-x)+B_x(\alpha +2,0)\right)}{\alpha +1}\right]^1_0$$

e.g. using

Limit[((Beta[x, 2 + \[Alpha], 0] + x^(1 + \[Alpha]) Log[1 - x])/(1 + \[Alpha])) PolyLog[2, x], x -> 1]

Mathematica can evaluate the integral,

$$-\int_0^1 \frac{x^{\alpha } \log ^2(1-x)}{\alpha +1} \, dx = -\frac{-6 \psi ^{(1)}(\alpha +2)+6 \left(H_{\alpha +1}\right){}^2+\pi ^2}{(\alpha +1) (6 \alpha +6)}$$

but then fails to evaluate:

$$\int_0^1 \frac{(-\log (1-x)) B_x(\alpha +2,0)}{(\alpha +1) x} \, dx.$$

However this is easily solved by substituting in the infinite series for the Incomplete Beta Function in this case

$$B_x(\alpha +2,0)=\sum _{n=0}^{\infty } \frac{x^{\alpha +n+2}}{\alpha +n+2}$$

and then by changing the order of the resulting integral and summation we have

$$\sum _{n=0}^{\infty } \int_0^1 \frac{x^{\alpha +n+1}(-\log (1-x)) }{(\alpha +1) (\alpha +n+2)} \, dx=\sum _{n=0}^{\infty } \frac{H_{n+\alpha +2}}{(\alpha +1) (\alpha +n+2)^2}$$

Then its just a matter of rearranging, changing the summations to start from $n=1$ and simplifying where necessary.


问:
Distribution of function of gaussians
Let $Z_i \sim \mathcal{N}(\mu_i,\Sigma_i)$, $i = 1,2$ be two Gaussian multivariate Random Variables.

Let $W = \max(Z_1,Z_2) - \min(Z_1,Z_2)$.

What is the distribution of $W$?

答:
As stated in the comment you can rewrite:
$$
W = \max(Z_1,Z_2) - \min(Z_1,Z_2) = |Z_1−Z_2|.
$$

$W$ is distributed as a Rectified Gaussian which has distribution

$$f(x;\mu ,\sigma ^{2})=\Phi (-{\frac  {\mu }{\sigma }})\delta (x)+{\frac  {1}{{\sqrt  {2\pi \sigma ^{2}}}}}\;e^{{-{\frac  {(x-\mu )^{2}}{2\sigma ^{2}}}}}{\textrm  {U}}(x), $$

and $U(x)$ is the unitary step function.

Title: Distribution of function of gaussians


问:
Calculate the probability of the first bulb to be burnt before the second one
I'm trying to understand and then generalize an exercise I saw. There were two lightbulbs. One has lifetime $X\sim \exp(1)$ months and second one has lifetime $Y\sim \exp(3)$ months. We turn on the first lightbulb and after one month we turn the second one (even if the first one is burnt until then). We want to calculate the probability of the first bulb to be burnt before the second one. It means we need to calculate $P(X<Y+1)$. What they did:
$$
\begin{align*}
P\left(X<Y+1\right)&=P\left(X<Y+1|X<1\right)\cdot P\left(X<1\right)+P\left(X<Y+1|X>1\right)\cdot P\left(X>1\right)\\&=(1-e^{-1})+\frac{1}{3+1}e^{-1}=0.724
\end{align*}
$$
First of all I don't understand how did they got the answer. I do understand that:
$$
P\left(X<1\right)=1-P\left(X\geq1\right)=1-e^{-1\cdot1}=1-e^{-1}\\P\left(X>1\right)=e^{-1\cdot1}=e^{-1}
$$
But this means that they think that $P\left(X<Y+1|X<1\right)=1$. Why? Also I see that $P\left(X<Y+1|X>1\right)=\frac{1}{3+1}$. So i'm guessing that they want me to say: $P\left(X<Y+1|X>1\right)=P\left(X<Y\right)$ (because then I understand why it's $\frac{1}{3+1}$). I also guess it has to do something with memorylessness. But I have never saw memorylessness which uses the two (or more) random variables. I'm familiar with $P(X>s+t|X>s)=P(X>t)$. But here we have another random variable $Y$.

To sum up, I have the following questions:

What is the memorylessness formula for two random variables? I think it will help me to understand second question.

Why $P\left(X<Y+1|X<1\right)=1$ and why $P\left(X<Y+1|X>1\right)=P\left(X<Y\right)$.

Given $X\sim\exp(\lambda_X)$ and $Y\sim\exp(\lambda_Y)$. Can we evaluate $P(X<Y+a)$ where $a\geq0$?

答:
But this means that they think that $\Pr[X < Y+1 \mid X < 1] = 1$. Why?

Because we require that $Y$ has a nonnegative lifetime, i.e. $Y \ge 0$; thus $Y + 1 \ge 1$, and given $X < 1$, we must have $X < Y + 1$ with certainty.

I also guess it has to do something with memorylessness.

That is correct.  Once the first bulb has lasted $1$ month, i.e. $X > 1$, its future lifetime is still exponential with mean $1$ month, due to memorylessness.  In other words, each moment that a bulb is still working, its future lifetime from that point forward is still exponential with the same parameter--as if it were a new bulb that had just been turned on.  So the conditional probability is simply $$\Pr[X < Y + 1 \mid X > 1] = \Pr[X < Y].$$  This has less to do with whether we have two random variables in the conditional probability than it has to do with what the notation actually is saying about what is happening with the bulbs.  If the first bulb has survived one month, it is as if it is new, so now the situation is equivalent to comparing the lifetimes of the two bulbs as if they had been turned on simultaneously, and seeing if the first one dies before the second.

I should point out that the solution you have provided implies that on average, the lifetime of the second bulb is shorter than the lifetime of the first; i.e., the notation $Y \sim \operatorname{Exp}(3)$ means $Y$ has mean lifetime $1/3$ months, not $3$ months, and $3$ is a rate parameter, not a scale parameter.

It is not difficult to extend the above to the general case described in your third question.  You want $$\begin{align}
\Pr[X < Y + a] &= \Pr[X < Y + a \mid X \le a]\Pr[X \le a] + \Pr[X < Y + a \mid X > a]\Pr[X > a] \\
&= 1(1 - e^{-\lambda_x a}) + \frac{\lambda_x}{\lambda_x + \lambda_y}(e^{-\lambda_x a}). \end{align}$$  A computational proof, however, is more involved:  one would need to evaluate $$
\Pr[X < Y + a] = \int_{y=0}^\infty \Pr[X \le y+a]f_Y(y) \, dy = \int_{y=0}^\infty (1 - e^{-\lambda_x(y+a)})\lambda_y e^{-\lambda_y y} \, dy.$$  One obtains the same result in both cases.


问:
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$
$x,y \in \mathbb{R}\setminus\{0\}$

Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.

My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$

But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.

答:
The following is equivalent to @NN2's solution, only written a bit differently.

First substitute $x=1/a, y=1/b$, so that the equation becomes
$$ \tag{*}
 a^3+b^3+3ab - 1 = 0 \, .
$$
We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then
$$
 0 = a^3+b^3+3ab - 1 = S^3+3PS + 3P -1 = (S-1)(S^2 + S-3P+1) \, .
$$

So $S=1$ is one possible value for $a+b$. Conversely, all $(a, b)$ with $a+b=1$ satisfy the equation $(*)$.

In order to determine the possible zeros of the second factor we use that $P \le S^2/4$ from the inequality between geometric and arithmethic mean:
$$
0 = S^2 + S-3P+1 \ge S^2 + S-\frac 34 S^2+1 = \frac 14 (S+2)^2 \ge 0
$$
is only possible if $S=-2$ and $P=1$, i.e. if $(a, b) = (-1, -1)$.

Therefore $S=a+b=1/x+1/y$ can only take the values $1$ and $-2$, and both values do actually occur.


问:
Optimizing a functional with initial value problem constraint
Let $y$ be a smooth function $[0,L] \to \mathbb{R}$. I am interested in finding the values of $\alpha\in[0,2\pi)$ such that the solution of the initial value problem
$$\tag{1}\label{eq1}
\begin{cases}
g(t,y,y')=0,\\
y(0)=\alpha
\end{cases}
$$
is a local extremum of the functional
$$J(y) = \int_{0}^{L}F(t,y)\,dt.$$

I thought I could solve this problem using calculus of variations. As far as I understood (and also looking at this question), if $y$ is both a solution of \eqref{eq1} and an extremum of $J$, then there exists a function $\lambda = \lambda(t)$ such that the following equation holds:
$$
\frac{\partial F}{\partial y} = \lambda \left(\frac{\partial g}{\partial y} - \frac{d}{dt} \frac{\partial g}{\partial y'}\right).
$$
Am I correct?

In particular, in my case the second term inside the parenthesis vanishes, so that I only have
$$\tag{2}\label{eq2}
\frac{\partial F}{\partial y} = \lambda \frac{\partial g}{\partial y}.
$$

Now, I can solve \eqref{eq2} for $y= y(\lambda)$. However, being $\lambda$ a function, substitution of $y(\lambda)$ into \eqref{eq1} gives a new initial value problem (this time in $\lambda$), and so I have not reduced the problem at all.

What am I missing?

答:
The calculus of variation generally solves problems of the form

What arbitrary function $y$ extremizes $I = \int {\cal F}[y](\mathbf{x}) \ {\rm d} \mathbf{x}$, possibly given some constraints on $y$ and fixed boundary values?

You have a problem of the form

What arbitrary boundary value $\alpha$ extremizes $I = \int {\cal F}[y_\alpha](\mathbf{x}) \ {\rm d} \mathbf{x}$, given that $y_\alpha(\mathbf{x})$ is a uniquely fixed function given by $\alpha$ (by a procedure involving $g$)?

Now you could rephrase your problem to involve an arbitrary function $y$ and impose the constraint $g(t,y,y')=0$ using a Lagrange multiplier -- but the constrained calculus of variations is only ever going to tell you that the only possible choice of $y$ is $y_\alpha$, the thing satisfying the $g$ constraint. And then you're still left wondering what $\alpha$ you should choose.

You need to actually have a varying function of $\alpha$ which you extremize at some point. Clearly, $\alpha$ determines $y_\alpha$ (via $g$ constraint) and hence $\alpha$ determines $j(\alpha) := J(y_\alpha)$. You need to vary $\alpha$ in this expression and optimize.

(Note that you may be able to rewrite $J(y_\alpha)$ in a useful way using the equation $y_\alpha$ obeys, in order to extract its $\alpha$ dependence, maybe without ever needing to actually work out $y_\alpha$.)


问:
Why is the following fraction an integer
I have $\sum_{k=0}^{p-1}\frac{(p-1)!}{(p-1-k)!k!}$. Whereby $p$ is prime.

how can I say that the fraction's result will be an integer?

I tried it with numbers and yes it works.
So I tried two approaches:

My first try was to write the factorial as a product,which means,
$\frac{\prod_{k=1}^{p-1}k}{\prod_{i=1}^{p-1-k}i\prod_{j=1}^{k}j}$ but I didn't get much further and the second one was to try and use some properties of the factorial and I wrote it as:
$\frac{1!(p-2)!(p-1)}{k!(p-1-k)(p-2-k)!}$.

In both cases I tried to expand them and stuff but didn't come to any conclusion. If I expand them I do understand why it must work, but I can't put it in some mathematical decent way.

I would like to solve it alone but I would be thankful for some tip, or maybe for some other property that I am missing, so that then I can continue it.

Thanks in advance

Annalisa

答:
This is the binomial expansion $$(a+b)^n = \sum\limits_{k=0}^n{n \choose k}a^kb^{n-k}$$
with $n=p-1$ and $a=b=1$. (Recall the symbol ${n\choose k} = \frac{n!}{k!(n-k)!}$ is an integer when $n$ and $k$ are, as it is the number of combinations of $n$ objects taken $k$ at a time.)

So your sum is $2^{p-1}$.


问:
derivative of $f\left(x_1\left(t\right),x_2\left(t\right)\right)$
What is the derivative of following function  $f\left(x_1\left(t\right),x_2\left(t\right)\right)$, where $x_1\left(t\right)$ and $x_2\left(t\right)$ are dependent on $t$?</p>

答: By the chain rule $\partial_t f = \partial_1 f\cdot \partial_t x_1 + \partial_2f\cdot\partial_t x_2.$</p>

ContentLicense: CC BY-SA 3.0


问:
limit of $\lim_{n\to \infty}-\frac{1}{n}(\log(\frac{1}{n}))$
How can I conclude that the limit of $\lim_{n\to \infty}-\frac{1}{n}(\log(\frac{1}{n}))$ is $0$? If $\log(\frac{1}{n})$ was bounded we can argue by the fact that:

A sequence which converges to $0$ times another bounded sequence converges to $0$ as well. But $\log(\frac{1}{n})$ is not bounded.

答:
$$-\frac{1}{n}\ln\left(\frac{1}{n}\right) = \frac{\ln(n)}{n} \to 0 ~~~~~~~ \text{as}~~~ n\to +\infty$$

For example you can use De L'Hôpital rule, or simply the hierarchy of infinities.

(The property I have used is simply $\ln\left(\frac{1}{n}\right) = -\ln(n)$).


问:
Do we have a version of Grönwall lemma for $t<b$?
Let $y(\cdot)$ be a continuous mappings defined on $[a,b]$ and $K\in (0,+\infty]$ a constant. An application of the Grönwall lemma is the following simple case: if we have for all $t\in[a,b]$

$$y(t)\leq K + \int_a^ty(s) ds ,$$
then we will have for all $t\in [a,b]$
$$y(t)\leq K e^{\int_a^tds}=Ke^{(t-a)}.
$$
My question is that do we still have such result if we change the order of the integral $\int_a^t ds$ to $\int_t^b ds$ in other words do we have the following implication:
$$\forall t\in [a,b]\; \;\;\; y(t)\leq K + \int_t^by(s) ds \implies \forall t\in [a,b]\; \;\;\;  y(t)\leq K e^{\int_t^bds}=Ke^{(b-t)}.$$

答:
By taking : $z(t)=y(-t)$ for all $t\in [-b,-a]$, and changing the measure $ds=-d\tau$ we obtain
$$z(t)\leq K+\int_{-b}^t z(\tau)d\tau $$
and then by applying the classic Gronwall lemma we get that :
$$z(t)\leq Ke^{t+b} \;\; \forall t\in [-b,-a],$$
in other words
$$y(t)\leq Ke^{b-t} \;\; \forall t\in [a,b]$$


问:
Find the number of different selections of two letters which can be made from the letters of the word PROBABILITY
I study maths as a hobby. I am tackling this problem:

Find the number of different selections of two letters which can be made from the letters of the word PROBABILITY. how many of these selections do not contain a vowel?

I first of all isolated the unique letters PROBAILTY.

I then said for the first selection the number = $\binom{11}{2} = 55$

For the second selection the number = $\binom{9}{2} = 36$

Third selection = $\binom{7}{2} = 21$

and so on to get $\binom{11}{2} \cdot \binom{9}{2} \cdot \binom{7}{2} \cdot \binom{5}{2} \cdot \binom{3}{2} = 55\times36\times21\times10\times3 = 1,247,400$

But now I am not sure of my working or how to proceed.

答:
Given that, we have to just 'select', this can be reduced to a simple combinations question.

Unique letters are $\text {{P, R, O, B, A, I, L, T, Y}}$

Therefore, the solution is simply $\binom{9}{2}=36$


问:
Maximum value of $(x−1)^2+ (y−1)^2+ (z−1)^2$ with constraint $x^2+y^2+z^2 ≤2 , z≤1$
So the problem is that I have $D(f)=\{(x,y,z), x^2+y^2+z^2 ≤2 , z≤1\}$ and I have to determine the maximum value for the function $(x−1)^2+ (y−1)^2+ (z−1)^2$ in $D$.

I'm just confused as I don't actually know if $z\le1$  counts as a constraint as well, or is it just for me to sketch the area, which is actually a part of the question.
Furthermore, I know that I have to use Lagrange multiplier method, but I honestly don't know how because $\le$ is making the question hard for me. Do I just calculate as usual and count $\le$ the same as $=$?

appreciate all the feedback

Edit: I have calculated the grad f =0 which is = $D(f)=(2(x-1), 2(y-1), 2(z-1))$ where I've got that $x=y=z= 1$ and $f(1, 1, 1)=0$. (I don't know what to do with this though). Then I calculated $L(x, y, x, λ) = (x−1)^2+(y−1)^2+(z−1)^2 +λ(x^2+y^2+z^2-2)$, then the four cases where I got the same value which is $-2λ= 2(x-1)/x = 2(y-1)/y = 2(z-1)/z$. Which means that $x=y=z$, put it in the $D$ function $x^2+ x^2+ x^2$ and ended up with $x=y=z= −+√2/√3$. I took the minus sign for the maximum distance from $(1,1,1)$. which means that the answer is $x=y=z= −√2/√3$. Is it correct?

答:
Hint. Find first the maximum value for the function $(x−1)^2+(y−1)^2+(z−1)^2$, which is the square of the distance of the point $(x,y,z)$ from $P=(1,1,1)$, in the larger set
$$D_1=\{(x,y,z): x^2+y^2+z^2 ≤2\}$$
which is the closed ball centered at $(0,0,0)$ of radius $\sqrt{2}$. This should be easy to find by using a geometric approach with very few calculations: the maximum point is on the boundary of $D_1$ along the line which joins the center $(0,0,0)$ and the point $(1,1,1)$.

Does this maximum point satisfy the constrain $z\leq 1$? What may we conclude?


答: With $x:=\ln n$ this becomes $\lim_{x\to\infty}xe^{-x}=0$, which is true because $\int_1^\infty xe^{-x}dx=2/e$ is finite, with a continuous non-negative integrand that monotonically decreases.

Body:
How can I conclude that the limit of $\lim_{n\to \infty}-\frac{1}{n}(\log(\frac{1}{n}))$ is $0$? If $\log(\frac{1}{n})$ was bounded we can argue by the fact that:

A sequence which converges to $0$ times another bounded sequence converges to $0$ as well. But $\log(\frac{1}{n})$ is not bounded.


问:
Prove that if $a$ is algebraic over $F$, then $F(a) = F(a^{-1})$.
Let $E$ be an extension of a field $F$ and let $a \in E$ be nonzero.

Given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.

Prove that if $a$ is algebraic over $F$, then $F(a) = F(a^{-1})$.

$\textbf{My Attempt:}$

Assume that $a$ is algebraic over $F$.

Notice that $F(a) = \{ c_0 + \cdots + c_{d-1} a^{d-1} ~|~ \text{each $c_i \in F$}\}$

where $d = [F(a) : F] = $ degree of minimal polynomial of $a$ over $F$.

Since, given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.

Then, $a^{-1}$ is algebraic over $F$.

Then, by definition $a^{-1} \in F(a)$.

Then, $a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$, for each $\frac{c_i}{c_0} \in F$.

Then, $c_0 a^{-1} = c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 \implies c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$.

So, $F(a^{-1}) = \{\frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0} ~|~ \text{each $\frac{c_i}{c_0} \in F$}\}$.

So, $F(a) = F(a^{-1})$.

$\textbf{Don't think the prove is correct, are there ways to fix to make it look better ?}$

答:
Hint

Your proof is almost correct. I say almost because one missing argument is that if

$$p(x) = a_d x^d +\dots +a_1 x +a_0$$ is irreducible over $\mathbb F$ then

$$\overline{p}(x)= a_0 x^d + \dots +a_d$$ is also irreducible.

With that, you can prove that

$$[\mathbb F(a): \mathbb F] = [\mathbb F(a^{-1}): \mathbb F] .$$


问:
Set of homomorphisms form a group?
Given vector spaces $V, W$ over field $F$, the set of all linear maps $V \to W$ forms a vector space over $F$ under pointwise addition.

Is there an analogue for groups?  Can the set of all homomorphisms from groups $G \to K$ be given a group structure?

答:
The set of all homomorphisms between two groups naturally forms a groupoid rather than a group; the objects of the groupoid are the homomorphisms and the morphisms are given by pointwise conjugation, so if $\varphi_1, \varphi_2 : G \to H$ are two homomorphisms then a morphism between them is an element $h \in H$ such that $\varphi_1(g) = h \varphi_2(g) h^{-1}$ for all $g \in G$.

This is a special case of the construction of functor categories, thinking of groups as one-object categories.

Some other relevant keywords from category theory: enriched category, preadditive category (what I would prefer to call an $\text{Ab}$-enriched category). The category $\text{Vect}$ of vector spaces is enriched over itself, as is the category $\text{Ab}$ of abelian groups, the category $\text{Cat}$ of (small) categories, and its subcategory $\text{Gpd}$ of groupoids.

I realize I've dodged the original question to some extent. There's a natural extra condition to ask for when you put additional structure on homsets, namely composition ought to respect that structure. So if you want to enrich the category of groups over itself, then it would be nice if composition

$$\text{Hom}(G, H) \times \text{Hom}(H, K) \to \text{Hom}(G, K)$$

were a group homomorphism. But this implies that $\text{Hom}(G, G)$ comes equipped with two monoid structures, one given by the enrichment and one given by composition, which satisfy the compatibility relation from the Eckmann-Hilton argument. It then follows that the monoid structures must be isomorphic and commutative, but now it follows that every group homomorphism $G \to G$ is an isomorphism and also that $\text{Aut}(G)$ is always commutative, both of which are clearly a contradiction.

(So why can we enrich the category of abelian groups over itself? The reason is that we give $\text{Ab}$ a different monoidal structure, namely the tensor product, which changes the condition we want on composition to

$$\text{Hom}(A, B) \otimes \text{Hom}(B, C) \to \text{Hom}(A, C)$$

being a group homomorphism, which can happen. Now the compatibility relation on $\text{Hom}(A, A)$ coming from its two monoid structures turns it into a ring. But the category of groups, unlike the category of abelian groups, doesn't have a natural notion of tensor product.)

Title: Set of homomorphisms form a group?


问:
How to prove that square root of $x$ times $y$ is square root of $x$ times square root of $y$
Simple question - how to prove that:

$\sqrt {x\times y} = \sqrt x \times \sqrt y$ ?

If I use the exponentation the answer seems easy, because

$(x\times y)^n = x^n \times  y^n$  because I get

$(x\times y) \times\cdots\times (x\times y)$ (where $x$ occurs $n$ times and $y$ occurs $n$ times) can be rewritten as: $x \times\cdots\times x \times y \times \cdots \times y$.

But in case of square roots that's not so obvious, because I can't rewrite the it the same way.
I can of course reason that $\sqrt x$ is $x^{\frac12}$ and $\sqrt y$ is $y^{\frac12}$ and think using induction, but that seems to not satisfy me.

答: If $z:=\sqrt{x}\sqrt{y}\ge0$ then $z^2=\sqrt{x}\sqrt{y}\sqrt{x}\sqrt{y}=\sqrt{x}\sqrt{x}\sqrt{y}\sqrt{y}=xy$ so $z=\sqrt{xy}$.


问:
Adding locally compact Hausdorff space into product besides two spaces of a quotient mapping.
I am strugling to prove the next theorem. More precisely, I have no clue how to even start solving it.

The problem states:

Let $f:X\rightarrow Y$ be a quotient mapping and let $K$ be a locally compact Hausdorff space. Prove that $f^*:X\times K\rightarrow Y\times K:(x,k)\mapsto(f(x),k)$ is one quotient mapping.

I know that the title is terrible. I am sorry, I just couldn't think of any better.

答:
Adding to the solution of Tipping Octopus:

There is one other way to construct the set $V'$ other than finding that chain of sets $V_i$.

Beggining from points $(y,k)$ and $(x,k)$ as in his original solution, one can firstly find an open set $W\ni k$ so that $\overline{W}$ is compact and $\{x\}\times\overline{W}$ is contained in $V$ (where $V=(f^*)^{-1}(U)$, as in his solution). Now one can take the set $A:=\{x'\in X|\{x'\}\times\overline{W}\subseteq V\}$. We have $x\in A$ and by using the tube lemma we can show that $A$ is open in $X$. To show that $A$ is saturated is easy. Now, $A$ has the same needed properties as $V'$ in his solution has.


答:
Since $n^{1/n} \to 1$ and $\ln$ is continuous, we get

$$\frac{1}{n}\ln\left(\frac{1}{n}\right)=\ln\left(\frac{1}{n^{1/n}}\right) \to \ln 1 =0.$$

Title: limit of $\lim_{n\to \infty}-\frac{1}{n}(\log(\frac{1}{n}))$


答:
Since $n$ isn't a natural number, we can approach the proof from a different angle.

Change the square root and rewrite it as a power of a $0.5 $, then use the power of a product rule to get:

$$\sqrt{xy} = (xy)^{0.5} = x^{0.5} \cdot y^{0.5} = \sqrt x \cdot \sqrt y$$

This is true, of course, as long as $x$ and $y$ aren't negative numbers.

Title: How to prove that square root of $x$ times $y$ is square root of $x$ times square root of $y$

Body:
Simple question - how to prove that:

$\sqrt {x\times y} = \sqrt x \times \sqrt y$ ?

If I use the exponentation the answer seems easy, because

$(x\times y)^n = x^n \times  y^n$  because I get

$(x\times y) \times\cdots\times (x\times y)$ (where $x$ occurs $n$ times and $y$ occurs $n$ times) can be rewritten as: $x \times\cdots\times x \times y \times \cdots \times y$.

But in case of square roots that's not so obvious, because I can't rewrite the it the same way.
I can of course reason that $\sqrt x$ is $x^{\frac12}$ and $\sqrt y$ is $y^{\frac12}$ and think using induction, but that seems to not satisfy me.


问:
For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more
I set out to solve the following question using the pigeonhole principle

Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.

My outline

[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.

[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.

My query

Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.

答:
Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$.

The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled.

The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.

Remark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.

Then $S_1+S_1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\cdots+10=55$.

But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}^{10}S_i=165$.

I prefer your proof, mildly modified.


问:
Given that the first lightbulb is turned for more than one month, what's the expected value until one of the two lightbulbs is burnt?
Following my previous question Calculate the probability of the first bulb to be burnt before the second one, which I'll add here so you won't have to jump back there:

There were two lightbulbs. One has lifetime $X\sim \exp(1)$ months and second one has lifetime $Y\sim \exp(3)$ months. We turn on the first lightbulb and after one month we turn the second one (even if the first one is burnt until then).

Given that the first lightbulb is turned for more than one month, what's the expected value of the time from turning the first lightbulb until one of the two lightbulbs is burnt?

As I understand, we want to calculate $E\left(\min\left\{X,Y\right\} |X>1\right)$. We know that $\min\left\{ T_{A},T_{B}\right\} \sim\exp\left(4\right)$. But now I'm stuck. In the solution they did:
$$
E\left(\min\left\{X,Y\right\} |X>1\right)=1+\frac{1}{4}=1.25
$$
Is it possible to explain how did they got to $1+\frac{1}{4}$?

答:
1- Intuitive explanation

Given that $X>1$, using memoryless property you can say that at the time 1 the bulbs are "good as new", thus the mean is $1+E(U)=1+\frac{1}{4}$

2- Analytic solution

Given that $X>1$ also $U=\min(X,Y)>1$ thus

$$f_{U|U>1}(u)=e^44e^{-4u}\mathbb{1}_{[1;+\infty)}(u)$$

Using the definition, the expectation is

$$e^4\int_1^{+\infty}4ue^{-4u}du=\frac{5}{4}$$


问:
Why does Munkres say "we shall seldom have occasion to use the Euclidean metric on $\mathbb{R}^n$"?
I'm not sure if opinion-based questions like this are appropriate here, but I'm a bit confused by something I read in Munkres in the section on the metric topology. Specifically, in the first edition on page 120, near where the Euclidean and square metrics are defined, it reads

"The proof that $d$ is a metric requires some work; it is probably already familiar to you. If not, a proof is outlined in the exercises. We shall seldom have occasion to use the Euclidean metric on $\mathbb{R}^n$."

It's this last sentence which is confusing to me. Surely the author doesn't mean in general, perhaps just in this book? You could argue that anytime they do anything with the usual topology on $\mathbb{R}^n$, we're using the Euclidean metric.

答:
Absolutely yes, it is not a general answer and the domain is the book itself. Munkres’ book is all about general topology (hence more general than a metric space) and algebraic topology, so it is not interested in the metric structure, that is more the topic of analysis or geometric analysis (differential geometry). In Real Analysis, when one generalizes the structure of the real numbers to a general metric space, the notion of distance (and the Euclidean is a particular one) is fundamental.

As regards the study of shapes and geometry overall, topology is the study of properties that are stable via homeomorphisms, regardless the metric structure. On the other hand, on more advanced topological spaces one wants to study functions that preserve the metric structure of the topological space (endowed with a particular metric) and so you jump into the field of differential geometry, Riemannian geometry and geometric analysis, that are a nice intersection of topology and analysis.


问:
Let $H$ be a subgroup of $S_7$ consisting of permutations that send $1,2,3$ to $1,2,3$ in some order. Is $H$ isomorphic to $S_3 \times S_4$?

Let $H$ be a subgroup of $S_7$ consisting of permutations that send $1,2,3$ to $1,2,3$ in some order. (The remaining values $4,5,6,7$ are still permuted just not to the positions $1,2$ or $3$.) Is $H$ isomorphic to $S_3 \times S_4$?

This turns out to be true, but I'm not sure how to get to the answer here. Shouldn't $H$ and $S_3 \times S_4$ share a bijective mapping since we're talking about permutations here? Clearly the cardinality of $S_3 \times S_4$ is greater than the cardinality of $H$ so there isn't a bijective map between them. What might be I missing here?

答:
1 In combinatorics

In how many ways can you make a seven-digit number from 1,2,...,6,7 without repetition (Group $S_7$) with 1,2,3 on the first three positions and 4,5,6,7 on the remaining four position? (description of $H$).

Well, you have six ways how to order the first three digits ($S_3$) and $4!=24$ ways how to arrange remaining digits ($S_4$).

In group theory

You define $\Phi: H\to S_3\times S_4$ by sending $\pi\in H$ on pair $(\pi_3, \pi_4)$ where by $\pi_3$ I mean permutation $\pi$ restricted on the subet $\{1,2,3\}$ and $\pi_4$ means permutation $\pi$ restricted on the subset $\{4,5,6,7\}$.

You can check that this is isomorphism.


答:
The question asks how many distinct selections of two letters.

There are only two ways to select two letters with repetition and they are  $\{$BB$\}$ and $\{$II$\}$. Now there are $9$ distinct letters after taking out $1$ B and $1$ I and that gives us $9 \choose 2$ selections of two letters without repetition.

So in total, we have ${9 \choose 2} + 2 = 38$ selections of two letters.

Now if there are no vowels. We have $7$ letters with two B's. We have only one selection with repetition which is $\{$BB$\}$. Taking out $1$ B, we have $6$ distinct letters without vowels and that gives us ${6 \choose 2}$ selections of two letters without repetition.

So in total, we have ${6 \choose 2} + 1 = 16$ selections of two letters without a vowel.

Body:
I study maths as a hobby. I am tackling this problem:

Find the number of different selections of two letters which can be made from the letters of the word PROBABILITY. how many of these selections do not contain a vowel?

I first of all isolated the unique letters PROBAILTY.

I then said for the first selection the number = $\binom{11}{2} = 55$

For the second selection the number = $\binom{9}{2} = 36$

Third selection = $\binom{7}{2} = 21$

and so on to get $\binom{11}{2} \cdot \binom{9}{2} \cdot \binom{7}{2} \cdot \binom{5}{2} \cdot \binom{3}{2} = 55\times36\times21\times10\times3 = 1,247,400$

But now I am not sure of my working or how to proceed.


问:
Prove that :$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq4$
I found this problem in an old textbook of mine and couldn't solve it:
Prove The following:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq4$$

$$a,b,c,d>0$$

This problem was from a mathematical olympiad related book and to my knowledge math olympiads use a lot of identities and tricks which are known by contestants

-If anyone can help prove it, I would be grateful

答: Just apply $AM \ge GM$ on $\frac{a}{b} , \frac{b}{c},\frac{c}{d},\frac{d}{a}$


问:
how to convert to a rational number with some decimal from base 10 to base 2
It is very eazy to convert an integer number to another base, say 2, using Maple. Can anyone tell me how to convert a rational numbers with several decimals from base 10 to other bases In maple? Well, I can multiply the rational number by 10th and get integer and then convert to another base, but this wont give me the true value for corresponding number in the other base. For example, I can multiply 0.25 by 10 and get 25 and after that convert 25 from base 10 to base 2 and get 10011 but this doesn't equal 0.25 in base 2.
Cam anyone tell me how to do that in Maple??
Thank you.</p>

答: Couldn't you just do the procedure you outlined (multiply say 0.25 to make it an integer=25) then divide by whatever you multiplied by? (in this case you would divide by 1100100)</p>


答: By AM. GM We have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4\sqrt[4]{\dfrac{a}{b}\cdot \dfrac{b}{c} \cdot \dfrac{c}{d}\cdot \dfrac{d}{a}}=4$$

Title: Prove that :$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq4$

Body:
I found this problem in an old textbook of mine and couldn't solve it:
Prove The following:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq4$$

$$a,b,c,d>0$$

This problem was from a mathematical olympiad related book and to my knowledge math olympiads use a lot of identities and tricks which are known by contestants

-If anyone can help prove it, I would be grateful


问:
Prove the space curve doest not have an asymptote
Prove that the function $\mathbb{R} \to \mathbb{R}^2: t \mapsto \left(t, at^2 + bt +c\right)$ doesn’t have an asymptote.

Here what I thought so far:

First of all Im using this definition for a line $L$ to be an asymptote of a curve:

$$
\lim\limits_{t \to A} d(f(t),L) = 0
$$

Where $A$ can be $\pm \infty$ and $\lim\limits_{t \to A} \left|f(t)\right| = \infty$.

The first solution I came up with is more conceptual, and since I want to practice my formal proof writing, I want to prove it using a more formal (delta epsilon) fashion.

Here’s the first solution:

By the definition of asymptote and the continuity of $f(t)$ and since $d(f(t), L) \geq \left|f(t) - L\right|$:

By contradiction, let’s assume the function does have an asymptote L. Then

$$
\lim\limits_{t \to A} \left|f(t) - L\right| = \left| \lim\limits_{t \to A} f(t) - L\right|= 0 \implies \lim\limits_{t \to A} f(t) = z \in L
$$

Which is, the sequence in $f(t)$ converges to a point in the line. So now, since the point $z$ lies in the line, we can use the fact that their derivative vector also matches the lines direction. But using the Lagrange Integral error bound for the parabola we can check that the function and the line at $z$ at the direction of it’s derivative diverges, which is, the derivative should match $L$’s derivative through any point, as well as the function $f(t)$ should be arbitrarily close to $L$, but the integral error shows the opposite. A contradiction.

Although this first solutions seems to make sense, it seems to lack formalism since Im using a point within the line on the limit. So I though of another way:

Since the function $f(t)$ has a well defined and simple expression, explicitly calculate the distance between $f(t)$ and its projection onto a generic parametric line $L(s) = p + vs $. This yields a quadratic expression that clearly does not converges to zero. So it shows no generic line can indefinite approach the parabola.

My questions are: are those 2 solutions reasonable or correct?

Second question is: is there a more formal delta epsilon way to solve it (possibly also more elegant?).

Thanks!

答:
If $z$ is a point of $L$, then by definition $$d(f(t),L) \le |f(t)−z|$$ and not the other way around, and this invalidates your first proof.

The second approach works, but of course requires some calculations. Anyway, there are some characterizations that allows for a somewhat simpler proof. For instance, with your notations, a function has an asymptote (at infinity) iff
$$ \exists \lim_{t\to\pm\infty} \frac{f(t)}{t} := m\in \mathbf{R}$$
and
$$ \exists \lim_{t\to\pm\infty} f(t)-mt := q\in \mathbf{R}.$$
In this case the line $t\mapsto mt+q$ is the asymptote.

On the contrary in your case $ \lim_{t\to\pm\infty} \frac{f(t)}{t} $ is not finite, hence there is no asymptote.


问:
Help with factoring $a^4(b-c)+b^4(c-a)+c^4(a-b)$
I want to factor $a^4(b-c)+b^4(c-a)+c^4(a-b)$. I found a similar question on the site, but still don't know how to factor it. from Macavity's answer I realized it has the factor of $(a-b)(a-c)(b-c)$. as mentioned in the answer I should find a quadratic that after multiplying on this factor getting the initial expression. I expanded these expressions:

$$a^4b-a^4c-b^4c-b^4a+c^4a-c^4b=(a^2b-a^2c-b^2a+b^2c+ac^2-bc^2)\times\underbrace{(\quad\quad\quad)}_\text{quadratic expression}$$
I don't know how to find it.

答:
Let $f(a,b,c)$ be the polynomial to factor. Write
$f(a,b,c)= g(a,b,c)(a-b)(a-c)(b-c)$ with $g(a,b,c)=r_1a^2+r_2ab+r_3ac+r_4b^2+r_5bc+r_6c^2$. Compare coefficients to obtain
$$
g(a,b,c)=a^2 + ab + ac + b^2 + bc + c^2.
$$
One can make this more efficient by arguing that the coefficients for $ac,ab,bc$ must be equal and the ones for $a^2,b^2,c^2$ as well.

So we have the factorization
$$
f(a,b,c)=(a^2 + ab + ac + b^2 + bc + c^2)(a - b)(a - c)(b - c)
$$


答:
You could prove it this way:

$0\neq a\in F(a)\Rightarrow a^{-1}\in F(a)$, which means $F(a^{-1})\subseteq F(a)$. Now do the same argument for $a^{-1}\in F(a^{-1})$ to get $F(a)\subseteq F(a^{-1})$ and you're done.

By the way, this result is true for any non-zero element you'd care to adjoin to $F$.

Body:
Let $E$ be an extension of a field $F$ and let $a \in E$ be nonzero.

Given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.

Prove that if $a$ is algebraic over $F$, then $F(a) = F(a^{-1})$.

$\textbf{My Attempt:}$

Assume that $a$ is algebraic over $F$.

Notice that $F(a) = \{ c_0 + \cdots + c_{d-1} a^{d-1} ~|~ \text{each $c_i \in F$}\}$

where $d = [F(a) : F] = $ degree of minimal polynomial of $a$ over $F$.

Since, given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.

Then, $a^{-1}$ is algebraic over $F$.

Then, by definition $a^{-1} \in F(a)$.

Then, $a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$, for each $\frac{c_i}{c_0} \in F$.

Then, $c_0 a^{-1} = c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 \implies c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$.

So, $F(a^{-1}) = \{\frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0} ~|~ \text{each $\frac{c_i}{c_0} \in F$}\}$.

So, $F(a) = F(a^{-1})$.

$\textbf{Don't think the prove is correct, are there ways to fix to make it look better ?}$


问:
integral of a discontinous periodic function?

Background

I am a biology student and I have been struggling with finding the integral of the function:

$$ a(t) = a_0\times e^{-k\times(t- \lfloor t/\tau \rfloor * \tau)}  \space\space\space \text{with}\space (k \neq 0, \tau \neq0)$$

Which (for some arbitrary values for $a_0, k$, $\tau $) looks like this. A fancy version of exponential decay.

My understanding of this function is

periodic (for a $\tau > 0$), therefore I only need to determine the integral over one period.

non-continuous /discontinuous.

What I did:
I found the integral of the "non-periodic" version of $a(t)$, lets call that one $\alpha(t)= a_0*e^{-kt}$. So,$$\int \alpha(t) dt = -\frac{a_0*e^{-kt}}k  $$
I then simply did $$\int_0^{\tau} \alpha(t) dt = -\frac{a_0*e^{-k \times \tau}}k +\frac{a_0}k  $$ but this way I am clearly missing that originally $a(\tau)= a_0$ (depicted by the black points in the image above).

My Question

Can I use  $\alpha(t)$ instead of $a(t)$ as to integrate over the interval fomr 0 to $\tau$? I am asking because obviously $a(tau)\neq\alpha(tau)$, but I really can't think of another way.

Bigger picture: $a(t)$ is a simplified model of the concentration of a "pollutant" in a waterbody. The concentration of this "pollutant" influences the growth of microorganisms. By finding $ \int_0^{\tau} a(t)  $ I ultimately want to determine the respective mean growth of microorganisms over $\tau$. But one step at a time. I want to do this analytically, not numerically.

答:
WLOG, $a_0=1$ and $\tau=1$. The integral is

$$\int_0^te^{-k\{u\}}du$$ (the braces denote the fractional part). Within the first period ($t<1$),

$$\int_0^te^{-ku}du=\frac{1-{e^{-kt}}}k.$$ For a complete period,

$$\int_0^1e^{-ku}du=\frac{1-{e^{-k}}}k.$$

Then for any value,

$$\int_0^t=\int_0^{\lfloor t\rfloor}+\int_{\lfloor t\rfloor}^t=\lfloor t\rfloor\int_0^1+\int_0^{\{t\}}.$$

You don't have to worry about the discontinuity $f(1)\ne f(0)$. It has null measure and does not influence the integral.

Below, $k=1$:


答:
From the inequality: $(a+b)^2 \le 2(a^2+b^2) $, we have $$(-(x+y))^2 \le 2(x^2+y^2) \implies -(x+y) \le\sqrt{2(x^2+y^2)}$$

And by applying the constraint $x^2+y^2+z^2 ≤2$,

\begin{align}
(x−1)^2+ (y−1)^2 &=  (x^2+y^2) -2(x+y)+2 \\
& \le (x^2+y^2) +2 \sqrt{2(x^2+y^2)}+2\\
& \le (2-z^2) +2 \sqrt{2(2-z^2)}+2\\
\end{align}

So, we can deduce that $$(x−1)^2+ (y−1)^2 +(z-1)^2  \le 5-2z + 2 \sqrt{2(2-z^2)}$$ with $-\sqrt{2} \le z\le 1$

Let's study the function $g(z)= -z+\sqrt{2(2-z^2)}$ in the interval $z\in(-\sqrt{2},1)$. We have $g'(z)=-1 -\frac{2z}{\sqrt{4-2z^2}}$ and reaches its maximum at $\sqrt{4-2z^2} = -2z$ or $z =-\frac{\sqrt{2}}{\sqrt{3}}$. We verify that this value is in the interval $z = -\frac{\sqrt{2}}{\sqrt{3}} \in(-\sqrt{2},1)$.

Note: we can use Cauchy Schwart inequality to find the maximum value of $g(z)$. This approach is more direct but not orthodox.

Finally, $(x−1)^2+ (y−1)^2 +(z-1)^2$ reaches the maximum value when ($z =-\frac{\sqrt{2}}{\sqrt{3}} $ and $x = y$ and $x^2+y^2 +z^2 =2$), or $x=y=z = -\frac{\sqrt{2}}{\sqrt{3}}$.

Title: Maximum value of $(x−1)^2+ (y−1)^2+ (z−1)^2$ with constraint $x^2+y^2+z^2 ≤2 , z≤1$

Body:
So the problem is that I have $D(f)=\{(x,y,z), x^2+y^2+z^2 ≤2 , z≤1\}$ and I have to determine the maximum value for the function $(x−1)^2+ (y−1)^2+ (z−1)^2$ in $D$.

I'm just confused as I don't actually know if $z\le1$  counts as a constraint as well, or is it just for me to sketch the area, which is actually a part of the question.
Furthermore, I know that I have to use Lagrange multiplier method, but I honestly don't know how because $\le$ is making the question hard for me. Do I just calculate as usual and count $\le$ the same as $=$?

appreciate all the feedback

Edit: I have calculated the grad f =0 which is = $D(f)=(2(x-1), 2(y-1), 2(z-1))$ where I've got that $x=y=z= 1$ and $f(1, 1, 1)=0$. (I don't know what to do with this though). Then I calculated $L(x, y, x, λ) = (x−1)^2+(y−1)^2+(z−1)^2 +λ(x^2+y^2+z^2-2)$, then the four cases where I got the same value which is $-2λ= 2(x-1)/x = 2(y-1)/y = 2(z-1)/z$. Which means that $x=y=z$, put it in the $D$ function $x^2+ x^2+ x^2$ and ended up with $x=y=z= −+√2/√3$. I took the minus sign for the maximum distance from $(1,1,1)$. which means that the answer is $x=y=z= −√2/√3$. Is it correct?


问:
Relating the Künneth Formula to the Leray-Hirsch Theorem
I am reading through Bott & Tu's Differential Forms in Algebraic Topology, which very early on discusses the Künneth formula and the Leray-Hirsch theorem for smooth principal bundles. The proof of Künneth is given explicitly, while Leray-Hirsch is said to follow by the same kind of argument but by adding the appropriate twists (no pun intended!).

It seems to me that the proofs Bott & Tu offer are independent of one another. Thus, in the presence of the Künneth formula, one might boldly restate Leray-Hirsch as

If a principal bundle contains global cohomology classes which freely generate the cohomology of the fibres, the bundle "looks" trivial in cohomology.

But this makes me wonder if I am putting Descartes before the horse (bad pun now intended); that is, perhaps Künneth is really just a simple corollary of Leray-Hirsch. Indeed, this would be the case if someone could prove something along the lines of the following:

Conjecture: Any smooth trivial principal bundle $\pi: G \times B \to B$ admits global cohomology classes on $G \times B$ which freely generate $H^*(G)$.

So here is where I'm not certain how to proceed. Triviality of the bundle guarantees we can actually define a projection onto the fibres: $\rho: G \times B \to G$, which I could then use to pullback cohomology classes from $G$ into the total space. Are these global?

答:
I'm not sure what 'global cohomology class' means, but for the Leray-Hirsch theorem I know of it's enough to ask that the inclusion of the fiber $G \rightarrow G \times B$ induces a surjection $H_{dR}^*(G \times B) \rightarrow H^*(G)$.

But this is clear since $G \rightarrow G \times B \rightarrow G$ is the identity, and so $G$ is a retract of $G \times B$. Since $H_{dR}^*$ is functorial it takes retracts to retracts (but flipping arrows), and so the hypotheses of the Leray-Hirsch theorem are satisfied, whence we have $H_{dR}^*(G) \otimes H_{dR}^*(B) \cong H_{dR}^*(G \times B)$.


问:
Problem in proving Cauchy sequence - how to tackle inequalities and changing indices [EDITED]
We are given this sequence:

$a_1 =1, a_{n+1} = 1 + \frac{1}{a_{n}}  \forall \;  n\geq 1$

Now before marking this question as "duplicate", consider that I have already referred these sources:

How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$

Convergence of sequences and limits

Now, I am relatively new to Real Analysis and for this particular question, I don't need to determine the limit, only to prove that it is Cauchy and hence convergent.

My approach:

Observe that $a_1 =1, a_2 =2, a_3 = \frac{3}{2}, a_4= \frac{5}{3}$ and so on.

Also observe,
$a_{n+1} = 1 + \frac{1}{a_n} \forall n\ge 1$

Multiplying throughout by $a_n$ we get,

$a_{n+1}a_{n} = a_{n} + 1 \; \forall  n\ge 1$

Claim: $a_n \; \ge 1 \; \forall \; n\ge 1$

Proof by Induction:
Base Case: $n=1:\; a_1 =1 \ge 1$ which is true.

Induction Hypothesis: $a_k \ge 1 \;$ for some $k \in \mathbb{N}, k \ge 1$

Induction Step: To prove: $a_{k+1} \ge 1$ for some $k \in \mathbb{N}, k\ge 1$

Now,

$a_k \ge 1$

$\therefore 0 \le \frac{1}{a_{k}} \le 1$

$\therefore 1 \le 1 + \frac{1}{a_{k}} \le 2$

$\therefore 1 \le a_{k+1} \le 2 $

Therefore by Priniciple of Mathematical Induction,

$a_n \; \ge 1 \; \forall \; n\ge 1$

Now Consider,

$a_{n+1}a_{n} = a_{n} + 1 \ge 1+1 =2  \; \; \forall \;  n\ge 1$

Changing indices, we get

$a_{n}a_{n-1} = a_{n-1} + 1 \ge 2 \; \; \forall \;  n\ge 2$

Now Consider:

\begin{aligned}
&\left|a_{n+1}-a_{n}\right| \\
=&\left|1+\frac{1}{a_{n}}-\left(1+\frac{1}{a_{n-1}}\right)\right| \\
=&\left|\frac{1}{a_{n}}-\frac{1}{a_{n-1}}\right| \\
=&\left|\frac{a_{n}-1-a_{n}}{a_{n} \cdot a_{n-1}}\right| \\
=&\left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right|
\end{aligned}

Since we have that
$a_{n}a_{n-1}  \ge 2 \; \; \forall \;  n\ge 2$

We get,
$\left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$

$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n}-1}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \forall n\ge 2$

$\therefore$

$$\left|a_{3}-a_{2}\right| = \left|\frac{a_{2}-a_{1}}{a_{2} \cdot a_{1}}\right| \leqslant \frac{\left|a_{2}-a_{1}\right|}{2} = \frac{1}{2}$$

$\therefore$

$$\left|a_{4}-a_{3}\right| = \left|\frac{a_{3}-a_{2}}{a_{3} \cdot a_{2}}\right| \leqslant \frac{\left|a_{3}-a_{2}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^2} $$

$\therefore$

$$\left|a_{5}-a_{4}\right| = \left|\frac{a_{4}-a_{3}}{a_{4} \cdot a_{3}}\right| \leqslant \frac{\left|a_{4}-a_{3}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^3} $$

$\therefore$

$$\left|a_{n+1}-a_{n}\right| = \left|\frac{a_{n}-a_{n-1}}{a_{n} \cdot a_{n-1}}\right| \leqslant \frac{\left|a_{n}-a_{n-1}\right|}{2} \leqslant \frac{\left|a_{2}-a_{1}\right|}{2^{(n+1)-2}}= \frac{\left|a_{2}-a_{1}\right|}{2^{n-1}} \forall n \ge 2 $$

Hence,

for some $m \ge n$

$$\left|a_m-a_n\right| \leqslant \left|a_{m}- a_{m-1}\right| + \left|a_{m-1}-a_{m-2}\right| + ... \left|a_{n+1}-a_{n}\right| $$
$\;$
$$\leqslant \frac{\left|a_2-a_1\right|}{2^{m-2}} + \frac{\left|a_2-a_1\right|}{2^{m-3}} + ... \frac{\left|a_2-a_1\right|}{2^{n-1}}$$

$$ =  \left| a_{2}-a_{1} \right| \cdot \frac{1 - {\frac{1}{2}}^{n-m}}{1 - \frac{1}{2}}$$

Setting $$n = N, m= 2N$$

We get for given $ \epsilon > 0 $

For $N> log_{2}{\frac{2-\epsilon}{\epsilon}}$

We have, $m>n \ge N$

$\left|a_m - a_n\right| < \epsilon $

Hence, this sequence is Cauchy.

Is my approach right?

答:
Step 1. By induction show that $$\alpha:=\frac32 \le a_n \le \frac{15}8.$$
(The upper-bound is the main trick!)

Step 2. We have
$$
|a_{n+1}-a_n| =
|\frac{1}{a_n}-\frac{1}{a_{n-1}}| =
\frac{|a_n-a_{n-1}|}{a_na_{n-1}} \le
\frac{|a_n-a_{n-1}|}{\alpha^2}.
$$
So by induction we have
$$|a_{n+1}-a_n|\le \alpha^{-2(n-1)}|a_2-a_1| = \frac12\alpha^{-2(n-1)}.$$

Step 3. For $n>m$, we have
$$|a_n-a_m| \le \sum_{k=m}^{n-1} |a_{k+1}-a_k|
\le
\frac12\sum_{k=m}^{n-1}\alpha^{-2(k-1)}
\le
\frac12\sum_{k=m}^{\infty}\alpha^{-2(k-1)} = \frac{\alpha^2}{2(1-\alpha^{-2})}\alpha^{-2m}.
$$

This shows that the sequence is Cauchy and convergent.


答:
I think the following method is better :we have  $$a^4(b-a+a-c)+b^4(c-a)+c^4(a-b)$$ $$=(a^4-c^4)(b-a)+(a-c)(a^4-b^4)$$ $$=(a-c)(b-a)(a^3+ac^2+a^2c+c^3)+(a-c)(a-b)(a^3+ab^2+a^2b+b^3)$$ $$=(a-c)(b-a)(\color{blue}{ac^2-ab^2}+\color{red}{a^2c-a^2b}+\color{green}{c^3-b^3})$$ can you end it now?

Note we just used well known identities $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...b^{n-1})$

Title: Help with factoring $a^4(b-c)+b^4(c-a)+c^4(a-b)$

Body:
I want to factor $a^4(b-c)+b^4(c-a)+c^4(a-b)$. I found a similar question on the site, but still don't know how to factor it. from Macavity's answer I realized it has the factor of $(a-b)(a-c)(b-c)$. as mentioned in the answer I should find a quadratic that after multiplying on this factor getting the initial expression. I expanded these expressions:

$$a^4b-a^4c-b^4c-b^4a+c^4a-c^4b=(a^2b-a^2c-b^2a+b^2c+ac^2-bc^2)\times\underbrace{(\quad\quad\quad)}_\text{quadratic expression}$$
I don't know how to find it.


问:
Algebra Equation with equals on left hand side
$6 = 4 - 2x =$

Show the answer with the mechanics of working out.

答:
Given the equation:
$$\boxed{6=4-2x}$$
We can subtract $-4$ from both sides:
$$6-4=4-4-2x$$
Simplifying we are left with:
$$2=-2x$$
Dividing by $-2$ on both sides to isolate x:
$$\dfrac{2}{-2}=\dfrac{-2x}{-2}$$
Leaving us with the solution:
$$\boxed{x=-1}$$

We can check we have the correct solution by plugging the $-1$ back into the original equation:

$$6=4-2(-1)$$

$$6=4+2$$

$$\boxed{6=6}$$

We find the answers are the same which proves we have the correct solution.


答: See both polynomials as polynomials in $a$; the numerator is$$(b-c)a^4+(-b^4+c^4)a+b^4c-bc^4\tag1$$and the denominator is$$(b-c)a^2+(-b^2+c^2)a+b^2c-bc^2.\tag2$$Now, apply long polynomial division to $(1)$ and $(2)$, and you will get$$a^2+(b+c)a+b^2+bc+c^2.$$


问:
Prove that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.
Let $E$ be an extension of a field $F$ and let $a \in E$ be nonzero.

Prove that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.

$\textbf{My Attempt:}$

prove of "if":

Assume that $a$ is algebraic over $F$.

Then, by definition there is a non-zero $f(x) \in F[x]$ such that $f(a) = c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$

where each $c_i \in F$ and for some $n \geq 0$.

Then, $c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0 \implies c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a = c_0$.

Then, $c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 = c_0 a^{-1} \implies a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$.

Since, each $c_i \in F$. So does each $\frac{c_i}{c_0} \in F$.

Notice that $F(a) = \{ b_0 + \cdot + b_{n-1} a^{n-1} ~|~ b_0, ..., b_{n-1} \in F \}$.

So, $a^{-1} \in F(a)$. Which means $a^{-1}$ is algebraic over $F$.

prove of "only if":

Assume that $a^{-1}$ is algebraic over $F$.

Then, there is a non-zero $f(x) \in F[x]$ such that $f(a^{-1}) = 0$.

Which $a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$, for each $\frac{c_i}{c_0} \in F$.

Then, $c_0 a^{-1} = c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 \implies c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$.

Where $F[a] = \{ c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 ~|~ c_n, ..., c_0 \in F \}$.

Which means $a \in F[a]$. So, $a$ is algebraic over $F$.

$\textbf{I don't think my proof is correct, but I don't know where to fix it!}$

答:
Well, the first thing to notice might be that the "if" and "only if" directions are the same: you prove one of them, and in that proof change all $a$ to $a^{-1}$, you will get a proof the other direction.

In your proof, you tried you divide by $c_0$. But $c_0$ can be zero, so that's why your proof is not correct. Nice try though!

Let us do your "only if" direction (which, in fact, is the "if" direction, but this doesn't matter.) Assume $a^{-1}$ is algebraic over $F$. Then, by definition there is a non-zero $f(x) \in F[x]$ such that $$f(a^{-1})=c_n{a}^{-n}+c_{n−1}a^{-n+1}+⋯+c_1a^{-1}+c_0=0$$
Just as you noted in your proof. Now multiply through by $a^n$, we get
$$c_n + c_{n-1}a + \cdots + c_1a^{n-1}+c_0a^n = 0$$
Now, we are still facing the problem that $c_0$ might be zero, but it doesn't matter now, since some $c_k$ is not zero: remember $f$ is a non-zero polynomial. Also it can't be the case that $c_n \neq 0$ but all others are $0$, because that would just mean $c_n = 0$. So, you disregard all the zero coefficients, and you are left with a non-zero polynomial relation in $a$ (although the degree might be lower), and you are done.


问:
Solving $\sin(\frac{2\pi x}{a}) = 0$ on the interval $[-\frac{3a}{2}, +\frac{3a}{2}]$
I have the equation $$\sin\left(\frac{2\pi x}{a}\right) = 0$$
and I have to solve for $x$ in the interval $\left[-\frac{3a}{2}, +\frac{3a}{2}\right]$.

Can someone help me?

答:
Hint:

$\sin \theta = 0$ for $\theta = n\pi, n\in \Bbb Z$.

So, $$\frac{2\pi x}{a} = n\pi \Rightarrow x = \frac{an}{2}$$

Now select $n\in \Bbb Z$ such that $x$ (one or more) lies in the required interval.


问:
Computing the integral $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$
Let $I=\int_{0}^{1}\sqrt{1-x^{\pi}}dx$ If we use $u=x^{\pi}$ then we get $dx=\frac{du}{\pi^{\frac{\pi-1}{\pi}}}$ are there any better identites or substitution that can make integral easier to compute?

Body:
Just for the fun of it

As said in comments and answers, the is no closed form expression for the antiderivative. However, we could have an infinite series representation of it since
$$\sqrt{1-x^{\pi}}=\sqrt{1-x^{3}}\Bigg[1+x^3\sum_{n=1}^\infty \frac { 1} {2^n\, n! }\left(\frac{\log (x)}{x^3-1}\right)^n P_n(x)\,(\pi-3)^n\Bigg] $$ where the first $P_n(x)$ are
$$\left(
\begin{array}{cc}
 n & P_n(x) \\
 1 & 1 \\
 2 & x^3-2 \\
 3 & x^6-2 x^3+4 \\
 4 & x^9-4 x^6-4 x^3-8 \\
 5 & x^{12}-4 x^9+36 x^6+56 x^3+16 \\
 6 & x^{15}-6 x^{12}-76 x^9-496 x^6-336 x^3-32 \\
 7 & x^{18}-6 x^{15}+288 x^{12}+3088 x^9+5328 x^6+1632 x^3+64
\end{array}
\right)$$

For any $n$, each term shows an explicit antiderivative in terms of hypergeometric function. For example, for $n=1$
$$\int \frac{ x^3 \log (x)}{ \sqrt{1-x^3}}\,dx=$$
$$-\frac{x^4}{16}  \left(\,
   _3F_2\left(\frac{1}{2},\frac{4}{3},\frac{4}{3};\frac{7}{3},\frac{7}{3};x^3\right)-
   4 \, _2F_1\left(\frac{1}{2},\frac{4}{3};\frac{7}{3};x^3\right) \log (x)\right)$$

Using only the elements given in this answer, computing the integral from $0$ to $t$ gives the following results

$$\left(
\begin{array}{cccc}
 t & \text{approximation} & \text{solution} & \text{error} \\
 0.1 & 0.099991285257983 & 0.099991285257855 & 1.29 \times 10^{-13}\\
 0.2 & 0.199846061391490 & 0.199846061391227 & 2.63 \times 10^{-13}\\
 0.3 & 0.299172693249054 & 0.299172693248698 & 3.56 \times 10^{-13}\\
 0.4 & 0.397263323235192 & 0.397263323234765 & 4.26 \times 10^{-13}\\
 0.5 & 0.493045151618189 & 0.493045151617706 & 4.83 \times 10^{-13}\\
 0.6 & 0.584997454713703 & 0.584997454713173 & 5.30 \times 10^{-13}\\
 0.7 & 0.670989777542907 & 0.670989777542338 & 5.70 \times 10^{-13}\\
 0.8 & 0.747926304221105 & 0.747926304220506 & 6.00 \times 10^{-13}\\
 0.9 & 0.810765563739254 & 0.810765563738633 & 6.20 \times 10^{-13}\\
 1.0 & 0.846941809210854 & 0.846941809208595 & 2.26 \times 10^{-12}
\end{array}
\right)$$

Otherwise and much simpler (but more difficult to compute)
$$I=\int\sqrt{1-x^{\pi}}\,dx=\frac{x \left(\pi  \, _2F_1\left(\frac{1}{2},\frac{1}{\pi };1+\frac{1}{\pi };x^{\pi
   }\right)+2 \sqrt{1-x^{\pi }}\right)}{2+\pi }$$


问:
Inverse image of prime ideal in noncommutative ring
We say an ideal $P\neq R$ in a ring $R$ is prime, if for any two ideals $I,J$ of $R$ the following implication holds: if $IJ\subseteq P$ then $I\subseteq P$ or $J\subseteq P$.

If $f\colon R \to S$ is a homomorphism of noncommutative rings and $P$ is a prime ideal in $S$, is $f^{-1}(P)$ a prime ideal in $R$?

Of course, the above statement is true for commutative $R$ and $S$. I managed to prove it for noncommutative $R$ and $S$ when $f$ is an epimorphism.

答: The answer to your question is no. Take a prime ring $R$ and $R'\subseteq R$ a subring which is not prime. Then the inverse image of $(0)$ is $(0)$ which is not a prime ideal of $R'$. An example is the following: $R=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ \mathbb Z & \mathbb Z\end{pmatrix}$ and $R'=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ 0 & \mathbb Z\end{pmatrix}$ with $n\ge 1$. (This example is taken from Lam, Exercises in  Classical Ring Theory, Exercise 10.7.)


问:
Kernel of a linear operator
Wikipedia:

If $V$ and $W$ are topological vector spaces such that $W$ is finite-dimensional, then a linear operator $ L: V → W$ is continuous iff ${\frak Ker}(L)$ is a closed subspace of $V$.

The theorem seems pretty general since it doesn't require any specific structure for $W$ and $V$.
So at first glance, it shouldn't hold if $\ \dim W,V = \infty\ $. But what if $V=W=\scr H$, a Hilbert space? Is there some hope?
Maybe not a double implication, but something like:

$L \in {\frak B}({\scr H}) \implies {\frak Ker}(L) \equiv \overline {{\frak Ker}(L)} $

答:
For any TVS $V,W$ (where $W$ is Hausdorff) we have $L: V\to W$ continuous implies $\text{Ker}(L)$ closed: the kernel of $L$ is the preimage of $\{0\}$, which is closed if $W$ is Hausdorff. Since $L$ is continuous, the preimage of $\{0\}$ is then closed in $V$.

For the other implication a simple counterexample goes as follows: let $(v_n)$ be an ONB (or at least an orthonormal sequence) and define $T v_n:= n v_n$. This operator is unbounded, hence discontinuous, but still has a closed kernel: the trivial subspace.

Title: Kernel of a linear operator


问:
The equation of a sphere of a given radius, whose center belongs to a given line and it is passing through a point
Determine the equation of a sphere of radius $\sqrt{5}$, whose center belongs to the line
$$d : {{x}\over1} = {{y-1}\over -1} = {{z+2}\over 1} $$
and it is passing through the point $A(0,2,-1)$

My solution:
If the center belongs to the line $d$, it means that the line is tangent to the sphere and so the distance from the center $C(x,y,z)$ to the line $d$ is equal with R = $\sqrt{5}$. After doing all the computation I got the following result:
$$2x^2+2y^2+2z^2-6x-6y+2xy-2xz+2yz+6 = \sqrt{15}$$
Now, this isn't really the equation of the sphere I'm looking for because the radius here is $\sqrt{15}$. Could you tell me where I went wrong in my solution?

答:
From the equations
$$d : {{x}\over1} = {{y-1}\over -1} = {{z+2}\over 1} $$
we get $$y=1-x;\;z=x-2$$
So the line $d$ has parametric equations
$$
\begin{cases}
x=t\\
y=1 - t\\
z= -2 + t\\
\end{cases}
$$
Thus the center of the sphere has coordinates $C(t,1-t,-2+t)$

The sphere passes through $A(0,2,-1)$ and radius is $r=\sqrt 5$

Therefore $AC^2=5$ that is
$$t^2+(t+1)^2+(t-1)^2=5$$
which gives $t=\pm 1$

The center of the sphere can be $$C_1(1,0,-1);\;C_2(-1,2,-3)$$


问:
Intuition about probability of choosing right key for a door

The key to opening is in a bunch of ten keys. A man attempts to open the door by trying the keys at random and discarding the wrong key. The probability that door is opened on fifth trial is:

The solution key involves using multiplication rule of probability for dependent events:

$$ P = \frac{9}{10} \frac{8}{9} \frac{7}{8} \frac{6}{8} \frac{1}{6} = \frac{1}{10}$$

Explanation, nine out of the ten are wrong and hence the probability of choosing wrong key in first choose must be $ \frac{9}{10}$, similarly once we have taken one wrong key , the probability of choosing the next key is given as $ \frac{8}{9}$ and so on till the fifth draw where we only have one 'right' key out of the six.

However, if we were to extend this result, it'd turn out that the chance of man opening the door on first try, second try, third try.. last try are all equal.. how is this possible?

I think that if we do a truly random experiment, then if we get wrong answer the first few times, it must increase our chances of getting the right answer later on.

答:
The probability that the door is opened on the first, second,....tenth trial is always constant and equal to $\frac{1}{10}$

The probability cannot change...

think at the following problem: you have only 1 good key among 10. the probablity to get it cannot depend by the fact that it is detected at first, second or tenth draw


问:
What is the "standard" way of proving that the transition map is $C^{\infty}$?
Sorry if this is a vague question.

I was handed a map between two manifolds, say, $M\to \mathbb{R^n}$. I was indeed able to find an open cover of $M$ and show that there is a homeomorphism between each open set and some open subset of $\mathbb{R^n}$. Proving homeomorphism usually takes me a long time but it's usually doable. But when it comes to showing that each chart is $C^{\infty}$-compatible, in other words, every transition map is smooth, I'm getting nowhere.

The same situation again happened on proving that Grassmannian in the case of $G(2,4)$ has a smooth structure. I identified each open set that covers $G(2,4)$ and proved homeomorphism, but then when it comes to proving the smoothness of transition maps, I get nowhere again. I get the map between two set of matrices and I had no idea how to argue. I think this happens because general theorems on homeomorphism are well-known but smoothness is really different for every situation. But since I am really having trouble proving smoothness, I'd like to know if there is any standard way or theorem of showing that the transition map is smooth. Thank you for your answers.

答: Well, if you're working with $\mathbb{R}^n$, smoothness just means that every component is smooth in all the others (due to some MVT argument, it even suffices to be smooth in every variable separately). And the next trick is that everything you are doing in the Grassmann example is rational (as in: your transition functions are ultimately rational functions), and rational functions are always smooth (where defined).


问:
Find the value of parameter a, so the polynomial has root with multiplicity
everyone!

Please help me understand what do I have to do here.

I have a function:

$$f(x) = x^3 + ax^2 +3x - 1$$

I need to find possible values of parameter a, such that the root of the function has multiplicity.

I have approached this problem like that:

$$f'''(x) = 6 + 2a$$
So I have found that

$$ a = -3$$

Are there any more values of parameters a, that correspond with the requirement above?

答:
For a double root, the discriminant
$$\Delta=(a+3)^2 \,(4 a-15)$$ must equal to $0$.

Have a look here.


问:
Example noetherian infinite dimensional topological space.
Definition. If $X$ is a topological space, we define the dimension of $X$ (denoted $\operatorname{dim} X$ ) to be the supremum of all integers $n$ such that there exists a chain $Z_{0} \subset Z_{1} \subset \ldots \subset Z_{n}$ of distinct irreducible closed subsets of $X.$

now let For $n \in \mathbb{Z}_{\geq 0},$ let $U_{n}=\{n, n+1, n+2, \ldots\} .$ Then the set $\tau=$ $\left\{\emptyset, U_{0}, U_{1}, \ldots\right\}$ is a topology of open sets on $\mathbb{Z}_{\geq 0}$. In this space, if $C$ and $C^{\prime}$ are closed sets, then it is easy to see that either $C \subset C^{\prime}$ or $C^{\prime} \subseteq C,$ that every nonempty closed set is irreducible, and that every closed set other then $\mathbb{Z}_{\geq 0}$ is finite. So this is an example of a Noetherian infinite dimensional topological space.

how we can see that either $C \subset C^{\prime}$ or $C^{\prime} \subseteq C,$ that every nonempty closed set ? why this topology is Noetherian ? why this topological space is infinite dimensional ?

答: As a hint, the closed set $C_n$ corresponding to $U_n$ is $\{0,\ldots,n-1\}$ und the sets $U_n$ form a basis of the topology.


问:
Formula for inner product in a RKHS
Given a kernel function $k$ for which a Reproducing Kernel Hilbert Space (RKHS) $H$ exists, can I write a formula how to compute the inner product of two functions in $H$? I am, of course, aware that because of the reproducing property, I can compute $\left<f, k(x,\cdot)\right>_H = f(x)$ for $f \in H$. But if I take any two arbitrary functions $f,g \in H$, how would the formula for computing $\left<f,g\right>_H$ look like? For example, if $k$ is a Gaussian kernel but also in general case? Can the inner product formula can be always explicitly written? Thank you!

答:
Hope it is not too late.. But maybe here are some intuitions on how to do it. The short answer is "it is difficult in general". Meaning for a given kernel it is not clear at first sight what $\langle f,g\rangle_{\mathcal{H}}$ does (where $\mathcal{H}$ is a RKHS). But with little assumptions we can do things.

A way to tackle this problem is to look at RKHS as a particular subset of square-integrable functions. With little rigor:

Suppose that you are in a space $(\mathcal{X},\mu)$ where $\mu$ is a finite measure. Suppose that $x\rightarrow k(x,x)$ is integrable w.r.t $\mu$ (which is not a "so" strong assumption). This implies that your RKHS $\mathcal{H}\subset L_2(\mu)$ meaning that your functions in the RKHS are square integrable. Indeed:
\begin{equation}
\begin{split}
\int |f(x)|^{2}d \mu(x)&=\int |\langle f,k(x,\cdot)\rangle_{H}|^{2}d \mu(x)\leq \int \|f\|^{2}_{H} \|k(x,\cdot)\|^{2}_{H}d \mu(x) \\
&\stackrel{*}{=}\|f\|^{2}_{H}\int |k(x,x)|d\mu(x)<+\infty
\end{split}
\end{equation}
where I used that $\|k(x,\cdot)\|^{2}_{H}=|\langle k(x,\cdot),k(x,\cdot)\rangle_{H}|=|k(x,x)|$ due to the properties of kernels.

Consider now the operator $\Sigma$ from $L_2(\mu)$ to $L_2(\mu)$ defined by:
\begin{equation}
\forall x \in \mathcal{X}, \ [\Sigma f](x)=\int f(y)k(x,y)d\mu(y)
\end{equation}
This operator is linear and finite. It acts like a kind of averaging on $f$ like a sensor. I am not going into details but it is also positive definite (due to Mercer's theorem) so we can talk about eigenvalues and eigenfunctions. So we can diagonalize it like for matrices and find its square root $\Sigma^{1/2}$.
If we moreover assume that $\mathcal{H}$ is dense in $L_2(\mu)$ then $\Sigma^{1/2}$ defines a bijection and more importantly an isometry from $\mathcal{H}$ to $L_2(\mu)$ and we have in particular:
\begin{equation}
\forall f,g \in \mathcal{H}, \ \langle f,g\rangle_{\mathcal{H}}=\langle \Sigma^{-1/2}f,\Sigma^{-1/2}g\rangle_{L_2(\mu)}
\end{equation}
This results states that the RKHS $\mathcal{H}$ is the subspace of square integrable functions $f$ such that $\|\Sigma^{-1/2}f\|_{L_2(\mu)}<+\infty$. It controls somehow the degree of smoothness of the functions.

So a way to compute the dot products using the kernel is the following: diagonalize the integral operator and just do a usual dot product of functions. The first part might be tricky but for some kernel and measure $\mu$ we know how to do it.

A good ref for this:

[1] Reproducing Kernel Hilbert Spaces in Probability and Statistics. Alain Berlinet et Christine Thomas-Agnan.

Title: Formula for inner product in a RKHS


问:
The order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$.
This is a follow-up to my previous problem. I think it will be clearer if I ask this as a stand-alone question. I apologise if I come up a bit hectic but I'm not entirely sure what my doubts are exactly.

Basically, at 16:30 here we change the order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$, separating this into series of real and imaginary parts. My question is - how do we know that it doesn't change the sum of the original series if it's not absolutely convergent? Is there a result that states this? The only theorem I've come across is:

The necessary and sufficient condition that the series of complex numbers
$$\alpha_1 + \alpha_2 + \alpha_3 + \dots + \alpha_n + \dots$$
converges to a limit $\alpha = a + bi$ is that
$$L_{n = \infty} A_n = a, \; L_{n = \infty} B_n = b.$$

However, I don't see how that helps. It seems to me like we just take it for granted that $1 - \frac{1}{3} + \frac{1}{5} - \dots$ (after rearrangement) sums up to $\frac{\pi}{4}$ (before rearrangement).

答:
If $\sum a_n$, $\sum b_n$ and $\sum a_n + b_n$ are all convergent, you have
$$\sum a_n + b_n = \sum a_n + \sum b_n$$

This has nothing to do with a rearrangement of the terms, there is no reordering here. This just uses the fact that when all the limits exist, one has
$$\lim (u_n + v_n) = \lim u_n + \lim v_n$$


答:
Let $z_n=a_n+ib_n$ be a sequence of complex numbers, where $a_n,b_n\in\mathbb{R}$ are the real and imaginary parts. In general, the (complex) series $\sum z_n$ is convergent if and only if the two (real) series $\sum a_n$ and $\sum b_n$ are both convergent.

Basically, this is because
$|a+bi|\le |a|+|b|\le 2|a+bi|$ and thus

$$
|\sum_{n=1}^N z_n-L|
\le |\sum_{n=1}^N a_n-\textrm{Re}(L)|+|\sum_{n=1}^N b_n-\textrm{Im}(L)|
\le 2|\sum_{n=1}^N z_n-L|
$$

If we know that $\sum z_n$ is convergent (and thus both the series of the real and imaginary parts convergent), the partial sum identity
$$
\sum_{n=1}^N z_n = \sum_{n=1}^N a_n+i\sum_{n=1}^N b_n
$$
implies that
$$
\sum_{n=1}^\infty z_n = \sum_{n=1}^\infty a_n+i\sum_{n=1}^\infty b_n
$$

Note that in the discussion above, we don't use anything about $\sum |z_n|$, i.e., we don't need to know whether $\sum z_n$ is absolutely convergent or not.

Title: The order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$.

Body:
This is a follow-up to my previous problem. I think it will be clearer if I ask this as a stand-alone question. I apologise if I come up a bit hectic but I'm not entirely sure what my doubts are exactly.

Basically, at 16:30 here we change the order of summation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} i^n$, separating this into series of real and imaginary parts. My question is - how do we know that it doesn't change the sum of the original series if it's not absolutely convergent? Is there a result that states this? The only theorem I've come across is:

The necessary and sufficient condition that the series of complex numbers
$$\alpha_1 + \alpha_2 + \alpha_3 + \dots + \alpha_n + \dots$$
converges to a limit $\alpha = a + bi$ is that
$$L_{n = \infty} A_n = a, \; L_{n = \infty} B_n = b.$$

However, I don't see how that helps. It seems to me like we just take it for granted that $1 - \frac{1}{3} + \frac{1}{5} - \dots$ (after rearrangement) sums up to $\frac{\pi}{4}$ (before rearrangement).


问:
Two questions about a special function
Define $$f(n):=\sum_{j=1}^n j^{n+1-j}=1+2^{n-1}+3^{n-2}+\cdots+(n-1)^2+n$$ for a positive integer $n$.

With PARI/GP, this function can be calculated with the self-defined function

```<code>f(n)=sum(j=1,n,j^(n+1-j))
</code>```

Question $1$ : Is $f(n)$ only a perfect power for $n=3$ ?

I checked upto $n=10^4$ and only found one perfect power, namely for $n=3$.
I do not even have an idea for the perfect squares. The PARI/GP-code for those who want to extend the search limit :

```<code>gp > for(m=1,10^4,if(ispower(f(m))>0,print1(m," ")))
3
gp >
</code>```

Question $2$ : If $f(n)$ only prime for $n=2$ and $n=34$ ?

Upto $n=7\ 400$ , there are no other primes $f(n)$.

It seems that $f(n)$ cannot be easily calculated with a formula. So, I think I cannot use fast primality testing tools like PFGW, but someone might have a faster software for testing general numbers.

The first such number from which I do not know a prime factor, is $f(138)$. So, small factors are not forced, hence I guess we can only search for further primes.

答:
To get a better intuition about the possible primality of $f(n)$ I've looked at periodicities in its primefactorizations.

Some short heuristics.

Primefactor $2$:

for $n_k=3+4k$ we have $2 \mid f(n)$

for $n_k=4+4k$ we have $2 \mid f(n)$

We analyze the sequences of p-adic valuations for that primefactor below.

Primefactor $p=3$:

with some vector $R$ of length $(p-1) \cdot p$ and residues $\pmod {(p-1) \cdot p^2}$ having

$\qquad R=[2,7,10,15,17,18]$

for $\qquad n_{j,k}=r_j+18k \qquad $ we have $ p \mid f(n)$

Primefactor $p=5$:

with some vector $R$ of length $p \cdot (p-1)$ and residues $\pmod {(p-1) \cdot p^2}$ having
$\qquad R=[5, 8, 13, 14, 16, 21, 44, 47, 49, 55, 62, 63, 70, 71, 78, 86, 92, 97, 99, 100]$

for $\qquad n_{j,k}=r_j+100k \qquad$ we have $ p \mid f(n)$

Primefactor $p=7,11,13,...$:

This is heuristically and in all tested cases analogue for the other small primefactors

We can even assume, that the higher p-adic valuations are periodic with $(p-1) \cdot p^A$ with stepwidth of $n$ like $(p-1) \cdot p^A$ . There is something more to say, but I can transfer the empirical tabular results to precise formulae only later.

Possibly we can say something about the covering of $f(n)$ by this periodicity of the primefactors, and how many holes per primefactor can occur allowing $f(n)$ to be composed by one prime factor alone.

Some initial impression:

```<code> n  primefactorization of f(n)
 --|--------------------------------------
 1   1
 2   3
 3       2^3
 4       2  .11
 5              5 .13
 6           11      .19
 7   3.  2^2            .61
 8       2^2   .5          .139
 9                           31.367
10   3      .11                      .1511
11       2^6                               .3637
12       2                                      .575957
13              5                                      .1203757
14              5                                              .41.103.1567
15   3  .2^2                                                              .7.29.78317
16       2^5   .5.13                                                                 .227.2437
17   3                                                                                .37.65240639
18   3                                                                                .73.216688337
19       2^3                                                                      .40394337023
20   3^2.2                                               .126821110583
</code>```

Title: Two questions about a special function

Body:
To get a better intuition about the possible primality of $f(n)$ I've looked at periodicities in its primefactorizations.

Some short heuristics.

Primefactor $2$:

for $n_k=3+4k$ we have $2 \mid f(n)$

for $n_k=4+4k$ we have $2 \mid f(n)$

We analyze the sequences of p-adic valuations for that primefactor below.

Primefactor $p=3$:

with some vector $R$ of length $(p-1) \cdot p$ and residues $\pmod {(p-1) \cdot p^2}$ having

$\qquad R=[2,7,10,15,17,18]$

for $\qquad n_{j,k}=r_j+18k \qquad $ we have $ p \mid f(n)$

Primefactor $p=5$:

with some vector $R$ of length $p \cdot (p-1)$ and residues $\pmod {(p-1) \cdot p^2}$ having
$\qquad R=[5, 8, 13, 14, 16, 21, 44, 47, 49, 55, 62, 63, 70, 71, 78, 86, 92, 97, 99, 100]$

for $\qquad n_{j,k}=r_j+100k \qquad$ we have $ p \mid f(n)$

Primefactor $p=7,11,13,...$:

This is heuristically and in all tested cases analogue for the other small primefactors

We can even assume, that the higher p-adic valuations are periodic with $(p-1) \cdot p^A$ with stepwidth of $n$ like $(p-1) \cdot p^A$ . There is something more to say, but I can transfer the empirical tabular results to precise formulae only later.

Possibly we can say something about the covering of $f(n)$ by this periodicity of the primefactors, and how many holes per primefactor can occur allowing $f(n)$ to be composed by one prime factor alone.

Some initial impression:

 n  primefactorization of f(n)
 --|--------------------------------------
 1   1
 2   3
 3       2^3
 4       2  .11
 5              5 .13
 6           11      .19
 7   3.  2^2            .61
 8       2^2   .5          .139
 9                           31.367
10   3      .11                      .1511
11       2^6                               .3637
12       2                                      .575957
13              5                                      .1203757
14              5                                              .41.103.1567
15   3  .2^2                                                              .7.29.78317
16       2^5   .5.13                                                                 .227.2437
17   3                                                                                .37.65240639
18   3                                                                                .73.216688337
19       2^3                                                                      .40394337023
20   3^2.2                                               .126821110583


问:
Help Understanding Complex Roots
I was reading a graphical explanation of complex roots, and between Figures 7 and 8 I became confused.

The roots appear in the imaginary plane, but I don't understand why the original function must be inverted before the graphical representation fits. I realize that inversion is necessary for the graphical representation of imaginary roots to fit, but the lack of explanation bothers me. I would like to understand this inverstion better than "we do it because it fits the data."

答:
Comment: In general, a nonzero quadratic has the form $y = A(x-a)^2+b^2$ where $A,a,b$ are real constants, but I set $A=1$ for the sake of clarity.

The graph is given to extend $y=f(x) = (x-a)^2+b^2$ to $y=Re(f(z)) = Re((z-a)^2+b^2)$ where $z = x+it$ is a complex variable. The graph given indicates all points in $(t,x,y)$ space such that the equation $y=Re(f(z))$ is satisfied. In particular,
$$ y = Re((x+it-a)^2+b^2) = (x-a)^2-t^2+b^2  $$
Considering $x=a$ which is a vertical plane in the pictures,
$$ y = -t^2+b^2 $$
This is a parabola which opens downward in the $x=a$ plane. It has $y=0$ when $t = \pm b$. This is why the graph opens downward. On the other hand, the plane $t=0$ (otherwise known as the $xy$-plane) we have $y = (x-a)^2+b^2$ which is a parabola with vertex $(a,b^2)$ which opens upward with no $x$-intercept.

The function is not inverted, it is extended to a complex domain, in this larger context the three-dimensional graph looks as pictured. However, beware the real picture casts $y = Re(f(x+it))$ as a surface, it is one-equation in three variables. This gives two parameters at most points on its solution set. The picture you post focuses on two interesting curves on this surface (it has parametrization $X(u,v) = (u,v,Re(f(u+iv)))$ )

Also, I was tempted to say it was the graph of $y = f(x+it)$, but that is more difficult to graph directly... think about it.

Title: Help Understanding Complex Roots