答:
4 is false. A counterexample is given with $G=\operatorname{PGL}(3)$ and with $H≅\operatorname{GL}(2)$ the subgroup of matrices of the form
$$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}\text{.}$$

Choose $x$ different from $0$ and $1$. Take
$$a=\begin{pmatrix}1&0&0\\0&x&0\\0&0&x\end{pmatrix}\text{.}$$
Then $a$ is in the center of $H$ so that $[a]_H = \{a\}$. But a conjugate in $G$ is
$$b = \begin{pmatrix}x&0&0\\0&x&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1/x\end{pmatrix}$$
which is not in the center of $H$, so that $|[b]_H| > 1$.

Title: The symmetry of conjugacy classes of a subgroup of a finite group

Body:
I have been solving some exercise in group theory and came up with two observations, the first and second below. Then I tried to generalize the fact from that exercise and got an idea of other two observations (the third and fourth below), but was unable to prove or disprove them. Any further observations about them is welcome, and a proof or disproof would resolve my problem.

Let $G$ be a finite group, let $H\leq G$ be its subgroup, and let's define the notation $[x]_A$ to be the conjugacy class of element $x\in A$ of group $A\in \{G,H\}$.

It holds that $(\forall x\in H)\Big(\frac{|[x]_G|}{|[x]_G\cap H|}\text{ divides }[G:H]\Big)$.

It holds that $(\forall x\in H)[x]_H\subseteq [x]_G$.

Does it hold that $(\forall x,y\in H)\big(\text{some of the numbers }\frac{|[x]_G|}{|[x]_G\cap H|}\text{ and }\frac{|[y]_G|}{|[y]_G\cap H|}\text{ divides the other one}\big)$?

Does it hold that $(\forall x\in H)(\forall a,b \in[x]_G\cap H)|[a]_H|=|[b]_H|$?

The exercise that I was solving was the next one:

Let $n\in\mathbb{N}$ be a natural number. And let $H\leq S_n$, $[S_n:H]<n$ be a subgroup of symmetric group $S_n$ of order less than $n$. If $n=5$, prove that $H$ is either $A_n$ or $S_n$.

To prove that, I used the first statement from above. Then I tried to generalize this exercise and replace the condition $n=5$ with $n\geq 5$ and came up with the third and fourth statements as possibly useful lemmas.


问:
What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? A hyperplane?
What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? Here $\theta\in\mathbb{R}^d$ and $v$ and $u$ fixed from $\mathbb{R}^d$, and $\langle a,b\rangle$ is dot product of two vectors $a$ and $b$.

My feeling is $\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is a hyperplane in $\mathbb{R}^d$ so that $\theta-u$ is orthogonal with $v\in\mathbb{R}^d$. The degree of freedom of $\theta\in\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is $d-1$. Is this correct?

答:
It is an affine hyperplane, i.e. a translate (coset) of a $1$-codimensional vector subspace.

In particular, it is $u+v^\perp$ where $v^\perp$ denotes the orthogonal complement of $v$.

ContentLicense: CC BY-SA 4.0


答:
Yes, it is a hyperplane. The equation for the plane is $$v\cdot\theta +b = 0,$$
where $b=-v\cdot u$. It indeed has $d-1$ degrees of freedom.

Title: What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? A hyperplane?

Body:
What is $\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\}$? Here $\theta\in\mathbb{R}^d$ and $v$ and $u$ fixed from $\mathbb{R}^d$, and $\langle a,b\rangle$ is dot product of two vectors $a$ and $b$.

My feeling is $\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is a hyperplane in $\mathbb{R}^d$ so that $\theta-u$ is orthogonal with $v\in\mathbb{R}^d$. The degree of freedom of $\theta\in\left\{\theta\in\mathbb{R}^d,\langle v,\theta-u\rangle =0\right\}$ is $d-1$. Is this correct?


问:
Logical definition with equalily and Law of Identity in Suppes' "Introduction to Logic"
Patrick Suppes' "Introduction to Logic" provides rules for formal definitions in chapter 8. The rules below are specified for a new operation symbol with equality:

An equivalence $D$ introducing a new n-place operation symbol $O$ is a proper definition in a theory if and only if $D$ is of the form:

$O(v_1, ..., v_n) = w \leftrightarrow S$

and the following restrictions are satisfied:

(i)   $v_1, ..., v_n, w$ are distinct variables.

(ii)  $S$ has no free variables other than $v_1, ..., v_n, w$.

(iii) $S$ is a formula in which the only non-logical constants are primitive symbols and previously defined symbols of the theory.

(iv)  The formula $\exists !w[S]$ is derivable from the axioms and preceding definitions of the theory.

There's also a prior mention of the Law of Identity:

If x is anything whatsoever, then $x=x$.

Now let's suppose that you have the following definition:

$$
\forall f,x,y[f_x = y \iff f \text{ is a function } \land \langle x,y \rangle \in f]
$$

Let's also assume that you have previously defined functions and ordered pairs such that you may prove $\exists !y[S]$ with extentionality, so it follows rule (iv).

Here's the problem: Within the bounds of this ruleset, it seems like one can use the Law of Identity with any variable, say $A$, to claim that $A_x=A_x$ and use that to claim that $A \text{ is a function } \land \langle x,A_x \rangle \in A$, and so, that $A$ is a function, even though we know nothing about it. That logic can be used with any variable, be it a normal relation, a simple set, or even a urelement, so this deduction must be wrong.

At first, I thought I was breaking rule (iii), as the statement "$A \text{ is a function } \land \langle x,A_x \rangle \in A$" has a not previously defined symbol in it, $A_x$, which is defined in the statement itself, so it would not be valid.

However, consider the following definition:
$$
\newcommand\liff{\leftrightarrow}
\newcommand\lif{\rightarrow}
\newcommand\lfi{\leftarrow}
\newcommand\ordp[2]{\langle #1,#2 \rangle}
\newcommand\mset[1]{\{ #1 \}}
\newcommand\isRel[1]{#1 \text{ is a relation}}
\newcommand\isFunc[1]{#1 \text{ is a function}}
\newcommand\isOneOne[1]{#1 \text{ is one-one}}
\mset{a} = p \iff \forall x[x \in p \liff x = a]
$$

It is unique by extentionality. It seems a clear consequence from it that $\mset{a} = \mset{b} \lif a = b$, but the only way I see to prove it is by using $\mset{a} = \mset{b}$ to get $\forall x[x \in \mset{b} \liff x = a]$, which would be disallowed if my interpretation was correct, so I don't think that is the answer.

My second instinct was that rule (i) is being broken, that $f_x = f_x$ doesn't count as being distinct variables. However, from the definition above it also seems that $a \in \mset{a}$ should follow. The only way I see to prove this is to use $\mset{a} = \mset{a}$ with the definition, which would be disallowed if this was the case, so I don't feel that's the solution either.

So my question is: What is the actual culprit of the fallacy?

Edit: After extended discussion, I'm adding some information to hopefully clarify what this question is and is not about.

This is not about set theory. My problem is with about the formal language of first-order logic provided by the book. To avoid the focus on set theory, I'll provide a second example. Let's suppose we have the following statements:

$$
\forall a,b,x,y[\text{isSingleChild}(x) \land \text{parentsOf}(a,b,x) \land \text{parentsOf}(a,b,y) \Rightarrow x = y] \\
\forall a,b,x[\text{son}\{a,b\} = x \iff \text{isAdult}(a) \land \text{isAdult}(b) \land \text{parentsOf}(a,b,x) \land \text{isSingleChild}(x)]
$$

The first statement guarantees that $x$ is unique in the definition of $\text{son}$.

The definition of $\text{son}\{a,b\}$ seems to follow all of the rules provided. It is not intended to state that any variable follows any specific predicate, but simply stating their logical relationship. However, if you use it together with the Law of Identity, you may derive:

$$
\newcommand{\fitch}[1]{\begin{array}{rlr}#1\end{array}}
\newcommand{\fcol}[1]{\begin{array}{r}#1\end{array}} %FirstColumn
\newcommand{\scol}[1]{\begin{array}{l}#1\end{array}} %SecondColumn
\newcommand{\tcol}[1]{\begin{array}{l}#1\end{array}} %ThirdColumn
\newcommand{\subcol}[1]{\begin{array}{|l}#1\end{array}} %SubProofColumn
\newcommand{\subproof}{\\[-0.25em]} %adjusts line spacing slightly
\newcommand{\fendl}{\\[0.037em]} %adjusts line spacing slightly
\small
\fitch{
 \fcol{1:\fendl 2:\fendl 3:\fendl\fendl 4:\fendl 5:\fendl 6:\fendl 7:\fendl 8:\fendl
 }
 & \scol {
   \forall a,b,x[\text{son}\{a,b\} = x \iff \text{isAdult}(a) \land \text{isAdult}(b) \land \text{parentOf}(a,b,x) \land \text{isSingleChild}(x)] \\
   \forall x[\text{son}\{a,b\} = x \iff \text{isAdult}(a) \land \text{isAdult}(b) \land \text{parentOf}(a,b,x) \land \text{isSingleChild}(x)] \\
   \text{son}\{a,b\} = \text{son}\{a,b\} \\\quad\iff \text{isAdult}(a) \land \text{isAdult}(b) \land \text{parentOf}(a,b,\text{son}\{a,b\}) \land \text{isSingleChild}(\text{son}\{a,b\}) \\
   \forall x[x = x] \\
   \text{son}\{a,b\} = \text{son}\{a,b\} \\
   \text{isAdult}(a) \land \text{isAdult}(b) \land \text{parentOf}(a,b,\text{son}\{a,b\}) \land \text{isSingleChild}(\text{son}\{a,b\}) \\
   \text{isAdult}(a) \\
   \forall a [\text{isAdult}(a)] \\
 }
 & \tcol{
  \text{P} \fendl
  1\ \forall\text{E}\ \fendl
  2\ \forall\text{E}\ \fendl\fendl
  \text{T}\ \fendl
  4\ \forall\text{E}\ \fendl
  3,5\ {\liff}\text{E}\ \fendl
  6\ {\land}\text{E}\ \fendl
  7\ \forall\text{I}\ \fendl
}}
$$

So from that definition, you may deduce that everyone is an adult. Note what I am not saying. I'm not saying that this argument is sound, nor defending it, I'm saying that the ruleset given in the book permits it (It probably doesn't, but I don't see any rule of logical deduction being broken). I know the argument is illogical, but the formal rules are being followed. My question is not about the soundness of the argument, but the soundness of the system provided in the book.

Also note that the assertion is not about set theory, nor "family theory", it is about the logic itself. My assertion is that (apparently) within the formal system given, any statement of the following form applies:

$$
\forall a,b,x[\text{entityFrom}\{a,b\} = x \iff \text{hasSomeProperty}(a) \land \text{uniqueRelation}(a,b,x)] \vdash \forall a[\text{hasSomeProperty}(a)]
$$

I understand that the definition doesn't entail the conclusion. Nonetheless, within the system, the conclusion seems to be deducible from it.

There are only three options. Either the formal system provided in not sound, the definition actually entails the conclusion, or I'm missing/misinterpreting some rule on the Law of Identity/Rules for Definition/Rules for Quantifiers.

The book and is more than 50 years old, any possible oversights in the system would have been noticed by this point (it also was written by Suppes, so I doubt there is any), thus I'm sure it's not the first one. The definitions also seem well formed and feel like they shouldn't lead directly to the conclusion, so it's probably not the second one as well. Leading to the conclusion that I am probably missing or misinterpreting some proviso/rule that would make that argument not valid. The question is, which one?

What will not answer the question:

"In set theory, functions have a specific domain and need to have [some set properties], so it is not possible for all variables to be functions."

"Your definition of parenthood does not describe the idea of parents correctly, as it does not imply that all children have parents and [some parenthood properties], so the definitions are not correct descriptions."

The solution cannot be about the unsoundness of the argument in one specific theory, that will not get to the root of the problem. A specific context may be used as an example, but the solution has to be on the level of the formal language.

What might answer the question:

"The ruleset given by the book is actually incomplete, because a definition with equality containing [some syntatic property] may lead to a fallacy. However, you may avoid it by adding a new rule that requires your definition to have [new definition contraint]"

"Your definitions logically entail the conclusion. Think about it, if your definition is [this], then [explanation of why the definition should logically lead to the conclusion], so the argument and conclusion are valid. I doubt that's what you intended to conclude with your definition though. I think what you actually mean is [well behaved definition]." $^{\dagger}$

"You've misinterpreted rule [n], perhaps you think it means [interpretation] when it actually says [different interpretation]. If you take that into account, line [x] of your argument is not valid."

"You are forgetting that you cannot substitute for defined terms like you do variables. You can only substitute for a defined term if [some syntatic condition] applies, so step $3$ of your deduction is invalid."

"The Law of Identity does not require only uniqueness, but also [some variable property], so you may not use it as in line $5$, since the variable in your definition doesn't follow this constraint."

Your answer need not be any of the above. I'm just presenting the types of answers I feel will most likely be useful: Answers that focus on the formal language.

Thank you for reading till the end, and I hope this makes clear enough the problem I'm wanting to solve.

$\dagger$ As pointed out by Mauro ALLEGRANZA, this case specially makes sense. As he put it:

Think about it: is there some axioms in your theory saying that not every object is an Adult?

Which I agree with. However, there's one problem: The ruleset shouldn't allow this.

Earlier in the same chapter, before the rules are established, their objective is laid out. The "Criteria for proper definitions". The objective is to separate an axiom from a definition. The first one (Criterion of Eliminability) is not important for this discusion, but the second is.

The Criterion of Non-Creativity states that a definition $S$ is non-creative if and only if:

There is no formula $T$ in which the new symbol does not occur such that $S \rightarrow T$ is derivable from the axioms and preceding definitions of the theory but $T$ is not so derivable.

The objective of the ruleset is to guarantee that our definitions follow both of these criteria. As stated in page 155: "[...] we turn to the task of stating rules of definition which will guarantee satisfacition of the two criteria of eliminability and non-creativity"

In my parenthood example, we have the first statement as an axiom, and the second as a definition. However, within that theory, the statement $\forall a [\text{isAdult}(a)]$ does not contain the new symbol and is derivable from the new definition, but not from the axioms alone, which would make the definition creative.

So in that case, my question then becomes: How come the definition is creative, when the ruleset is supposed to guarantee non-creativity?

答:
The ruleset given by the book is not incomplete. The example derivation you give holds up to scrutiny as well. You get (seemingly) paradoxical conclusions because restriction (iv) does not actually hold in any of your examples.

In your first example, the formula $S$ denotes the following: "$v_2 \text{ is a function } \wedge \langle v_1,w \rangle \in v_2$". So restriction (iv) is not satisfied unless the following is a theorem of the theory under consideration:

$$\exists! w. v_2 \text{ is a function } \wedge \langle v_1,w \rangle \in v_2 $$

which, since $v_1,v_2$ are distinct free variables, holds precisely if

$$\forall v_1. \forall v_2. \exists! w. v_2 \text{ is a function } \wedge \langle v_1,w \rangle \in v_2 $$

is a theorem of your theory as well. Needless to say, this latter statement is very much not a theorem of any reasonable set theory. In particular it would imply "$\forall v. v \text{ is a function }$" by itself.

In your second example, the formula $S$ denotes the following: "$\text{isAdult}(v_1) \wedge \text{isAdult}(v_2) \wedge \text{parentsOf}(v_1,v_2,w) \wedge \text{isSingleChild}(w)$". As above, restriction (iv) is not satisfied unless the following is a theorem of the theory under consideration:

$$ \forall v_1. \forall v_2. \exists! w. \text{isAdult}(v_1) \wedge \text{isAdult}(v_2) \wedge \text{parentsOf}(v_1,v_2,w) \wedge \text{isSingleChild}(w) $$

But if the sentence given above is a theorem of your theory, then you can already prove (directly, starting from the sentence above as a premise, and using $\forall E$, $\wedge E$ and $\forall I$) that $\forall v_1.  \text{isAdult}(v_1)$ is a theorem of your theory.


问:
natural isomorphism for modules
This is Proposition D.26 in Jeffrey Lee's book Manifolds and Differential Geometry. If $V$ is an $R$-module, then we have natural isomorphisms $$V\otimes{}R\rightarrow{}V\rightarrow{}R\otimes{}V$$given on decomposable elements as $v\otimes{}r\rightarrow{}rv\rightarrow{}r\otimes{}v$ (assuming that $R$ is commutative).

I am probably just being stupid here, but how is this not a counterexample? Let $V=\mathbb{R}^{2}$ and $R=\mathbb{R}$. Then $$\left(\begin{array}{c}3\\1\end{array}\right)\otimes{}2\rightarrow{}\left(\begin{array}{c}6\\2\end{array}\right)$$and$$\left(\begin{array}{c}6\\2\end{array}\right)\otimes{}1\rightarrow{}\left(\begin{array}{c}6\\2\end{array}\right)$$

答: Thank you, Kenta. For modules, $r(v\otimes{}w)=rv\otimes{}w=v\otimes{}rw$, therefore the "counterexample" is no counterexample. All good now.

Title: natural isomorphism for modules


问:
Doubt about uniqueness of limit proof
My lecturer gave us a proof of uniqueness of limit by contradiction, the context is: let $X \subset \mathbb{R}$ be an open set, let $x_0$ be a limit point for $X$ and let $f:X \to \mathbb{R}$ a function. If the limit of $f$ exists then it must be unique.

I have a doubt in the proof, which is the following: suppose by the sake of contradiction that $f$ has two limits $y_1 \ne y_2$ as $x \to x_0$. Let $V_1$ be a neighbourhood of $y_1$ and $V_2$ be a neighbourhood of $y_2$, with $V_1 \cap V_2 = \emptyset$. Since $x_0$ is a limit point, we can find two neighbourhoods $U_1$ and $U_2$ of the point $x_0$ such that for all $x \in U_1 \cap X \setminus \{x_0\}$ it is $f(x) \in V_1$ and for all $x \in U_2 \cap X \setminus \{x_0\}$ it is $f(x) \in V_2$. Since $U-1$ and $U_2$ are neighbourhoods of $x_0$ it is $U_1 \cap U_2 \ne \emptyset$ and since $x_0$ is a limit point there exists at least a point $\bar{x} \in X$ such that $\bar{x} \in U_1 \cap U_2$. So if $x \in U_1 \cap U_2$ it is $f(x) \in V_1$ and $f(x) \in V_2$, meaning $V_1 \cap V_2 \ne \emptyset$ and this contradicts the fact that $V_1 \cap V_2 = \emptyset$.

My doubt is: what if $V_1$ and $V_2$ aren't disjoint? I will try to explain my doubt: proofs must be general, and it seems to me that when he says "let $V_1$ and $V_2$ be such that $V_1 \cap V_2 =\emptyset$" we are restricting to a case that it isn't general because it imposes a condition on $V_1$ and $V_2$, that is that they are disjoint. Why this proof works then? I have a suspect that it is because this restriction is "apparent", in the sense that we can always lead back to the case when $V_1$ and $V_2$ are disjoint (for example, if $V_1$ and $V_2$ aren't disjoint I can consider another neighbourhoods $V_1'$ and $V_2'$ with radius, for example, a fraction the radius of $V_1$ and $V_2$ big enough to assure that $V_1'$ and $V_2'$ are disjoint), and so obtain a general proof because the condition we impose can always be imposed and this makes $V_1 \cap V_2 =\emptyset$ an assumption without loss of generality. This has some sense or am I missing the point? Thanks.

答:
I think that you are misunderstanding the meaning of the idea that proofs must be general. We are assuming that both $y_1$ and $y_2$ are limits of $f$ at $x_0$. Asserting that $y_1$ is limit of $f$ at $x_0$ means that for every neighborhood $V_1$ of $y_1$, there is a neighborhood $U_1$ of $x_0$ such that $x\in U_1\implies f(x)\in V_1$. And asserting that $y_2$ is limit of $f$ at $x_0$ means that for every neighborhood $V_2$ of $y_2$, there is a neighborhood $U_2$ of $x_0$ such that $x\in U_2\implies f(x)\in V_2$. Since this is true for every two neighborhoods $V_1$ and $V_2$ of $y_1$ and $y_2$ respectively, then, in particular, it must still be true when we choose those neighborhoods in such a way that they don't intersect. And doing that leads to a contradiction.

A proof being general means that it has to apply to every single case. But, in this context, this does not mean that it must work if we pick any two neighborhoods $V_1$ and $V_2$. If one pair of such neighborhoods leads to a contradiction, that's fine and we have then a proof.

Title: Doubt about uniqueness of limit proof


问:
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$

I did:

I take $S=\{x\in \mathbb{R} \mid x^3+x<3\}$

So, this set is bounded and by completeness of the real numbers exists $a=\sup(S)$

Then I want to prove that $a^3+a=3$

And then in order to get a contradiction I suppose that $a^3+a>3$

and how $a $ is the supremum then $\forall\epsilon\in\mathbb{R} $ exists $b \in S$ such that
$a-\epsilon<b$ then $(a-\epsilon)^3+b<b^3+b<3$

But I'm stuck from here. I'm not able to use derivatives, or continuity so I can't use IVT

答:
Hint
If
$$a^3+a >3$$
and
$$0< \epsilon < \min \{ 1, \frac{a^3+a -3}{3a^2+3a+2} \}$$

[ I figured out the value by finding first the upperbound $3a^2\epsilon-3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon+2\epsilon$ below and then solving $3a^2\epsilon+3a\epsilon+2\epsilon< a^3+a-3$. I like to point out to my students that in many proofs in real Analysis you do the computations first to figure out the inequalities you need, but write the proof in reverse order]

Then
$$
3a^2\epsilon-3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon^2+\epsilon^3+\epsilon <3a^2\epsilon+3a\epsilon+2\epsilon< a^3+a -3
$$
This gives
$$(a-\epsilon)^3+a-\epsilon=a^3+a-3a^2\epsilon+3a\epsilon^2-\epsilon^3-\epsilon>3$$

Now, use the fact that there exists some $b \in S$ such that $b >a-\epsilon$. But this leads to the contradiction
$$
3< (a-\epsilon)^3+a-\epsilon \leq b^3+3 \leq 3
$$


问:
Prove that Cameron–Martin space is set of elements with finite Cameron–Martin norm
Let $\mu$ be a centered Gaussian measure on a separable Banach space $\mathscr{U}$. The construction of the Cameron-Martin Hilbert space $H_\mu \subset \mathscr{U}$ typically goes as follows: Let $C_\mu: \mathscr{U}^* \to \mathscr{U}$ be defined by
$$ C_\mu(\ell) = \int u \ell(u) \mu(du).$$
This is well-defined for $\ell \in \mathcal{U}^*$ since $\int_\mathscr{U} \|u\|^2 \mu(du) < \infty$ by the Fernique theorem. Moreover, there is a natural embedding $\mathscr{U}^* \subseteq L^2(\mu)$. Let $R_\mu$ denote the closure of $\mathscr{U}^*$ in $L^2(\mu)$. Then $R_\mu$ is a Hilbert space with respect to the usual inner product. The Cameron-Martin Hilbert space $\mathcal{H}_\mu$ is then defined as $C_\mu(R_\mu)$ and endowed with the inner product
$$ \langle h_1, h_2 \rangle_\mu = \langle C_\mu \ell_1, C_\mu \ell_2 \rangle_\mu := \langle \ell_1, \ell_2 \rangle_{L^2(\mu)}. $$

Question: One can show that for any $h \in \mathcal{H}_\mu$ there holds
$$\|h\|_{\mu} = \text{sup}_{\|\ell\|_{L^2(\mu)}\le 1} |\ell(h)|.$$
How does one show that $h \in \mathscr{U}$ is in the Cameron-Martin space if and only if the previous supremum is finite?

答:
First, let's note that for all $\ell, k \in \mathscr{U}^*$, we have the identity
$$\ell(C_\mu(k)) = \ell\left(\int u k(u)\,\mu(du)\right) = \int \ell(u) k(u) \,\mu(du) = \langle \ell, k \rangle_{L^2(\mu)}$$
and so by continuity and density the same identity holds for all $\ell \in \mathscr{U}^*$ and $k \in R_\mu$.

Suppose the supremum is finite.  This means the map $\ell \mapsto \ell(h)$ is a bounded linear functional on $\mathscr{U}^*$ equipped with the $L^2(\mu)$ inner product.  It thus extends to a bounded linear functional on the Hilbert space $R_\mu$, which by Riesz representation can be identified with some $k \in R_\mu$. This means in particular that $\langle \ell, k \rangle_{L^2(\mu)} = \ell(h)$ for all $\ell \in \mathscr{U}^*$.  By the note above, we also have $\ell(C_\mu(k)) = \langle \ell, k \rangle_{L^2(\mu)}$ for all $\ell \in \mathscr{U}^*$.  By the Hahn-Banach theorem, we conclude that $h = C_\mu(k)$ and so $h \in \mathcal{H}_\mu$.


问:
Confusion over joint probability and joint expectation
This question is likely mainly a notational issue, but I would greatly appreciate some clarification.

Consider two (discrete) random variables X and Y. It makes sense to think about the joint probability distribution between the two rv’s, which can be written as $f(x,y)=P(X=x,Y=y)$. Note the commas. Does it make sense to also think about a joint expectation, denoted as $E(X,Y)$? Is there a requirement that the expectation operator only takes one “argument”? Why?

I often see $E(XY)$ but I don’t think this is the same thing, as this is the expectation of a new random variable $Z=XY$ which the product of X and Y.

答:
The probability is a value attributed to an event which is a subset of a sample space, $\Omega$, e.g, $P\{A\}$ is the probability of the occurrence of $A\subseteq \Omega$. A joint probability, something like $P\{A,B\}$, is the probability of the intersection of $A$ and $B$, i.e., the probability of both $A$ and $B$ occur. Using such a notation, the comma constitutes the logical binary operation: and.

A random variable usually attributes a quantity to an outcome or an event. It is actually a function as $X(\omega): \Omega \rightarrow \mathbb{R}^n$, or other plausible numeric domains. The expectation operation returns the quantity that we expect (on average) for that random variable to be, according to the probability distribution of the random variable, or equivalently, the probability distribution of the events. Using such a notation, $E\{X\}$ is actually $E\{X(\omega)\}$ over all $\omega\in\Omega$. So, the expectation is an operation on a random variable with a numerical support. In that case, $E\{X,Y\}$ is not quite meaningful.

As unusual as it is, if $X(\omega)$ represents a set (a subset of some other universal set), as well as some $Y(\omega)$, then by the above-mentioned conventions, $E\{X,Y\}$ could be interpreted as the expectation of the intersection of $X$ and $Y$. The result in that case should be a set.


问:
Differential equation (with Laplace transform)
Here is the text of the problem:

(Solve the following initial value problem.)
$$y' + 2y + 10\int_{0}^{t}y(\tau)\,d\tau = 255[1 - u(t - \pi)]\sin t, \quad y(0) = 0.$$

I tried to use Laplace which seems the best method to use and this is what I did:

$$sY(s) - y(0) + 2Y(s) + 10Y(s)/s = 255\frac{1+e^{-\pi s}}{s^2+1}$$

I can reorder a bit left but then I'm stuck here:

$$Y(s)(s+2+10/2) = 255\frac{1+e^{-\pi s}}{s^2+1}$$

How can I go on from here?

答:
I think there's two ways to approach this problem. The first is in the vein of my suggestion to first differentiate in order to get a simpler ODE. For now, though, I'll instead focus the Laplace-transform method for the original ODE as presented by the OP.

First, as I noted in comments, it suffices to take the inverse Laplace transform of
$$G(s)=\frac{255}{s+2+10/s}\frac{1}{s^2+1}=\frac{255s}{(s^2+2s+10)(s^2+1)}$$
which is a rational function. Such a rational function is not altogether attractive, but we can make it look much nicer if we expand it in partial fractions. To that end, we look for an expansion of the form

$$G(s)=\frac{255s}{(s^2+2s+10)(s^2+1)}=\frac{A(s+1)+B}{s^2+2s+10}+\frac{Cs+D}{s^2+1}$$ for appropriate coefficients $A,B,C,D$. (The choice of $A(s+1)+B$ is convenient later.) Multiplying both sides by the denominator, we have

\begin{align}
255s
&=(As+A+B)(s^2+1)+(Cs+D)(s^2+2s+10)\\
&=(A+C)s^3+(A+B+2C+D)s^2+(A+10C+2D)s+(A+B+10D).
\end{align}
We now need to match these coefficients term-by-term and solve for $A,B,C,D$. For instance, the lack of a cubic term on the right-hand side means that $A+C=0$. The coefficients thankfully prove to be integers, so I'll leave the computation of such as an exercise.

Having obtained such an expansion, what do we do with it? What we should notice now is that $G(s)$ may be expressed as

$$G(s) = A\frac{s+1}{(s+1)^2+9}+B\frac{1}{(s+1)^2+9}+C\frac{s}{s^2+1}+D\frac{1}{s^2+1}.$$

Each of these terms may be handled by use of your Laplace transform table. For instance, $\frac{s}{s^2+1}$ is the Laplace transform of $\cos t$. You'll also be well-served to remember how to generate Laplace transforms of the form $F(s+a)$. Hence we can determine $g(t)$ explicitly, and thereby determine $y(t)$ as desired. As above, the details are simple enough that I'll leave them as an exercise.

Alright, what about the differentiate-first approach? Before doing so, I'll focus attention on the interval $0\leq t\leq \pi$. Over this domain, we have

$$y' + 2y + 10\int_{0}^{t}y(\tau)\,d\tau = 255\sin t, \quad y(0) = 0.$$
By assumption, $y(0)=0$. Therefore if we evaluate this ODE at $t=0$ we have

$$y'(0)+2y(0)+10\int_0^0 y(\tau)\,d\tau =2y(0)=0$$
Hence $y'(0)=0$ as well. We now differentiate both sides of the ODE, obtaining the following IVP:

$$y''+2y'+10y=255\cos t,\quad y(0)=y'(0)=0.$$

This is an inhomogenous second-order ODE. The homogeneous part has the general solution
$$y_h(t)=A e^{-t}\cos(3t)+Be^{-t}\sin(3t).$$
For the particular solution, I'll take an ansatz of the form $y_p=A_p \sin t+B_p\cos t$. Inserting this into the ODE, we find

$$(9A_p+2B_p)\sin t+(9B_p-2A_p)\cos t=255\cos t$$
which yields the appropriate $A_p,B_p$. We thus have

$$y(x)=A e^{-t}\cos(3t)+Be^{-t}\sin(3t)+A_p \sin t+B_p\cos t.$$
where $y(0)=y'(0)=0$ uniquely determines $A,B$.

So much for $0\leq t\leq \pi$. For $t\geq \pi$, the relevant ODE (upon differentiation) is $$y''+2y+10y=0.$$ This is just a homogeneous ODE, so the solution over this domain is simply $$y(t)=A'e^{-t}\cos(3t)+B'e^{-t}\sin(t).$$ However, the boundary conditions at $y(\pi),y'(\pi)$ must be obtained from those of the previous solution. Fortunately, the form of this solution is particularly nice at $t=\pi$ and so this doesn't present much hurdle. Therefore we can indeed construct the solution for all $t$ without resorting to the Laplace transform. As with the prior solution, the determination of the relevant coefficients is left as an exercise.


问:
Double partial derivative proof
Let $z=f(x,y)$  where $x:=r \cos \theta$ and $y := r \sin \theta$.
I had to prove that $\frac {\partial ^{2}z}{\partial \ x^{2}} + \frac {\partial ^{2}z}{\partial \ x^{2}} = \frac {\partial ^{2}z}{\partial \ r^{2}} \ + \dots$ The whole equation was given.

My doubt:

I applied Jacobian and tried to solve this but I ended up in a loop where I was substituting the same things back and forth.

The solution provided said this.

$\frac {\partial ^{2}z}{\partial r^{2}} \ = \frac {\partial}{\partial r}(\frac {\partial z}{\partial r})  = \frac {\partial}{\partial r}(\cos \theta \frac {\partial z}{\partial x} + \sin \theta \frac {\partial z}{\partial y})$

$ = \cos \theta \ [ \cos \theta \frac {\partial}{\partial x} + \sin \theta \frac {\partial }{\partial y} ](\frac {\partial z}{\partial x}) \ + \sin \theta [ \cos \theta \frac {\partial }{\partial x} + \sin \theta \frac {\partial }{\partial y} ] (\frac {\partial z}{\partial y}).$

I do not understand the last step. What exactly is happening? Why do we use the partial derivative of $z$ with respect to $r$ (and skip the $z$ in the numerator) and multiply it with partial derivative of z with respect to $x$ and $y$? Since $x$ and $y$ are functions of $r$ and $\theta$ and since in $\frac {\partial z}{\partial x}$ we have kept $y$ constant are we accounting for that? I am really confused as this relevant theory wasn't taught in class and the professor hasn't been helpful to clear this part. Any insight is highly appreciated.

答:
Here we look at first at the general situation and develop the second partial derivative by consequently applying the chain rule. This way we can better see  symmetries which is helpful when calculating the special case.

We consider the functions
\begin{align*}
z=f(x,y)\qquad\qquad x&=x(r,\theta)=r\cos\theta\\
y&=y(r,\theta)=r\sin\theta
\end{align*}

First partial derivative:$\quad\Large{\frac{\partial z}{\partial r}}$

We obtain by applying the chain rule
\begin{align*}
\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial r}
+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial r}\tag{1}
\end{align*}

Second partial derivative:$\quad\Large{\frac{\partial^2 z}{\partial r^2}}$

We obtain by applying $\frac{\partial}{\partial r}$ to (1) using the chain rule again as well as the product rule of derivatives
\begin{align*}
\frac{\partial^2 z}{\partial r^2}
&=\frac{\partial }{\partial r}
\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial r}
+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial r}\right)\tag{2}\\
&=\frac{\partial }{\partial r}
\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial r}\right)
+\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial r}\right)\tag{3}\\
&=\color{red}{\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial x}\right)}\cdot\frac{\partial x}{\partial r}
+\frac{\partial z}{\partial x}\cdot\frac{\partial }{\partial r}\left(\frac{\partial x}{\partial r}\right)\\
&\qquad+\color{blue}{\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial y}\right)}\cdot\frac{\partial y}{\partial r}
+\frac{\partial z}{\partial y}\cdot\frac{\partial }{\partial r}\left(\frac{\partial y}{\partial r}\right)\tag{4}\\
&=\color{red}{\left(\frac{\partial^2 z}{\partial x^2}\cdot\frac{\partial x}{\partial r}
+\frac{\partial^2 z}{\partial y\partial x}\cdot \frac{\partial y}{\partial r}\right)}\cdot\frac{\partial x}{\partial r}
+\frac{\partial z}{\partial x}\cdot\frac{\partial^2 x}{\partial r^2}\\
&\qquad+\color{blue}{\left(\frac{\partial^2 z}{\partial x\partial y}\cdot\frac{\partial x}{\partial r}
+\frac{\partial^2 z}{\partial y^2}\cdot \frac{\partial y}{\partial r}\right)}\cdot\frac{\partial y}{\partial r}
+\frac{\partial z}{\partial y}\cdot\frac{\partial^2 y}{\partial r^2}\tag{5}\\
&=\frac{\partial^2 z}{\partial x^2}\cdot\left(\frac{\partial x}{\partial r}\right)^2
+\frac{\partial^2 z}{\partial y\partial x}\cdot \frac{\partial x}{\partial r}\cdot\frac{\partial y}{\partial r}
+\frac{\partial z}{\partial x}\cdot\frac{\partial^2 x}{\partial r^2}\\
&\qquad\frac{\partial^2 z}{\partial y^2}\cdot \left(\frac{\partial y}{\partial r}\right)^2
+\frac{\partial^2 z}{\partial x\partial y}\cdot\frac{\partial x}{\partial r}
\cdot\frac{\partial y}{\partial r}
+\frac{\partial z}{\partial y}\cdot\frac{\partial^2 y}{\partial r^2}\tag{6}
\end{align*}

Comment: This is due to presentation issues a rather lengthy calculation. Usually we might skip (3) and (4) and obtain (5) directly from (2).

In (2) we apply the chain rule as we did in (1)

In (3) we use the linearity of the differential operator

In (4) we apply  the product rule

In (5) we apply the chain rule again

In (6) we multiply out and do a few rearrangements

Special case: $x(r,\theta)=r\cos \theta,\ y(r,\theta)=r\sin \theta$

The partial derivatives of the special case give
\begin{align*}
&x(r,\theta)=r\cos \theta&&\frac{\partial}{\partial r}x(r,\theta)=\cos \theta&&\frac{\partial^2}{\partial r^2}x(r,\theta)=0\\
&y(r,\theta)=r\sin \theta&&\frac{\partial}{\partial r}y(r,\theta)=\sin \theta&&\frac{\partial^2}{\partial r^2}y(r,\theta)=0
\end{align*}
Putting these results in (6) we finally obtain
\begin{align*}
\color{blue}{\frac{\partial^2 }{\partial r^2}}&\color{blue}{f(x(r,\theta),y(r,v\theta))}=\frac{\partial^2 }{\partial r^2}f(r\cos \theta,r\sin \theta)\\
&=\frac{\partial^2 z}{\partial x^2}\cdot \cos^2\theta
+\frac{\partial^2 z}{\partial y\partial x}\cdot \cos \theta\cdot\sin\theta
+\frac{\partial z}{\partial x}\cdot 0\\
&\qquad+\frac{\partial^2 z}{\partial y^2}\cdot \sin^2 \theta
+\frac{\partial^2 z}{\partial x\partial y}\cdot \cos\theta\cdot\sin\theta
+\frac{\partial z}{\partial y}\cdot0\\
&\,\,\color{blue}{=\frac{\partial^2 z}{\partial x^2}\cdot\cos^2\theta
+\left(\frac{\partial^2 z}{\partial y\partial x}+\frac{\partial^2 z}{\partial x\partial y}\right)\cdot\cos \theta\cdot\sin\theta
+\frac{\partial^2 z}{\partial y^2}\cdot\sin^2 \theta}\tag{7}
\end{align*}
in accordance with OPs formula.

Title: Double partial derivative proof


问:
$\lim_{n\to\infty}\left(\sum_{k=1}^n\left(\frac{1}{3}\prod_{l=1}^k \frac{2+3l}{ 4\cdot 3(l+1)}\right)\right)$
Limit of the series

$$\left(\frac{5}{3\cdot6}\right)\cdot\frac{1}{4} + \left(\frac{5\cdot8}{3\cdot6\cdot9}\right)\cdot\frac{1}{4^2} + \left(\frac{5\cdot8\cdot11}{3\cdot6\cdot9\cdot12}\right)\cdot\frac{1}{4^3} +.....$$

Try: Actually I was trying to do it by multiplying by $\frac{1}{4}$ and then substract by shifting one place but then I can't get any useful.....

A small hint is warmly appreciated .

答:
Hint

Note that after multiplying the series by $2\times 1/4$, you can reorganize each term into the form
$$
\frac{1}{n!} \left(\prod_{k=1}^n \frac{3k-1}{3}\right) \frac{1}{4^n}\ .
$$
This should indicate to you that this could be the power series of some function evaluated at $\frac{1}{4}$.

Now the product of the factors $(3k-1)/3 = k - 1/3$ should be the $n$-th derivative of some function. What kind of function has an $n$-th order derivative of this form? More specifically, we want a function whose $n$-th order derivative at $0$ can be obtained from multiplying its $(n-1)$-th order derivative by some factor, and this factor increase by $1$ each time you increase the order by $1$.

 Generically, $(1-x)^{k}$ for $k<0$ behaves like this. In this case, $k = - 2/3$. Of course you still need to subtract off the first few terms to get the exact form.


问:
Showing that $\sum_{n=1}^{\infty} a_n c_n$ converges for all $a=(a_n)\in l^1(\mathbb{N})$ then $c$ is bounded.
Let $c=(c_n)\subset \mathbb{C}$ and suposse that $\sum_{n=1}^{\infty} a_n c_n$ converges  for all $a=(a_n)\in l^1(\mathbb{N})$. Show that $c$ is bounded.

Hi. I cannot conclude on this problem. I attach the possible ideas I have.

Idea: For all $k$, $|c_k|\leq \sum_{n=1}^{k} |a_n||c_n|\leq \sum_{n=1}^{\infty} b_nc_n$ with $a=(\frac{\overline{c_1}}{|c_1|},\ldots, \frac{\overline{c_k}}{|c_k|},0,\ldots, )\in l^1(\mathbb{N})$
and I would like to consider $b$ as $b=(\frac{\overline{c_1}}{|c_1|},\ldots, \frac{\overline{c_k}}{|c_k|},\ldots)$ but this $b\not\in l^1(\mathbb{N})$... and that does not allow me to obtain a bound for $| c_k|$

答:
If $c$ is not bounded then there must be a subsequence $(c_{n_k})$ such that $|c_{n_k}|>k^2$ for each $k\in\mathbb{N}$. Now define $(a_n)$ like this:

$a_n=\begin{cases} \frac{1}{k^2} &,  n=n_k \\ 0 & \text{,otherwise} \end{cases}$

Clearly $(a_n)\in\ell^1$. However, note that for each $k\in\mathbb{N}$ we have $|a_{n_k}c_{n_k}|>1$. So the series $\sum_{n=1}^\infty a_nc_n$ can't converge, as its general term doesn't even tend to zero. So we have a contradiction.


问:
Gradient of a multiple variable function

I am a bit confused here. This question from Advance calculus by Folland, asks me to calculate prime of H(x,y) but its a multiple variable function and when we derivate a multiple variable function we calculate the gradient vector as far as I know but how am I supposed to find the prime of H ? is there some mistake in the question or am I missing something here? any ideas on how to solve this ?

答:
Using Folland's notation we have

$$G'(a,b,c) = \begin{pmatrix} 1 & 1 & 0 \\ 0 & \frac{1}{2\sqrt{b+c}} & \frac{1}{2\sqrt{b+c}} \\ \end{pmatrix} $$

$$F'(x,y) = \begin{pmatrix} y & x \\ 2xy^2 & 2x^2y \\ 3x^2y^3 & 3x^3y^2 \\ \end{pmatrix} $$

Which means

$$H'(x,y) = G'(xy,(xy)^2,(xy)^3)\cdot F'(x,y) = \begin{pmatrix} y+2xy^2 & x + 2x^2y \\ \frac{2xy^2+3x^2y^3}{2\sqrt{(xy)^2+(xy)^3}} & \frac{2x^2y+3x^3y^2}{2\sqrt{(xy)^2+(xy)^3}} \\ \end{pmatrix}$$

Be careful with absolute values when simplifying.


问:
What is the dimension of the orthogonal complement of an one-dimensional subspace in $\mathbb R^n$?
$V$ is a subspace of $\mathbb R^n$ and $\mathbf u$ is a vector in $\mathbb R^n$. If $V=\{c\mathbf u: c \in \mathbb R\}$ then $\dim V^\bot \ge n-1.$ Why is this statement true? For orthogonal complements the following relationship holds: $\dim V + \dim V^\bot = n,$ and since $V$ is a line it is one dimensional, so shouldn't $\dim V^\bot = n-1$? Why is it greater than or equal to?

答:
If $\dim V^\perp = n-1$ then $\dim V^\perp \ge n-1.$

But in one instance you have $\dim V^\perp = n,$ namely when $\mathbf u = \mathbf 0.$


问:
Counting problem in probability
How many 7-digit phone numbers are possible assuming that the first 3 digits can't start with 911?

I should add, I just started studying probability and watching lectures on youtube. The first part of this question is to assume the first digit can't be a 0 or 1 which I solved; 8 possibilities for the first number( ignoring 0 and 1), 10 possibilities for the rest. I know the answer to the problem I posted (8 x 10^6-10^4) but I don't understand the answer and I'd like an explanation which is why I posted it here. You don't have to respond to this post if you don't want to, there's no need to be condescending. It may be elementary to some of you but I don't understand it.

答:
The multiplication principle, paraphrased, states that if you want to count the number of outcomes a scenario has... if you can describe each and every outcome uniquely via a sequence of steps such that the number of available options for a particular step is consistent and does not rely on previously made choices at earlier steps (though the list of options may vary)... then the total number of outcomes is the product of the number of options for the steps.

In terms of set theory, this is commonly stated as:

$$|A\times B| = |A|\times |B|$$

Next, the addition principle... if you want to count the number of outcomes a scenario has and you can partition the set of outcomes into two smaller disjoint exhaustive sets, you can count those smaller sets and add their respective results to get the larger set.

In terms of set theory, this is commonly stated as:

$$|A\sqcup B| = |A|+ |B|$$

(Where $\sqcup$ is the union symbol but carries with it the connotation that it is understood that $A$ must be disjoint from $B$ to be used)

 Note, the more general identity is $|A\cup B|=|A|+|B|-|A\cap B|$ where we drop the requirement that $A$ and $B$ must be disjoint.  The idea here is that we might have double-counted those outcomes which were in the overlap of $A$ and $B$ and the overall count needs to then be corrected to account for this.

We can rearrange this last one by subtracting to the other side to get the related identity that to count the number of "good" outcomes, we can count the overall number of outcomes and subtract the number of "bad" outcomes, that $|A\sqcup B|-|B|=|A|$

So... using these... we count the number of overall outcomes for the number of 7-digit phonenumbers if we don't care about it starting with 911.

Given the solution, it is assumed that a phone number can not start with $0$ or with $1$.

Pick the first number.  There are $8$ choices

Pick the second number.  There are $10$ choices

Pick the third number.  There are $10$ choices

Pick the fourth number.  There are $10$ choices

$\vdots$

Applying multiplication principle, there are then $8\times 10\times 10\times \dots \times 10 = 8\times 10^6$ different 7-digit phone numbers.

Now, to count the "bad" phone numbers from those which were those who started with 911.

The first three numbers are 9,1,1 respectively.  There is only the one choice available for these for the number to have started with 9,1,1.

Pick the fourth number.  There are $10$ choices

Pick the fifth number.  There are $10$ choices

$\vdots$

Multiplying these number of options together, that gives a total of $1\times 10\times 10\times 10\times 10 = 10^4$ "bad" phone numbers that we had accidentally included in our count earlier.

Taking the difference gives the final result as being:

$$8\times 10^6 - 10^4$$

Title: Counting problem in probability


问:
Polar decomposition of invertible operators
I'm looking for references wich treat the following properties:

Let H be a hilbert space. Let $T \in B(H)$ be an invertible operator. Suppose that $T= U |T|$ is the polar decomposition of T. Then we have:

$T^{-1} = U^*|T^{-1} |$ is the polar decomposition of $T^{-1}$.

$|T^{-1}|= {|T^*|}^{-1}.$

${({|T|}^{-1})}^\alpha = U^* {(|T^{-1}|)}^\alpha U$ ,  for each $\alpha >0$.

Could you please suggest me some references on this please.

Thank you !

Body:
In most textbooks, these would be exercises at best. Mostly, you need to notice that
$$\tag1
|T^{-1}|^2=(T^{-1})^*T^{-1}=(T^*)^{-1}T^{-1}=(TT^*)^{-1}=|T^*|^{-2},
$$
which is 2.

From the fact that $T$ is invertible, it follows that $U$ is a unitary:
$$
U^*U=(T|T|^{-1})^*T|T|^{-1}=|T|^{-1}T^*T|T|^{-1}=|T|^{-1}|T|^2|T|^{-1}=I,
$$
and $U$ is surjective because $T$ is.

Now, from $T=U|T|$ we get $T^{-1}=|T|^{-1}U^*$; and then
$$
|T^{-1}|^2=(|T|^{-1}U^*)^*|T|^{-1}U^*=U|T|^{-2}U^*=(U|T|^{-1}U^*)^2,
$$
so $$\tag2|T^{-1}|=U|T|^{-1}U^*.$$ Then
$$
T^{-1}=|T|^{-1}U^*=U^*|T^{-1}|UU^*=U^*|T^{-1}|
$$
is the polar decomposition.

Finally, item 3 follows directly from $(2)$.


问:
How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$?
I'm given a generating function $f(x)$:
$$
f(x)=\frac{1}{1-x^2}
$$
Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.

So far I figured out what follows:
$$
f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac{1}{2}}{1-x} + \frac{\frac{1}{2}}{1+x}=\frac{1}{2}\cdot\left( \sum_{n=0}^\infty x^n+\frac{1}{1-(-x)} \right)
$$$$
\frac{1}{1-(-x)}=1+(-x)+x^2+(-x^3)+x^4+\cdots
$$$$
\frac{1}{1-(-x)}=(1+x^2+x^4+\cdots)-(x+x^3+x^5+\cdots)=\sum_{n=0}^\infty x^{2n}-x^{2n+1}
$$
Therefore:
$$
f(x)=\frac{1}{2} \left( \sum_{n=0}^\infty x^n+x^{2n}-x^{2n+1} \right)
$$
Now I am lost. Is my train of thought good? Am I anywhere near finding the series which is generated by function $f(x)$?

Body: $\frac 1 {1-x^2}=1+x^2+x^4+x^6+\cdots$ for $|x| <1$ since $\frac 1 {1-r}=1+r+r^2+\cdots$ for $|r|<1$. So $a_n=1$ for $n$ even and $a_n=0$ for $n$ odd.


问:
Conditional expectations of $X$ with respect to a filtration converges to $X$
Let $\mathbb F = \left( \mathcal F_n\right)_{n \in \mathbb N}$ be a filtration, and let $\mathcal F_\infty = \sigma\left(\mathcal F_n : n \geq 1\right)$. I want to try proving the following result:

If $X \in \mathcal L^1\left(\mathcal F_\infty\right)$, then $\mathbb E\left[X|\mathcal F_n\right] \to X$ as $n \to \infty$ almost surely.

It’s easy to show that $\left(\mathbb E\left[X|\mathcal F_n\right]\right)_{n \in \mathbb N}$ is a martingale with respect to $\mathbb F$, and since $X$ is integrable, one can show that
$$
\sup_{n \geq 1} \mathbb E\left[\mathbb E\left[X|\mathcal F_n\right]^+\right] \leq \mathbb E[|X|] < \infty
$$
So by the martingale convergence theorem, $\left(\mathbb E\left[X|\mathcal F_n\right]\right)_{n \geq 1}$ converges almost surely, say to the random variable $\tilde X$. But why does $X = \tilde X$ a.s.? How can I show that $\left| X - \mathbb E\left[X|\mathcal F_n\right]\right| \to 0$? I’m wondering if there’s a functional analytic reason, considering the projections $\mathbb E\left[\cdot|\mathcal F_n\right] : \mathcal L^1\left(\mathcal F_\infty\right) \to \mathcal L^1\left(\mathcal F_n\right)$, but I haven’t gotten very far with this approach. Or is there an obvious reason for this that I’m just overlooking?

答:
Let $X_n=\mathbb{E}\left[X\mid\mathcal{F}_n\right]$. This is a closed martingale so it converges in $L^1$ and almost surely to $\tilde{X}$. For all $n\geq 1$ and $A\in\mathcal{F}_n$, the convergence in $L^1$ together with the continuity of the conditional expectation give
$$\mathbb{E}\left[\mathbf{1}_AX\right]=\mathbb{E}\left[\mathbf{1}_AX_n\right]=\mathbb{E}\left[\mathbf{1}_A\tilde{X}\right]$$
This equality holds for any $A\in\bigcup_{n=1}^\infty\mathcal{F}_n$. If we can show that it holds for $A\in\mathcal{F}_\infty$ we are done because both $X$ and $\tilde{X}$ are $\mathcal{F}_\infty$-measurable. Now, the set $\bigcup_{n=1}^\infty\mathcal{F}$ is stable by finite intersection and the class $$\mathcal{G}=\left\{A\in\mathcal{F}_\infty:\mathbb{E}\left[\mathbf{1}_AX\right]=\mathbb{E}\left[\mathbf{1}_A\tilde{X}\right]\right\}$$
is a monotone class. Since $\bigcup_{n=1}^\infty\mathcal{F}_n\subset\mathcal{G}$ the monotone class lemma implies $\mathcal{F}_\infty\subset\mathcal{G}$.


问:
What is the correct definition of the $clip(\cdot)$ function?
I was wondering how to properly (or correctly if only one correct definition exists) define the $clip(x,min\_value, max\_value)$ function, which clips $x$ to the range $[min\_value, max\_value]$.

For convenience, take $f(\cdot) = clip(\cdot)$, $a = min\_value$, and $b = max\_value$. I am considering two candidate definitions:

1:

$f(\cdot) = \left\{\begin{matrix}
a\ \text{if}\ x \leq a\\
b\ \text{if}\ x \geq b\\
x\ \text{else}
\end{matrix}\right.$

or 2:

$f(\cdot) = \left\{\begin{matrix}
a\ \text{if}\ x  b\\
x\ \text{else}
\end{matrix}\right.$

Hence, as I see it, the important detail here is whether to use the $\leq$ and $\geq$ operators or the $<$ and $>$ operators alternatively.

I am wondering which of the two versions above is correct/preferred and if there is any paper I can cite for its definition.
The difference might seem trivial at first glance, but it of course has implications for defining the derivative of the function in the next step.

答: Both are correct and exactly identical. In particular $f(a)=a$ and $f(b)=b$ for both your proposed definitions.


问:
Proving the inequality $(1+x)^n \geq 1 + nx$ with induction
I have to prove this equation with induction. $$\text{if } x > -1 \text{ then } (1+x)^n \geq 1 + nx\;\; \forall\; n \in \mathbb{N}$$

I've solved a base case (when $n = 0$), but I don't know how to prove the induction step.

答:
As @HagenvonEitzen has mentioned, the base case $n = 0$ is true because $(1+x)^{0} = 1 \geq 1 + 0\times x$.

Now, assume $(1+x)^{n} \geq 1 + nx$. Then we have that
\begin{align*}
(1+x)^{n+1} = (1+x)^{n}(1+x) \geq (1+nx)(1+x) = 1 + (n+1)x + nx^{2} \geq 1 + (n+1)x
\end{align*}

and we are done.

Hopefully this helps!


问:
limsup and liminf and the product of sequences
I'm trying to show that if $ \limsup s_n = +\infty$ and $\liminf t_n > 0$, then $\lim\sup s_n t_n = +\infty$.

Could someone check my proof/give feedback?

Since $\lim\inf t_n > 0$, we know that there is a natural $N_1$ such that $m = \inf \{t_n \ | \ n > N_1 \} > 0$. Also because $\lim\sup s_n = +\infty$, there exists $N_2 \in \mathbb{N}$ and $M > 0 $ such that $\displaystyle \sup s_n > \frac{M}{m}$ for $n > N_2$. Then for $n > \max \{ N_1, N_2 \}$, we have $s_n t_n \geq s_n m$. Now we can write $\displaystyle \lim\sup s_n t_n \geq \lim\sup s_n m = m \lim\sup s_n  > m \left( \frac{M}{m} \right) = M$. Since $M$ was any arbitrary number greater than 0, we see that $\lim\sup s_n t_n = +\infty$.

Edit

Since $\limsup s_n = +\infty$, there exists a subsequence $s_{n_j} \to +\infty$ as $j \to \infty$. Then we can also take a subsequence $t_{n_j}$ of $t_n$. Now we have $s_{n_j}t_{n_j} \geq s_{n_j}m$ and $\limsup s_{n_j}t_{n_j} \geq m \limsup s_{n_j} = +\infty$. Since $\limsup s_{n_j}t_{n_j} = +\infty$, we must have $\limsup s_n t_n = +\infty$

答:
This is red herring, but there are a few mistakes in this line:

"...because $\limsup s_n=+\infty$, there exists $N_2\in\mathbb{N}$ and $M>0$ such that $\sup s_n>\frac{M}{m}$ for $n>N_2$."

First of all, if by $\sup s_n$ you mean $\sup\{s_n: n \in \mathbb{N}\}$ then clearly this number is $\infty$, so your statement is vacuous. If by $\sup s_n$ you are referring to $\sup\{s_n:n>N_2\}$, then this number is still $+\infty$.

Second, this $N_2$ will depend on $M$; the correct order of quantification is "for any $M>0$ there exists $N_2\in\mathbb{N}$ such that..."

I say this is a red herring because this line is irrelevant to the proof. You are done once you are able to show that $\limsup s_nt_n \geq m\limsup s_n$, since the right hand side is equal to $+\infty$. You have not quite done this yet. Instead of doing this strange business with the $\limsup$, try going back to the original definition:

By $\limsup a_n=+\infty$ we mean that there exists a subsequence $(a_{n_k})_{k=1}^\infty$ such that $a_{n_k}\to+\infty$ as $k \to \infty$.

Getting the inequality $s_nt_n\geq s_nm$ is a good idea (in fact, this is the only hard part of the proof), but you need to ensure $s_n \geq 0$ for this inequality to work. Since you're working with $\limsup$, you don't need the entire sequence to be nonnegative, but only a subsequence. Can you extract such a subsequence from the definition?

ContentLicense: CC BY-SA 3.0


问:
How to prove inequality including natural log?
Let $0<p<1$ and $0<q<1$ be real numbers and $0 < p + q < 1$.
How can I prove the following inequality is correct or not?

$$q > \frac{\ln(\frac{p}{1-q})}{\ln(\frac{p}{1-q}) + \ln(\frac{q}{1-p})}$$

$\ln$ is natural logarithm.

答:
Recall the definition of Relative Entropy. In this case:

$D(q||(1-p))=\sum\limits_{x\in\mathcal{X}} q(x)\ln\left(\frac{q(x)}{1-p(x)}\right)$.

In the binary case:

$D(q||(1-p)) = q\ln(q)-q\ln(1-p)+(1-q)\ln(1-q)-(1-q)\ln(p)\geq 0$

So,

$D(q||(1-p)) = q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) \geq 0$

EDIT
Use the fact that relative entropy is non-negative. The equality holds iff the two distributions are identical. The last condition actually means that two distributions 1-p and q are not identical, so the relative entropy in this case is positive.

Correction

Since $p+q<1$, then $p<1-q$ and $q<1-p$, so $\ln(\frac{p}{1-q})$ and $\ln(\frac{q}{1-p})$ are both negative, thus

$q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) > 0$

$\Rightarrow q\left[ \ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})\right] > \ln(\frac{p}{1-q}) $

$\Rightarrow q < \dfrac{\ln(\frac{p}{1-q})}{\ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})} $


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
If we want to be absolutely clear, then the argument for the derivative should be the following: $\lim\limits_{h\to0}\frac{(x+h)^2-x^2}{h}$ and $\lim\limits_{h\to0}2x+h$ both exist and are equal if and only if at least one of them exists. Since $\lim\limits_{h\to0}2x+h$ does in fact exist and is $2x$, so too must the other limit (that's $\lim_{h\to0}\frac{(x+h)^2-x^2}{h}$) exist and be $2x$.

This does not work for your logarithm example: You can argue that $\log0$ and $\log0+\log0$ exist and are the same if at least one of the two exists. But neither exists, so the point is moot.


答:
You're correct that it doesn't really make sense to write $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$ unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written $\frac{f(x+h)-f(x)}{h}=2x+h$, and then use the fact that $\lim\limits_{h\to 0}x=x$ and $\lim\limits_{h\to 0}h=0$ as well as the constant-multiple law and the sum law for limits.

Adding to the last sentence: most of the familiar properties of limits are written "backwards" like this. I.e., the "limit sum law" says
$$\lim\limits_{x\to c}(f(x)+g(x))=\lim\limits_{x\to c}f(x)+\lim\limits_{x\to c}g(x)$$
as long as $\lim\limits_{x\to c}f(x)$ and $\lim\limits_{x\to c}g(x)$ exist. Of course, if they don't exist, then the equation we just wrote is meaningless, so really we should begin with that assertion.

In practice, one can usually be a bit casual here, if for no other reason than to save word count. In an intro analysis class, though, you would probably want to be as careful as you reasonably can.

Title: The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place

Body:
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?


问:
Calculating the area of the sphere
I'm trying to learn about calculus on manifolds and read that $ \omega = x dy \wedge dz - y dx \wedge dz + z dx \wedge dy$ gives the area on the sphere. I tried to find the integral $\int_{S^2}\omega$ by using the parametrization $\phi : [0, 2 \pi] \times [0, \pi] \rightarrow S^2$ given by $\phi(u, v) = (\sin(v)\cos(u),\,\sin(v)\sin(u),\,\cos(v))$.

I calculated:

$dx \wedge dy  = -\sin(v)\cos(v)du \wedge dv $,

$dx \wedge dz = \sin^2(v)\sin(u) du \wedge dv$ and

$dy \wedge dz = -\sin^2(v)\cos(u) du \wedge dv$.

When I integrated I got:

\begin{align} \int_S \omega &= \int_{S^2}x dy \wedge dz - y dx \wedge dz + z dx \wedge dy \\\\ &=  \int_0^{2 \pi} \int_0^{\pi} \sin(v)\cos(u)[-\sin^2(v)\cos(u)] - \sin(v)\sin(u)[\sin^2(v)\sin(u)] + \cos(v)[-\sin(v)\cos(v)] dvdu \\\\ &= - 2 \pi \int_0^{\pi}\sin(v)dv \\\\ &= -4 \pi  \end{align}
Since the area should be $4 \pi$, I was wondering what my mistake was.

答: You need to write the pullback as a function multiplying $dv\wedge du$, not $du\wedge dv$, and then do the integration. Your order of variables is not the usual one to get the outward-normal pointing actually outward. When you switch the order of your $2$-forms as I suggest, that will take care of your negatives. The crucial point is that your must have an orientation-preserving parametrization, and the given $2$-form $\omega$ has it built in that $(x,y,z)$ is the unit outward-pointing normal. (You might find some of my YouTube lectures on differential forms and surface integrals helpful. See the link in my profile.)

Title: Calculating the area of the sphere


答:
To repair the approach you were taking:
$$f(x) = \frac{1}{1-x^2} = \frac{1/2}{1-x} + \frac{1/2}{1+x} = \frac{1}{2} \sum_{n=0}^\infty x^n + \frac{1}{2} \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{1+(-1)^n}{2} x^n,
$$
so
$$
a_n = \frac{1+(-1)^n}{2} = \begin{cases} 1 &\text{if $2\mid n$} \\0 &\text{otherwise}\\ \end{cases}
$$

Title: How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$?

Body:
I'm given a generating function $f(x)$:
$$
f(x)=\frac{1}{1-x^2}
$$
Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.

So far I figured out what follows:
$$
f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac{1}{2}}{1-x} + \frac{\frac{1}{2}}{1+x}=\frac{1}{2}\cdot\left( \sum_{n=0}^\infty x^n+\frac{1}{1-(-x)} \right)
$$$$
\frac{1}{1-(-x)}=1+(-x)+x^2+(-x^3)+x^4+\cdots
$$$$
\frac{1}{1-(-x)}=(1+x^2+x^4+\cdots)-(x+x^3+x^5+\cdots)=\sum_{n=0}^\infty x^{2n}-x^{2n+1}
$$
Therefore:
$$
f(x)=\frac{1}{2} \left( \sum_{n=0}^\infty x^n+x^{2n}-x^{2n+1} \right)
$$
Now I am lost. Is my train of thought good? Am I anywhere near finding the series which is generated by function $f(x)$?


答: No property of the limit was used in the first argument before the last step so actually what we have done inside the limit is just rewriting and when we reach the last step we can show the existence using the epsilon-delta definition which apparently deals with the existence issue , same thing applies to the chain rule thing since every thing in the proof before the last steps is just rewriting and the final steps which use the properties of limits which is justified since the epsilon delta definition deals with the existence issue, hope this helps</p>


问:
How to get the value of $16\cot\theta$ from a half circle?
The problem is as follows:

The figure from below represents the half of an apple pie. Assuming
that each portion costs $16\cot\theta$ cents. How many people would
have eaten if it was collected $3264$ cents?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{148 kids}\\
2.&\textrm{136 kids}\\
3.&\textrm{124 kids}\\
4.&\textrm{156 kids}\\
\end{array}$

From the given expressions I could spot that:

$\triangle PCR$ is a right isosceles. Because $PR=CR$

The same holds for $\triangle CQR$ because $PR=RQ$ as $\triangle POQ$ is isosceles.

But that's where I'm stuck. The whole key in this problem is how to get the PO in terms of what I've found?.

Can someone help me here?.

Because if I'm able to get that result the number of kids can be found from:

$K=\frac{3264}{16\cot\theta}=\frac{204}{\cot\theta}$

But again, what should be done here?. Can someone help me please?.

答:
The radius of the pie is $CR+RO$.

Given $CR=MP=PR$, consider the right-angled triangle $\triangle MRO$,

$$\begin{align*}
MO &= \text{radius}\\
MO^2 &= RO^2 + MR^2\\
(PR+RO)^2 &= RO^2 + (2PR)^2\\
PR^2 + 2PR\cdot RO + RO^2 &= RO^2 + 4PR^2\\
2PR\cdot RO &= 3PR^2\\
\frac{RO}{PR} &= \frac{3}{2}\\
\cot\theta &= \frac 32
\end{align*}$$


问:
Probability related to min and max of Brownian motion
Given the Brownian motion $X(t)$ and $s$ with value greater than $0$.

I want to ask about the probability:

\begin{align*}
\Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) .
\end{align*}

Intuitively, I believe $\Pr(\min_{0 \leq u \leq t} X(u) > 0)$ and $\Pr(\max_{0 \leq u \leq t} X(u) > s)$ are independent when $s > 0$. So
\begin{align*}
\Pr(\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s) =  \Pr(\min_{0 \leq u \leq t} X(u) > 0)\Pr(\max_{0 \leq u \leq t} X(u) > s).
\end{align*}
But I just could not prove it logically.

Can someone give me a hint? I really appreciate it. Thank you very much.

答:
If $B_t$ is Brownian motion, then with probability one, for all $\delta>0$, there will exists $0<t<\delta$ for which $B_t<0$. Therefore, $\min_{0\le t\le u}B_t$ is almost surely negative, so your probability is zero.

One way to prove this is to use the standard fact that $W_t:= tB_{1/t}$ is also a Brownian motion, and then apply the law of the iterated logarithm to $W_t$.


问:
Heat equation with initial condition $f(x)=\sin(2\pi x)-\sin(5\pi x)$
Given the following heat equation problem
$$u_{t}=u_{xx},\quad u(0,t)=u(1,t)=0,\quad u(x,0)=\sin(2\pi x)-\sin(5\pi x)$$
when I calculate
$$c_{n}=2\int_{0}^{1}(\sin(2\pi x)-\sin(5\pi x))\sin(n\pi x)dx=0$$
I get
$$u(x,t)=0$$
what's wrong here?

Body:
$$ A_n = \dfrac{\left<  \sin(n\pi x),\sin(2\pi x) - \sin(5\pi x) \right>}{\left<  \sin(n\pi x , \sin(n\pi x)\right>} $$

$$ \int_0^1 \sin^2(n\pi x) \mathrm{d}x  = \dfrac{ 2\pi n - \sin(2\pi n)}{4\pi n} = \dfrac{1}{2} \ , \ \sin(2\pi n) = 0 \ \forall n \in \mathbb{N}  $$

$$ \int_0^1 \sin(n\pi x) ( \sin(2\pi x) - \sin(5\pi x) )\mathrm{d}x = \frac{7 \, {\left(n^{2} - 10\right)} \sin\left(\pi n\right)}{100 \, \pi + \pi n^{4} - 29 \, \pi n^{2}}$$

But considers $ h(x) = 100 \, \pi + \pi x^{4} - 29 \, \pi x^{2} \ , \ h(x) = 0 \iff x=-5 , x=-2, x=2, x=5 $ then $ 100 \, \pi + \pi n^{4} - 29 \, \pi n^{2} \ = 0  \iff n=2 , n=5$

Therefore $A_n = 0 \hspace{0.5cm} \forall n \in \mathbb{N} \backslash \{ 2,5 \} $

Then

$$ \int_0^1 \sin(2\pi x) ( \sin(2\pi x) - \sin(5\pi x) )\mathrm{d}x = \dfrac{1}{2}$$

$$ \int_0^1 \sin(5\pi x) ( \sin(2\pi x) - \sin(5\pi x) )\mathrm{d}x = \dfrac{-1}{2}$$

$$ A_2  = 1 \ , \ A_5 = -1$$

$$ u(x,t) = e^{-4\pi^2t} \ \sin(2\pi x) - e^{-25\pi^2t} \ \sin(5\pi x) $$


问:
Why should we use the fundamental definition of derivative while checking differentiability?
I have seen in many examples that using the fundamental definition of derivative only gives the correct results while checking for differentiability at a point. But why do we get wrong results while we directly apply the rules of differentiation and taking the limit at the point. Here is an example

$$y = \begin{cases}
x^2, &-\infty < x < 1\\
2x, &1\leq x < \infty
\end{cases}
$$

If we check the differentiability at the point $x=1$ then when we apply the rules of differentiation left hand and right hand derivative becomes same. But by using fundamental definition of derivative it will be different.

But if I am right the rules of differentiation are in fact derived from the fundamental definition of derivative? So both of these should give correct result. Or it seems like I am missing some basic concept.

I have seen same question like this one here. The answer to this question only says it gives wrong result.But it doesn't answer the question of why?

答:
Hint: Note that in order that $y=y(x)$ is differentiable at $x=1$ it has to be continuous at $x=1$. Here we have
\begin{align*}
\lim_{x\to 1^{-}}y(x)=\lim_{x\to 1^{-}}x^2&\color{blue}{=1}\\
\lim_{x\to 1^{+}}y(x)=\lim_{x\to 1^{+}}2x&\color{blue}{=2}=y(1)
\end{align*}
Since the left-side limit of $y$ is not equal to the right-side limit of $y$ at $x=1$ the function is not continuous at $x=1$ and can't be differentiable at $x=1$. There is no need to apply the fundamental definition of derivatives.


问:
Derivative equal to original function
How many different functions $f(x)$ exist such that $f'(x) = f(x)$?

The ones I know of right now are $f(x) = 0$ and $f(x) = ne^x$, for any real number $n$. What other functions satisfy this property?

答:
Write the equation as $\frac{dy}{dx} = y$.

Take the reciprocal of both sides, integrate and make $y$ the subject:

$\frac{dx}{dy} = \frac{1}{y}$

$x = \ln y + C$

$x - C = \ln y$

$e^{x-C} = y$

$y = C_1 e^x$

Title: Derivative equal to original function


问:
Find the period of $f(x)=\sin^2(x)$
$f(x+p)=f(x)$

$\sin^2(x+p)=\sin^2(x)$

$\sin^2(x+p)-\sin^2(x)=0$

$[\sin(x+p)-\sin(x)][\sin(x+p)+\sin(x)]=0$

$\sin(x+p)-\sin(x)=0$

$\sin(x+p)=\sin(x)$

This is the part iI don't get:

$x+p=x+2kπ$

$p=2kπ$

My question: how did we get $x+p=x+2kπ$ from $\sin(x+p)=\sin(x)$

And yeah I know the exercise doesn't end there, but I only want to understand this part first.

答:
That is not quite correct, as we have the basic equivalence for equations in sines:
$$\sin x=\sin y\iff \begin{cases}x\equiv y\mod 2\pi &\text{or} \\x\equiv \pi -y\mod 2\pi.\end{cases}$$

The initial equation can also be solved  linearising both sides:

$$\sin^2(x+p)=\sin^2x\equiv \frac{1-\cos 2(x+p)}{2}=\frac{1-\cos2x}{2}\iff =\cos 2(x+p)=\cos 2x\\\iff 2(x+p)\equiv \pm 2x\mod 2\pi.$$

Title: Find the period of $f(x)=\sin^2(x)$


问:
Do Regular Functions Separate Closed Sets in Affine Variety?
I'm focusing on the language of classic algebraic geometry.

Let $X$ be an affine variety and $A(X)$ be its coordinate ring. Let $E,F\subseteq X$ be $2$ disjoint closed subsets of $X$. Does there exist an $f\in A(X)$ such that $f(E) = 0$ and $f(F) = 1$?

If this is true, is there any generalization of this or in the language of scheme?

答:
Let $I(E)$ and $I(F)$ be the regular functions vanishing on $E$ and $F$,

$I(E)$ is an ideal of $A(X)$ so its image in $A(X)/I(F)$ is an ideal.

If it is smaller than the whole ring $A(X)/I(F)$ then it must be contained in some maximal ideal $m$ containing both $I(F)$ and $I(E)$, which is impossible because $F$ and $E$ are disjoint.

Thus, there is some $f\in I(E)$ such that $f=1\in A(X)/I(f)$ ie. $f-1\in I(F)$.

A consequence is that $$A(X)/I(E)\times A(X)/I(F)\cong A(X)/(I(E\cup F)), \qquad (g,h)\to (1-f)g+fh$$


答:
The other answers are perfectly fine; just a perspective that can save your day in situations in which the existence of the limit is actually a critical point.

The crucial definition is the one of limsup and liminf: these are always well defined, and all you have to know at the moment are the following two properties:

$\liminf_{x \to x_0} f(x) \le \limsup_{x\to x_0} f(x) $

The limit of $f$ exist if and only if $\liminf_{x \to x_0} f(x) = \limsup_{x\to x_0} f(x) $, and in this case the limit agree with this value.

Now imagine you do your computation twice: firstly, you compute the liminf; then you compute the limsup. In both computations, as soon as you arrive to something that actually has limit (like $2x+h$), because of property (2) you can forget about the inf/sup story and just compute the limit.

Since with some manipulations you arrive to something that actually has limit, both calculations will give the same result and, because of property (2) again, the limit exist and coincide with the value you just computed.

Now this is not really the thing you should do if you are doing introductory analysis and you don't know liminf and limsup: formal properties of these two are slightly different from the formal properties of lim, and you could end up with an error. But as long as you don't "touch" the limit, and you just make some manipulation inside theimit, the same argument will carry on: if you end up with a well defined result, it is the limit :)


答: Let CR = $\frac12$MR = a and RO = b. Then, OM = a + b and the Pythagorean’s rule for MRO establishes $b^2+4a^2 =(a+b)^2$, yielding $\cot \theta = \frac ba =\frac32$ and the answer (2).

Title: How to get the value of $16\cot\theta$ from a half circle?

Body:
The problem is as follows:

The figure from below represents the half of an apple pie. Assuming
that each portion costs $16\cot\theta$ cents. How many people would
have eaten if it was collected $3264$ cents?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{148 kids}\\
2.&\textrm{136 kids}\\
3.&\textrm{124 kids}\\
4.&\textrm{156 kids}\\
\end{array}$

From the given expressions I could spot that:

$\triangle PCR$ is a right isosceles. Because $PR=CR$

The same holds for $\triangle CQR$ because $PR=RQ$ as $\triangle POQ$ is isosceles.

But that's where I'm stuck. The whole key in this problem is how to get the PO in terms of what I've found?.

Can someone help me here?.

Because if I'm able to get that result the number of kids can be found from:

$K=\frac{3264}{16\cot\theta}=\frac{204}{\cot\theta}$

But again, what should be done here?. Can someone help me please?.


问:
Coin that lands in a plane tessellated by triangles..
A coin of radius 1 cm is tossed onto a plane surface that has been tessellated by right triangles whose sides are 8 cm, 15 cm, and 17 cm long. Find the probability that the coin lands within a triangle.

I know that this has to do with similarity because the inner triangle that is formed by the area where the coin can land is similar to the outer triangle. Therefore, I know the angles of this triangle, but I am not sure how to find one side of this triangle.

答:
You don't need to compute the sides of the new triangle, you only need to know the change in the inradius.

Given any triangle with area $\Delta$ and perimeter $\ell$, its inradius $r$ equals to $\frac{2\Delta}{\ell}$.

If one parallel shift all three edges inwards for a small distance $\rho$, the shrinked triangle will be similar to original triangle, sharing same incenter and has inradius $r - \rho$. The area of the shrinked triangle is given by:

$$\Delta_{\rm shrink} = \Delta \left( 1 - \frac{\rho}{r}\right)^2 = \Delta\left(1 - \frac{\rho\ell}{2\Delta}\right)^2$$

For the problem at hand, the area and perimeter of a triangle tile are:
$$\Delta = \frac{8\cdot 15}{2} = 60\quad\text{ and }\quad \ell = 8+15+17 = 40$$

In order for the coin to falls completely inside the interior of a triangle tile, its center needs to lie inside a shrinked triangle of the tile obtained by a shift with $\rho = 1$. This means the probability for the coin to fall complete inside a tile is

$$\frac{\Delta_{\rm shrink}}{\Delta} = \left(1 - \frac{\rho\ell}{2\Delta}\right)^2 = \left(1 - \frac{1\cdot 40}{2\cdot 60}\right)^2 = \left(1 - \frac13\right)^2 = \frac{4}{9}$$


问:
Clarity Regarding Dependent Events in Probability
Given the Scenario in Communication:
Let $R^∗_f$ and $R^∗_n$ denote the target rates of the far and near user. Then, the probability of outage of the near user data is given by,
$P_n=P_r(R_n<R^∗_n)$
The far user's signal can be decoded successfully only if the near user's data is decoded correctly. Therefore, the outage probability of the far user is given as follows:
$P_f=P_r(R_n<R^∗n)+P_r(R_n>R^∗_n,R_f<R^∗_f)$

Is this correct expression or we need to multiply the probabilities in dependent events.
(Note: The relations within the bracket are verified regarding achievable rates, the question is specific to $+$ (Plus) sign.

Thankyou.

答:
The two events are disjoint, so the expression for the probability of failure as you mentioned is correct. Among four possible outcome, {SS, SF, FS, FF}, only one, {SS}, constitutes a successful end-to-end connection. If the first link fails, it doesn't matter whether the second link is failed or successfully connected, hence the first term in $P_f$. But if the first link is connected successfully, the overall link fails if the second link fails, hence the second term in $P_f$.

You are good to go. :)


答:
The equation $f(x+p) = f(x)$ is the definition of a periodic function, which you have used on $f(x) = \sin^2(x)$ to reach the conclusion that $\sin(x+p) = \sin(x)$ must hold. If it was another function that we knew nothing about, we couldn't really go any further but say that $\sin^2(x)$ and $\sin(x)$ have the same period. But because this is the sine function which we understand very well, we know that its period is $2\pi$. Thus, the output of the function is repeated at $2\pi,4\pi,6\pi$... Hence we can say that the input of the function must satisfy $x+p=x+2k\pi$.

To give a different example.. say we arrived to the case where we had $\tan(x+p) = \tan(x)$ instead. Here, we know that the period of $tan$ is $\pi$, so we can conclude that $x+p=x+k\pi$.

Title: Find the period of $f(x)=\sin^2(x)$

Body:
$f(x+p)=f(x)$

$\sin^2(x+p)=\sin^2(x)$

$\sin^2(x+p)-\sin^2(x)=0$

$[\sin(x+p)-\sin(x)][\sin(x+p)+\sin(x)]=0$

$\sin(x+p)-\sin(x)=0$

$\sin(x+p)=\sin(x)$

This is the part iI don't get:

$x+p=x+2kπ$

$p=2kπ$

My question: how did we get $x+p=x+2kπ$ from $\sin(x+p)=\sin(x)$

And yeah I know the exercise doesn't end there, but I only want to understand this part first.


答: $$|CR||RC'|=|MR||RN| \quad\to\quad u\cdot(u+2u\cot\theta)=2u\cdot2u \quad\to\quad\cot\theta=\frac32$$


问:
Maximum on circle through normal p.d.f in $\mathbb{R}^3$
Given a normal distribution on $\mathbb{R}^3$

$$p(\mathbf{x})\propto\exp(-\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu}))$$

and a circular trajectory through $\mathbb{R}^3$

$$\mathbf{x}(\theta)=r(\mathbf{\hat{i}}\cos\theta+\mathbf{\hat{j}}\sin\theta)$$

where

$r$ is the radius of curvature

$\mathbf{\mu}$ is the distribution's mean

$\Sigma$ is the distribution's variance matrix (which is symmetric)

Can you suggest how to solve (or alternatively approximate) $\theta$ maximizing $p(\mathbf{x}(\theta))$?

答:
Note that $x \mapsto \exp(- \frac 12 x)$ is a decreasing function. Thus, it suffices to find the $\mathbf x(\theta)$ that minimizes
$$
f(\mathbf x) = (\mathbf x - \boldsymbol \mu)^T\Sigma^{-1}(\mathbf x - \boldsymbol \mu).
$$
One approach is to simply plug in. Denote $\mathbf x = (x_1,x_2,x_3)$, $\boldsymbol \mu = (\mu_1,\mu_2,\mu_3)$, $M = \Sigma^{-1}$ and the entries of $M$ are $m_{ij}$. We have
$$
f(\mathbf x) = \sum_{i,j = 1}^3 m_{ij}(x_i - \mu_i)(x_j - \mu_j).
$$
Plugging in $\mathbf x(\theta)$ yields
$$
f(\mathbf x(\theta)) = m_{11}(\cos \theta - \mu_1)^2 + 2m_{12}(\cos \theta - \mu_1)(\sin \theta - \mu_2) + m_{22}(\sin \theta - \mu_2)^2.
$$
This function can be maximized in the usual fashion; just set $f'(\theta) = 0$ and solve.


问:
How to obtain the maximum integer value for $\frac{\tan \theta}{10}$?
The problem is as follows:

The diagram from below shows a cruiser which has departed from an
allied shore. A nearby lighthouse is located on point $A$. Assume the
height between the bottom of the lighthouse on $A$ to the top of the
lighthouse $P$ is equal to $\textrm{10 m}$. The cruiser is located at
$1.2\,km$ from the lighthouse ($MA=1.2\,km$). After advancing $1\,km$
in a straight line the cruiser makes a stop on point $N$ to do
maintenance and repairs. Assuming the amount of sailors which needs to
be rescued is given by $\frac{12E}{7}$ where $E$ is the maximum
integer value, which it can take $\frac{\tan \theta}{10}$ and $\angle
> APN=\theta$, find the amount of people who will be rescued.

The alternatives given on my book are as follows:

$\begin{array}{ll}
1.&\textrm{48 sailors}\\
2.&\textrm{36 sailors}\\
3.&\textrm{24 sailors}\\
4.&\textrm{42 sailors}\\
\end{array}$

This problem I don't know how to approach it correctly. The thing which confuses me the most is how to make sure \tan\theta is maximum?

I do remember that:

If it mentions that \theta is acute then this means it lies in the first quadrant. Hence:

$0\leq \tan \theta < +\infty$

In short what it is asking is how to get maximimum integer value for:

$E=\frac{\tan\theta}{10}$

But given such condition it isn't very logical to find the maximum because it is on the positive infinity side. What to do next?.

I've attempted using derivatives, but it wouldn't make much sense as it would confirm that given this situation you can find the minimum:

$\tan\theta'=\sec^2\theta=\frac{1}{\cos^2\theta}$

And to equate the previous equation to zero would mean that the denominator has to be infinity.

So what to do here?. I don't know how to relate the other information provided. In other words the height of the lighthouse and the distance covered by the cruiser.

Can someone help me here please?.

答:
Triangle $ANM$ gives us: $AN < 2200$.

Triangle $PNA$ gives us: $\tan \theta <220$.

$E = \frac{\tan \theta}{10}< 22$.

$E_{max}=...$

$\text{Amount}=...$

Can you finish?


问:
Ratio of areas of similar triangles given SS
What is the ratio $ A_I:A_{II} $ ?

I know that the given angles and common angle prove the triangles are similar. Using proportionality, I found the length of the middle lengthed side of $\triangle_I$ to be $\frac 34 \times 7$. The length of the long side of the quadrilateral is $ 8 - (\frac 34 \times 7) $.

I am stuck here. I cannot think of how to compute the area only given two sides of a scalene triangle, although I imagine the next step is finding the last side and then using Hero's formula. Thanks for any help.

This problem is from Geometry for Enjoyment and Challenge, Chapter 11: Areas of triangles, parallelograms, trapezoids, kites, etc. It does not yet cover the Law of Sines or Cosines, although I couldn't imagine how one could find the angle either way.

答:
The answer's $\frac97$.

In similar triangles, $(\text{side}:\text{side})^2 = \text{area}:\text{area}$. Thus $\left(\frac{6}{8}\right)^2= \frac{9}{16}$, and $\text{tri}_1 : \text{tri}_{\text{whole}} = \frac9{16}$. Then $\text{tri}_2$ must equal $\frac7{16}$, thus the ratios are $\frac97$.


问:
IVP - Differential equation
Working on this. Have to find a solution. Very new to DE:

$$\begin{cases}
 2yx'(y)-1=x^2(y) \\
 x(y=1)=1 \\
 \end{cases}
$$

Writing this as:

\begin{align}
2y \frac{dx}{dy}-1&=x^2y\\
\frac{1}{x^2}2y\,dx-1&=y\,dy\\
\frac{1}{x^2}\,dx &= 1+y\,dy\\
\int \frac{1}{x^2}\,dx &= \int 1+y\,dy\\
-\frac{1}{x}+C &= \frac{y^2}{2}+y+C
\end{align}

Is this correct? Feel like I'm just doing algebra, but don't really understand if it's correct. Would love feedback.

Title: IVP - Differential equation

Body:
$$ 2yx'(y)-1=x^2(y) \\$$
$$ 2y-y'=x^2y' $$
$$ 2y=y'(x^2+1) $$
This is separable:
$$2\int \dfrac {dx}{x^2+1}=\int \dfrac {dy}{y}$$
$$2 \arctan x =\ln y +C$$


问:
Determinant: signed area of parallelogram question
I'm trying to understand the geometrical aspects of the determinant and have questions.  Wiki states that:

$\begin{vmatrix} a \ b \\ c \ d \end{vmatrix} = ad - bc$

is the area of this parallelogram:

How is this area established?  I'm reading Wiki on parallelograms and do not see how the determinant equals the area.

I see in another post someone suggests using Heron's Formula to prove $A_p = ad - bc$.  Wiki later says:

Signed Area = $|u||v| sin \theta = |u^\Gamma||v| cos \theta' = \begin{bmatrix} -b \\ a \end{bmatrix} \cdot \begin{bmatrix} c \\ d \end{bmatrix} = ad - bc$, where $u = (a, b)$ and $u^\Gamma = (-b, a)$ is of the complementary angle to a perpendicular vector.

Does anyone know why $u^\Gamma = (-b, a)$?  I'm unable to understand this.  I think the coordinates are swapped to provide the same result with $cos$ as $sin$, but am confused by the $-$ sign.

答:
The iamge is the original image with the original
parallelogram completed into an
$(a+c)\times(b+d)$ rectangle by adding two $b\times c$
rectangles, two $a\times b$ right triangles, and two
$c\times d$ right triangles.
The area of the parallelogram is thus
$$ (a\!+\!c)(b\!+\!d) - 2(b\,c) -
 2\Big(\frac12 a\,b+\frac12 c\, d\Big)
 = a\, d-b\, c.\tag{1}$$

Your question regarding $\,u^\Gamma = (-b,a)\, $
is answered by noting that it is
$\,u\,$ rotated by $\,90^\circ\,$ which is confirmed
by $\,u^\Gamma\cdot u = -b\,a+a\,b =  0.\,$


答:
Before we start consider this preliminary question.

If $\sin a = \sin b$ solve for $b$ in terms of $a$.

Well if $(\cos a, \sin b) = (x,y)$ there is only one other point of a circle with the same $y$ value.  That would be $(-x,y)$ and that occurs if $b = \pi - a$.

so if $0\le a,b < 2\pi$ then only options are $b= a$ or $b = \pi -a$.  And if we remove the restrictions we have $b = a +2k\pi$ or $b= (2k+1)\pi -a$ for so integer $k$.

....

Now we do your question.

Suppose $\sin (x+p) = \sin x$ for all $x$.

Let $y = x+p$ and we have $\sin(y) = \sin x$ and we just did that.  $y = x +p = x +2k\pi$ or $y=x+p= (2k+1)\pi - x$.

In the first case we have $p =2k\pi$.  And in the second case $p=(2k+1)\pi - 2x$ but that makes $p$ dependent upon $x$ and won't have a single value for every possible $x$.

so if $p$ is a period it must be the first case and $p = 2k\pi$ for some $k$.  And to find "the" period we need the smallest positive value of $2k\pi$ so that $\sin^2(x+2k\pi) = \sin^2 x$ and as $k=1$ works that's the smallest.

Not $k=0$ and $p =0$ is not an acceptable value for a period.  Claiming a function "has period $0$" would be to so "$f(x) = f(x+0)$ for all $x$" and that's....  well, that's just pointless.

Title: Find the period of $f(x)=\sin^2(x)$


问:
Prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ is cyclic of order $5$
I have been asked to prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ (presumably over $\mathbb{Q}$ is cyclic of order $5$.  Unfortunately, I have no idea where to start.

The hint says to show that this is the minimal polynomial of $\zeta_{11} + \zeta_{11}^{-1}$.  Which I guess I could do by plugging it into the polynomial, but there is most likely a better way to do it.  Anyway, even if I could establish that, I wouldn't know any of the other roots so it's not like I could find any of the automorphisms.  I guess it would kind of help because it would show that the Galois group had order divisible by 5.  Except a lot of them are.  I think I am completely missing the point of the hint.

The section was about how you could reduce polynomials mod $p$, and the Galois group would have to contain a permutation with cycle type $(n_1 n_2 ... n_k)$, where $n_1$, ..., $n_k$ are the degrees of the irreducible factors (mod $p$).  But this is not very relevant because since the group is cyclic of prime order, all the permutations would be 5-cycles.  And lots of other subgroups of $S_5$ have 5-cycles too.

So at this point I am kind of stuck.  Can you give me a hint as to how to solve the problem?  Thanks!

P.S. We are allowed to use the computer program SAGE for this assignment, but not the command that calculates the Galois group.  Also, I tried to use SAGE to calculate the roots but it gave me numerical answers.

答:
The simplest way to show that the minimal polynomial is this one is just to plug it in and develop using the binomial theorem ; just remember that $\zeta_n \zeta_m = \zeta_{n+m}$ and that $\zeta_{n} \zeta_m^{-1} = \zeta_{n-m}$ when you're computing, and then prove its irreducibility by noticing that $1$ and $-1$ are not roots of it (rational root theorem), plus it cannot factor as a product of degree $2$ and $3$ (use linear algebra to do this part).

Now $[\mathbb Q(\zeta_{11}+\zeta_{11}^{-1}) : \mathbb Q] = 5 = |\mathrm{Gal}(\mathbb Q(\zeta_{11}+\zeta_{11}^{-1}) \backslash \mathbb Q)| $ and the $5$ comes from the fact that the dimension of the extension divides $5$ (obviously) but is not $10$ since $\mathbb Q(\zeta_{11}) \neq \mathbb Q(\zeta_{11} + \zeta_{11}^{-1})$ . Since the only group of order $5$ is the cyclic one, you're done. (Any non-trivial element of the group has order dividing $5$, hence $5$, so that the group is cyclic.)


问:
Drawing dynamic circles based on input value
Is there a formula that will allow me to calculate the radius of a circle based on an input value?  The input value could be as small as zero or as large as $10^7$, or larger.  The circle is restricted to a minimum radius of $10$ and a maximum radius of $100$.

Does anyone know how to calculate something like this?

UPDATE

The input values correspond to state/country population.  I want to calculate the radius (how big the circle should be) of the circle based on the input value.

答: For this application, I would just make the area of the circle proportional to the population.  Let $R=\sqrt{\text{maximum population of a state/city}}$  Then plot each circle as $r=100\sqrt{\frac {\text{population}}R}$, boosting the ones you want to show that are below $r=10$ to $r=10$ to satisfy your minimum.  But I don't understand the minimum.</p>


问:
Is there a reasonably accurate and easy way to approximate lbs-stones or kg-stones (and vice versa)?
For example, when I need to talk with my American friends about body weight:

$\mathrm{KG}\times2+10\%$

($100\times2=200$, $200+10\%=220$... Pretty accurate and easy to do mentally, both ways)

However, when I need to talk with my British friends about body weight, how do I approximate from KG to Stones?

Additionally, if I have a similar formula for Stones-lbs, that will be useful too!

I mean, I know a Stone is 14 lbs, but how to approximate 247lbs in Stones without pulling out a calculator?

For your reference, a Stone is about 6.35 KG

I am thinking that a "good, practical" approximation follows these rules:

5% accuracy is kinda bad... Imagine weighing 100kg and wanting to lose 5kg - the error alone is making discussion about it difficult! I am thinking up to 2% error

Maximum of 3 steps

Only addition, subtraction, multiplication and division allowed (% also allowed)

Multiplication only by "easy" factors: 2, 3, 4, 5, 10

Division only by "easy divisors": 2, 10

Only "easy" percentages like: 1%, 10%...

答:
I worked out another mental method for converting kilograms to stone that takes a little more work, but gives a more accurate answer. First, assuming you're not going to deal with weights much beyond 150 kg, you'll need to know your multiples of 19 from 1 to 8:

$\\19 \times 1 = 19 \ \ \ \ 19 \times 5 = 95\\19 \times 2 = 38 \ \ \ \ 19 \times 6 = 114\\19 \times 3 = 57 \ \ \ \ 19 \times 7 = 133\\19 \times 4 = 76 \ \ \ \ 19 \times 8 = 152$

In the steps below, I'll use the conversion of 86 kg to stone in the examples.

Step 1: Take the number of kilograms and divide it by 19, thinking of the answer as a mixed fraction. Since $86 = (19 \times 4) + 10$, we can quickly see that $\frac{86}{19} = 4\frac{10}{19}$

Step 2: Multiply the total you get in step 1 by 3. $3 \times 4\frac{10}{19} = 12\frac{30}{19} \Rightarrow 13\frac{11}{19}$

The total from step 2 is your estimate!

How good is this estimate? This method of estimation is equivalent to dividing by 6.33, and thus yields a good approximation of dividing by 6.35. Your estimate will be off by only about $\frac{3}{10} \%$!

Example: The estimate yields $13\frac{11}{19} \approx 13.57 \ stone$, while the actual conversion gives $86 \ kg = 13.54 \ stone$.

Bonus: If you want to be able to give your answer in decimals, learn to mentally divide by 19 with decimal precision here.


问:
Unique question about packing problem
I added the related pages from part 3 of the book: combinatorial geometry by János Pach,Pankaj K.Agarwal (1995) (which is not available on net so I added them as pictures).

A. Prove that one can always find a packing $Ç$ of the plane with congruent copies of a convex disc $C$ whose density $d(Ç,R^2)$ exists and is equal to $\delta(C)$.
Similarly show that there is a Lattice packing $Ç$ with $d(Ç,R^2)=\delta_L(C)$.

B.construct a packing $Ç$ of unit discs and two convex discs $P,P'$ such that $lim_{r\to \infty} d(Ç,P(r)) \neq lim_{r\to \infty} d(Ç,P'(r))$.

Hints:
Part A:To prove the first statement, construct a packing $C_n$ of congruent copies of $C$ in the disc $D(n)$ with density $d(C_n,D(n))\geq \delta(C)-O(1)/n$

For every n choose a subsequence $n_1,n_2,...$ such that $C_{n_i}$ converges when restricted to $D(1)$ as $i\to \infty$ .
From this choose a subsequence for which $C_n$ converges when restricted to $D(2)$ and so on.
Show that the limit packing meets the requirements.

For part B: Let $Ç$ be the densest packing of unit discs in a cone.
Choose $P$ and $P'$ to be circular disc and a regular triangle respectively.

Can you kindly help me to understand this (I'm a beginner in this field).

1.(https://i.stack.imgur.com/2k1mz.jpg)

2.(https://i.stack.imgur.com/ZEdCp.jpg)

3.(https://i.stack.imgur.com/fhhvM.jpg)

4.(https://i.stack.imgur.com/WIVsz.jpg)

5.(https://i.stack.imgur.com/B2O4m.jpg)

6.(https://i.stack.imgur.com/ZuXvX.jpg)

Many thanks.

答:
I'm adding another answer, since my first is already pretty dense and I think it would be contextually useful to leave it up--- at least to make the comments make sense. If this is a StackExchange faux pas, apologies.

Lemma: Fix some convex disc $C$ and $R \subseteq \mathbb{R}^2$. Then there is a packing $P$ which achieves an arbitrary density between $0$ and $\delta(C)$. That is to say, there is a packing $P$ so that $d(P,R) = r \in \mathbb{R}$, where $0 \leqslant r \leqslant \delta(C)$.

There are many ways to prove this, but I think if you want to make this rigorous a separation into cases and then IVT via translating a specific copy of $C$ through the boundary to increase density.

Proof of A:

Fix some convex disc $C$, and create an infinite family of packings $C_n$ so that:

$$d(C_n, D(n)) \geqslant \delta(C) - \frac{O(1)}{n}$$

for each $n$. Such a construction is justified by the lemma, since $O(1)/n > 0$. Moreover, the interval $[0,A(D(n))]$ is compact, so we choose a subsequence which converges in $D(1)$, then one which converges in $D(2)$ from that subsequence, and so on forever. Now let:

$$P = \lim_{n \rightarrow \infty} C^\ast_n$$

Where $C^\ast_n$ is the limiting sequence. I claim that $P$ is our desired packing. Indeed:

$$\delta(C) \geqslant \lim_{n \rightarrow \infty} d(C^\ast_n, D(n)) \geqslant \delta(C) - \lim_{n \rightarrow \infty} \frac{O(1)}{n}$$

since the limiting sequence converges in all $D(n)$ and has the bounds in all $D(n)$. As $\lim_{n\rightarrow \infty} O(1)/n$ goes to $0$, we have what we desired.

See if you can formulate something for the lattice packing.

Proof of B:

Fix our region $Q$ a convex cone of angle $\pi/3$ whose vertex is at the origin, and let $P$ be the densest packing of $Q$ with unit discs.

Then letting $P(r)$ be the disc $d(0,x) \leqslant r$, we can see immediately that $d(P,P(r)) \leqslant 1/6$ for large $r$.

Now let $P'(r)$ be an equilateral triangle with vertex at the origin and two legs on the boundary of $Q$. Since the angle of $Q$ is $\pi/3$, $P'(r)$ will always be in $Q$, and as this is the densest packing, it follows that $$\lim_{n \rightarrow \infty} d(P,P'(r)) = \frac{\pi}{\sqrt{12}}$$

by the results mentioned in the textbook.

As $\pi/\sqrt{12} > 1/6$, their densities as $r \rightarrow \infty$ are different.

EDIT: Here is an image of the cone and the disc.

Title: Unique question about packing problem


问:
Is the heat semigroup $P_t$ smooth for any function $f$ for which $P_tf$ is well-defined?
Consider the heat semigroup (or heat kernel) $P_t$ which can be define on the space of non-negative functions, i.e. for any non-negative measurable function $f$
$$P_tf(x):=\frac{1}{\sqrt{2\pi t}^d}\int_{\mathbb R^d}f(y)e^{-|y-x|^2/(2t)}\,dy.$$
It can be infinite. Of course it can be defined for a much larger class of functions. My question, however, is the following:

Problem. Assume for some non-negative function $f$ we have that $P_tf(x)$ is finite for all $t\in (0,1]$ and a.e. $x\in\mathbb R^d$, i.e. one can evaluate the integral from the definition of $P_tf(x)$ and it is finite. Is it true that $P_tf(x)$ is differentiable (in a weak sense in both variables).

Actually I wanted to show that the function $u(x, t) =P_tf(x) $ solves (weakly) $$\tag{1}\partial_t u =\frac 12\Delta_x u$$
But I thought let's study the differentiability before even bothering whether the PDE (1) is satisfied. One knows that for any (regular enough $f$) the solution to the PDE (1) with boundary condition $u(x, 0)=f(x)$ is given by $P_tf(x) $. My question is like the reverse: what if I know $P_tf(x) $ is a well-defined function, then does it satisfy the PDE?

The issue is that we do not know the regularity of $f$, but we only know that it can be integrated against the heat kernel. This is related to a problem coming from stochastic processes, minimizing entropy where one gets that such quantities $P_tf(x)$ are well defined.

My attempts.  I really have no clue, I know the heat semigroup has a smoothing effect for integrable functions, or even functions growing fast, because then we can differentiate inside the integral. Here we are only given that $P_tf$ is well defined. I deep inside think that a counterexample exists. But also since the derivative of the integrand adds a polynomial term, I think that that polynomial term will not affect anything, so differentiating under the integral would be allowed.

Also, what if we add that the a.e. limit as $t\to 0^+$ exists and is equal to $f$ a.e.? Does it make the problem easier?

For the ones wondering, I added the tag 'stochastic analysis', because I would also be satisfied if we can show (or disprove) the following:

Assume $B$ is a Brownian motion. Recall the function $u(x, t) =P_tf(x) $. Do we have
$$u(B_t, 1-t) =u(B_0, 1)+\int^t_0 \nabla_x u(B_s, 1-s)\, dB_s. $$
Note that we cannot expect pointwise evaluation of $u$ so the stochastic equation may be understood equality in $L^2_{loc}(B) $. This would follow from Ito's formula if we would have $u$ satisfy the PDE (1) in a classical sense (or weakly in a nice sobolev space). As you can see both are related. I'm not sure which one is easier. The second one may use more methods from probability theory avoiding the whole differentiating, maybe? I'm clueless.

Edit. (Based on John Dawkins' answer). So the martingale in John Dawkins' answer, namely $\mathbb E[f(B_1)\mid \mathcal F_t]$ is where this question came from. Note we do not have integrability because the starting value of the Brownian motion is random. But if we are allowed to apply martingale representation theorem, then noting that $u(B_t,1-t)=\mathbb E[f(B_1)\mid \mathcal F_t]$ allows us to write
$$u(B_t,1-t)=u(B_0,0)+\int^t_0 H_s\,dB_s,$$
for some process $H$. The question is are we able to conclude that $H_s=\nabla_x u(B_s,1-s)$ in some sense?

答:
(I will stick to the 1-dimensional case, for simplicity.)

If $f$ is bounded then $u(x,t):=P_t f(x)$ is smooth on $\Bbb R\times(0,\infty)$, and you even have $\lim_{t\downarrow 0} u(x,t) = f(x)$ provided $f$ is also continuous. In this case, as you note,
$$
{\partial u\over\partial t}={1\over 2}{\partial^2u\over\partial x^2}.
$$
This in turn implies that if you fix $T>0$ then $u(B_t,T-t)$, $0\le t\le T$, is a martingale; in fact. by Ito's formula,
$$
u(B_t,T-t) = P_Tf(x)+\int_0^t u'_x(B_s,T-s)\,dB_s,
$$
where $u'_x$ is alternate notation for ${\partial u\over\partial x}$.

A more direct way to see the martingale property is to notice that $u(B_t,T-t)=\Bbb E[f(B_T)\mid\mathcal F_t]$ for $0\le t\le T$.  This last statement persists under the conditions you envision: If $f\ge 0$ is Borel measurable and $u(x,t):=P_tf(x)<\infty$ for each $t>0$ and $x\in\Bbb R$, then for each $T>0$ the process
$$
u(B_t,T-t),\qquad 0\le t\le T,
$$
is a continuous martingale, a.s. $\Bbb P^x$. (Here $\Bbb P^x$ is the law of the Brownian motion  started at $x$.) I'd guess that $u'_x$  exists in a good enough sense that the stochastic integral expression fo $u(B_t,T-t)$ above is still true in the more general setting, but a proof eludes me.