问:
Problem 17(v) Topology Without Tears (pg 96)

The Question

My Understanding(final edit)

(My proof will probably suck,but it is the best I can do)

We are required to prove that $R\setminus C \in \tau$  and  $R\setminus A\in\tau$ are Hausdorff but not regular.

Assumptions

1.Let X be the set,of all real numbers and
$ S=\{1/n:n \in N\}$

2.Define a set $C\subseteq\Bbb R$ to be closed if $C=A \cup T$. is $A$ is closed in the Euclidean topology on $\Bbb R$. $T\subset S$

If C is closed by Def 1.2.4 R\C is open. Similarly if A is closed
in Euclidean space A\C is open

But T$\subset S $. Then every point of S is an isolated point, since a sufficiently small interval about 1/n doesn’t contain 1/m for any integer m$\ne$ n, and S has no interior points. The set of boundary points ofS, S ∪ {0}. The point 0 $\notin$ S is the only accumulation point ofS, since every open interval about 0 contains 1/n for sufficiently large n.

 Source
This implies that $\bigcup_{n\in N} S\cup {0}=[0,1]$
Thus R\S=$(-\infty,0) \cup (1,\infty$)

Next l will show it is a topology

$\emptyset$,X $\in\tau$, since R\C and R\A are open
and byProp 1.2.2 (ii) & (iii) are satisfied cause both sets
are open Thus R\C,R\A $\in\tau$ Finally R\S by prop 1.2.2 (ii)
is satisfied.

Thus C=A $\cup T \in \tau$

Then we can find x$\in (-\infty,0)\subset$ R\S
and y$\in. (1,\infty)\subset $R\S so that
-$\infty$<x<0 and 1<y<$\infty$

.

Let U=(-$\infty$,0) V=(1,$\infty$) Then U $\cap V=\emptyset$
So $\tau $ on R is Hausdorff and since O$\notin S$ we can’t
separate T$\in\tau$ so it is not regular.

答:
We are required to prove that $X\setminus C \in \tau$  and  $R\setminus A\in\tau$ are Hausdorff but not regular.

This is incorrect. What you are supposed to prove is that $\tau$ is a topology, and that the space $\langle\Bbb R,\tau\rangle$ is Hausdorff but not regular. You should begin by showing that $\tau$ really is a topology on $\Bbb R$, so I’ll skip down to that part of your answer.

Next l will show it is a topology

$\emptyset$,X $\in\tau$, since R\C and R\A are open

What are $C$ and $A$? What do they have to do with $\varnothing$ and $X$? You have asserted that $\varnothing$ and $X$ are in $\tau$, but what follows since has no identifiable connection with this assertion and therefore does not justify it.

Let $\mathscr{C}$ be the collection of sets that are defined in the problem to be closed: a set $C\subseteq\Bbb R$ is in $\mathscr{C}$ if and only if there are a set $A$ that is closed in the Euclidean topology and a set $T\subseteq S$ such that $C=A\cup T$. To show that $\Bbb R\in\tau$, you must show that it is the complement of some member of $\mathscr{C}$, i.e., that $\Bbb R\setminus\Bbb R\in\mathscr{C}$. $\Bbb R\setminus\Bbb R=\varnothing$; is $\varnothing\in\mathscr{C}$? Yes: $\varnothing$ is closed in the Euclidean topology, $\varnothing\subseteq S$, and $\varnothing=\varnothing\cup\varnothing$, so $\varnothing\in\mathscr{C}$, and therefore $\Bbb R=\Bbb R\setminus\varnothing\in\tau$. And $\varnothing=\Bbb R\setminus\Bbb R$, so to show that $\varnothing\in\tau$, you must show that $\Bbb R\in\mathscr{C}$. This is also true: $\Bbb R$ is closed in the Euclidean topology on $\Bbb R$, $\varnothing\subseteq S$, and $\Bbb R=\Bbb R\cup\varnothing$, so $\Bbb R$ meets the criteria to belong to $\mathscr{C}$, and $\varnothing=\Bbb R\setminus\Bbb R$ is therefore in $\tau$.

and byProp 1.2.2 (ii) & (iii) are satisfied cause both sets
are open Thus R\C,R\A $\in\tau$ Finally R\S by prop 1.2.2 (ii)
is satisfied.

This is nonsense, I’m afraid. Proposition $\mathbf{1.2.2}$ says that if you already have a topological space, then it satisfies these three conditions, but you don’t yet know that $\langle\Bbb R,\tau\rangle$ is a topological space: that’s what you’re trying to prove. To do so, you must show that $\tau$ satisfies the requirements of Definitions $\mathbf{1.1.1}$. The first two of these are identical to (i) and (ii) in Proposition $\mathbf{1.2.2}$, and the third requires just that the intersection of any two open sets be open. I proved above that $\tau$ satisfies (i), but (ii) and (iii) remain completely unjustified.

Thus C=A $\cup T \in \tau$

Since you never specified what $C,A$, and $T$ are, this is meaningless. We can guess, however, that you’re referring to the definition of closed sets given in the problem, in which case it is in general false: if $C\in\mathscr{C}$, then by definition $\Bbb R\setminus C$ is in $\tau$, not $C$, and whether or not $C$ is in $\tau$ is irrelevant to the task of showing that $\tau$ is a topology.

To prove that $\tau$ satisfies (ii), you must start with an arbitrary family $\mathscr{U}\subseteq\tau$ and show that $\bigcup\mathscr{U}\in\tau$. If $U\in\mathscr{U}$, then by definition $\Bbb R\setminus U\in\mathscr{C}$, so there are a Euclidean closed set $A_U$ and a $T_U\subseteq S$ such that $\Bbb R\setminus U=A_U\cup T_U$, and hence $U=\Bbb R\setminus(A_U\cup T_U)$. Then

$$\begin{align*}
\bigcup\mathscr{U}&=\bigcup_{U\in\mathscr{U}}\big(\Bbb R\setminus(A_U\cup T_U)\big)\\
&=\Bbb R\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\,,
\end{align*}$$

so to show that $\mathscr{C}\in\tau$, you must show that $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\in\mathscr{C}$. Let $A=\bigcap_{U\in\mathscr{U}}A_U$; this is an intersection of Euclidean closed sets, so it is Euclidean closed. Clearly $A\subseteq\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$; what other points might be in $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$? Any other point of $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$ must belong to $S$, so $A\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\subseteq S$. Thus, if we let $T=A\setminus\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)$, we have $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)=A\cup T$, where $A$ is Euclidean closed, and $T\subseteq S$, and hence $\bigcap_{U\in\mathscr{U}}(A_U\cup T_U)\in\mathscr{C}$, as desired. This shows that $\tau$ is closed under arbitrary unions, i.e., that it satisfies condition (ii).

Finally, you have to show that if $U,V\in\tau$, then $U\cap V\in\tau$. By definition there are Euclidean closed sets $A_U$ and $A_V$ and sets $T_U,T_V\subseteq S$ such that $U=\Bbb R\setminus(A_U\cup T_U)$ and $V=\Bbb R\setminus(A_V\cup T_V)$. Thus,

$$\begin{align*}
U\cap V&=\big(\Bbb R\setminus(A_U\cup T_U)\big)\cap\big(\Bbb R\setminus(A_V\cup T_V)\big)\\
&=\Bbb R\setminus\big((A_U\cup T_U)\cup(A_V\cup T_V)\big)\\
&=\Bbb R\setminus\big((A_U\cup A_V)\cup(T_U\cup T_V)\big)\,.
\end{align*}$$

$A_U\cup A_V$ is a Euclidean closed set, and $T_U\cup T_V\subseteq S$, so $$(A_U\cup A_V)\cup(T_U\cup T_V)\in\mathscr{C}\,,$$ and its complement $U\cap V$ is therefore in $\tau$. Thus, $\tau$ satisfies condition (iii) of Definitions $\mathbf{1.1.1}$ and is therefore a topology on $\Bbb R$.

The next step is to prove that $\langle\Bbb R,\tau\rangle$ is Hausdorff. For this you must let $x$ and $y$ be arbitrary distinct points of $\Bbb R$ and show that there are disjoint $U,V\in\tau$ such that $x\in U$ and $y\in V$. You can do this by considering cases — both points in $S$, exactly one of $x$ and $y$ in $S$, or both points in $\Bbb R\setminus S$ — but this is unnecessary. Any two distinct points of $\Bbb R$ have disjoint Euclidean open nbhds, and as Henno Brandsma proves in the second paragraph of his answer, every Euclidean open set is in $\tau$, so automatically any two distinct points of $\Bbb R$ have disjoint $\tau$-open nbhds.

The last step is to show that $\langle\Bbb R,\tau\rangle$ is not regular. For this you write:

since O$\notin S$ we can’t
separate T$\in\tau$ so it is not regular.

If by ‘O’ you mean $0$, you may possibly have identified the counterexample to regularity, but what you’ve written doesn’t make any sense. Here is an actual argument.

First, by definition $S=\varnothing\cup S$ is $\tau$-closed, since $\varnothing$ is Euclidean closed and $S\subseteq S$. And $0\notin S$, so if $\langle\Bbb R,\tau\rangle$ were regular, there would be disjoint $U,V\in\tau$ such that $0\in U$ and $S\subseteq V$. Thus, you can prove that $\langle\Bbb R,\tau\rangle$ is not regular by proving that no such $U$ and $V$ exist.

Suppose, then, that $U,V\in\tau$, $0\in U$, and $S\subseteq V$; you need to prove that $U$ and $V$ cannot be disjoint. By definition there are Euclidean closed sets $A_U$ and $A_V$ and sets $T_U,T_V\subseteq S$ such that

$$U=\Bbb R\setminus(A_U\cup T_U)=(\Bbb R\setminus A_U)\cap(\Bbb R\setminus T_U)\,,$$

and similarly,

$$V=\Bbb R\setminus(A_V\cup T_V)=(\Bbb R\setminus A_V)\cap(\Bbb R\setminus T_V)\,.$$

$\Bbb R\setminus A_U$ is a Euclidean open set; call it $W_U$. Then

$$U=W_U\cap(\Bbb R\setminus T_U)=W_U\setminus T_U\,.$$

Similarly $\Bbb R\setminus A_V$ is Euclidean open and if we call it $W_V$, we have $V=W_V\setminus T_V$. But $V\supseteq S$, so $T_V=\varnothing$, and $V=W_V$. In other words, $V$ is already a Euclidean open nbhd of $S$.

$0\in U\setminus S\subseteq U\setminus T_U$, so $0\in W_U$. $W_U$ is a Euclidean open nbhd of $0$, so there are $a,b\in\Bbb R$ such that $0\in(a,b)\subseteq W_U$. Note that $(a,b)\setminus S\subseteq W_U\setminus T_U=U$.

Clearly $0<b$, so there is a positive integer $n$ such that $\frac1n<b$. $\frac1n\in S$, and $V$ is a Euclidean open nbhd of $S$, so there are $c,d\in\Bbb R$ such that $\frac1n\in(c,d)\subseteq V$. Let $r=\max\{\frac1{n+1},c\}$, and let $x=\frac12\left(r,\frac1n\right)$; then $\frac1{n+1}<x<\frac1n$, so $x\notin S$, and therefore $x\in(a,b)\setminus S\subseteq U$. Thus, $x\in U\cap V$, and $U$ and $V$ are therefore not disjoint. Since $U$ and $V$ were arbitrary $\tau$-open nbhds of $0$ and $S$, we’ve shown that $0$ and the $\tau$-closed set $S$ cannot be separated by disjoint open sets and hence that $\langle\Bbb R,\tau\rangle$ is not regular.

That completes the proof, but I want to comment a few other things that you wrote. Early on you wrote this:

But T$\subset S $. Then every point of A is an isolated point, since a sufficiently small interval about 1/n doesn’t contain 1/m for any integer m$\ne$ n, and A has no interior points. The set of boundary points of A is A ∪ {0}. The point 0 $\notin$ A is the only accumulation point of A, since every open interval about 0 contains 1/n for sufficiently large n.

 Source
This implies that $\bigcup_{n\in N} S\cup \{0\}=[0,1]$
Thus R\S=$(-\infty,0) \cup (1,\infty$)

It implies nothing of the kind. $\bigcup_{n\in\Bbb N}S$ is simply $S$, and $S\cup\{0\}$ certainly is not the entire closed unit interval. Then you say that we can find

x$\in. (1,\infty)\subset $R\S so that
-$\infty$<x<0 and 1<x<$\infty$

Clearly x is contained in both intervals

This is absurd: $x$ cannot be contained in both rays, as it cannot be both less than $0$ and greater than $1$.


问:
Which is the axiom: well ordering principle, principle of induction, both, or none?
From analysis 1 by Terence Tao, I learn that the principle of induction is a peano axiom. In many other analysis books, like analysis by Bartle and Sherbert, the well ordering principle is used to derive the principle of induction. I am thinking now: either both are equivalent statements or one of them is the axiom.</p>

答:
The answer, unfortunately, is that it depends on the circumstances. In some cases, the two statements are equivalent (and therefore either one could properly function as the fifth Peano axiom). In other cases, induction is stronger than the well-ordering principle in the sense that induction implies well-ordering, but not vice versa.

This is expounded upon more fully in the wonderful article:

Öhman, L. Are Induction and Well-Ordering Equivalent?. Math Intelligencer 41, 33–40 (2019). https://doi.org/10.1007/s00283-019-09898-4

which can briefly be summarized by his two conclusions:

(A) The induction principle and the well-ordering principle are not equivalent relative to the common first 4 axioms of the Peano system, since the resulting axiomatic systems admit different models.

(B) In the axiomatic system consisting of the common first 4 axioms together with the well-ordering principle, induction cannot be a theorem, since there is a model Ord ($\omega+\omega$) in which all five of these axioms are satisfied, but induction is not true.

He goes on to state:

We may also note that the axiomatic system consisting of the Peano axioms with induction admits only models that are isomorphic to the natural numbers, and since the natural numbers are well-ordered, in this system well-ordering principle is in fact a theorem. Induction is therefore stronger than well-ordering in this context, in that it has the power to rule out more possible models.

ContentLicense: CC BY-SA 4.0


问:
Lower-bound analysis of a polynomial with square roots.
So I'm trying to prove that $f = \Omega(g)$ with the following:

$f(n) = \sqrt{7n^2 + 2n - 8}$, so I simplify this out and get: $f(n) = n\sqrt{7} \sqrt{1+\frac{2}{7n}-\frac{8}{7n^2}}$. So going from there I can see that my $c$ = $\frac{a_k}{2}=\frac{\sqrt{7}}{2}$, and since I have negative coefficients my $n_0=\frac{2|A|}{\sqrt{7}}=\frac{2\sqrt{8}}{\sqrt{7}}=4\sqrt{\frac{2}{7}}$.

I have $g(n) = n\sqrt{7}$ in this case.

But I get stuck here because I know that $\sqrt{1+\frac{2}{7n}-\frac{8}{7n^2}} \ge 0$. But then to go further along with the inequality I can't multiply both sides by $n\sqrt{7}$ since one side is still $0$. So I don't get how to prove it from there? What am I missing, I've read through my textbook multiple times at this point.

答:
In the definition "$f = \Omega(g)$ iff there are $C, n_0$ such that $f(n) \ge C g(n)$ for all $n \ge n_0$" you get to choose $n_0$. From looking at where the square root is defined, you've determined that $n_0$ should be at least $4 \sqrt{\frac27}$, but if that's not working out, you should feel free to choose a bigger $n_0$.

I want to emphasize that there's never any requirement to find the best $C$ or the best $n_0$ to use. The $\Omega$ asymptotic notation specifically exists to handwave that away.

In this case, if we take $n_0 = 10$ (for example) then we can say that for $n \ge n_0$,
$$
   7 + \frac2n - \frac8{n^2} \ge 7 - \frac8{n^2} \ge 7 - \frac8{10^2} = 6.92.
$$
By taking $C = \sqrt{6.92}$, this is the start of a proof that $f = \Omega(n)$. Alternatively, by taking $C = \sqrt{\frac{6.92}{7}}$, this is the start of a proof that $f = \Omega(n \sqrt7)$. (Since we are free to choose the constant $C$ however we like, we usually don't talk about $\Omega(n \sqrt 7)$, since it's not any different from $\Omega(n)$.)

You should also be aware that an alternative definition of $f = \Omega(g)$ is via limits: $f = \Omega(g)$ iff $\lim_{n \to \infty} \frac{f(n)}{g(n)} > 0$ (including the case where the limit is $+\infty$). This might be easier to think about if you're comfortable with limits, though when you do homework for a specific class you might be stuck using the definition the class uses.


问:
Example for $\operatorname{dist}(x_0,Y)<1$ in Riesz' lemma
Let $$X:=\left\{f\in C([0,1]):f(1)=0\right\}$$ be equipped with the supremum norm and $$Y:=\left\{f\in X:\int_0^1f(t)\:{\rm d}t=0\right\}.$$

I want to show that if $f_0\in X$ with $\left\|f_0\right\|_\infty=1$, then $$\operatorname{dist}(f_0,Y)<1\tag1.$$ How can we do that?

Note that, since $0\in Y$, $$\operatorname{dist}(f,Y)\le\left\|f\right\|_\infty\tag2\;\;\;\text{for all }f\in X.$$

答:
Define the functional $x^* : X \to \mathbb R$ via $x^*(f)= \int_0^1 f$. Clearly $||x^*|| \leq 1$. In fact, $||x^*|| =1$. To see this, consider the functions $f_n \in X$ defined by $f_n(t)=1 $ on $[0,1-1/n]$ and $f_n(t) =\frac 1n (1-t)$ on $(1-1/n,1]$. Evidently $ x^*(f_n)=1-\frac{1}{2n}$ and so $||x^*|| \geq 1 - \frac{1}{2n} $ for all $n$.

Notice that $Y = \ker x^*$. It is  well known that $dist(f,\ker x^*) = \frac{|x^*(f)|}{||x^*||} = |x^*(f)|$. Assume that there exists $f$ with norm one such that $|x^*(f)|=|\int_0^1 f |=1$. Since
$$ 1= | \int_0^1 f | \leq \int_0^1 |f| \leq 1$$
we see that $ \int_0^1 |f| =1$ or equivalenty that $ \int_0^1( 1- |f|)=0$. However, $1- |f| \geq 0 $ and so $|f|\equiv 1$. This is impossible since $f(1)=0$.

A nice byproduct of this is that X is not reflexive. Indeed, as we showed $x^*$ does not attain its norm and so by James's theorem we infer that X is not reflexive.

$\textbf{Edit:}$ Generally, if $X$ is a non reflexive space, then by James theorem there exists $x^* \in X^*$ that is not norm attaining. The above shows that $ dist(x, \ker x^*) <1$ for all $x \in  X$ with norm one.


问:
Probability density function of a Gamma random variable plus a constant
Suppose $X$ has a Gamma distribution $\Gamma(a,b)$. What is the distribution of $X+c$? where $c$ is a positive constant.

EDIT: I know how extract the expression of the pdf and cdf. What I wanted to know is if the resulting distribution is a know one.

Body:
Simply

$$f_Y(y)=\frac{b^a}{\Gamma(a)}(y-c)^{a-1}e^{-(y-c)b}$$

$y\in[c;+\infty)$


问:
Rescaling and reciprocal a set with empty interior
Assume $A\subset \Bbb R$ with empty interior, that is, there is no $a, b\in \Bbb R$ with $a<b$ such that $$(a,b)\subset A$$ let $r\in\Bbb R\setminus\{0\}$ then $r A$ still has an empty interior. For contradiction, assume there exists an open interval $(a,b)$ such that $(a,b)\subset r A$ but this implies $$\frac{1}{r} (a,b)\subset A$$ notice that $\frac{1}{r} (a,b)$ is an open set  which is a contradiction since $A$ has an empty interior.

Also, assume $0\notin A$, the set $B:=\{\frac{1}{a}\colon a\in A\}$ has an empty interior. The argument is as above.

Now, assume what I said above it is correct. if $A=\{a_\xi\colon \xi<\mathfrak c\}$ has an empty interior how about  $$\{r_\xi a_\xi\colon \xi<\mathfrak c\}?$$  Does it still have an empty interior? where $r_\xi$ is a real number for every $\xi<\mathfrak c$

Could you please check my work and tell if it is correct or not? Or there is another way to show it.

Any help will be appreciated greatly.

答:
Not necessarily. Let $\Bbb R\setminus\{0\}=\{x_\xi:\xi<\mathfrak{c}\}$, and without loss of generality assume that $0\notin A$. For $\xi<\mathfrak{c}$ let $r_\xi=\frac{x_\xi}{a_\xi}$; then

$$\{r_\xi a_\xi:\xi<\mathfrak{c}\}=\{x_\xi:\xi<\mathfrak{c}\}=\Bbb R\setminus\{0\}\,.$$


问:
Product measure of complete measure and counting measure
Let $(X,B, \mu)$ be a complete measure and $(\mathbb{N},P(\mathbb{N}),\nu)$ be counting measure.

Are the following true?

a) $A \subset X \times \mathbb{N}$ if and only if $A^n \in B$ for all $n \in \mathbb{N}$.

b)$f(x,n)$ is $\mu \times \nu$- measurable if and only if $f^n(x)=f(x,n)$ is $\mu$-measurable for all $n\in \mathbb{N}$.

c)If $f\in L^+(\mu \times \nu)$, then $\int f d\mu \times \nu = \sum_{n \geq 1} \int f^n(x) d\mu(x)$, where $\mu$ is not necessarily $\sigma$-finite.

My attempt:

a) true, by Proposition 2.34 in Folland.

b) true, by Proposition 2.34 in Folland.

c) false, since we need $\sigma$- finiteness in Fubini-Tonelli.

I'm not sure if the above is correct, please let me know if something is wrong... Thank you!

答:
(a) Consider measurable spaces $(X,\mathcal{F})$ and $(\mathbb{N},\mathcal{P}(\mathbb{N}))$.
Let $(X\times\mathbb{N},\mathcal{F}\otimes\mathcal{P}(\mathbb{N}))$
be their product. Let $A\subseteq X\times N$. We go to prove that
$A\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$ iff for each $n\in\mathbb{N}$,
$A^{n}\in\mathcal{F}$, where $A^{n}=\{x\in X\mid(x,n)\in A\}$.

Proof: Let $\pi_{X}:X\times\mathbb{N}\rightarrow X$ and $\pi_{\mathbb{N}}:X\times\mathbb{N}\rightarrow\mathbb{N}$
be the canonical projections, i.e., $\pi_{X}(x,n)=x$ and $\pi_{\mathbb{N}}(x,n)=n$.
For each $x\in X$ and $n\in\mathbb{N}$, define $i_{x}:\mathbb{N}\rightarrow X\times\mathbb{N}$
by $i_{x}(n)=(x,n)$ and $j_{n}:X\rightarrow X\times\mathbb{N}$ by
$j_{n}(x)=(x,n)$. Since the product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$
is the smallest $\sigma$-algebra on $X\times\mathbb{N}$ such that
$\pi_{X}$ and $\pi_{\mathbb{N}}$ are measurable, it can be proved
easily that $\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$ has the following
universal property:

For any measurable space $(Y,\mathcal{M})$ and map $g:Y\rightarrow X\times\mathbb{N}$,
$g$ is $\mathcal{M}/\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$-measurable
iff $\pi_{X}\circ g$ and $\pi_{\mathbb{N}}\circ g$ are measurable.

Now $\pi_{X}\circ j_{n}=id_{X}$ (the identity map on $X$) and $\pi_{\mathbb{N}}\circ j_{n}:X\rightarrow\mathbb{N}$
is the constant function $x\mapsto n$, which are both measurable.
Therefore, $j_{n}$ is measurable. Similarly, $i_{x}$ is also measurable.

$\Rightarrow:$ Let $A\subseteq X\times\mathbb{N}$. Suppose that
$A\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$. Let $n\in\mathbb{N}$
be arbitrary. Note that $A^{n}=j_{n}^{-1}(A)$ and hence $A^{n}\in\mathcal{F}$
because $j_{n}$ is measurable.

$\Leftarrow:$ Let $A\subseteq X\times\mathbb{N}$. Suppose that for
each $n\in\mathbb{N}$, $A^{n}\in\mathcal{F}$. For each $n\in\mathbb{N}$,
$A^{n}\times\mathbb{N}=\pi_{X}^{-1}(A^{n})\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N}).$
Observe that $X\times\{n\}=\pi_{\mathbb{N}}^{-1}(\{n\})\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$.
Therefore, $\left(A^{n}\times\mathbb{N}\right)\cap\left(X\times\{n\}\right)\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$.
Finally, $A=\cup_{n}\left(A^{n}\times\mathbb{N}\right)\cap\left(X\times\{n\}\right)\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$.

Remark: In the above, if we replace $(\mathbb{N},\mathcal{P}(\mathbb{N}))$
by an arbitrary measurable space $(Y,\mathcal{G})$, $\Rightarrow$
still holds. However, in general, $\Leftarrow$ fails to hold. (Notice
that, in the proof of $\Leftarrow$, we use the fact that $\mathbb{N}$
is countable in the last step.)


问:
Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6
I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.

First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:

$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$

Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.

Equation for violations becomes W+X+Y+Z = 15-7 = 8

As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)

⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4

Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.

When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1

⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4

When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.

Therefore, the number of integer solutions for $W+X+Y+Z = 15$ where $W, X, Y, Z \leq 6$ is $$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$

答:
There is a mistake in your last term. The answer should be $180$ and not $172$. Also you can simplify the working.

There are multiple ways to tackle the problem -

Solve as is using P.I.E what you did

Simplify using change of variable and solve

Solve using generating function

a) Using first method (how you did; I have used boxes and balls analogy in my working),

the answer should be $\displaystyle {18 \choose 3} - {4 \choose 1} {11 \choose 3} + {4 \choose 2} {4 \choose 1} = 180$

The second term is where we choose one of the boxes to have $7$ balls and then distribute rest $8$ balls in any of the boxes.

The third term is where we choose two of the boxes to have $7$ balls each and then put one remaining ball in any of the boxes. This term is wrong in your working.

b) Using second method,

We substitute $a = 6 - w, b = 6 - x, c = 6- y, d = 6 - z$.

So we have, $a + b + c + d = 9 \ $ where $0 \leq a, b, c, d \leq 6$.

Only one of them can have more than $6$ balls. So the answer is

${12 \choose 3} - {4 \choose 1} {5 \choose 3} = 180$

c) Using generating function

Find coefficient of $x^{15}$ in $(1+x+x^2+x^3+x^4+x^5+x^6)^4$ which is indeed $180$. I will leave it to work through it if this is a method which is of interest to you.


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

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

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

Thank you very much guys

答:
This can only be done numerically.

This is how it looks:

The plot shows $x$ and the integral. Blue is the integral calculated numerically and the sand colored graph is $x$.

This solution is graphical methodology, plot and adopted the plot interval. This is done with Mathematica and the built-in NIntegrate with default settings.

In Mathematica a numerical solution input is

```<code>FindRoot[NIntegrate[Cos[1/t] Log[3 - 1/t], {t, 1, x}] == x, {x,
   27.22}, AccuracyGoal -> 8, PrecisionGoal -> 8][[1, 2]]
</code>```

The output is $27.2244$. This result is properly bigger than one. The difference between x and the integral is smaller and $10^{-8}$ at this point.

At the start the value of the integral is less than $x$. The increase is already faster than $x$. The value of the integral grow faster than x for positive value bigger than 1.

This uses a famous algorithm called fixpoint iteration. The rewrite into the fixpoint form and the iteration are done internally and optimized. This does not find a more exact value than the fiven one. This fixpoint iteration does involve the first derivative.

This is only a numerical prove open to verification. But this verification can only be done numerically or graphically.

Plot of the both function on a larger interval:

Plot of the crossing of both functions represented by the difference:
.

Body:
This can only be done numerically.

This is how it looks:

The plot shows $x$ and the integral. Blue is the integral calculated numerically and the sand colored graph is $x$.

This solution is graphical methodology, plot and adopted the plot interval. This is done with Mathematica and the built-in NIntegrate with default settings.

In Mathematica a numerical solution input is

FindRoot[NIntegrate[Cos[1/t] Log[3 - 1/t], {t, 1, x}] == x, {x,
   27.22}, AccuracyGoal -> 8, PrecisionGoal -> 8][[1, 2]]

The output is $27.2244$. This result is properly bigger than one. The difference between x and the integral is smaller and $10^{-8}$ at this point.

At the start the value of the integral is less than $x$. The increase is already faster than $x$. The value of the integral grow faster than x for positive value bigger than 1.

This uses a famous algorithm called fixpoint iteration. The rewrite into the fixpoint form and the iteration are done internally and optimized. This does not find a more exact value than the fiven one. This fixpoint iteration does involve the first derivative.

This is only a numerical prove open to verification. But this verification can only be done numerically or graphically.

Plot of the both function on a larger interval:

Plot of the crossing of both functions represented by the difference:
.


答:
\begin{align}
& \Pr(X+c\le x) = \Pr(X\le x-c) \\[8pt]
= {} & \int_0^{x-c} f(u)\, du \text{ if } x-c\ge0 \\[8pt]
= {} & \int_c^x f(v-c)\, dv \\[8pt]
& \text{where} \\
& v = u+c \\
& dv = dx \\
\text{so that } & \text{when } u=x-c \text{ then } v= c \\
\text{and } & \text{when } u=0 \text{ then } v=c.
\end{align}
So the probability density of $X+c$ is
$$
g(x) = \begin{cases} f(x-c) & \text{for } x>c, \\ 0 & \text{for }x<c. \end{cases}
$$

Title: Probability density function of a Gamma random variable plus a constant

Body:
Suppose $X$ has a Gamma distribution $\Gamma(a,b)$. What is the distribution of $X+c$? where $c$ is a positive constant.

EDIT: I know how extract the expression of the pdf and cdf. What I wanted to know is if the resulting distribution is a know one.


问:
I need to prove that $\{e_n\ |\ n \in \Bbb Z \}$ is an orthonormal basis for $L^2 (\Bbb S^1).$

Let $\mathcal H = L^2 (\Bbb S^1).$ Define $e_n(z) = z^n,\ n \in \Bbb Z.$ Show that $\{e_n\ |\ n \in \Bbb Z \}$ is an orthonormal basis for $L^2 (\Bbb S^1) \simeq L^2 [0,1].$

I need to first show that $\displaystyle \int_{0}^{1} e^{2 \pi (m-n)it}\ dt = 0,$ for $m \neq n,$ which is very easy to show. Secondly I need to show that $\{e_n\ |\ n\in \Bbb Z \}$ is total. Can somebody please shed some light on how to show totality of the orthonormal system? Is it due to Fourier series expansion? Can every square summable function on $[0,1]$ admits Fourier series expansion? Any help or suggestion in this regard will be highly appreciated.

Thanks for your time.

答: Hint: Note that $C(S^1)$ is dense in $L^2(S^1)$. Then, use this fact together with the Stone-Weierstrass theorem to conclude.


问:
I'm using the formula below and want to find the length of γ
I want to find the length of the smooth curve $γ(t)$ from $t = 0$ to $t = \pi/4$

I'm using the formula for γ:
$$l(γ) = \int_a^b|\gamma'(t)|\,dt$$
apologies I can't work out how to format that correctly.

I have $ \gamma'(t) $
I am now aware this is incorrect but am struggling to amend it myself
I'm not sure how to apply this into the formula without it just integrating back into its original form?

Any help or identification of areas where I've made a mistake would be really useful.

答:
Note that if $$\gamma(t)=e^{it}+t(\sin(t)-i\cos(t))=e^{it}+t\sin(t)-it\cos(t)$$
so, you have that $$\gamma'(t)=\color{blue}{ie^{it}}+\sin(t)+t\cos(t) +it\sin(t)-i\cos(t). \quad [1]$$
Now, since that for all $t\in \mathbb{R}$, we have that $$e^{it}=\cos(t)+i\sin(t)$$
The, you can re-write $[1]$ as $$\gamma'(t)=\sin(t)+t\cos(t)\color{blue}{-\sin(t)}+it\sin(t)-i\cos(t)\color{blue}{+i\cos(t)}.$$
Using that for all $z\in \mathbb{C}$ we have that $$|z|=\sqrt{\operatorname{Re}(z)^{2}+\operatorname{Im}(z)^{2}}$$
So, you have that \begin{eqnarray*}
|\gamma'(t)|&=&\sqrt{(\sin(t)+t\cos(t)-\sin(t))^{2}+(t\sin(t)-\cos(t)+\cos(t))^{2}}\\
&=&\sqrt{t^{2}\cos^{2}(t)+t^{2}\sin^{2}(t)}\\
&=&\sqrt{t^{2}(\cos^{2}(t)+\sin^{2}(t))}\\
&=&\sqrt{t^{2}}, \quad \text{since that} \quad \forall x\in \mathbb{R}: \quad \cos^{2}(x)+\sin^{2}(x)=1.
\end{eqnarray*}

Finally, you need calculate $$\ell(\gamma(t))=\int_{0}^{\frac{\pi}{2}}|\gamma'(t)|dt.$$


问:
What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

For the case when $x=y$ it is clear, what about the case when $x\ne y$?

答:
"OP" asked the query:

$x^y=y^x$

Above has numerical integer solution:

$(x,y)=(2,4)$


问:
How many different identifiers can a fictitious programming language have?
How many different identifiers can a fictitious programming language have, when the following rules must be obeyed:

Only upper case letters are allowed, so elements of $\{A, ..., Z\}$,

The identifier must be at least of length $1$ and at most length $5$.

The keywords AND, OR, IF, THEN and GOTO are not allowed to appear in any form in the identifier.

I approached this by looking at the allowed lengths individually and then add all the combinations up.

Length = 1:
Here we simply have only $26$ possibilities.

Length = 2:
This time we have $26^2 - 2$ combinations, since IF and OR can't appear.

Length = 3:
This one is also quite simple. We have $25^3 - 1 - 2*26 - 2*26$ combinations, since AND is not allowed to appear and all combinations that contain IF (IF. and .IF) or OR (OR. and .OR) are invalid.

Length = 4:
This is where I start to have my problems. I tried to do the same thing here, but I am not quite sure. There are $26^4$ total combinations. This time THEN and GOTO can't appear, so $-2$ to accommodate this. AND also can't appear, so $-2*26$. But for IF and OR I am not certain how to calculate it. At first I thought it would simply be $-3*26^2 - 3*26^2$ (IF.. , .IF. , ..IF , OR.. , .OR. , ..OR), but then I thought this would count certain combinations more than once. The combination IFOR for example would be counted twice, so I would have to add $1$ to make up for that. Then there are the combinations IFIF and OROR, which also would be subtracted twice so I would have to add $+1+1$ again. And I am not sure if I missed some of these.

So for Length > 3 I am not quite sure how to tackle this, so I don't miss any combinations and everyone is counted only once.

答:
For $n=4$, you also need to add back in IFOR and ORIF. In other words, all of the doubly subtracted codes look like
$$
AABB,
$$
where $AA$ and $BB$ are both bad words of length $2$. There are two choices for $AA$, and two choices for $BB$, so you need to add back in $2\times 2$.

When $n=5$, there are more possibilities for doubly subtracted words: they can be
$$
AABB*,\quad AA{*}BB,\quad*AABB,\\
AACCC,\quad CCCAA
$$
Again, $AA$ and $BB$ are arbitrary two-letter bad words, and $CCC$ is an arbitrary three letter bad word (okay, there is only one possibility for $CCC$, but I am trying to show how this will generalize).

That seems to be all the ways to cram two of the bad words into a five letter string. But these are only the ways where the two words do not overlap. Since the end of GOTO is the same as the beginning of OR, there is one final doubly subtracted word, namely,
$$
GOTOR.
$$
You can check this is the only possibility for overlap. In general, if you had lots of overlapping possibilities, things would be more more complicated. Furthermore, if $n\ge 6$, you would have to worry about triply subtracted words, like IFOROR. This problem, however, is quite tractable.

In summary, you initially subtract the words which contain each possible bad word in each possible position, not worrying about double counting, resulting in $$26^5-4\cdot 26^3-4\cdot 26^3-3\cdot 26^2-2\cdot 26-2\cdot 26.$$ But then you must add in all of the doubly subtracted words in the previous paragraphs. For example, for words like $AABB*$, you would add back in $2\times 2\times 26$. I will see you if can finish the rest.

This type of counting is called the inclusion exclusion principle, or in and out method.


问:
Can you think of a set $A$ for which $A \subseteq \mathcal{P}(A)$?
It's an exercise from Velleman's How to Prove It. I can only think of the empty set ($A=\emptyset$), but we're asked to think of another, if there is such a set $A$.

Edit: $\mathcal{P}(A)$ being, of course, the powerset of $A$.

答: $A=\{\emptyset\}$ or $\{\emptyset,\{\emptyset\}\}$ or $\{\emptyset,\{\emptyset\},\{\{\emptyset\}\}\}$ and so on.


问:
Why are the payoffs of this family of maximization problems convex?
This is an exercise for the reader from my optimization course. A simple profit-model is
$$\max_x p\cdot q(x) - c^Tx\quad\text{ s.t. }\quad x\ge 0$$
for some price $p$, costs $c$ and a continuous function $q:\mathbb R^n\rightarrow \mathbb R$. Now for fixed $c>0$ and an arbitrary continuous function $q$, denote the solution of the problem by $x(p)$ (we may assume it exists). Show that the function
$$g(p)=p\cdot q(x(p))-c^Tx(p)$$
of the optimum payoff for given price $p$ is convex on $(0,\infty)$.

Since $q$ is not assumed to be differentiable, I don't think calculus can be used.
On the other hand, if I work with $g(\lambda p_1+(1-\lambda)p_2)$, I don't know what to do with$x(\lambda p_1+(1-\lambda)p_2)$.

Is there some relationship/property of $x(p)$ that I'm missing here that would allow me to do a direct convexity proof? Or is there an indirect way to show that $g$ is convex?

答:
$$g(p) = \max_{x : x\ge 0} \{ p\cdot q(x) - c^Tx\}$$
For fixed $x$, $p\cdot q(x) - c^Tx$ is affine in $p$, so $g$ is the maximum of a collection of convex functions, and is therefore convex on $\mathbb{R}$.

Note that it is not relevant that $q$ is continuous, that $c>0$ or that $x(p)$ is unique.


问:
How to restructure Compound interest formula (with regular contributions) to solve for the periodic payment amount
I need help restructuring this formula to solve for the payment PMT rather than the Total:

$$ \text{Total} = \text{Compound interest for principal} + \text{Future value of a series} , $$

$$ \text{Total} = P \left( 1+\frac{r}{n} \right)^{nt} + \text{PMT} * \frac{\left(1 + \frac{r}{n}\right)^{nt} - 1) }{\frac{r}{n}} . $$

Where:

$A$ = the future value of the investment/loan, including interest

$P$ = the principal investment amount (the initial deposit or loan amount)

$\text{PMT}$ = the monthly payment

$r$ = the annual interest rate (decimal)

$n$ = the number of times that interest is compounded per unit t

$t$ = the time (months, years, etc) the money is invested or borrowed for

答:
If you know the total, you can just do a bit of algebraic rearranging:

$$ \text{Total} = P \left( 1+\frac{r}{n} \right)^{nt} + \text{PMT} * \frac{\left( 1 + \frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} $$

$$ \Rightarrow \text{Total} - P\left( 1+\frac{r}{n} \right)^{nt} = \text{PMT} * \frac{\left( 1 + \frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} $$

$$ \Rightarrow \frac{r}{n} * \left( \text{Total} - P\left( 1+\frac{r}{n} \right)^{nt} \right) = \text{PMT} * \left(\left( 1 + \frac{r}{n} \right)^{nt}-1\right) $$

$$ \Rightarrow \text{PMT} = \frac{r}{n} * \frac{\text{Total} - P\left( 1+\frac{r}{n} \right)^{nt}}{\left( 1 + \frac{r}{n} \right)^{nt}-1} $$


问:
Problem 7.3 I. Martin Isaacs's Character Theory Of Finite Groups

Let $H\subseteq G$ and $\xi\in Irr(H)$. Suppose $(\xi-\xi(1)1_H)^G=\theta$ and $[\theta,\theta]=1+\xi(1)^2$. Show that there exists $N\lhd G$ with $N\cap H=\ker\xi$ and every $x\in G-N$ conjugate to some element of $H$.

My attempt:

I am trying to imitate the proof of the Frobenius theorem(Theorem 7.2).

Obviously $ \xi\ne 1_H $ since $ [\theta,\theta]>1 $.

Let $ \chi=\theta+\xi(1)1_G $, it is a generalized character of $ G $, we claim that $ \chi\in Irr(G) $ by showing that $ [\chi,\chi]_G=1 $:

\begin{align*}
  [\chi,\chi]_G&=[\theta,\theta]+2\xi(1)[\theta,1_G]+\xi(1)^2\\
  &=1+\xi(1)^2+2\xi(1)[\xi^G-\xi(1)(1_H)^G, 1_G]+\xi(1)^2\\
  &=1+2\xi(1)^2+2\xi(1)[\xi^G, 1_G]-2\xi(1)^2[(1_H)^G, 1_G]\\
  &=1+2\xi(1)^2+0-2\xi(1)^2\\
  &=1.
 \end{align*}
Also, $ \chi(1)=\theta(1)+\xi(1)=\xi(1)>0 $, it is indeed an irreducible character of $ G $.

Suppose we have shown that $ \chi|_H=\xi $.

Let $N=\ker\chi$.

We have $ N\lhd G $ and we claim that $ N\cap H=\ker\xi $. Since if $ h\in N\cap H $, then $ \xi(h)=\chi(h)=\chi(1)=\xi(1) $, we know that $ h\in\ker\xi $. Conversely, if $ \xi(x)=\xi(1) $, then of course $ x\in H $ and $ x\in\ker\xi\subseteq N $. Hence $ N\cap H=\ker\xi $.

To show that every $ x\in G-N $ conjugate to some element of $ H $, it suffices to show that $ G\setminus N\subseteq\bigcup_{g\in G} H^g $ which is equivalent to show that $ G\setminus\bigcup_{g\in G} H^g\subseteq N $. For any $ y\in G\setminus\bigcup_{g\in G} H^g  $, and $ \chi|_H=\xi\in Irr(H)\setminus\{1_H\} $, we have $ \chi(y)-\chi(1)=\theta(y)=(\xi-\xi(1)1_H)^G(y)=0 $ by the definition of induced characters.

But the problem is, I do not know how to show that $ \chi|_H=\xi $. We do not have any TI set here and I have tried to compute $[\chi|_H,\xi]=[\xi^G,\xi^G]-\xi(1)[(1_H)^G,\xi^G]$. And it suffices to show that $[\chi|_H,1_H]=0$.

答:
Here is a more streamlined solution with notation that sticks closer to the text (not the problem) in Isaacs' book. It also includes the solution to Problem (7.4).

Notation If $\varphi $ is a generalized character of some subgroup $H \leq G$, we write $\varphi_0=\varphi-\varphi(1)1_H$. Observe that $\varphi_0(1)=0$, and in the notation of M.I. Isaacs CTFG page 107, $\varphi \in \mathbb{Z}[Irr(H)]^\circ$. Since $\varphi$ is a generalized character, also $\varphi_0$ and $\varphi_0^G$ are generalized characters of $H$ and $G$ respectively. If $\varphi \neq 1_H$ then $[\varphi_0,\varphi_0]=1+\varphi(1)^2$.

Proposition Let $H \lt G$ and let $\varphi \in Irr(H)$, with $[\varphi_0^G,\varphi_0^G]=1+\varphi(1)^2$. Then the following hold.

$(a)$ There exists an $N \unlhd G$, with $H \cap N=ker(\varphi)$

$(b)$ $(G-\bigcup_{g \in G}H^g) \cup \{1\}\subseteq N$, in other words, every element outside $N$ is conjugate to some element of $H$.

Proof The start of the proof is similar to the proof of how a normal complement to a Frobenius complement can be constructed. Observe that $[\varphi_0^G,1_G]=[\varphi_0,1_H]=-\varphi(1)$. Hence $\varphi_0^G=\varphi^* - \varphi(1)1_G$, for some generalized character $\varphi^*$ of $G$ with $[\varphi^*,1_G]=0$. Observe that $\varphi_0^G(1)=|G:H|\varphi_0(1)=0=\varphi^*(1)-\varphi(1)$, so $\varphi^*(1)=\varphi(1) \gt 0$. Using this we get
$$1+\varphi(1)^2=[\varphi_0^G,\varphi_0^G]=[\varphi^*-\varphi(1)1_G,\varphi^*-\varphi(1)1_G]=[\varphi^*,\varphi^*]+\varphi(1)^2$$
whence $[\varphi^*,\varphi^*]=1$ and since $\varphi(1) \gt 0$ we conclude that $\varphi^* \in Irr(G)$.

To summarize: $\varphi_0^G=\varphi^G-\varphi(1)1_H^G=\varphi^*-\varphi(1)1_G$, so $\varphi^G-\varphi^*=\varphi(1)(1_H^G-1_G)$. Now let us take a closer look at the character $1_H^G-1_G$. Since $[1_H^G,1_G]=[1_H,1_H]=1$, we can write $1_H^G=1_G+\Delta$, with $\Delta$ a (possibly reducible) non-zero character of $G$ with $[\Delta,1_G]=0$. In addition (see also lemmas (2.21) and (5.11) in CTFG), observe that $ker(1_H^G)=core_G(H)=ker(1_G) \cap ker(\Delta)=ker(\Delta)$. We can rewrite $$\varphi^G=\varphi^*+\varphi(1)\Delta$$ and from this it follows at this point that $core_G(ker(\varphi))=ker(\varphi^*) \cap ker(\Delta)=ker(\varphi^*) \cap core_G(H)$. But we are going to prove something stronger, namely that even $ker(\varphi)=ker(\varphi^*) \cap H$, and then $N=ker(\varphi^*)$ is the normal subgroup we are looking for.

Now $\Delta=\sum_{\chi \in Irr(G)}a_\chi\chi$ for some non-negative integers $a_{\chi}$, where $a_{1_G}=0$ but not all $a_{\chi}$'s can be zero. We are going to show that $\varphi^*$ extends $\varphi$, that is $\varphi^*_H=\varphi$. First of all
$$[\varphi^*_H,\varphi]=[\varphi^*,\varphi^G]=[\varphi^*,\varphi^*+\varphi(1)\Delta]=1+\varphi(1)[\varphi^*,\Delta]=1+\varphi(1)a_{\varphi^*}$$
yielding $$\varphi^*_H=(1+\varphi(1)a_{\varphi^*})\varphi+\Gamma$$ where $\Gamma$ is a character of $H$ with $[\Gamma,\varphi]=0$ or $\Gamma=0$. It follows that $\varphi^*(1)=\varphi(1)=\varphi(1)+\varphi(1)^2a_{\varphi^*}+\Gamma(1)$, so $\varphi(1)^2a_{\varphi^*}+\Gamma(1)=0$. Since this last equation concerns non-negative integers and of course $\varphi(1) \gt 0$, this can only happen when $a_{\varphi^*}=0$ and $\Gamma=0$. Hence $\varphi^*_H=\varphi$, which implies $ker(\varphi^*_H)=H \cap ker(\varphi^*)=ker(\varphi)$. This proves (a).

To prove (b) pick an $x \in G-\bigcup_{g \in G}H^g$ (note that this set is non-empty since $H$ is proper). Then by the definition of induced characters we have $\varphi_0^G(x)=0=\varphi^*(x)-\varphi(1)$. Since $\varphi^*$ and $\varphi$ have the same degree it follows that $\varphi^*(x)=\varphi^*(1)$, that is $x \in ker(\varphi^*)$ and we are done.

Corollary (Isaacs Problem (7.4))
Let $H \lt G$ then the following are equivalent.

$(a)$ $H$ is a Frobenius complement.

$(b)$ Induction to $G$ is an isometry on $\mathbb{Z}[Irr(H)]^\circ$.

Proof (a) $\Rightarrow$ (b). This is clear from Lemma (7.7) in Isaacs' CTFG with $X=H$ (note that in this case $N_G(X)-X=\emptyset$, since $H=N_G(H)$).

(b) $\Rightarrow$ (a) Let $\varphi \in Irr(H)$, $\varphi \neq 1_H$. Then $\varphi_0 \in \mathbb{Z}[Irr(H)]^\circ$ and $[\varphi_0,\varphi_0]=1+\varphi(1)^2$. Since $[\varphi_0^G,\varphi_0^G]=[\varphi_0,\varphi_0]=1+\varphi(1)^2$, we can apply Problem (7.3). So each $\varphi$ has an extension to $G$ called $\varphi^*$, where it is understood that $1_H$ extends to $1_G$. Now put $M=\bigcap_{\varphi \in Irr(H)}ker(\varphi^*)$. Then $M \unlhd G$. Since $ker(\varphi)=H \cap ker(\varphi^*)$ for all $\varphi \in Irr(H)$ , we get $M \cap H=1$. We now use a counting argument to show that in fact $G=MH$ and then we can apply Problem (7.1) (d) $\Leftrightarrow$ (f) in combination with the Problem(7.3). Problem (7.3) yields the set $G - \bigcup_{g \in G}H^g \subseteq ker(\varphi^*)$ for all $\varphi \in Irr(H)$. It follows that $G-\bigcup_{g \in G}H^g \subseteq M$, that is $G-M \subsetneq \bigcup_{g \in G}H^g$ (note $1 \notin G-M$). Now
$$\#( \bigcup_{g \in G}H^g) \leq |G:N_G(H)|(|H|-1)+1 \leq |G:H|(|H|-1)+1=|G|-|G:H|+1$$
We conclude that $\#(G-M)=|G|-|M| \lt |G|-|G:H|+1$, so $|G:H| \lt |M|+1$, whence $|G:H| \leq |M|$. Then $|G| \leq |M| \cdot |H|=|MH|$, since $M \cap H=1$. This implies $G=MH$ and we conclude that $H$ is indeed a Frobenius complement.


问:
Problem in a Taylor expansion of Runge-Kutta method
I'm stuck with this problem:

Determine all three-stage, thre-order RK methods

I'm reading some papers about this and they do:

$$ y_{i} = y_{i-1} + \sum_{i=0}^m b_{j}k_{j} $$

$$
\begin{cases}
k_{1} = hf(t_{i-1},y_{i-1}) \\
k_{2} = hf(t_{i-1}+c_{2}h,y_{i-1}+a_{21}k_{1}) \\
k_{3} = hf(t_{i-1}+c_{3}h,y_{i-1}+a_{31}k_{1}++a_{32}k_{2}) \\
\end{cases}
$$

Now, they do a third order Taylor expansion of $y(t;t_{i-1},y_{i-1})$. The point is here:

$$
y_{i} = y_{i-1}
 + f(t_{i-1},y_{i-1})h
 + \cfrac{1}{2}\left[  \frac{\partial }{\partial t}  f(t_{i-1},y_{i-1})
    + f(t_{i-1},y_{i-1}) \frac{\partial }{\partial y}  f(t_{i-1},y_{i-1})  \right]h^2
 + ...
$$

I'm trying to get this expression but I don't know to do it:

$$
\frac {d}{dh} f(t_{i-1},y_{i-1})
= \frac {\partial }{\partial t} f(t_{i-1},y_{i-1})\frac {dt}{dh}
+ \frac {\partial }{\partial y} f(t_{i-1},y_{i-1})\frac {dy}{dh}
$$

So that, it means that:

$$
\begin{cases}
\frac {dt}{dh} = 1 \\
\frac {dy}{dh} = f(t_{i-1},y_{i-1}) \\
\end{cases}
$$

But I don't understand it (in case this is right).

Please, anyone could help me to understand it?
Thanks in advance.

答:
In the derivative $\frac{dy}{dh}$, you are considering the function $h\mapsto y(t_{i-1}+h)$ for an exact solution of the ODE with initial condition $y(t_{i-1})=y_{i-1}$. In consequence you get indeed
$$
\frac{dy}{dh}(t_{i-1})=y'(t_{i-1})=f(t_{i-1},y_{i-1}).
$$

Any ODE problem can be transformed into an autonomous first order system $u'(t)=F(u(t))$. In application this can then be reduced back to the non-autonomous case. In the autonomous case the propagation of the exact solution has the Taylor expansion
$$\begin{multline}
u(t+h)=u(t)+h·F(u(t))+\frac{h^2}2F'[F]+\frac{h^3}6\Bigl(F''[F,F]+F'[F'[F]]\Bigr)
\\+\frac{h^4}{24}\Bigl(F'''[F,F,F]+3F''[F'[F],F]+F'[F''[F,F]]+F'[F'[F'[F]]]\Bigr)+O(h^5)
\end{multline}$$
For the stages of the method one gets the expansions
$$\begin{align}
k_1&=hF\\
k_2&=hF+a_{21}h^2F'[F]+\tfrac12 a_{21}^2h^3 F''[F,F]+O(h^4)\\
k_3&=hF+hF'[a_{31}hF+a_{32}h(F+a_{21}hF'[F]]+\tfrac12hF''[a_{31}hF+a_{32}hF]^2+O(h^4)\\
&=hF + (a_{31}+a_{32})h^2F'[F] + a_{32}a_{21}h^3F'[F'[F]] + \tfrac12(a_{31}+a_{32})^2h^3F''[F,F]+O(h^4)
\end{align}$$
and assembled
$$\begin{multline}
u_{+1} = u+(b_1+b_2+b_3)hF+(b_2a_{21}+b_3(a_{31}+a_{32}))h^2F'[F] \\+ \tfrac12(b_2a_{21}^2+b_3(a_{31}+a_{32})^2)h^3F''[F,F] + b_3a_{32}a_{21}h^3F'[F'[F]] + O(h^4)
\end{multline}$$
Now using $a_{21}=c_2$ and $a_{31}+a_{32}=c_3$ and comparing both expansions gives the conditions
\begin{align}
\text{order $1$}&:& b_1+b_2+b_3&=1\\
\text{order $2$}&:&  b_2c_2+b_3c_3&=\tfrac12\\
\text{order $3$}&:&  b_2c_2^2+b_3c_3^2&=\tfrac13\\
 & &  b_3a_{32}c_2&=\tfrac16\\
\end{align}
See also Butcher 2008: RK introduction and low order methods for a more systematic approach and examples of methods.


问:
sum and product of random variables

Let $X$ and $Y$ be two (general) random variables with finite means and let $Z=X+Y$ and $Z' = XY.$ Define for a random variable $X,X^+$ to be the random variable equal to $\max\{X(\omega), 0\}$ and $X^-$ so that $X^-(\omega) = \max\{-X(\omega), 0\},$ where $\omega \in \Omega,$ the sample space.

Express $Z^+$ and $Z^-$ in terms of $X^+, X^-, Y^+, Y^-,$ with justification.

If $X$ and $Y$ are independent, express $Z'^+$ and $Z'^-$ in terms of $X^+, X^-, Y^+, Y^-,$ with justification.

I think $E(Z^+) - E(Z^-) = E(X^+) - E(X^-) + E(Y^+) - E(Y^-).$ But I'm not sure how to express $Z^+$ and $Z^-$ from this. I tried considering all the possible linear combinations of $X^+, X^-, Y^+, Y^-,$ with coefficients of absolute value less than or equal to $1$, but that seems very tedious. For instance, clearly $Z^+\neq X^+ + Y^+ - X^- - Y^-$ because they're unequal when $X(\omega), Y(\omega) < 0.$ Is there some sort of way for me to deterministically find the right coefficients?

Also, if $X$ and $Y$ are independent, then $E(Z'^+) - E(Z'^-) = E(Z')= E(XY) = E(X)E(Y) = (E(X^+) - E(X^-))(E(Y^+)- E(Y^-))$, and so the values of $Z'^+$ and $Z'^-$ should reflect this.

Could someone give some hints as to how to find the required relationships?

答:
Using all the possibilities that I believe are possible, e.g. when X, Y are both positive, when both negative, when X is positive and Y is negative but greater in absolute value or negative but less in absolute value, and with X and Y's roles switched for this last statement

```<code>X    Y    Z     X+   Y+   X-    Y-    Z+    Z-
1    2    3     1    2    0     0     3     0
1    -1/2 1/2   1    0    0     1/2   1/2   0
-1/2 1    1/2   0    1    1/2   0     1/2   0
1    -2   -1    1    0    0     2     0     1
-2   1    -1    0    1    2     0     0     1
-1   -3   -4    0    0    1     3     0     4
</code>```

I found that

$$Z^+=\max\{X^+-Y^-, 0\}+\max\{Y^+-X^-, 0\}\\
Z^-=\max\{X^--Y^+, 0\}+\max\{Y^--X^+, 0\}$$

and e.g. both positive, exactly one of them negative, and both negative,

```<code>X    Y    Z'    X+   Y+   X-    Y-    Z'+    Z'-
1    2    2     1    2    0     0     2     0
1    -1/2 -1/2  1    0    0     1/2   0     1/2
-1    2   -2    0    2    1     0     0     2
-1   -3   3     0    0    1     3     3     0
</code>```

gives

$$Z'^+=X^+Y^++X^-Y^-\\
Z'^-=X^+Y^-+X^-Y^+$$

Title: sum and product of random variables

Body:
Using all the possibilities that I believe are possible, e.g. when X, Y are both positive, when both negative, when X is positive and Y is negative but greater in absolute value or negative but less in absolute value, and with X and Y's roles switched for this last statement

X    Y    Z     X+   Y+   X-    Y-    Z+    Z-
1    2    3     1    2    0     0     3     0
1    -1/2 1/2   1    0    0     1/2   1/2   0
-1/2 1    1/2   0    1    1/2   0     1/2   0
1    -2   -1    1    0    0     2     0     1
-2   1    -1    0    1    2     0     0     1
-1   -3   -4    0    0    1     3     0     4

I found that

$$Z^+=\max\{X^+-Y^-, 0\}+\max\{Y^+-X^-, 0\}\\
Z^-=\max\{X^--Y^+, 0\}+\max\{Y^--X^+, 0\}$$

and e.g. both positive, exactly one of them negative, and both negative,

X    Y    Z'    X+   Y+   X-    Y-    Z'+    Z'-
1    2    2     1    2    0     0     2     0
1    -1/2 -1/2  1    0    0     1/2   0     1/2
-1    2   -2    0    2    1     0     0     2
-1   -3   3     0    0    1     3     3     0

gives

$$Z'^+=X^+Y^++X^-Y^-\\
Z'^-=X^+Y^-+X^-Y^+$$


问:
Trouble deriving the value of $⟨x_i|x_i⟩$ in an infinite dimensional vector space as the Dirac Delta $δ(x_i- x_i)$
On pg. 57 of Principles of Quantum Mechanics, the author considers a vector space of "infinite dimensional vectors", conceived of as a vector space of functions defined on some closed interval.  The inner product $⟨f,g⟩$ is defined as

$$
⟨f|g⟩ = ∫_a^b f^*(x)g(x)dx
$$

while a basis $|x_i⟩$ is defined as a function $x_i(x)$ s.t. $x_i(x) = 0$ everywhere except for $x_i(x_i) = 1$ (the author doesn't actually explicitly state this, but this I'm assuming is what he means?).  At this point, it seems to me that

$$
⟨x_i|x_i⟩ = ∫_a^b x_i^*(x)x_i(x)dx = 0
$$

since the integral of any function 0 almost everywhere is 0. But instead the author goes on to derive that the value is actually $δ(x_i-x_i) = ∞$:

This seems to me to be highly problematic since since an inner product is supposed to be a scalar function which returns scalar values, and $∞ ∉ \mathbb{C}$ is not scalar. In fact even the notion of $⟨x_i|f⟩$ seems to me to be confused, since

$$
⟨x_i|f⟩ = ∫_a^b x_i^*(x)f(x)dx = 0
$$

since again the integral of a function 0 everywhere except for $x^*_i(x_i)f(x_i) = 1*f(x_i) = f(x_i)$ is still just $0$. Yet the author wants to say that this integral evalutes to $f(x_i)$?

答: Further to @CosmasZachos's answer, nothing in the excerpt says $x_i(x_i)=1$, i.e. that $x_i(x)=\delta_{x,\,x_i}$ is a Kronecker delta. This is a misconception on your part. It really should be a Dirac delta, so$$\langle x_i|x_j\rangle=\int_a^b\delta(x-x_i)\delta(x-x_j)dx=\delta(x_i-x_j)$$ provided $x_i\in(a,\,b)$ (otherwise the RHS is $0$ instead).


问:
If $A\in M_{n}(\mathbb{C})$ is a Hermitian matrix  so $I-iA$ is Invertible
I'd love your help with this.

Let $A\in M_{n}(\mathbb{C})$ be a Hermitian matrix $(A^{*}=A)$.

How can I prove that $I-iA$ is Invertible?

Thank you

答:
Define $$B= (I+A^2)^{-1}(I +i A).$$ Note that $I+A^2$ is invertible, as $\det(I+A^2) = \det(I+A^\dagger A) \geq 1$ (a sum of a positive definite and a positive semidefinite matrix is invertible). Then
$$B(I-iA) = (I+A^2)^{-1}(I +i A)(I-iA) = (I+A^2)^{-1} (I+ A^2) =I,$$
so $B$ is the inverse of $A$.</p>

ContentLicense: CC BY-SA 3.0


问:
Decomposing a rectangle in $\mathbb{R}^d$
Let $K=[a_1,b_1]\times\dots \times [a_d,b_d]$ be a closed rectangle in $\mathbb{R}^d$ and suppose that $K\subset\cup_{i=1}^n R_i$, where each $R_i$ is an open rectangle. In his measure theory book Cohn writes:

$K$ can be decomposed into a finite collection $\{K_j\}$ of rectangles that
overlap only on their boundaries and are such that for each $j$ the interior of $K_j$ is
included in some $R_i$.

If $\overline{R}_i$ denote the closure of $R_i$, then $K\cap\overline{R}_i$ is a closed rectangle. For each $1\leq m \leq d$, let $P_m$ denotes all the left and right $m$th coordinates appearing among the $K\cap\overline{R}_i$ $ (1\leq i\leq n)$. Then $P_m\subset [a_m,b_m]$, and if we order the elements of $P_m$ from left to right we obtain a partition of $[a_m,b_m]$ into a finite collection if closed intervals that overlap only at the boundary. Let $\{K_j\}$ denote the finite collection of all closed rectangles in $\mathbb{R}^d$ which are formed by taking $d$-products of those closed intervals (by choosing one closed interval in the partition of $[a_m,b_m]$ for each $m$). These rectangles intersect only at their boundaries and

$$K=\cup K_j$$

Indeed if $x\in K$, then $x_m \in [a_m,b_m]$ for each $m$, and so $x\in K_j$ for some $j$. Conversely if $x\in K_j$ for some $j$, then $x_m \in [a_m,b_m]$ for each $m$ and so $x\in K$.

Am stuck at proving that the interior of each $K_j$ is included in some $R_i$. It would suffice to show that $K_j\subset K\cap\overline{R}_i$ for some $i$. But I don't see how to show it.

Any help is greatly appreciated.

答:
Use the $\sup$ norm for convenience.

For $k \in K$, let $\delta(k) = \sup \{ r | B(k,r) \subset R_i \text{ for some } i\}$. Since the $R_i$ are open we see that $\delta(k) > 0$ for all $k$.

Let $\delta= \inf_{k \in K} \delta(k)$, it is straightforward to show that $\delta>0$ and for any $k$ there is some $R_i$ such that $B(k,\delta) \subset R_i$.

Let $l_i = b_i-a_i$ and take the rectangles $$\Delta_{n,k_1,...,k_d} = [a_1+{k_1 \over n} l_1, a_1+{k_1+1 \over n} l_1] \times \cdots \times [a_d+{k_d \over n} l_d, a_d+{k_d+1 \over n} l_d] $$
with  ${1 \over n} < \delta$.
Note that for each $n$ we have  $K = \cup_{k_i=0,...,n-1} \Delta_{n,k_1,...,k_d}$.

Since the center of each rectangle lies in some $R_i$, we see that each rectangle lies in some $R_i$.

Addendum:

To see why $\delta>0$: Suppose $\delta=0$ then there is some sequence $k_i$ such that $\delta(k_i) \to 0$. Without loss of generality we can assume $k_k \to k$. We know that $\delta(k) >0$ and hence for $k_i$
sufficiently close to $k$ we have $\delta(k_i) \ge {1 \over 2} \delta(k)$ which is a contradiction.


问:
If $X$ and $Y$ are fundamental matrices then exists $C$ invertible such that $Y=XC$
If $X(t)$ and $Y(t)$ are fundamental matrices then exists $C\in M_n$ invertible such thta $Y(t)=X(t)C$
Both for the homogeneous linear system $x'=A(t)x$.

attempt: Since $X$ is fundamental, it spans the solution space i.e.,  $$X'(t)=A(t)X(t)$$ but doing the same for $Y$ I go nowhere on finding $C$ invertible. I now that since $X$ is fundamental there is $t_0$ such that $\det(X(t_0))\neq 0$, even more $\det(X(t))\neq 0$ for all $t$, but again go nowhere.

答:
You determine $C=X(t_0)^{-1}Y(t_0)$ at one point and then use that both solve the same linear differential equation to confirm that this relation holds with a constant $C$, that is, outright define $Y(t)=X(t)C$ and conclude from
$$
X'=AX\\
(XC)'=X'C=AXC\\
Y'=AY
$$
that $Y$ is indeed a solution.


问:
What algorithm is Newton using in the "De analysi" to extract the square root of a polynomial?
I'm reading a 1745 English translation of Newton's De analysi (apparently the most up-to-date there is, surprisingly). The Latin is here.

In this tract he shows how to use the integral power rule for rational exponents (together with a sum rule: the area under a sum of curves being the sum of the areas) to find quadratures, and he gives examples in three types of how to deal with the case of integrating $y$ when it is not a polynomial: examples using long division e.g. $y=a^2/(b+x)$, examples using the extraction of square roots e.g. $y=\sqrt{a^2+x^2}$, and examples using "the resolution of affected equations" when $y$ is defined implicitly via a polynomial $f(x,y)=0$.

For the second example, $y=\sqrt{a^2+x^2}$, he uses an algorithm to extract the root, but I don't recognize the algorithm. Here's an image:

Note that Newton is not applying his binomial series: he is doing something else.

I realize this is equivalent to the following procedure. Assume $y=\sum b_ix^i$ and we are given $y^2=\sum a_ix^i$; the goal is to solve for the $b_i$. Using the Cauchy product, squaring the first equation gives $y^2=\sum c_ix^i$ where $c_i=\sum_{k=0}^ib_kb_{i-k}$. Pattern matching the coefficients $c_i=a_i$ allows us to recursively solve for the $b_i$, because $c_i$ is the first term to use $b_i$. Thus

$$c_0=b_0^2$$
$$c_1=2b_0b_1$$
$$c_2=2b_0b_2+b_1^2$$
$$c_3=2b_0b_3+2b_1b_2$$
$$c_4=2b_0b_4+2b_1b_3+b_2^2$$
etc.

But Newton is clearly doing something else: drawing on some established algorithm with a visual representation. I don't think it's his algorithm. So:

Three Questions

How does the algorithm work?

Does the algorithm have a name?

Who first used the algorithm?

EDITED

After puzzling it out a little, I at least see the answer to my first question. Here is the procedure applied to $\sqrt{a^2+x^2}$.

Step 1: Guess $a$. Store this as the latest estimate $S_0$.

Step 2: Square, giving $a^2$, then subtract from the radicand: $x^2$. Store this as the latest remainder $R_0$.

Step 3: Update $S_i$ to $S_{i+1}$ by adding a term $y_i$ to $S_i$ so that $2y_iS_0$ agrees with the lowest-degree term of $R_i$. In this case, we want to add $y_0$ so that the lowest-degree term of $2y_0a$ agrees with $x^2$. Hence $y_0=x^2/2a$. The estimate $S_0$ is thus updated to $S_1=a+x^2/2a$.

Step 4: Multiply $y_i$ by $(2S_i+y_i)$, then subtract from the last remainder and store the result as the latest remainder $R_{i+1}$. In this case, we multiply $x^2/2a$ by $(2a+x^2/2a)$, giving $x^2+x^4/4a^2$. Subtracting from the last remainder gives $-x^4/4a^2$.

Step 5: Repeat steps 3/4.

Just to illustrate for the next term, we want to add $y$ to that $2ya=-x^4/4a^2$. Thus we add $y=-x^4/8a^3$ and the estimate is updated to $a+x^2/2a-x^4/8a^3$. Now we multiply

$$-\frac{x^4}{8a^3}\left(2\left(a+\frac{x^2}{2a}\right)-\frac{x^4}{8a^3}\right)=-\frac{x^4}{4a^2}-\frac{x^6}{8a^4}+\frac{x^8}{64a^6}$$

Subtracting from the previous remainder $-x^4/4a^2$ yields the new remainder

$$\frac{x^6}{8a^4}-\frac{x^8}{64a^6}$$

and so on.

答:
Partial answer with historical context

After googling "square root of a polynomial," I found that the algorithm is lucidly described in many standard 19th and early 20th century algebra textbooks for high schools and colleges: e.g. here, here, and here. (How poorly modern textbooks fare by comparison!)

Encouraged by this evidence, I decided to glance at George Chrystal's famous textbook Algebra (first published 1886). He gives a thorough discussion in section 17 ("Square root of an integral function of $x$") in Chapter XI ("Arithmetical theory of surds") in volume 1. He also leaves a historical footnote, dating the algorithm to 1557 in English. I reproduce this passage here for ease of reference:

When an integral function of $x$ is a complete square as regards $x$,
its square root can be found by a method analogous to that explained
in $\S12$, for finding the square root of a number. Although the
method is of little interest, either theoretically or practically, we
give a brief sketch of it here, because it illustrates at once the
analogy and the fundamental difference between arithmetical and
algebraical operations.*

I. We may restate Proposition I. of $\S 12$, understanding now $A$ and
$B$ to mean integral functions of $x$.

II. If
$F=p_ox^{2n}+p_1x^{2n-1}+\cdots+p_{2n}$, and if
$$\sqrt{F}=(q_nx^n+q_1x^{n-1}+\cdots+q_{n-p+1}x^{n-p+1})+(q_{n-p}x^{n-p}+\cdots+q_0)=P+Q$$ say; and if we suppose the first $p$ terms, namely
$P=q_nx^n+q_1x^{n-1}+\cdots+q_{n-p+1}x^{n-p+1}$, of $\sqrt{F}$ to be
known, then the next $p$ terms will be the first $p$ terms in the
integral part of $(F-P^2)/2P$.

For we have $$F=P^2+2PQ+Q^2$$ hence
$$\frac{F-P^2}{2P}=Q+\frac{Q^2}{2P}$$ Now the degree of the integral
part of $Q^2/2P$ is $2(n-p)-n=n-2p$. Hence $Q^2/2P$ will at most
affect the term in $x^{n-2p}$. Hence $(F-P^2)/2P$ will be identical
with $Q$ down to the term in $x^{n-2p+1}$ inclusive. In other words,
the first $n-p-(n-2p)=p$ terms obtained by dividing $F-P^2$ by $2P$
will be the $p$ terms of the square root which follow $P$.

We may use this rule to obtain the whole of the terms one at a time,
the highest being of course
found by inspection as the square root of
the highest term of the radicand; or we may obtain a certain number in
this way, and then obtain the rest by division.$\dagger$

$*$ The method was probably obtained by analogy from the arithmetical process. It was first given by Recorde in The Whetstone of Witte
(black letter, 1557), the earliest English work on algebra.

$\dagger$ Just as in division, we may, if we please, arrange the
radicand according to ascending powers of $x$. The final result will
be the same whichever arrangement be adopted, provided the radicand is
a complete square. If this is not the case the operation may be
prolonged indefinitely just as in continued division. We leave the
learner to discover the meaning of the result obtained in such cases.
The full discussion of the matter would require some reference to the
theory of infinite series.


问:
Prove that the set of positive integers can be expressed as a countably infinite union of disjoint countably infinite sets.
Consider the set $\mathbb Z^+$ of positive integers. Prove that $\mathbb Z^+$ can be expressed as a countably infinite union of disjoint countably infinite sets.

I know that countably infinite sets means it has the same cardinality as $\mathbb N$ (the set of natural numbers).

However, I am not sure how to go about proving this.

答:
An easy and cool aproach to this question is using the infinity of the prime numbers.

Let $\mathbb{P}$ be the set of prime numbers. For each $p \in \mathbb{P}$ consider $A_{p} = \lbrace p^{n} : n \in \mathbb{N}\rbrace $ which is an countably infinite set. Consider $ R = \mathbb{Z^{+}}$\ $ \bigcup_{p \in \mathbb{P}}A_{p}$ which is an countably infinite set.

You have that $\mathbb{Z^{+}}= \bigcup_{p \in \mathbb{P}}A_{p} \cup R$ so you have expressed $\mathbb{Z^{+}}$ as a countably infinite union of disjoint countably infinite sets.


问:
Show $f(x)=x^TBx$ is not uniformly continuous if $B$ is symmetric
I am in the initial fase of learning analysis. I am working on the following problem:

Consider the quadratic form $f \, : \, \mathbb{R}^k \to \mathbb{R} $ given by $f(x)=x^TBx$ for $x \in \mathbb{R^k}$ and a $k \times k$ matrix $B$.

Show that if there exists $x_0 \in \mathbb{R}^k$ such that $x_0^TBx_0 \neq 0$, then $f$ is not uniformly continuous.

Hint: Consider $t \mapsto f(tx_0)$.

Show that if $B$ has a diagonal element $b_{ii}$ such that $b_{ii} \neq 0$, then $f$ is not uniformly continuous.

Show that if $B$ is symmetric and different from the zero matrix, then $f$ is not uniformly continuous.

Find an example of a matrix $B$ different from the zero matrix such that $f$ is uniformly continuous.

Regarding question 4, I figured out that the uniform continuity is not ruled out by 1-3 in the example where B is a 2x2 matrix with zeros in the diagonal and the other two elements equal to $a$ and $−a$. And then the function $f$ maps every element to zero, so this function is clearly uniformly continuous.

But I am having trouble with the first 3 questions. Any hints would be greatly appreciated!

答:
(1) Prove by contradiction. Suppose the contrary that $f$ is uniformly
continuous. For $\epsilon=1$, there exists $\delta>0$ such that
$|f(x)-f(y)|<\epsilon$ whenever $||x-y||<\delta$. Note that $f(tx)=t^{2}f(x)$
for any $t\in\mathbb{R}$ and $x\in\mathbb{R}^{k}$. Choose $t_{1}>\frac{2||x_{0}||}{|f(x_{0})|\delta}.$
Define $t_{2}=t_{1}+\frac{\delta}{2||x_{0}||}$ (Note that $x_{0}\neq0$,
so $||x_{0}||>0).$ Observe that $||t_{1}x_{0}-t_{2}x_{0}||=||x_{0}||\cdot|t_{2}-t_{1}|=\frac{\delta}{2}<\delta$,
so we have $\left|f(t_{1}x_{0})-f(t_{2}x_{0})\right|<\epsilon=1.$
On the other hand, by direct calculation,
\begin{eqnarray*}
\left|f(t_{1}x_{0})-f(t_{2}x_{0})\right| & = & |f(x_{0})||t_{1}^{2}-t_{2}^{2}|\\
 & = & |f(x_{0})|\cdot(t_{2}-t_{1})(t_{2}+t_{1})\\
 & \geq & |f(x_{0})|\cdot\frac{\delta}{2||x_{0}||}\cdot2t_{1}\\
 & \geq & |f(x_{0})|\cdot\frac{\delta}{2||x_{0}||}\cdot\frac{4||x_{0}||}{|f(x_{0})|\delta}\\
 & = & 2,
\end{eqnarray*}
which is a contradiciton.

(3) follows from the fact : If $B$ is a symmetric $k\times k$
matrix and $\langle Bx,x\rangle=0$ for all $x\in\mathbb{R}^{k}$,
then $B=0$.

Proof: Let $x,y\in\mathbb{R}^{k}$. Since $B$ is symmetric, we have
that $\langle By,x\rangle=\langle y,B^{t}x\rangle=\langle y,Bx\rangle=\langle Bx,y\rangle.$
Therefore
\begin{eqnarray*}
0 & = & \langle B(x+y),(x+y)\rangle\\
 & = & \langle Bx,x\rangle+\langle By,y\rangle+\langle Bx,y\rangle+\langle By,x\rangle\\
 & = & 0+0+2\langle Bx,y\rangle.
\end{eqnarray*}
That is, $\langle Bx,y\rangle=0$. It follows that $B=0$.

Therefore, if $B$ is symmetric and $B\neq0$, there exists $x_{0}$
such that $\langle Bx_{0},x_{0}\rangle\neq0$.

(4) Simply find a non-zero $k\times k$ matrix such that $B+B^{t}=0.$
Such matrix exists. For example, let $B=(b_{ij})$ with $b_{12}=-1$,
$b_{21}=1$, and $b_{ij}=0$ for other entries. In this case $\langle Bx,x\rangle=\langle x,B^{t}x\rangle=\langle x,-Bx\rangle=\langle-Bx,x\rangle=-\langle Bx,x\rangle$
and it will follow that $\langle Bx,x\rangle=0$ for all $x$.


问:
period of a function $\lvert\sin x\rvert+\lvert\cos x\rvert$
I have read that $$y=\lvert\sin x\rvert+ \lvert\cos x\rvert $$ is periodic with fundamental period $\frac{\pi}{2}$.

But Wolfram says it is periodic with period $\pi$.

Please tell what is correct.

答:
The results coming from any software should be checked and considered from multiple angles.  Part of the reason the graph is provided is precisely to help you do that.  In this case, the graph makes it crystal clear that the result is twice the smallest period.

This is a bug and you can feel free to report it as such using the "Give us your feedback" window at the bottom of the page.

Ultimately, the problem is the following Mathematica computation:

```<code>Periodic`PeriodicFunctionPeriod[Abs[Sin[x]] + Abs[Cos[x]], x]
(* Out: Pi *)
</code>```

Body:
The results coming from any software should be checked and considered from multiple angles.  Part of the reason the graph is provided is precisely to help you do that.  In this case, the graph makes it crystal clear that the result is twice the smallest period.

This is a bug and you can feel free to report it as such using the "Give us your feedback" window at the bottom of the page.

Ultimately, the problem is the following Mathematica computation:

Periodic`PeriodicFunctionPeriod[Abs[Sin[x]] + Abs[Cos[x]], x]
(* Out: Pi *)


答: $x^x=y^y$ is obviously equivalent to $(1/y)^{1/x}=(1/x)^{1/y}$, and that's a very popular problem at MSE (e.g. $x^y = y^x$ for integers $x$ and $y$ containing also the general parametric solution).

Body:
What are all real solutions of equation $x^x = y^y$? What about integer solutions and rational solutions?

For the case when $x=y$ it is clear, what about the case when $x\ne y$?


问:
If $\operatorname{rank}(A) = \operatorname{rank}(A^2)$, thn $\operatorname{rank}(A) = \operatorname{rank}(A^n)$, if $A$ has $2$ rows and $2$ columns?
How to prove that if $\operatorname{rank}(A) = \operatorname{rank} (A^2)$, then $\operatorname{rank}(A) = \operatorname{rank}(A^n)$, if $A$ has $2$ rows and $2$ columns?

答:
There are only three cases, $rank(A)=0,1,2$. Note that for $2\times 2$ matrices $A$ the following holds true;

$rank(A)=0$ if and only if $A=0$

$rank(A)=1$ if and only if $A=fg^\top$ for nonzero column vectors $f,g$

$rank(A)=2$ if and only if $\det A\not=0$.

Thus if $rank(A)=rank(A^2)=0$ then $A=0$, and clearly so is $A^n$, hence $rank(A^n)=0$.

If $rank(A)=rank(A^2)=2$ then $\det A\not = 0$, hence $\det(A^n)=(\det A)^n\not =0$ and also $rank(A^n)=2$.

Finally, if $rank(A)=rank(A^2)=1$, then $A=fg^\top$, as mentioned above. We note that $g^\top f\not=0$; otherwise, $A^2=0$, which contradicts the hypothesis. Thus $A^n=(fg^\top)^n=(g^\top f)^{n-1}f g^\top$ which is also rank $1$ for all $n\geq 1$. Hence $rank(A^n)=1=rank(A)$ for all $n\geq 1$.


问:
Absolute convergence of $\sum_{n=0}^{\infty}a_nb_n$ if $\sum_{n=0}^{\infty}b_n^2$ and $\sum_{n=0}^{\infty}a_n^2$ converge
Let $(b_n)$ and $(a_n)$ sequences of real numbers. If $\sum_{n=0}^{\infty}b_n^2$ and $\sum_{n=0}^{\infty}a_n^2$ converge, prove that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely.

I would like to have a feedback on my proof and know if everything is correct, please.

Proof.

As $\sum_{n=0}^{\infty}b_n^2$ converges, $\lim_{n\to\infty}b_n^2=0$. Therefore, $b_n^2$ is bounded and we have that: $\exists M>0$: $|b_n|<\sqrt{M}\ \forall n\in\mathbf{N}$ .

By analogous reasoning we obtain that $\forall \epsilon>0\ \exists N \ \forall n\ge N'$: $|a_n|<\frac{\epsilon}{\sqrt{M}(n-N)}$.

Let's show that $\sum_{n=0}^{\infty}|a_nb_n|$ is Cauchy. By definition we have to show the following: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N'$:

$|\sum_{k=N'}^{n}|a_kb_k||<\epsilon$

But,

$|\sum_{k=N'}^{n}|a_kb_k||=\sum_{k=N'}^{n}|a_kb_k|=\sum_{k=N''}^{n}|a_k||b_k|\le \sqrt{M}\sum_{k=N'}^{n}|a_n|\le \sqrt{M}\cdot \frac{\epsilon}{\sqrt{M}(n-N)}\cdot (n-N')$

If we set $N'=N$, we obtain that

$|\sum_{k=N'}^{n}|a_kb_k||\le \epsilon$. We conclude then that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely.

答:
Your proof cannot be correct, as in the end you only used that (i) $(b_n)_n$ was bounded, and (ii) $\lim_{n\to \infty} a_n =0$. But those those assumptions alone cannot imply convergence (let alone absolute convergence) of $\sum_n a_n b_n$, as otherwise we would get convergence of $\sum_n \frac{1}{n}\cdot 1$.

Now, why is it false? Well, you write

By analogous reasoning we obtain that $\forall \epsilon>0\ \exists N \ \forall n\ge N$: $|a_n|<\frac{\epsilon}{\sqrt{M}(n-N)}$.

This step, for instance, is not true. It may holds for fixed $N,M$ (with a fixed $n$ on the RHS), but how do you prove it holds for all $n\geq N$? Here is a counterexample:
$$
a_n = \frac{1}{n^{2/3}}
$$
Note that $\sum_n a_n^2 < \infty$, but the statement you wrote does not hold.


答:
I suggest one possible solution is
$$ \begin{matrix}
      1 &  2 &  4 &  7 & \cdots \\
      3 &  5 &  8 & 12 & \cdots \\
      6 &  9 & 13 & 18 & \cdots \\
     10 & 14 & 19 & 26 & \cdots \\
    \vdots & \vdots &\vdots & \vdots  & \ddots
\end{matrix}$$
This infinite two dimensional array is given by two
integer sequences. Namely,
OEIS sequence A002260
and OEIS sequence A004736.
Each positive integer $\,n\,$ appears exactly once in row
number $\,t_1(n)\,$ and column number $\,t_2(n).\,$
The formulas for these functions are
$$ t_1(n) := n - C(\lfloor 1/2+\sqrt{2n}\rfloor,2) $$
and
$$ t_2(n) := C(\lfloor 3/2+\sqrt{2n}\rfloor,2) $$
where $\, C(n,2) := n(n-1)/2.\,$
Each row  represents a countably
infinite set of  positive integers and there
are a countably infinite set of rows. The positive integer
in row $\,n\,$ and column $\,k\,$ is
$$ T(n,k) := C(n+k-1,2)+n. $$ It needs to be proven
that $\, T(t_1(n),t_2(n)) = n\,$ for all positive integer
$\,n\,$ and then this implies that the rows are disjoint
and hence their union is all positive integers.

Title: Prove that the set of positive integers can be expressed as a countably infinite union of disjoint countably infinite sets.

Body:
Consider the set $\mathbb Z^+$ of positive integers. Prove that $\mathbb Z^+$ can be expressed as a countably infinite union of disjoint countably infinite sets.

I know that countably infinite sets means it has the same cardinality as $\mathbb N$ (the set of natural numbers).

However, I am not sure how to go about proving this.


问:
Find portion of Earth between line of longitude.
Find the portion of the Earth between 10 degrees west and 40 degrees west.

I am not sure how to approach this problem. I found the radii of the circles created by the longitudinal lines, but I can not find the surface area of the spherical cap.

答:
hint

All the Earth corresponds to $ 360 ^° $, and your portion to $ 40-10=30^°=\frac{360}{12}$


问:
Why do we believe the Church-Turing Thesis?
The Church-Turing Thesis, which says that the Turing Machine model is at least as powerful as any computer that can be built in practice, seems to be pretty unquestioningly accepted in my exposure to computer science.  Why?  Do we have any more evidence for its truth than "nobody has been able to think of a more powerful model than Turing's yet?"  If so, why do we accept that argument as truth?  It seems to me that in any other mathematical discipline, a claim like the Church-Turing thesis whose support amounts to "we haven't yet found a counterexample" would not be taken seriously by the community.</p>

答:
The Church-Turing thesis asserts that the informal notion of a function that can be calculated by an (effective) algorithm is precisely the same as the formal notion of a recursive function. Since the prior notion is informal, one cannot give a formal proof of this equivalence. But one can present informal arguments supporting the thesis. For example, every known attempt at formally modeling this informal notion of  computability has led to precisely the same class of recursive functions, whether it be via lambda-calculus,
Post systems, Markov algorithms, combinatory logic, etc. This remarkable confluence lends strong support for the importance of this class of functions.

It is worth emphasizing that the Turing Machine was devised by Turing not as
a model of any type of physically realizable computer but, rather, as an
ideal limit to what is computable by a human calculating in a step-by-step
mechanical manner (i.e. without any use of intuition). This point
is widely misunderstood -- see Sieg [1] for an excellent exposition
on this and related topics.

The finiteness limitations postulated by Turing for his Turing Machines
are based on postulated limitations of the human sensory apparatus.
A Turing style analysis of physically realizable computing devices and
analogous Church-Turing theses did not come until much later (1980)
due to Robin Gandy -- with limitations based on the laws of physics.
As Odifreddi says on p. 51 of [2] (bible of Classical Recursion Theory)

Turing machines are theoretical devices, but have been designed with
    an eye to physical limitations. In particular, we have incorporated
    in our model restrictions coming from: (a) ATOMISM, by ensuring that
    the amount of information that can be coded in any configuration of
    the machine (as a finite system) is bounded; and (b) RELATIVITY, by
    excluding actions at a distance, and making causal effect propagate
    through local interactions. Gandy [1980] has shown that the notion
    of Turing machine is sufficiently general to subsume, in a precise
    sense, any computing device satisfying similar limitations.

and on p. 107: (A general theory of discrete, deterministic devices)

The analysis (Church [1957], Kolmogorov and Uspenskii [1958],
    Gandy [1980]) starts from the assumptions of atomism and
    relativity. The former reduces the structure of matter to a finite
    set of basic particles of bounded dimensions, and thus justifies the
    theoretical possibility of dismantling a machine down to a set of
    basic constituents. The latter imposes an upper bound (the speed of
    light) on the propagation speed of causal changes, and thus
    justifies the theoretical possibility of reducing the causal effect
    produced in an instant $t$ on a bounded region of space $V$, to actions
    produced by the regions whose points are within distance $ct$ from
    some point $V$. Of course, the assumptions do not take into account
    systems which are continuous, or which allow unbounded action-at-a-distance (like Newtonian gravitational systems).

Gandy's analysis shows that the THE BEHAVIOR IS RECURSIVE, FOR ANY
    DEVICE WITH A FIXED BOUND ON THE COMPLEXITY OF ITS POSSIBLE
    CONFIGURATIONS (in the sense that both the levels of conceptual
    build-up from constituents, and the number of constituents in any
    structured part of any configuration, are bounded), AND FIXED
    FINITE, DETERMINISTIC SETS OF INSTRUCTIONS FOR LOCAL AND GLOBAL
    ACTION (the former telling how to determine the effect of an action
    on structured parts, the latter how to assemble the local
    effects). Moreover, the analysis is optimal in the sense that, when
    made precise, any relaxing of conditions becomes compatible with any
    behavior, and it thus provides a sufficient and necessary
    description of recursive behavior.

Be forewarned that Gandy's 1980 paper [3] is regarded as difficult
even by some cognoscenti. You may find it helpful to first peruse
the papers in [4] by J. Shepherdson, and A. Makowsky.

[1] Sieg, Wilfried.  Mechanical procedures and mathematical experience.

pp. 71--117 in Mathematics and mind.  Papers from the Conference on the
Philosophy of Mathematics held at Amherst College, Amherst, Massachusetts,
April 5-7, 1991. Edited by Alexander George.
Logic Comput. Philos., Oxford Univ. Press, New York, 1994. ISBN: 0-19-507929-9
MR 96m:00006 (Reviewer: Stewart Shapiro) 00A30 (01A60 03A05 03D20)

[2] Odifreddi, Piergiorgio.  Classical recursion theory.

The theory of functions and sets of natural numbers. With a foreword
by G. E. Sacks. Studies in Logic and the Foundations of Mathematics, 125.
North-Holland Publishing Co., Amsterdam-New York, 1989. xviii+668 pp.
ISBN: 0-444-87295-7  MR 90d:03072 (Reviewer: Rodney G. Downey)
03Dxx (03-02 03E15 03E45 03F30 68Q05)

[3] Gandy, Robin.  Church's thesis and principles for mechanisms.

The Kleene Symposium. Proceedings of the Symposium held at the
University of Wisconsin, Madison, Wis., June 18--24, 1978.
Edited by Jon Barwise, H. Jerome Keisler and Kenneth Kunen.
Studies in Logic and the Foundations of Mathematics, 101.
North-Holland Publishing Co., Amsterdam-New York, 1980. xx+425 pp.
ISBN: 0-444-85345-6  MR 82h:03036 (Reviewer: Douglas Cenzer) 03D10 (03A05)

[4] The universal Turing machine: a half-century survey. Second edition.

Edited by Rolf Herken. Computerkultur [Computer Culture], II.
Springer-Verlag, Vienna, 1995. xvi+611 pp. ISBN: 3-211-82637-8
MR 96j:03005 03-06 (01A60 03D10 03D15 68-06)


问:
Existence of a finite set of sentences $\Delta$

Let $\Gamma_1, \Gamma_2$ be sets of sentences s.t. for any structure $M$ it holds that:

$M\vDash \Gamma_1 \iff M\not\vDash \Gamma_2 $

Show that exists a finite set of sentences $\Delta$ s.t. for any sentence $A$, it holds that $\Gamma_1 \vDash A \iff \Delta \vDash A$.

I got lost with all the iff's. I tried assuming towards contradiction but I could only contradict one case. I can't see how the information about $\Gamma_2$ is helpful, and I certainly can't come up with such $\Delta$ that have that property (and any $\Delta$ which is a subset of either $\Gamma_1$,$\Gamma_2$, or $\Gamma_1 \cup \Gamma_2$ seems not suitable). Would appreciate any hint.

答:
In fact the desired $\Delta$ is indeed a subtheory of $\Gamma_1$ - which is reasonable, since being a subtheory of $\Gamma_1$ gives us one of our three "goal conditions" immediately, namely being entailed by $\Gamma_1$ itself.

I've broken my answer into a series of exercises:

As you say, the main difficulty with this problem comes from its phrasing, and the "nested implications" it contains. It helps to break things down as we go along. Let's start with the hypothesis in question, namely $$\forall M(M\models\Gamma_1\iff M\models\Gamma_2).$$

This has two pieces which are much easier to think about on their own:

No structure can satisfy both $\Gamma_1$ and $\Gamma_2$ - that is, $\Gamma_1\cup\Gamma_2$ is unsatisfiable.

On the other hand, every structure does have to satisfy one or the other.

Besides being individually simpler than the whole hypothesis itself, these sub-hypotheses can be considered one at a time, which helps us see how our argument should flow. Specifically, while the latter is a bit subtle, the first is very nice in that it suggests an immediate first step - namely, whenever we have an unsatisfiable theory (in this case $\Gamma_1\cup\Gamma_2$), we can get a subtheory $\Theta$ which is itself unsatisfiable but is "simpler" in a particular sense:

 The compactness theorem then says that there is some finite $\Theta\subseteq\Gamma_1\cup\Gamma_2$ which is itself unsatisfiable.

OK, so we have a "simple" (in the above sense) theory $\Theta\subseteq\Gamma_1\cup\Gamma_2$ which is unsatisfiable. This $\Theta$ has a "$\Gamma_1$-part" and a "$\Gamma_2$-part," namely $\Theta\cap \Gamma_1$ and $\Theta\cap \Gamma_2$ respectively. Now right off the bat we see from this decomposition a theory which is pretty close to what we want, and so which we'll choose as our $\Delta$:

 The theory $\Theta\cap\Gamma_1$ is finite, and we have $\Gamma_1\models\Theta\cap\Gamma_1$. So we'll set $\Delta=\Theta\cap\Gamma_1$.

Per the above, we only need to show that this $\Delta$ entails $\Gamma_1$ itself. The second bulletpoint gives an equivalent rephrasing of this, and consequently a different way to phrase our goal:

 Suppose no model of $\Delta$ satisfies $\Gamma_2$. Then every model of $\Delta$ must satisfy $\Gamma_1$, since every structure - and so a fortiori every model of $\Delta$ - satisfies one of $\Gamma_1$ and $\Gamma_2$. So all we need to prove is that $\Delta\cup\Gamma_2$ is unsatisfiable.

But this follows immediately from the definition of $\Delta$:

 We have $\Delta\cup\Gamma_2\supseteq\Theta$.


问:
Restriction of a covering map on a closed set having finite fibers is a proper map?

Let $p\colon X\to Y$ be a covering map between two connected
topological manifolds. Let $C$ be a closed and path-connected
subset of $X$ such that each fiber of the restriction map $p\big |C\to
 Y$ is finite. Is $p\big|C\to Y$ a proper map?

I am trying to show $p$ is a closed map so that I can use this: A continuous closed map defined on a Hausdorff space having compact fibers is a proper map.

Also, note that a finite-fold covering map is a closed map.

答: Hint: This is false even for the universal covering $p$ of the once-punctured plane and $C$ such that $p|_C$ is 1-1.


问:
Equation of a circle passing through the intersection of 'two circle' or 'one circle and one line'.
In our mathmatics textbook , there is a question of finding out the Equation of a circle passing through the intersection of 'two circle' or 'one circle and one line'. They do it the following way.Lets say:

Equation of a circle            : $f(x,y) = 0$

Equation of another circle/line : $g(x,y) = 0$

Then the equation of a circle passing through the intersection of $f$ and $g$ is:

$f(x,y) + k.g(x,y) = 0 $ (Here k is any constant)

But I didn't really understood this last part because it didn't really have any explanation or proof along with it.What is the proof of this theorem?

In an another Analytic Geometry paper.I Saw this quote with no proof:

"If $S = 0$ and $S'= 0$ are equations of two curves, then $S + KS'= 0$ is the general equation of all curves, passing through all the points common to the two curves"

The above quote is clearly a theorem.But what does it intuitively mean , what is the proof of the theorem and Does it have anything to do with the original problem I had with circle? (It is okay if calculus is involved)

答:
The two results you quote do indeed involve the same idea and they certainly need to be rigorously proved. As an example, let's look at the intersection of a circle and a line. This will illustrate the general idea in a simple context.

The proof is elementary but needs a few prerequisites:

(1) The general equation of a circle is $ax^2+ay^2+bx+cy+d=0, a\ne 0$.

(2) The general equation of a line is $lx+my+n=0$.

(3) There is precisely one circle through any given three non-collinear points.

Theorem

Let $f(x,y)=0$ be the equation of a circle $\mathcal{C}$ and let $g(x,y)=0$ be the equation of a line $\mathcal{L}$ which intersects $\mathcal{C}$ at two points, $P$ and $Q$.

(a) For a constant $k$, $f(x,y)+kg(x,y)=0$ is the equation of a circle through $P$ and $Q$.

(b) Every circle through $P$ and $Q$ has  equation $f(x,y)+kg(x,y)=0$  for some constant $k$.

Proof

(a) From prerequisites (1) and (2), $f(x,y)+kg(x,y)=0$ is the equation of a circle, $\mathcal{D}$ say.

Now let $(a,b) $ be the coordinates of either $P$ or $Q$, then $f(a,b)=g(a,b)=0$.  Therefore $f(a,b)+kg(a,b)=0$ and so $(a,b)$ is a point on $\mathcal{D}$. Thus (a) is proved.

(b) Let $\mathcal{E}$ be any circle through $P$ and $Q$. Let $R(c,d)$ be a third point on $\mathcal{E}$ and then $R$ is automatically not on the line $\mathcal{L}$.

Since $R$ is not on $\mathcal{L}$, $g(c,d)\ne 0$. We can therefore find $k$ such that  $f(c,d)+kg(c,d)=0$. Then $f(x,y)+kg(x,y)=0$ represents a circle through $P,Q$ and $R$ and, by prerequisite (3), it is the circle  $\mathcal{E}$, as required.


问:
Negation of "For all x ∈ R, if ⌊5x⌋ = 5⌊x⌋, then x ∈ Z"
Negation of "For all x ∈ R, if ⌊5x⌋ = 5⌊x⌋, then x ∈ Z"

Is there only one correct negation?

I know "There exists x ∈ R, so that ⌊5x⌋ = 5⌊x⌋, but x ∉ Z" is correct.

But what about: "For some x ∈ R, ⌊5x⌋ = 5⌊x⌋ and x ∉ Z" ?

答:
What you wrote is equivalent. Here is another equivalency, but a shorter one:
$$
\exists x \in \mathbb{R} \backslash \mathbb{Z} \text{ such that }
\lfloor 5x \rfloor = 5\lfloor x\rfloor
$$


答:
Formally negation for $$(\forall x \in R)(⌊5x⌋ = 5⌊x⌋ \Rightarrow x \in Z)$$ is $$(\exists x \in R)(⌊5x⌋ = 5⌊x⌋ \land x \notin Z )$$
You can use several variants how to say it, but while you mean what is written, then they are same.

Body:
Negation of "For all x ∈ R, if ⌊5x⌋ = 5⌊x⌋, then x ∈ Z"

Is there only one correct negation?

I know "There exists x ∈ R, so that ⌊5x⌋ = 5⌊x⌋, but x ∉ Z" is correct.

But what about: "For some x ∈ R, ⌊5x⌋ = 5⌊x⌋ and x ∉ Z" ?


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

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

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

Thank you very much guys

答:
Let $I(x) = \int_1^x \cos(\frac1t) \ln(3 - \frac1t) dt$. As Ben suggested, we will use the intermediate value theorem.

First, $I(1) = 0$. Thus, $I(1) - 1 < 0$.

Then, for $x > 1$ :

$$I(x) = \ln(3) \int_1^x \cos(1/t) dt + \int_1^x \cos(1/t) \ln(1 - 1/3t) dt$$

Both integrals diverge, and the integrand are positive, thus :

$$I(x) = \ln(3) \int_1^x 1 \times dt + o(x) - \int_1^x 1 \times \frac1{3t} dt + o(\ln(x)) = \ln(3) x + o(x) \quad [x \to \infty]$$

Finally :

$$I(x) - x = (\ln(3)-1) x + o(x) \geq 0.005 x + o(x) \to \infty \quad [x \to \infty]$$

Thus, there exist $x_0 > 1$ such that $I(x_0) - x_0 > 0$. Moreover $x \mapsto I(x)$ is continuous (antiderivative of a continous function). The intermediate value theorem gives the existence of $x$ such that $I(x) = x$.


问:
Fitting a set of points in a plane to a smooth curve obtained by joining a half-line and an arc of a circle
I have a set of points in the plane and I want to find a curve that best fits these points (e.g., in a least squares manner, or using some other convenient "measure"). I want that the curve be either:

a line

a circle

a circle whose center lies on a given line

a curve that looks like a hook, that is obtained from:

(a) an arc of a circle whose center lies on a given line

(b) a half-line tangent to one of the endpoints of the arc circle (in this way I can get a curve that is smooth at the point where the arc of the circle and the half-line meet).

I managed to get an answer for all but the last problem. Any suggestions are welcomed.

答:
Here is a numerical solution. The parameters are taken as $A$,$B$,$C$,$M$,$K$,$c$. The equation of the arc is given by
$$(y-C)^2+(x-B)^2=A^2$$
and for the line
$$y=M\,x+K$$
The break-point is taken as $c$. Therefore the objective function to be minimized becomes
$$F=\sum_1^n\bigg((y_i-F_{1,i})^2+(y_i-F_{2,i})^2\bigg)$$
$$F_{1,i}=-\sqrt{A^2-(x_i-B)^2}+C\qquad \text{if }x_i\le c\text{ and }0\text{ elsewhere}$$
$$F_{2,i}=M\,x_i+K\qquad \text{if }x_i>c\text{ and }0\text{ elsewhere}$$
The first constraint function assures that the arc and line meet at the break-point $c$
$$-\sqrt{A^2-(c-B)^2}+C-M\,c-K=0$$
The second constraint that the line is tangent to the arc at $c$
$$\frac{c-B}{\sqrt{A^2-(x-B)^2}}-M=0$$
And the last constraint is that the center of arc lies on a line given by $y=m_1x+k$
$$C-m_1\,B+k_1=0$$
To test the algorithm I generated 120 points and added error terms of $N(0,0.1)$ as given in below graph. The black dots are for circle with $A=3$, $B=1$, $C=4$; and the blue dots for line with $M=0.8944$, $K=-0.9193$. The break-point is given as $c=3$

In Matlab I used "fmincon" with "interior-point" algorithm. The initial guess is randomly generated as $[A=0.9045;B=0.8663;C=0.9225;M=0.5637;K=0.1863;c=0.6766]$. After 170 iterations the algorithm stopped convergent with objective function value of $1.16322$ and maximum constraint violation of $1.66\cdot 10^{-16}$. The parameters are found to be
$$A=2.97\qquad B=0.99\qquad C=3.96\qquad M=0.88\qquad K=-0.88\qquad c=2.95$$
As you can see the calculated parameters are very close to the original parameters even though random error is added to the data. The final grap shows the data (red) and the calculated fitting function (blue)

-------------------ADDITION OF CODE------------------------

Please note that $A\rightarrow x(1)$, $B\rightarrow x(2)$, $C\rightarrow x(3)$, $M\rightarrow x(4)$, $K\rightarrow x(5)$, $c\rightarrow x(6)$

First the function file for objective function (save it as a seperate file as objfun.m)

```<code>function f = objfun( x )
global n
global Y_er X
sum=0;
for i=1:2*n
    if X(i)<=x(6)
        f1=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
        sum=sum+(Y_er(i)-f1)^2;
    else
        f2=x(4)*X(i)+x(5);
        sum=sum+(Y_er(i)-f2)^2;
    end
end
f=sum;
end
</code>```

As the second constraint equations (save it as a seperate file as constfun.m)

```<code>function [ c ceq ] = constfun( x )
c=[];
f1=-sqrt(x(1)^2-(x(6)-x(2))^2)+x(3);
f2=x(4)*x(6)+x(5);
g1=x(6)-x(2);
g2=x(4)*(sqrt(x(1)^2-(x(6)-x(2))^2));
ceq(1)=f1-f2;
ceq(2)=g1-g2;
ceq(3)=x(3)-3*x(2)-1;
end
</code>```

And the final one is main file (save it as m.file with any name in the same folder with objfun.m and constfun.m)

```<code>clc
clear
global Y_er X n
global beta
%------------------- GENERATION OF SAMPLE POINTS -------------------------
% Generate 60 uniformly distributed random points between 0 and 3
n=60;
a=0;
b=3;
X1=a+(b-a)*rand(n,1);
% Generate y=-sqrt(A-(x-B)^2) + C function for random data
A=3;
B=1;
C=4;
for i=1:n
    Y1(i,1)=-sqrt(A^2-(X1(i)-B)^2)+C;
end
plot(X1,Y1,'k.');

% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y1_er=Y1+Er;
hold on;
plot(X1,Y1_er,'ro');

% Now generate 60 more samples between 3 and 6
clear a b
a=3;
b=6;
X2=a+(b-a)*rand(n,1);

% Generate y=m*x+k function for random data
m=(-B+a)/sqrt(A^2-(a-B)^2);
k=-sqrt(-(a-A-B)*(a+A-B))-a*m+C;
for i=1:n
    Y2(i,1)=m*X2(i)+k;
end
plot(X2,Y2,'b.');

% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y2_er=Y2+Er;
hold on;
plot(X2,Y2_er,'ro');

%Construct the final data set
X=[X1;X2];
Y_er=[Y1_er;Y2_er];
figure(2)
plot(X,Y_er,'b.');

%--------------------------------OPTIMIZATION---------------------------
x0=rand(6,1);
options = optimset('Display','iter-detailed','Algorithm','interior-point','TolFun',1e-12,'TolX',1e-12);
x = fmincon(@objfun,x0,[],[],[],[],[],[],@constfun,options);
figure(3)
hold on
for i=1:2*n
    if X(i)<=x(6)
        r=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
    else
        r=x(4)*X(i)+x(5);
    end
    plot(X(i),r,'b.')
end
</code>```

PS: If you have your own data set put x-data in variable X and y-data in Y_er. In addition put n= your number of data points. Hope it works for you.

PS2: Please modify your constraint "center of arc on a given line" in "constfun.m" in ceq(3).

Body:
Here is a numerical solution. The parameters are taken as $A$,$B$,$C$,$M$,$K$,$c$. The equation of the arc is given by
$$(y-C)^2+(x-B)^2=A^2$$
and for the line
$$y=M\,x+K$$
The break-point is taken as $c$. Therefore the objective function to be minimized becomes
$$F=\sum_1^n\bigg((y_i-F_{1,i})^2+(y_i-F_{2,i})^2\bigg)$$
$$F_{1,i}=-\sqrt{A^2-(x_i-B)^2}+C\qquad \text{if }x_i\le c\text{ and }0\text{ elsewhere}$$
$$F_{2,i}=M\,x_i+K\qquad \text{if }x_i>c\text{ and }0\text{ elsewhere}$$
The first constraint function assures that the arc and line meet at the break-point $c$
$$-\sqrt{A^2-(c-B)^2}+C-M\,c-K=0$$
The second constraint that the line is tangent to the arc at $c$
$$\frac{c-B}{\sqrt{A^2-(x-B)^2}}-M=0$$
And the last constraint is that the center of arc lies on a line given by $y=m_1x+k$
$$C-m_1\,B+k_1=0$$
To test the algorithm I generated 120 points and added error terms of $N(0,0.1)$ as given in below graph. The black dots are for circle with $A=3$, $B=1$, $C=4$; and the blue dots for line with $M=0.8944$, $K=-0.9193$. The break-point is given as $c=3$

In Matlab I used "fmincon" with "interior-point" algorithm. The initial guess is randomly generated as $[A=0.9045;B=0.8663;C=0.9225;M=0.5637;K=0.1863;c=0.6766]$. After 170 iterations the algorithm stopped convergent with objective function value of $1.16322$ and maximum constraint violation of $1.66\cdot 10^{-16}$. The parameters are found to be
$$A=2.97\qquad B=0.99\qquad C=3.96\qquad M=0.88\qquad K=-0.88\qquad c=2.95$$
As you can see the calculated parameters are very close to the original parameters even though random error is added to the data. The final grap shows the data (red) and the calculated fitting function (blue)

-------------------ADDITION OF CODE------------------------

Please note that $A\rightarrow x(1)$, $B\rightarrow x(2)$, $C\rightarrow x(3)$, $M\rightarrow x(4)$, $K\rightarrow x(5)$, $c\rightarrow x(6)$

First the function file for objective function (save it as a seperate file as objfun.m)

function f = objfun( x )
global n
global Y_er X
sum=0;
for i=1:2*n
    if X(i)<=x(6)
        f1=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
        sum=sum+(Y_er(i)-f1)^2;
    else
        f2=x(4)*X(i)+x(5);
        sum=sum+(Y_er(i)-f2)^2;
    end
end
f=sum;
end

As the second constraint equations (save it as a seperate file as constfun.m)

function [ c ceq ] = constfun( x )
c=[];
f1=-sqrt(x(1)^2-(x(6)-x(2))^2)+x(3);
f2=x(4)*x(6)+x(5);
g1=x(6)-x(2);
g2=x(4)*(sqrt(x(1)^2-(x(6)-x(2))^2));
ceq(1)=f1-f2;
ceq(2)=g1-g2;
ceq(3)=x(3)-3*x(2)-1;
end

And the final one is main file (save it as m.file with any name in the same folder with objfun.m and constfun.m)

clc
clear
global Y_er X n
global beta
%------------------- GENERATION OF SAMPLE POINTS -------------------------
% Generate 60 uniformly distributed random points between 0 and 3
n=60;
a=0;
b=3;
X1=a+(b-a)*rand(n,1);
% Generate y=-sqrt(A-(x-B)^2) + C function for random data
A=3;
B=1;
C=4;
for i=1:n
    Y1(i,1)=-sqrt(A^2-(X1(i)-B)^2)+C;
end
plot(X1,Y1,'k.');

% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y1_er=Y1+Er;
hold on;
plot(X1,Y1_er,'ro');

% Now generate 60 more samples between 3 and 6
clear a b
a=3;
b=6;
X2=a+(b-a)*rand(n,1);

% Generate y=m*x+k function for random data
m=(-B+a)/sqrt(A^2-(a-B)^2);
k=-sqrt(-(a-A-B)*(a+A-B))-a*m+C;
for i=1:n
    Y2(i,1)=m*X2(i)+k;
end
plot(X2,Y2,'b.');

% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y2_er=Y2+Er;
hold on;
plot(X2,Y2_er,'ro');

%Construct the final data set
X=[X1;X2];
Y_er=[Y1_er;Y2_er];
figure(2)
plot(X,Y_er,'b.');

%--------------------------------OPTIMIZATION---------------------------
x0=rand(6,1);
options = optimset('Display','iter-detailed','Algorithm','interior-point','TolFun',1e-12,'TolX',1e-12);
x = fmincon(@objfun,x0,[],[],[],[],[],[],@constfun,options);
figure(3)
hold on
for i=1:2*n
    if X(i)<=x(6)
        r=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
    else
        r=x(4)*X(i)+x(5);
    end
    plot(X(i),r,'b.')
end

PS: If you have your own data set put x-data in variable X and y-data in Y_er. In addition put n= your number of data points. Hope it works for you.

PS2: Please modify your constraint "center of arc on a given line" in "constfun.m" in ceq(3).