问:
Why is present value of repayment of principal in an amortization schedule constant?
It is a common observation that if a loan is paid off in equal installments, and if the number of installments is large, then the early installments consist mostly of payment of interest, while, as the final installment nears, the repayment of principal increases dramatically.

I'm looking for a conceptual explanation of the observation that the present value of the repayment of principal in an amortization schedule is constant.

Details: A loan in the amount $P$ is to be repaid in $n$ monthly installments, starting at the end of the first month.  The monthly interest rate is $r$.  The monthly repayment $A$ can be computed as follows:
$$
\begin{aligned}
\text{present value of loan}&=\text{present value of all future payments}\\
P&=\sum_{i=1}^nA(1+r)^{-i}\\
&=A\frac{(1+r)^{-n-1}-(1+r)^{-1}}{(1+r)^{-1}-1}\\
&=\frac{A}{r}[1-(1+r)^{-n}],
\end{aligned}
$$
with the result that
$$
A=\frac{Pr(1+r)^n}{(1+r)^n-1}.
$$
Let $B_i$ be the balance on the loan at the end of $i$ months.  So $B_0=P$, $B_1=(1+r)P-A$, and in general
$$
B_{i+1}=(1+r)B_i-A.
$$
The repayment of principal on the $i^\text{th}$ payment, $R_i$, is the reduction in the balance resulting from that payment, $R_i=B_{i-1}-B_i$.
One can show by induction that
$$
\begin{aligned}
R_i&=(1+r)^{i-1}(A-rP)\\
&=(1+r)^{i-1}rP\frac{1}{(1+r)^n-1},
\end{aligned}
$$
which increases geometrically with growth factor $1+r$.  It follows immediately that the present value of the repayment of principle is constant:
$$
\begin{aligned}
\text{PV}(R_i)&=(1+r)^{-i}(1+r)^{i-1}rP\frac{1}{(1+r)^n-1}\\
&=\frac{rP}{1+r}\cdot\frac{1}{(1+r)^n-1}.
\end{aligned}
$$
I can't escape the feeling that, if only I had the right conceptual framework, I would have been able to predict this (not the precise value, but the fact that it is constant) without doing the calculation above.  Is there a way of looking at things that makes this clear?

答:
One month before the final payment the amount outstanding must be $\frac{A}{1+r}$ since then this amount plus interest is $A$ and is paid off completely by the final payment.

Another month earlier, the amount outstanding must be $\frac{A}{1+r}+\frac{A}{(1+r)^2}$ since then this amount plus interest is $A+\frac{A}{1+r}$ and is reduced to $\frac{A}{1+r}$ by the next payment.

In general, the amount outstanding with $k$ months to go is $\frac{A}{1+r}+\frac{A}{(1+r)^2}+...+\frac{A}{(1+r)^k}$. The next repayment therefore reduces the outstanding amount by $\frac{A}{(1+r)^k}$ and this repayment of principal has a present value of  $\frac{A}{(1+r)^{n+1}},$ a constant.


问:
Prove that the set of words (from a countable infinite alphabet) is countable
Here is the question:

Define a word as a finite string of letters from a countable infinite alphabet $\{ a_{1}, a_{2}, a_{3},...\}$

a) Prove that the set of three letter words is countable.
b) Prove that the set of all finite-letter words is countable.

I'm having a little trouble with part a). My first instinct was to define a function $f: \mathbf{W} \to \mathbf{N}$ (where $\mathbf{W}$ is the set of three letter words) as $f(w) = ijk$ where $w \in \mathbf{W}, w = a_{i}, a_{j}, a_{k}$ for some $i,j,k \in \mathbf{N}$. But this wouldn't be a bijection, since natural numbers can have multiple sets of factors so more than one word would map to the same natural number.

I've seen a similar proof for words in the English language being countable, but the thing that's tripping me up here is the fact that the alphabet is infinite, so I don't think I could use the same method of proof as I would for a finite (English) alphabet. Am I on the right track here? Some guidance would be appreciated.

答:
Let us denote by $W_n$ the set of all $n$-letter words over your alphabet $\Sigma$ and further, let $W = \bigcup_{n \in \mathbb N} W_n$ be the set of all finite words over $\Sigma$.

(a) The idea is actually really close to one standard proof of this fact which goes like this: As $\Sigma$ is countable, there exists some bijection $f \colon \Sigma \to \mathbb N$.
Now consider the map $g \colon W_3 \to \mathbb N$ with $g(\alpha \beta \gamma) = 2^{f(\alpha)} \cdot 3^{f(\beta)} \cdot 5^{f(\gamma)}$ where $\alpha, \beta, \gamma \in \Sigma$.
Note that for any $w \in W$ we know that $g(w) = 2^i \cdot 3^j \cdot 5^k$ and by doing a prime factorization of $g(w)$ we can recover $i, j$ and $k$, allowing us to find $w = f^{-1}(i)f^{-1}(j)f^{-1}(k)$.
Hence $g$ is an injection and hence, $W_3$ must be countable.

(b) As we have infinitely many primes we can reuse the idea from the first part with the simple change of
$$h \colon W \to \mathbb N, g(\sigma_1...\sigma_m) \mapsto \prod_{i \in [1, m]} p_i^{f(\sigma_i)},$$
where $p_i$ denotes the $i$-th prime number.
By essentially the same argument we get that $h$ is an injection, showing that $W$ is countable (in fact, one can show that $h$ is a bijection).

ContentLicense: CC BY-SA 4.0


问:
$M\subset\Bbb R^n$ contains an open ball iff its Hausdorff measure is $n$
$\Rightarrow$ is proved here.

I suspect that if a subset $M\subset\Bbb R^n$ has Hausdorff dimension $n$, then it must contain an open ball. Is that true?

答:
No, it is not necessarily true that if $M$ has Hausdorff dimension $n,$ then $M$ contains an open ball. I will provide you with a construction of such a set. Consider for simplicity the case $n = 2$ (you can construct similar examples in higher dimensions) by the same technique. Consider the square $[-2, 2] \times [-2, 2].$ We will fill it with disjoint scaled Cantor dusts with slightly tweaked parameters such that its dimension is $2,$ but it doesn't contain any open balls by construction.

In $[0,1] \times [0,1]$, we construct a Cantor dust $D_1$ with parameter $f_1 = \frac13.$ Its dimension will be $\dim_H D_1 = \log_3 4.$ In $[1 + \frac12, 1 + \frac12 + \frac14] \times [1 + \frac12, 1 + \frac12 + \frac14],$ we construct a Cantor dust $D_2$ with parameter $f_2 := \frac{1}{2 + \frac12}$ and its dimension will be $\dim_H D_2 = - \log_{\frac{2}{5}} 4.$ Inductively, in $[\sum_{0 \leq l \leq 2k - 1} \frac{1}{2^l}, \sum_{1 \leq l \leq 2k} \frac{1}{2^l}] \times [\sum_{0 \leq l \leq 2k - 1} \frac{1}{2^l}, \sum_{1 \leq l \leq 2k} \frac{1}{2^l}]$ we construct a Cantor dust $D_k$ with parameter $f_k := \frac{1}{2 + \frac1k}$ and $\dim_H D_k = - \log_{f_k} 4.$ Denote $D := \bigcup_{1 \leq k} D_k.$ Then $\dim_H D = \sup_{k \geq 1} \dim_H D_k = 2,$ but it doesn't contain any open ball by construction. I hope this helps. :)


问:
Calculating this implicit differentiation for $\frac{dy}{dx}$
I'm trying to calculate this differential equation using implicit differentiation for $\frac{dy}{dx}$ in terms of x and y:

$2x^{3}y + 5x + 6y - sin(\pi y) = 8$

What I have tried:

By differentiating with respect to x

$6x^{2}y+\frac{dy}{dx}2x^{3}+5+\frac{dy}{dx}6-\frac{dy}{dx}\pi \cos(\pi y) = 0$

I'm not sure where to go from here.

答:
You have already obtained a relation between $x,y,y'$. Now, you need to "solve" with respect to $y'$, getting
$$
y' = \dfrac{-6x^2 y-5}{2x^3+6-\pi \cos(\pi y)}.
$$


问:
Can the Fubini theorem be applied to a trapezoid?
I have two questions:

How does one apply the Fubini theorem?

and

Can it be applied to a trapezoid?

Thank you.

答:
Fubini's theorem says a double integral (where the integral of the absolute value is finite) is equal to either of two iterated integrals.  One concrete case:
$$
\iint\limits_{[a,b]\times[c,d]} f(x,y)\,d(x,y) = \int_a^b\left(\int_c^d f(x,y) \, dy \right) \, dx = \int_c^d \left( \int_a^b f(x,y)\, dx \right) \, dy,
$$
provided that
$$
\iint\limits_{[a,b]\times[c,d]} |f(x,y)|\,d(x,y) <\infty.
$$

One way to use this theorem is to exploit the fact that one of the two iterated integrals may be readily evaluated.

As for trapezoids, here is an example:
$$T=\{(x,y) : 1\le y\le2,\  0\le x\le y \}\tag{1}$$
Suppose one wants
$$
\iint\limits_T e^{y^2} \,d(x,y).
$$
Applying $(1)$, this becomes
$$
\int_1^2 \left( \int_0^y e^{y^2} \, dx \right)\,dy.
$$
This is easily evaluated, whereas if one had integrated first with respect to $y$ and afterward with respect to $x$, one would face the intractable integral $\int e^{y^2}\, dy$.

Does Fubini's theorem justify this? The answer is "yes" because one can view the integral as
$$
\iint\limits_{[0,2]\times[1,2]} f(x,y) \, d(x,y)
$$
where
$$
f(x,y)=\begin{cases} e^{y^2} & \text{if }x\le y, \\[10pt] 0 & \text{otherwise}.  \end{cases}
$$

ContentLicense: CC BY-SA 3.0


问:
Show that a set $ F $ is uncountable
Let $$ \{a,b,c\}^n = \big\{(x_1,x_2,\dots,x_n): x_j \in \{a,b,c\} \big\}$$ and

$$ \{a,b,c\}^{*} = \bigcup_{n=1}^{\infty} \{a,b,c\}^n.$$

Consider the set $ F = \big\{ f: \{a,b,c\}^{*} \to \{a,b,c\} \big\}$ and show that $ F $ is not countable.

I've tried a couple of solutions:

i) No bijection between $ F \to \mathbb{N} $. Although I've tried to use Cantors diagonalizing argument I'm getting nowhere, since the arguments of $ f$ is such a mess.

ii) $ F $ is clearly an infinite set. Then it's sufficient to show that $ \mid F \mid \neq \mathbb{N}^0$ (aleph-null) or $ \mid F \mid > \mathbb{N}^0 $.

How do I solve this?

答: The set $A=\{a,b,c\}^{*}$ is countable. Thus $P(A)$, the power set of $A$, is uncountable. Now we see $P(A)\cong\big\lbrace f:A\to \lbrace 0,1\rbrace\big\rbrace\subset F$, so $F$, too, is uncountable.

Title: Show that a set $ F $ is uncountable


问:
$(0,1]$ is not homeomorphic to $(a,b)$ - Carothers Example 5.6(g)
I'm reading Carothers' Real Analysis, and I'm currently looking at homeomorphisms. The author says "two intervals that look different, are different" - i.e. they are not homeomorphic. The proof is done for the case $(0,1]$ and $(a,b)$, where the first interval is semi-open so we expect that the two are not homeomorphic.

I have some difficulty following the book's argument which I paraphrase here (for convenience) with inline questions.

Proof by contradiction. Suppose $(0,1]$ and $(a,b)$ are homeomorphic. Then by removing $1$ from $(0,1]$ we get $(0,1)$, and by removing its image $f(1) = c$ from $(a,b)$ we have that $(0,1)$ is homeomorphic to $(a,c)\cup (c,b)$.

Wait, why is that? I understand that we are deleting an element each from the domain and codomain, so the bijection remains a bijection. What about the continuity of $f$ and $f^{-1}$ though? We literally threw away an element, how do we know that continuity still holds?

But $(0,1)$ is homeomorphic to $\Bbb R$, so $\Bbb R$ would have to be homeomorphic to $(a,c)\cup (c,b)$ too. So $\Bbb R$ can be written as the disjoint union of two non-trivial open sets, which is impossible.

Two questions:

Why is $(0,1)$ homeomorphic to $\Bbb R$? We really need to produce a homeomorphism, that's all I'm asking.

How to prove that $\Bbb R$ cannot be written as the disjoint union of two non-trivial open sets? The idea that comes to my mind is - we already know that any open subset of $\Bbb R$ can be written as the disjoint union of unique open maximal intervals. Since $\Bbb R$ is open in $\Bbb R$, we can obviously do the same for it as well, and clearly, our choice of intervals would not be maximal in the event that we could write $\Bbb R$ as the union of two disjoint open intervals. Does this make sense?

Thank you!

答:
In $\Bbb R^2$ let $C$ be the circle centered at $(0,2)$ with radius $1.$ Let $D=\{(x,y)\in C: y<2\}$...Now $D$ is a semi-circle without its end-points, and $D$ is homeomorphic to $(0,1).$

For $p\in D$ let $l_p$ be the line thru $p$ and $(0,2)$ and let $l_p$ intersect the $x$-axis at $(f(p),0)$... (Draw a diagram of this.)

$f: D\to \Bbb R$ is a homeomorphism. A composition of homeomorphisms is a homeomorphism, so if $g: (0,1)\to D$ is a homeomorphism, then so is $(fg):(0,1)\to\Bbb R.$


问:
Number theory with sequences
The sequence $x_0,x_1,x_2,x_3,x_4,.....$ is defined as:

\begin{cases}
x_0 = 2 \\
x_k =2\cdot x_{k-1}^2-1 &\text{for every }k>0\\
\end{cases}
Prove that if an odd prime number $p$ divides $x_n$, then $2^{n+3}$ divides $p^2-1$

My idea was to firstly treat simple cases like this:

If $n=0$, then $x_0=2$ doesn't have any odd divisors.

If $n=1$, then $x_1=7$. It has only one odd prime divisor, $7$ and $2^{n+3}=2^4=16$ divides $p^2-1=7^2-1=48$

.........

From here, my idea was to apply some sort of induction but induction doesn't work too well with prime numbers so this is where I got stuck with this idea.

Another idea was to write $x_n$ in its polynomial form and making it easier to prove what we want. I think that induction might work here but I still didn't manage to do it. Can you help me, please?

答:
The general solution to the recursion is:

$$x_n = \frac{\alpha^{2^n} + \alpha^{-2^n}}{2}$$

Which can be shown by induction.

By the initial condition, we get $\alpha = 2 + \sqrt3$. Then:

$$x_n = \frac{(2 + \sqrt3)^{2^n} + (2 - \sqrt3)^{2^n}}{2}$$

Now things start to become weird. If you expand both terms in parenthesis, you can eliminate square roots because they are negative in the second expansion. Take a prime that divides this integer and use $r$ as an integer square root of 3 in the modulus p. I still can't prove it exists. But assuming it is true, we get:

$$\frac{(2 + r)^{2^n} + (2 - r)^{2^n}}{2} \equiv 0\pmod p$$

$$(2 + r)^{2^n} + (2 - r)^{2^n} \equiv 0\pmod p$$

$$(2 + r)^{2^n} \equiv - (2 - r)^{2^n} \pmod p$$

Multiplying both sides by $(2 + r)^{2^n}$ and then squaring:

$$(2 + r)^{2^{n + 1}} \equiv -1\pmod p$$

$$(2 + r)^{2^{n + 2}} \equiv 1\pmod p$$

We see an exponent for which we get residue 1 modulus p, and if it is the minimum one (the order), it must divide every other exponent which also gives 1 modulus p for the same base, in particular p - 1, which works for every base.

It is really the minimum one, because, if any other is the minimum, it needs to be in the form $2^k$, $k < n + 2$, because it must then divide $2^{n+2}$.

This is not possible because we would have $(2 + r)^{2^{n + 1}} \equiv 1\pmod p$ by squaring both sides of $(2 + r)^{2^k} \equiv 1\pmod p$ multiple times, and we know the truth is $(2 + r)^{2^{n + 1}} \equiv -1\pmod p$.

We now get:

$$2^{n + 2} | p - 1$$

Since p + 1 is also even, the product $p^2 - 1 = (p-1)(p+1)$ must be divisible by $2^{n+3}$.

Title: Number theory with sequences


问:
Let $A$ be an $n \times n$ matrix. If $A$ is invertible, then $|A| \neq 0$?
Let $A$ be an $n \times n$ matrix. If $A$ is invertible, then $|A| \neq 0$.

How I can justifying the answer?

答: If $A$ is invertible, then there exists a matrix $B$ with $AB=I$, the identity. Then $\det(A)\cdot\det(B)=\det(AB)=\det(I)=1$, so $\det(A)\ne 0$.


问:
Condition number of matrix proof
The condition of a differentiable function $f:X\rightarrow Y$ is defined as $$\kappa = \frac{\Vert x\Vert_X}{\Vert f(x)\Vert_Y}\Vert Df(x)\Vert_{\leadsto Y},$$
where $\Vert\cdot\Vert_X$ is the norm defined on $X$, $\Vert\cdot\Vert_Y$ the norm defined on $Y$, and $\Vert\cdot\Vert_{\leadsto Y}$ the norm induced by $\Vert\cdot\Vert_Y$.

The condition of a regular matrix $A$ is defined as $k = \Vert A^{-1}\Vert_{\leadsto X} \Vert A\Vert_{\leadsto X}$. I want to show that this is a special case of the aforementioned general case if $f(x) = Ax$. I have managed to show that $$\kappa\leq k$$
using the submultiplicativity of norms: $$\kappa = \frac{\Vert x\Vert}{\Vert Ax\Vert}{\Vert A\Vert}\leq \frac{\Vert A^{-1}\Vert \Vert Ax\Vert}{\Vert Ax\Vert}\Vert A\Vert  = \Vert A^{-1}\Vert \Vert A\Vert = k$$
(ignoring the norm subscripts in the notation). However, I fail to work out the other ineqality: $$\kappa = \frac{\Vert x\Vert}{\Vert Ax\Vert}\Vert A\Vert\geq \qquad?\qquad\geq k$$

Does this inequality even hold? I am aware of Matrix Condition Number Definition, however, in that answer it seems that the two notions of condition numbers $k$ and $\kappa$ are just related and not the same.

Title: Condition number of matrix proof

Body:
Proof

$$\kappa=\frac{\|x\|}{\|Ax\|}\|A\|_{ip}\ge\frac{\|x\|}{\|A\|_{ip}\|x\|}\|A\|_{ip}\\
=1\\
=\|I\|_{ip}\\
=\|A^{-1}A\|_{ip}\\
\ge \|A^{-1}\|\|A\|\\
=k$$


问:
Fourier series of $sgn(\cos(x))$
Knowing formulas for Fourier series when I'm given interval $(0,2l)$ & $(-l,l)$ I can't seem to find Fourier series of $sgn(\cos(x))$. I know that  $a_n = \frac{2}{l} \int_0^{l}f(x)\cos(\frac{n\pi x}{l})dx$, $b_n$ same but with $\sin$. If I'm given interval $(0,2l)$ then $a_n = \frac{1}{l} \int_{\lambda}^{\lambda + 2l}f(x)\cos(\frac{n\pi x}{l})dx$, $b_n$ same where we usually put $\lambda = 0$.

My question is if I have that my function $sgn(cos(x))$ is:

$\begin{cases}
 1&\text{if}\, x \in [\frac{\pi}{2},\frac{3\pi}{2}]\\
 -1&\text{if}\, x \in [\frac{-\pi}{2},\frac{\pi}{2}]\\
\end{cases}
$

What interval I should use for my Fourier series and this is the part where my understanding of this fails. Any help would be appreciated.

答:
since your function is $2\pi$ periodic you only need any intervall of the length $2\pi$. As the following shows:
Let $a\in\mathbb R$ then we can write $a=2\pi k+r$ with $r\in[0,2\pi)$ and $k\in \mathbb Z$.

$$\pi a_n=\int_a^{a+2\pi}f(x)\text{cos}(nx)dx=\int_{2\pi k+r}^{2\pi(k+1)+r}f(x)\text{cos}(nx)dx=\int_r^{2\pi  +r}f(x+2\pi k)\text{cos}(nx+2\pi nk)dx$$
Where the last step is just an easy substitution. Now we use that $f$ is $2\pi$ periodic, so we get:
$$\pi a_n=\int_r^{2\pi +r}f(x)\text{cos}(nx) dx=\int_r^{2\pi}f(x)\text{cos}(nx)dx+\int_{2\pi}^{2\pi+r}f(x)\text{cos}(nx)dx$$
Now we use again the periodicity on the right hand term:
$$\int_{2\pi}^{2\pi+r}f(x)\text{cos}(nx)dx=\int_{0}^{r}f(x+2\pi)\text{cos}(nx+2\pi n)dx=\int_{0}^{r}f(x)\text{cos}(nx)dx$$
Combined it follows:
$$\pi a_n=\int_a^{a+2\pi}f(x)\text{cos}(nx)dx=\int_0^{2\pi}f(x)\text{cos}(nx)dx$$

As you can see, the acutal interval ist not that important, you only need an interval of leingh $2\pi$.

Title: Fourier series of $sgn(\cos(x))$


问:
Confused about something that has to do with uniform continuity

I am confused about something that has to do with uniform continuity.

The rectangles in the image above represent the regions formed by the following intervals, by the definition of uniform continuity

$$c-\delta<x<c+\delta \Longrightarrow f(c)-\epsilon<f(x)<f(c)+\epsilon$$

The intervals above are illustrated in the graph - makes sense.

However, I am confused about the extremes of the interval in the $y$-axis. Namely, $f(c)-\epsilon \text{ and } f(c)+\epsilon$.

They do not seem like actual outputs that are mapped to the line of the function. Consider the first rectangle. If you start at the top edge of the rectangle ($f(c)+\epsilon$) and try to move horizontally to the point that it is mapped to on the line, there isn't one. There is no point on the line for ($x+\delta$, $f(c)+\epsilon$).

This is what I am confused about. How does this work? I know there is something I am either overthinking or just confused about. I would appreciate an explanation, elaboration, or clarification.

Thank you.

Image from WIKIMEDIA COMMONS by Claudia4 " Gleichmäßig stetige Funktion, visualisiert durch epsilon-delta-Umgebungen gleicher Größe" 22 Sept. 2016

答:
The graph of the function, when restricted on $(x-\delta,x+\delta)$, is a subset of one of the gray rectangles in your picture. In each rectangle, only the green part is part of the graph of the function.

The inequality $f(c)-\epsilon<f(x)< f(c)+\epsilon$ simply says that $f(x)$ falls in the interval  $(f(c)-\epsilon, f(c)+\epsilon)$. It is not necessarily true that $f(x)$ achieves the value, say, $f(c)+\epsilon$.

Consider the example when $f$ is a constant function $f(x)=1$. For any $x$, in any rectangle, $f(x)$ takes only one value in the interval $(f(c)-\epsilon, f(c)+\epsilon)$.


答:
This proof is basically the hint that @Hanul_Jeon gave in the comments as well as a follow on to your attempt number $1$. First, note that the set $\{a,b,c\}^{*}$ is countable. This is because there exists an injective function between this set and a subset of the rational numbers. One such function can be formally written as

$$g(\{x_1,x_2,...,x_n\})=\sum_{i=1}^n \frac{h(x_i)}{4^i}$$

where

$$h(x_i)=\begin{cases}
      1 & x_i=a\\
      2 & x_i=b \\
      3 & x_i=c
   \end{cases}$$

Now that we have shown that $\{a,b,c\}^{*}$ is countable, we can show that $F$ is uncountable. Let $r:\{a,b,c\}^{*}\to\mathbb{N}$ be a bijection between $\{a,b,c\}^{*}$ and the natural numbers and suppose by way of contradiction that $F$ is countable. Then we can list the elements of $F$ in a bijection with $\mathbb{N}$:

$$f_1:\{a,b,c\}^{*}\to\{a,b,c\}$$

$$f_2:\{a,b,c\}^{*}\to\{a,b,c\}$$

$$f_3:\{a,b,c\}^{*}\to\{a,b,c\}$$

$$\vdots$$

We will now construct a new function $f^{*}$ that is not on this list (Technically, the diagonal arguement is a constructive arguement, not contradiction. But its easier to understand as a proof by contradiction so I'll leave that terminology from above). For each element in $\{a,b,c\}^{*}$, define

$$f^{*}(\{x_1,x_2,...,x_n\})=K[f_{r(\{x_1,x_2,...,x_n\})}(\{x_1,x_2,...,x_n\})]$$

where

$$K(x)=\begin{cases}
      a & x=b\\
      b & x=c \\
      c & x=a
   \end{cases}$$

There is a lot to unpack in that function, but the basic idea is this: To find what $f^{*}$ does to an arbitrary element $s\in\{a,b,c\}^{*}$ do the following

Use the bijection $r$ to get the natural number corresponding to the $s$

Go to the function $f_{r(s)}$ in the list (the one we assumed to exist)

Change the output of $f^{*}$ to be different from $f_{r(s)}$

In this way, we are assured that $f^{*}$ is different from all functions on the list. As this is a contradiction, we conclude that $F$ is not countable.

Title: Show that a set $ F $ is uncountable

Body:
Let $$ \{a,b,c\}^n = \big\{(x_1,x_2,\dots,x_n): x_j \in \{a,b,c\} \big\}$$ and

$$ \{a,b,c\}^{*} = \bigcup_{n=1}^{\infty} \{a,b,c\}^n.$$

Consider the set $ F = \big\{ f: \{a,b,c\}^{*} \to \{a,b,c\} \big\}$ and show that $ F $ is not countable.

I've tried a couple of solutions:

i) No bijection between $ F \to \mathbb{N} $. Although I've tried to use Cantors diagonalizing argument I'm getting nowhere, since the arguments of $ f$ is such a mess.

ii) $ F $ is clearly an infinite set. Then it's sufficient to show that $ \mid F \mid \neq \mathbb{N}^0$ (aleph-null) or $ \mid F \mid > \mathbb{N}^0 $.

How do I solve this?


问:
Probability of getting 2 tails out of 5 tosses of a coin
Consider three coins having the probability of obtaining a single head in a single trial as $\frac{1}{4}$,$\frac{1}{2}$,$\frac{3}{4}$. A player selects one of these three coins at random(each coin is equally likely to be selected). If the player tosses the selected coin five times independently, then the probability of obtaining two tails in five tosses is qual to.?

probability of selected each coin is $\frac{1}{3}$

so $\frac{1}{3}\left( {5 \choose2}\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^{3}\right)+\frac{1}{3} \left( {5 \choose2}\left(\frac{1}{2}\right)^{5}\right)+\frac{1}{3}\left( {5 \choose2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3}\right)=\frac{85}{384}$

I dont have answer for it can anybody confirm my way of doing this question?

Body:
I got this

> 1/3*dbinom(2, 5, 1/4)+1/3*dbinom(2, 5, 1/2)+1/3*dbinom(2, 5, 3/4)
[1] 0.2213542
> 85/384
[1] 0.2213542

It looks correct.


问:
Minimum area bounded between two curves
What is the minimum area bounded between the curve $ y = \dfrac{x^2}4 $ and $y=ax+9$, where $a$ is some real number.

I tried to use application of derivative but not able to reach a final solution.

答:
The line $y=ax+9$ will always lie above the parabola $y=\frac{x^2}4$, intersecting when

$$ax+9=\frac{x^2}4\implies x^2-4ax-36=0$$

at two points, $x_1$ and $x_2$, both dependent on $a$. (It's given that $a\in\mathbb R$, so the line is never vertical, hence there are two points of intersection.)

The area between the curves is then a function of $a$,

$$\operatorname{Area}(a)=\int_{x_1(a)}^{x_2(a)}\left((ax+9)-\frac{x^2}4\right)\,\mathrm dx$$

which can be minimized by applying the fundamental theorem of calculus. Can you continue from here?

Title: Minimum area bounded between two curves


问:
Proving Statement with Diagonalizable Matrices
In Serge Lang’s Linear Algebra, a diagonalizable linear map is defined to be a linear map $F$ such that it has a basis $B$ where $M^B_B(F)$ is a diagonal matrix.

So the theorem I’ve been stumped on proving is stated on p.93 as follows:

Let V be a finite dimensional vector space over $K$, let $F:V\to V$ be a linear map, and let $M$ be its associated matrix relative to a basis $B$. Then $F$ (or $M$) can be diagonalized (in $K$) if and only if there exists an invertible matrix $N$ in $K$ such that $N^{-1}MN$ is a diagonal matrix.

I proved the forward direction but I have no idea how to start the proof for the backward direction. I am pretty confused because the theorem Lang stemmed off of to justify this theorem was that for bases $B,B’$ for $V$, and $F:V\to V$ a linear map, there exists invertible $N$ such that $M_{B’}^{B’}(F)=N^{-1}M^B_B(F)N.$ Not sure how to use this.
Any help would be appreciated.

答: For linear transformation $F$ to be diagonalizable, it need to have a full set of eigenvectors. This is because, the eigenvectors corresponding to a linear transformation is always fixed and if we look at the action of a diagonal matrix on the corresponding basis vectors (we view action on a basis by left multiplication of the matrix with the correspo0nding basis matrix), it just scales the vectors. So, if a matrix can be written as $P^{-1}DP$, $D$ being a diagonal matrix, the basis has to be eigenvectors of $F$, and we get that $F$ has full set of eigenvectors, hence diagonalizable.


问:
Circular argument for infinitude of primes from Euler product?
Many textbooks will explain how the Euler product formula can be used as a basis for proving the infinite of primes:

$$\sum_{n}\frac{1}{n^{s}}=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1}$$

My concern is that this is a circular argument because the derivation of the product formula uses the fact there are infinite primes.

Maths history texts will suggest that Euler's derivation of the product formula was done by a process of "sieving the zeta function", outlined next:

We start with the zeta function, having proven (elsewhere) $s>1$ for convergence.

$$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\ldots$$

We can divide this series by $2^{s}$.

$$\frac{1}{2^{s}}\zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\frac{1}{12^{s}}\ldots$$

These denominators are multiples of $2^{s}$. By subtracting these terms from $\zeta(s)$, we sieve out terms with these multiples of $2^{s}$.

$$(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\ldots$$

Repeating for $3^s$ gives us:

$$(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

And again for $5^s$:

$$(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

Repeating the process, we can only remove multiples of successive primes. We can't remove multiple of a number C if it is non-prime, C=AB, because we would have already sieved them when we removed multiples of its factors A or B.

In the limit of this process (having removed multiples of an infinitude of primes), only 1 remains on the RHS, and on the left we have an infinite product over all primes:

$$\ldots\cdot(1-\frac{1}{11^{s}})\cdot(1-\frac{1}{7^{s}})(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1$$

Back to my question.

This derivation assumes there are an infinitude of primes in order to attain the single 1 on the RHS. Therefore this Euler product can't be used as basis to say that there is an infinitude of primes.

Where is the flaw in my logic?

Comment: I am aware of the other derivation which uses the FTA and doesn't assume an infinitude of primes, but this is the method used by Euler, and he used it to demonstrate the infinitude of primes, for the first time since Euclid, according to the maths history books.

I am not a university trained mathematician so would appreciate replies with minimal technical terminology.

答:
You are not using that there are infinitely many primes when expanding the Euler product, just that every integer has a unique factorization in prime powers.

$$\prod_{p\in \text{ set of all primes}} (1+\sum_{k\ge 1}(p^k)^{-s})= \sum_{n\ge 1}n^{-s}$$

For $s > 1$ the LHS is bounded by $\prod_{p\in \text{ set of all primes}} 2$.

That the RHS $\to \infty$ as $s\to 1^+$ implies that there are infinitely many primes.


问:
Confused about inductive proof that any Prüfer sequence has a corresponding tree

Let $(a_1, \dots, a_{n-2})$ be a sequence of integers with $1 \leq a_i \leq n$. Prove that $(a_1, \dots, a_{n-2})$ is the Prufer sequence of some tree with vertex set $\{1, \dots, n\}$.

The proof is a proof by induction, and is as follows: Let $k$ be the smallest integer in $\{1, \dots, n\}$ that does not appear in $(a_1, \dots, a_{n-2})$. Then by the inductive hypothesis, $(a_2, \dots, a_{n-2})$ is a Prüfer sequence of a tree with vertex set $\{1, \dots, k-1, k+1, \dots, n\}$. We form a tree $T$ by adding the vertex $k$ and connecting it to $a_1$. Then $T$ has Prufer sequence $(a_1, \dots, a_{n-2})$.

I'm confused about a few things about this proof:

If $n$ appears in $(a_2, \dots, a_{n-2})$, then we cannot apply the inductive hypothesis as there would be a restriction on $a_i \leq n-1$. How does this work?

When applying the inductive hypothesis, doesn't the vertex set of the generated tree need to be of form $(1, \dots, n-1)$? Why can we arbitrarily choose its vertex set to be $\{1, \dots, k-1, k+1, \dots, n\}$?

Any help would be appreciated!

答:
What you're asking for is the decoding of a Prüfer sequence.

I'm going to describe this decoding on one example: start with the Prüfer sequence $P=(5,2,1,6)$ (hence on vertex-set $V=\{1,2,3,4,5,6\}$).

We are going to construct the edge list $T$ of the corresponding tree.

The least number appearing in $V$ but not in $P$ is $3$, so we add the edge $\{3,5\}$ to  $T$, and remove $5$ from $P$ and $3$ from $V$. $T$ is currently $\{ \{3,5\} \}$.

We now have $P=(2,1,6)$ (hence on vertex-set $V=\{1,2,4,5,6\}$.
The least number appearing in $V$ but not in $P$ is $4$, so we add the edge $\{4,2\}$ to  $T$, and remove $2$ from $P$ and $5$ from $V$. $T$ is now $\{ \{3,5\};\{2,4\} \}$.

We now have $P=(1,6)$ and remaning vertices are $V=\{1,2,5,6\}$.
The least number appearing in $V$ but not in $P$ is $2$, so we add the edge $\{2,1\}$ to  $T$, and remove $1$ from $P$ and $2$ from $V$. $T$ is now $\{ \{3,5\};\{2,4\} ; \{1,4\} \}$.

Now $P=(6)$ a nd $V=\{1,5,6\}$.
The least number appearing in $V$ but not in $P$ is $1$, so we add the edge $\{1,6\}$ to  $T$, and remove $6$ from $P$ and $1$ from $V$. $T$ is now $\{ \{3,5\};\{2,4\} ; \{1,4\} ; \{1,6\} \}$.

Now $P$ is empty, and vertices $5$ and $6$ are left in $V$, so we add the edge $\{5,6\}$ to $T$, which is finally full (size $6-1 = 5$) and equal to $\{ \{3,5\};\{2,4\} ; \{1,4\} ; \{1,6\} ; \{5,6\} \}$.

This can be used to show that Prüfer encoding is a bijection from trees with $n$ vertices to sequences in $\{1,...,n\}^{n-2}$ because:

1- from any sequence $P$ in  $\{1,...,n\}^{n-2}$ you get a tree $T$ on $n$ vertices

2- the Prüfer sequence of $T$ is $P$ (we haven't quite proved this but is holds, you can test this on the example)


问:
Why don't introductory calculus textbooks usually introduce the notion of differentiability on closed intervals?
The title says it all. I find the apparent neglect of this idea rather unfortunate because (1) the notion of a function being differentiable on a closed interval is intuitively reasonable (and ought to be discussed), and (2) several theorems from calculus can be strengthened if they're stated in terms of closed interval differentiability.

Before proceeding further, it's worth taking time to make sure we agree on what we mean we say that a function is "differentiable over a closed interval". Here's the definition I had in mind:

A function $f:[a,b]\to\mathbb{R}$ is differentiable over $[a,b]$ if it is differentiable (in the ordinary sense) over $(a,b)$, right-differentiable at $a$, and left-differentiable at $b$.

As you can probably guess, a function $f$ is said to be left-differentiable at a number $x_0$ if the limit

$$\lim_{h\to 0^-}\frac{f(x_0+h)-f(x_0)}{h}$$

exists, with a similar definition applying to right-differentiability. The limits are then defined to be the left-hand and right-hand derivatives of $f$, respectively.

With that said, here's an example of a theorem (FTC1) that, under its usual hypotheses, can be strengthened (albeit slightly) if the notion of closed-interval differentiability is used:

If $f:[a,b]\to\mathbb{R}$ is continuous, then the function $F:[a,b]\to\mathbb{R}$ defined by $F(x)=\int_{a}^{x}f(t)\text{ }dt$ is differentiable over $[a,b]$. Moreover, for all $x\in[a,b]$, $F'(x)=f(x)$ (at the endpoints, $F'(x)$ is understood to be a left or right-hand derivative)

The return you get by applying this notion to this example is obviously minimal. That said, I truly believe that making these extra definitions is worthwhile anyway, namely because the resulting concepts strongly align with our intuitions.

To bring everything back together, I'll restate my question one last time: why don't introductory calculus textbooks usually introduce the notion of differentiability on closed intervals? Any and all responses are greatly appreciated.

Note: the "theorems" I mentioned earlier do not include existence theorems like Rolle's theorem or the MVT. I understand why these only require differentiability in the interior and why we shouldn't strengthen that requirement.

答:
Your question is not new in this forum. See for example Why does the concept of differentiation have the meaning only on open sets? and Most general $A \subseteq \mathbb R$ to define derivative of $f: A \to \mathbb R$?

For each $f : J \to \mathbb R$ living on an arbitrary interval $J \subset \mathbb R$ you can of course define that $f$ is differentiable in $x_0 \in J$ if the limit
$$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} $$
exists and denote this limit by $f'(x_0)$. This means, as expected, that
$$\forall \epsilon > 0 \phantom {x} \exists \delta > 0 \phantom {x} \forall x \in J : 0 < \lvert x - x_0 \rvert < \delta \implies \left\lvert \frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) \right\rvert < \epsilon .$$
If $x_0$ is a right / left boundary point of such an interval $J$, then you can say that $f$ is left /right differentiable at $x_0$ with left /right derivative $f'(x_0)$ because you can approach $x_0$ only from the left / right side. But this is implicit in the definition and in my opinion there is no need to emphasize it.

Doing so, the concepts of left and right derivative of $f$ nevertheless make sense for an interior point $x_0$ of $J$ because you can approach $x_0$ from both sides.

So why do most calculus textbooks introduce the notion of differentiability on open intervals only? In my opinion it is just tradition. See my answer to Why does the concept of differentiation have the meaning only on open sets? But note that there are also textbooks with a more general approach.


问:
A good introduction to gradient flows?
I would like to learn more about gradient flows and their application to the theory of PDEs. I was wondering if anyone had any favourite texts that they thought provided a good and solid introduction to the topic?

There is a similar question here which was asked in 2012. They seemed to recommend the text Partial Differential Equations by L. Evans.

I ask the question again because it hasn't received a great deal of attention and also there may be more modern texts available now which give an introduction to gradient flows.

Thanks for any recommendations given.

EDIT: I realise there is the book by Gigli et al. which discusses gradient flows on metric spaces. I would prefer to work with something describing the theory on $\mathbb{R}^n$ if such a text was available. However it may be the case that this book simply is the best resource out there.

答: I suggest the Bible of Gradient Flows, the book "Gradient Flows: In Metric Spaces and in the Space of Probability Measures" by Luigi Ambrosio, Nicola Gigli, and Giuseppe Savare</p>

Title: A good introduction to gradient flows?


问:
$λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.
Determine the value of the constant $λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.

I tried equating it to n =  1 and found that $\lfloor \lambda \rfloor = 0 , 2$

Then I equated it to n =  2 and found out that $\lfloor \lambda\rfloor = 2$

I use the original equation to get (let x =  fractional part of lambda)

$2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$

=  $2\lfloor{(2+x) n}\rfloor = 1 - n + \lfloor (2+x) \lfloor(2+x) n \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor (2+x)(2n+ \lfloor nx \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor 4n +2\lfloor nx \rfloor + 2nx +x\lfloor nx \rfloor\rfloor$

->  $n-1 = \lfloor 2nx + x\lfloor nx \rfloor \rfloor$.

I do not know how to proceed, I tried doing $n-1<y< n$ but it didn't give me anything

答:
Take $n=1$, we have $2\lfloor{\lambda }\rfloor =  \lfloor\lambda \lfloor\lambda  \rfloor\rfloor\Rightarrow \lfloor{\lambda }\rfloor=2,\ \lfloor{2\lambda }\rfloor=4$ (another situation is impossible).

Take $n=2$, we have $2\lfloor{2\lambda }\rfloor =-1+ \lfloor\lambda \lfloor2\lambda  \rfloor\rfloor\Rightarrow \lfloor{4\lambda }\rfloor=9$.

Let $a_1=2,a_2=4,a_3=9$, and $a_{n+1}=\lfloor{a_n\lambda }\rfloor$. Then the condition becomes
$$2a_n=1-a_{n-1}+a_{n+1}\Rightarrow a_n=\frac{1}{2}+\frac{(1-\sqrt2)^{n+1}+(1+\sqrt2)^{n+1}}{4}.$$

Hence,
$$\frac{a_{n+1}}{a_n}\le \lambda<\frac{a_{n+1}+1}{a_n},\quad\forall n\in\mathbb{N}.$$
As a consequence, $\lambda$ has the unique value $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1+\sqrt2.$


问:
Is there a "global" convexity locally around a minimum?
$$\newcommand{\til}{\tilde}$$

Let  $F:(0,\infty) \to [0,\infty)$ be a continuous function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$. Suppose also that $F|_{[1-\epsilon,1+\epsilon]}$ is strictly convex, for some $\epsilon>0$.

Question: Does there exist a $\delta>0$ such that $F$ is convex at every point $y \in (1-\delta,1)$?

By convexity at $y\,\,$ I mean that for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =y$,
$$
F(y)=F\big(\alpha x_1 + (1- \alpha)x_2 \big) \leq \alpha F(x_1) + (1-\alpha)F(x_2). \tag{1}
$$
Equivalently, $F$ admits a supporting line at $y$, i.e. $\exists m \in \mathbb{R}$ such that
$$
F(x) \ge F(y)+m (x-y) \, \, \, \text{ for every } \, \, x \in (0,\infty). \tag{2}
$$

Edit: I have a proof for a positive answer
if $F \in C^1$, and an incomplete argument for the case where we do not assume $F \in C^1$. I am interested to know the answer when $F$ is merely continuous.

(Feel free to skip over my proofs/arguments below. The question stands as it is.)

A (hopefully correct) proof assuming $F \in C^1$:

If there is no such $\delta$, there exists $y_n \in (0,1)$, $y_n \to 1$ such that $F$ is not convex at $y_n$. Thus, $\exists x_n \in (0,\infty)$ such that
$$
F(x_n) < F(y_n)+F'(y_n) (x_n-y_n) \tag{3}
$$
The tangent at $y_n$ is below the $x$-axis for $x>1$
(since $F$ is convex at $(y_n,1]$ it decreases more slowly than its tangent after the tangency point). Thus, $F>0$ is above its tangent which is negative for $x>1$, which implies $x_n \in (0,1)$.

We may assume that $x_n \to x_0$; taking limits on both sides of $(3)$, we obtain
$F(x_0) \le F(1)=0$, since $|F'(y_n) (x_n-y_n)| \le |F'(y_n)| \to 0$.

This implies $x_0=1$, so both $x_n,y_n \to 1$. Looking again at inequality $(3)$, this contradicts the convexity of $F|_{[1-\epsilon,1+\epsilon]}$.

We have used $F'(y_n) \to F'(1)=0$. If we do not assume $F \in C^1$, we need to replace $F'(y_n)$ with slopes of supporting lines $m_n$, which do not necessarily converge to zero.

An incomplete proof without assuming $F \in C^1$:

Assume there is no such $\delta$. Then $\exists s_n \in [0,1]$, $s_n \to 1$ such that $F$ is not convex at $s_n$.

Thus $\exists x_n,y_n \in (0,\infty), \alpha_n \in [0,1]$, $x_n \le s_n \le y_n$ such that
$$
s_n=\alpha_n x_n + (1- \alpha_n)y_n, \, \, \text{ and } \, \,
F\left(s_n \right) > \alpha_n F(x_n) + (1-\alpha_n)F(y_n). \tag{4}
$$

$x_n \le s_n \le 1$ so $x_n$ is bounded. W.L.O.G we may assume that $y_n \le 1$. Indeed, if $y_n >1$, we can replace it with $\til y_n=1$, and choose $\til x_n$ such that $\alpha_n \til x_n + (1- \alpha_n)\til y_n=s_n$. Then $x_n \le \til x_n \le s_n \le 1=\til y_n \le y_n$, hence $F(\til x_n) \le F(x_n), F(\til y_n) \le F(y_n)$. Then inequality $(4)$ holds with $x_n,y_n$ replaced by $\til x_n,\til y_n$.

We now have $x_n \le s_n \le y_n \le 1$, and $s_n \to 1$ so we may assume that $x_n \to x, y_n \to 1, \alpha_n \to\alpha$. Taking limits of inequality $(4)$ we get
$$
0=F(1) \ge \alpha F(x) + (1-\alpha)F(1) = \alpha F(x)\ge 0,
$$
so $\alpha F(x)=0$. If $\alpha>0$, then $F(x)=0$ and $x=1$, so
$x_n,y_n,s_n \to 1$, thus they eventually lie at $(1-\epsilon,1]$, which together with inequality $(4)$ contradicts the convexity of $F|_{[1-\epsilon,1+\epsilon]}$.

The problem is that if $\alpha=0$ we cannot deduce that $x=1$!

答:
A counterexample:
$$
 F(x) = \begin{cases}
\frac 74 - x & \text{ for } 0 \le x \le \frac 12 \, ,\\
x^2-4x+3  & \text{ for } \frac 12 \le x \le 1 \, ,\\
(x-1)^2 & \text{ for } 1 \le x \, .
\end{cases}
$$
$F$ is strictly decreasing on $[0, 1]$ and strictly convex on $[1/2, \infty)$.

Now assume that there are supporting lines
$$
 F(x) \ge F(y) + m_y(x-y)
$$
for all $y \in (1-\delta, 1)$. Then $m_y \le F_-'(1) = -2$ so that for $0 < x < 1/2$
$$
 F(x) \ge F(y) - 2(x-y) \implies F(x) \ge  F(1) -2(x-1) = 2-2x
$$
and that does not hold for $x$ close to zero.

Here is a plot with the function $F$ (blue) and the lower bound for all supporting lines (red):


问:
A an nxn matrix. P a permutation matrix that permutes columns of A. How many operations does P*A involve?
Essentially, I am supposed to count how many operations a particular computational algorithm involves, and I've gotten stuck on this one part. My understanding is that for two nxn matrices, matrix multiplication requires of order $0(n^3)$ operations.

If P is a permutation matrix, and we also know the particular permutation that it represents, how many operations does P*A involve?

I'm assuming that it's of order $0(n)$? Or maybe even none (say, if the permutation simply involves storing the values differently - switching which values are in which columns)?

Can someone confirm this?

答:
For an $n \times 1$ vector $v$ and $n \times n$ matrix $A$, to get the first entry of the resulting vector $Av$ you require $n$ multiplications and $n-1$ additions. Doing this for each ot the $n$ rows (entries), you get $n(n+n-1) \sim O(n^2)$ operations. Matrix multiplication is just $Av$ done for each of the $n$ columns $v$ of the second matrix, so the resulting operation count is $O(n^3)$, as you said.

However, in your case, multiplying on the left by $P$ corresponds to permuting the rows of $A$. In other words, if $P$ represents the permutation $\sigma \in S_n$, $\sigma : \{1,\dots,n\} \rightarrow \{1,\dots,n\}$, then the resulting matrix $PA$ has entries $a(\sigma(i),j)$, where $a(i,j)$ is the $i,j$ entry of $A$. Depending on the language you are using and how you code this, this should take no time at all.


问:
Help understanding Garling's proof that closed and bounded is equivalent to compactness
I'm self-studying real analysis using A course in mathematical analysis by Garling

In the proof that, in $\Bbb{R}$, compactness $\Leftrightarrow$ closed and bounded I don't understand how they deduce that  $V_n = \{x \in \Bbb{R} : |x-a| \gt \frac{1}{n} \}$ covers the arbitrary compact set B if $a \not\in B$

I don't believe any more of the proof is relevant to this but to this point they have proven compactness $\Rightarrow$ boundedness and that closed and bounded $\Rightarrow$ compactness

答:
Let $x\in B$ then $|x-a|\ne 0$ and so there is a positive integer $n$ such that $|x-a|>\frac{1}{n}.$

Therefore $x\in V_n.$


问:
$gcd(a,b)=lcm(a,b)$ if and only if $a=\pm b$
I proceeded like this.
Let $a \leq b$.

Since $gcd(a,b)=lcm(a,b)$, then $gcd(a,b)^2=ab$

Now $gcd(a,b)=\pm\sqrt ab$ which is an integer. So b must be of the form $am^2$

so the numbers are a and $am^2$ whose gcd is clearly a. hence $gcd(a,b)=a=\pm am$ which implies $m=\pm 1$

so $b=am^2=\pm a$ which is proved.
I do not know how to prove the converse.

答:
Let $g$ be the gcd, then there are integers $u$ and $v$ with no common factors such that $$a=gu, b=gv.$$

Then the lcm is $|guv|$ and so the lcm is $g$ if and only if $|uv|=1$ i.e. $|a|=|b|$.


问:
A definition of mathematical expression that can distinguish between expressions that denote the same object
I asked a similar question on philosophy stack exchange, but was told to ask it here. In an elementary school math class, we sometimes are asked "What is 2+3?". It would be marked wrong if we simply wrote "2+3". What the teacher is after is a certain mathematical expression. To formalize this idea, we need a definition of mathematical expression that can distinguish even between two expressions that denote the same object. Is there such a definition somewhere, perhaps in some computer science textbook?</p>

答:
I think at a fundamental level, your confusion comes from definitions of mappings. Take your example of addition. Then we could more formally create a mapping:
$$
\phi : \mathbb{N}\times \mathbb{N}\to \mathbb{N}
$$
given by
$$
\phi(a,b) = a+b
$$
now - as with your example - you could ask "what is ${\phi(2,3)}$?", and someone could keep answering over and over again "well it's just ${\phi(2,3)}$", and they would of course be right. But this still hasn't given an explicit definition of ${\phi(2,3)}$. This answer is as about helpful as saying nothing.

You could say "well, doesn't this mean we are missing something? Shouldn't we make some sort of new definitions to avoid these nonsense answers?" and I would argue no. To explain why, think of the following: when we define addition, do we literally define the answer to every possible addition? No, of course not. It's impossible, we only have finite amounts of time. So really you could say we have actually never explicitly defined the map ${\phi(a,b)}$ - we have only ever written down rules that, given any two specific inputs, allow you to explicitly find an output. And this is the best we can do. (I guess you could actually ask "how do we even know that addition is well-defined then? How do we know the rules we set for addition are consistent and make sense?" - this is a rather difficult question to answer). Even with this, however, given the rules we set out for addition we can still figure out further properties of addition, and as I said - we have the toolkit to actually figure out for example that ${\phi(2,3)=5}$.

Hopefully that helps a little.

EDIT: to give you a taste of the sorts of rules we define ${\phi}$ by, it's essentially the following:
$$
\phi(n,0) = n
$$
$$
\phi(a,\phi(b,1))=\phi(\phi(a,b),1)
$$
We could define ${2 := \phi(1,1)}$, ${3 := \phi(2,1)}$, ${4 := \phi(3,1)}$, ${5 := \phi(4,1)}$. Going to your example, say we want to calculate ${\phi(3,2)}$. Then this is
$$
=\phi(3,\phi(1,1)) = \phi(\phi(3,1),1) = \phi(4,1) = 5
$$


问:
Determine the number of elements of $\text{Hom}(\mathbb{Z}/5 \mathbb{Z}, D_5)$ and of $\text{Hom} (\mathbb{Z}/6 \mathbb{Z} ,D_6)$.
I'm trying to an exam of last year, to practice. I find this exercise quite difficult:

Determine the number of elements of $\text{Hom}(\mathbb{Z}/5
 \mathbb{Z}, D_5)$ and of $\text{Hom} (\mathbb{Z}/6 \mathbb{Z}  ,D_6)$.

I don't know where to start. A hint would be appreciated to guide me in the right direction.

答: Suppose $\phi: \mathbb{Z}_5 \to D_5$ is a group homomorphism. Then $\phi$ is determined by $\phi(1)$ since $\phi(3) = \phi(1)^3$ etc. So we just need to decide where $\phi(1)$ goes. Where can it go? Well we must have $\phi(1)^5=1_{D_5}$ so $\phi(1)$ is either rotation by $0, 2\pi/5, 4\pi/5, 6\pi/5$ or $8\pi/5.$ It can be any of those, so there are $5$ elements in the first Hom group. Use similar ideas for the second one. </p>


问:
Measurable functions and $\sigma$-algebras generated by a collection of subsets.
Let $(X,\Sigma_X)$ be a measurable space. Let $Y$ be a set and $\mathcal{S}$ be a collection of subsets of $Y$. Consider the measurable space $(Y,\Sigma_Y)$ where $\Sigma_Y=\sigma(\mathcal{S})$ and consider a function $f:X\to Y$. What is are (minimal) sufficient conditions on the collection $\mathcal{S}$ so that the following holds
$$f^{-1}(E)\in\Sigma_X,\;\forall E\in\mathcal{S}\implies f^{-1}(E)\in\Sigma_X,\;\forall E\in\Sigma_Y.\;\;\;\;\;(*)$$

For example, for $X=Y=\mathbb{R}$ and $\mathcal{S}=\{(-\infty,b]:b\in\mathbb{R}\}$, and $\Sigma_X=\Sigma_Y=\sigma(\mathcal{S})$, we have that (*) actually holds. What is special about $\mathcal{S}$ in this case? What conditions did it satisfy so that $(*)$ holds? And how far can the argument used here be generalized?

答:
$\sigma(\mathcal S)=\Sigma_Y$ is already sufficient. In fact, the family $U=\{A\in\mathcal P(Y)\,:\,f^{-1}[A]\in \Sigma_X\}$ is a $\sigma$-algebra:

$\emptyset\in U$

if $A\in U$, then $f^{-1}[Y\setminus A]=X\setminus f^{-1}[A]\in \Sigma_X$

if $A_n\in U$ for all $n\in\Bbb N$, then $f^{-1}\left[\bigcup_{n\in\Bbb N}A_n\right]=\bigcup_{n\in\Bbb N}f^{-1}[A_n]\in \Sigma_X$.

And therefore $U\supseteq \sigma(\mathcal S)$.


问:
Intensional identity
I have some basic questions about intensional identity. Two types are said to be (intensionally) identical iff “they have the same objects and identical
objects of one of the types are also identical objects of the other'' (p. 141 in Nordstrom et al. (1990)).

Does this mean that two types are (intensionally) identical iff they imply each other (or they are equivalent)?

If this is the case, then if we negate an intensional identity type such as the following: $\neg (A=_UB)$, does it mean that $A$ and $B$ does not imply each other (or alternatively, that $A$ and $B$ do not have the same objects)?

Then apart from their syntactic distinction, what's the difference between $\neg (A=_UB)$ and $\neg(A\leftrightarrow B)$?

答:
First, I would note that what you're quoting doesn't appear to be defining the identity type. It appears to be talking about judgmental equality of types, which is distinct from types that represent proofs that two things are equal/identical. As for your questions:

No.

The notion of having 'the same' objects is not just that there are functions between them. It means that they classify exactly the same things. However, this explanation is not really something you can use to derive an equality between $A$ and $B$. You don't prove that for all $e$, $Γ ⊢ e : A$ if and only if $Γ ⊢ e : B$ and get $Γ ⊢ A = B$ or similar, even for judgmental equality. This explanation is, I think, mainly used to motivate the rules that are subsequently introduced from a meta-reasoning perspective.

(In fact, I'm a little skeptical that the meta-reasoning is actually true. I think it might be possible to have types $A$ and $B$ that classify all the same values, but they are not able to be shown equal by the rules of type theory. However, it's all right as a motivation for how equality should work.)

No.

For this I would bring up again that what you quoted was talking (I think) about judgmental equality, not the identity type. So it doesn't really make sense to negate it or hypothesize it, at least as part of reasoning within type theory.

But also, the answer is still no for the actual identity type over all. It is correct to say that if $¬(A =_U B)$ holds, then they must not classify all the same terms (because by the previous answer's meta-reasoning, they'd be judgmentally equal in that case). However, that is not equivalent to your other condition.

First, two non-equivalent types can imply one another. It's easy to define functions with types $⊤ → ℕ$ and $ℕ → ⊤$, but the natural numbers have many more values than the singleton type. But also, Martin-löf type theory doesn't even require that equivalent types (where those functions would be inverses to one another) are equal. You could extend MLTT with rules that imply $¬ (A =_U B)$ when those types have distinct appearing elements. For instance, $¬ (ℕ =_U ⊤×ℕ)$, because values of the former are like $5$ and values of the latter are like $(\mathsf{tt}, 5)$. That isn't derivable in plain MLTT, but it isn't refuted either (just like it doesn't refute equivalence implying identity, allowing for homotopy type theory).

The second is much stronger in general.

The first assumption just says that $A$ is not identical to $B$. The latter says that there are no pairs of functions between them. The only time those really come close to coinciding is for 'propositional' types. But even then, #2 shows that MLTT doesn't rule out e.g. $¬(⊤ =_U ⊤×⊤)$, while it exhibits $⊤ \leftrightarrow ⊤×⊤$

Editing to give a too-long answer to the comment:

The answer to this is going to change somewhat based on the context. In HoTT, there is a principle that ensures that two propositions are equal if and only if they imply each other. However, this principle is not part of MLTT (it is univalence for $\mathsf{Prop}$). Strictly speaking, it is an extensionality principle. 'Intensional' refers to only recognizing things with the same definition as equal.

MLTT does this in a weak way: things with the same definition* are provably equal, and things without the same definition are not [provably equal]. It is fairly non-committal about types being provably disequal. For instance $¬ (⊥ =_U ⊤)$ is derivable, and something like $¬(\mathsf{Fin}(n) =_U \mathsf{Fin}(\mathsf{suc}\ n))$ is derivable. However, $⊤ =_U ⊤×⊤$ is only not-provable, not provably false.

However, just as one can add the extensionality principle of univalence, one could add principles that make equality more committally intensional. Actually, I'm recalling that the book you are reading does this, because it includes an induction principle for the universe. So, it is possible to distinguish values of the universe that are defined differently, even if they denote equivalent types. So I believe $¬\mathsf{Id}(U,\hat ⊤,\hat ⊤ \hat × \hat ⊤)$ is derivable (using the notation of the book). However, this is an atypical definition of the universe of MLTT.

[*] One could quibble about induction principles and computation rules making things with distinct definitions provably equal, though.

Title: Intensional identity


问:
Sum of an indeterminate limit: $\lim_{x\to0}\;\left(2x\sin\left(\frac{1}{2x^2}\right)-\frac{1}{x}\cos\left(\frac{1}{2x^2}\right)\right)$

$\lim\limits_{x\to 0}\;\left(2x\sin\left(\frac{1}{2x^2}\right)-\frac{1}{x}\cos\left(\frac{1}{2x^2}\right)\right)$

How do I solve this limit? If I evaluate, then the left part is something that goes to zero multiplied by something bounded, so zero. But the right part, I have seen it gives an indetermination (which I quite don't understand). I would appreciate an explanation or a hint of why that's the case, and how do I evaluate the sum of both. Thank you.

答: Upon further thought,  yes, the limit doesn't exist.  The first part goes to 0, the second oscillates between $\pm \infty$,  so the whole thing oscillates


答:
Since you asked about computer science as one option, note the line of code

```<code>x = 2+3
</code>```

causes the l-value x to be $5$, whereas the r-value from which that was obtained was $2+3$.

Title: A definition of mathematical expression that can distinguish between expressions that denote the same object

Body: I asked a similar question on philosophy stack exchange, but was told to ask it here. In an elementary school math class, we sometimes are asked "What is 2+3?". It would be marked wrong if we simply wrote "2+3". What the teacher is after is a certain mathematical expression. To formalize this idea, we need a definition of mathematical expression that can distinguish even between two expressions that denote the same object. Is there such a definition somewhere, perhaps in some computer science textbook?</p>

Body:
Since you asked about computer science as one option, note the line of code

x = 2+3

causes the l-value x to be $5$, whereas the r-value from which that was obtained was $2+3$.


答: A variant approach is to just show that for natural numbers, using the fundamental theorem of arithmetic the lcm is just the product of the primes to the max of the exponents present in either number,   and the gcd is the product of the primes to the min of the exponents present in each number.  So if lcd=gcd,  the exponent for each prime in the prime factorization is the same.   Extending this to integers gets you the $\pm$

Title: $gcd(a,b)=lcm(a,b)$ if and only if $a=\pm b$

Body:
I proceeded like this.
Let $a \leq b$.

Since $gcd(a,b)=lcm(a,b)$, then $gcd(a,b)^2=ab$

Now $gcd(a,b)=\pm\sqrt ab$ which is an integer. So b must be of the form $am^2$

so the numbers are a and $am^2$ whose gcd is clearly a. hence $gcd(a,b)=a=\pm am$ which implies $m=\pm 1$

so $b=am^2=\pm a$ which is proved.
I do not know how to prove the converse.


问:
Find integers $a,b,n$ such that the polynomial $x^2+1$ divide the polynomial $a+bx+x^{n}$
I came across with following problem:

Find integers $a,b,n$ such that the polynomial $x^2+1$ divide the polynomial $a+bx+x^{n}$. Or some conditions to verify this property. However, I have no idea to start. We can restrict $a,b$ in the numbers $-1,0,1$. So, we have a set of polynomials of the form: $x^{n}$, $1+x^{n}$,$-1+x+x^{n}$, etc...

答: If $x^2+1$ divides your polynomial, then $\pm i$ are roots of your polynomial. You can take it from there.


答:
You can prove it reducing to absurdity.

Suppose $ \lvert a \rvert \neq \lvert b \rvert$.

Then $ gcd(a,b) \leq min(\lvert a \rvert,\lvert b \rvert) < max(\lvert a \rvert,\lvert b \rvert) \leq lcm(a,b) $.

We get $gcd(a,b) < lcm(a,b)$ so that would mean $gcd(a,b) \neq lcm(a,b) \ !!$

The other implication is obvious given that $gcd(a,-a) = a = lcm(a,-a)$


问:
simple natural log calculation
I'm trying to calculate for

$y = x^2 \ln(x)$ for $x \ge 1$

I'm getting confused with the correct approach to this.

What I have tried:

$\frac{dy}{dx}= 2x \cdot \frac{1}{x}$

Or:

$\exp{y}=x^{2}x$; $\frac{dy}{dx}\exp{y} = x^{3}$; = $\frac{x^{3}}{\exp{y}}$?

Essentially, I want to calculate $x$ at $1.2$, what is the appropriate expansion to figure out the result?

答:
Assume you left out the words $\color{red}{\text{the derivative}}$ in the sentence "$\text{I'm trying to calculate }\color{red}{\text{the derivative}}\text{ for}$"

You seem to some very fundamental "freshman dreams" about distribution across multiplication.

$[f(x)g(x)]' \ne f'(x)g'(x)$

And $\exp (f(x)g(x))\ne f(x)exp(g(x))$.

To be blunt these are very naive errors, and there is no reason you should have ever have made them.

In calculus (why on earth is the tag precalculus) you should have learned the product rule and you should know it backwards and forward in your sleep.

$[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$.

So you should be able to rattle off:

$[x^2\ln x]= (x^2)'\ln x + (x^2)\ln' x = 2x\ln x + x^2\frac 1x = 2x\ln x +x$

if you like you can (probably should) factor

$= x(2\ln x + 1)$.

Title: simple natural log calculation


问:
I’m doing an introduction to propositional logic course and am stuck on finding a proof for that s statement
Let $\Sigma \subseteq L$. Suppose $\Sigma$ is truth-consistent. Show that there exists a set $\Gamma \subseteq L$ such that $\Gamma$ is truth-consistent, $\Sigma \subseteq \Gamma$ and, for all $Q \in L$, $\Gamma \models Q$ or $\Gamma \models \neg Q$.

How do you prove this?

答:
I will assume that by truth-consistent you simply mean consistent: that is, for no formula $\varphi$ one has both $\Sigma\vDash\varphi$ and $\Sigma\vDash\neg\varphi$. Then, as Mauro ALLEGRANZA mentions in a comment, this is the existence of a maximally consistent set of formulas.

You define $\Gamma$ by transfinite induction: assume $\{\varphi_{n} : n\in\mathbb{N}\}$ is an enumeration of $L$ (you usually assume the number of connectives and propositional variables is at most enumerable, so this is a reasonable request). Then:

define $\Gamma_{0}=\Sigma$;

if $\Gamma_{n}\cup \{\varphi_{n}\}$ is consistent, define $\Gamma_{n+1}=\Gamma_{n}\cup\{\varphi_{n}\}$; if the union is inconsistent, then define $\Gamma_{n+1}=\Gamma_{n}$.

Notice, and this is very important, that $\Gamma_{n}$ is always consistent: $\Gamma_{0}$ is consistent, and if $\Gamma_{n}$ is consistent, either $\Gamma_{n}\cup\{\varphi_{n}\}$ is consistent and thus so is $\Gamma_{n+1}=\Gamma_{n}\cup\{\varphi_{n}\}$; or $\Gamma_{n}\cup\{\varphi_{n}\}$ is inconsistent, $\Gamma_{n+1}=\Gamma_{n}$ is consistent given our induction hypothesis.

We finally define $\Gamma=\bigcup_{n\in\mathbb{N}}\Gamma_{n}$: it certainly satisfies $\Sigma=\Gamma_{0}\subseteq \Gamma$. We see that $\Gamma$ is consistent by compactness: suppose, otherwise, that $\Gamma\vDash \varphi$ and $\Gamma\vDash \neg\varphi$; by the compactness theorem, there exist $\alpha_{1}, ... , \alpha_{n}, \beta_{1}, ... , \beta_{m}\in \Gamma$ such that
$$\alpha_{1}, ... , \alpha_{n}\vDash\varphi\quad\text{and}\quad\beta_{1}, ... , \beta_{m}\vDash\neg\varphi.$$
Since $\{\varphi_{n} : n\in\mathbb{N}\}$ is an enumeration of $L$, there exist $i_{1}, ... , i_{n}, j_{1}, ... , j_{m}\in\mathbb{N}$ such that $\alpha_{1}=\varphi_{i_{1}}, ... , \alpha_{n}=\varphi_{i_{n}}, \beta_{1}=\varphi_{j_{1}}, ... , \beta_{m}=\varphi_{j_{m}}$, and by taking $N=\max\{i_{1}, ... , i_{n}, j_{1}, ... , j_{m}\}$, you can see that $\alpha_{1}, ... , \alpha_{n}, \beta_{1}, ... , \beta_{m}\in \Gamma_{N}$, what implies $\Gamma_{N}\vDash\varphi$ and $\Gamma_{N}\vDash\neg\varphi$, contradicting the aforementioned result that all $\Gamma_{n}$ are consistent! $\Gamma$ must then be also consistent.

The final step is proving $\Gamma$ is maximal, meaning that, for any formula $\psi\notin \Gamma$, $\Gamma\cup\{\psi\}$ is inconsistent. Since $\{\varphi_{n} : n\in\mathbb{N}\}$ is an enumeration of $L$, there must exist $n\in\mathbb{N}$ such that $\varphi_{n}=\psi$; now, $\Gamma_{n}\cup\{\varphi_{n}\}$ must be inconsistent, since otherwise we would have $\varphi_{n}\in \Gamma_{n+1}\subseteq \Gamma$, and therefore $\Gamma\cup\{\varphi_{n}\}\supseteq\Gamma_{n}\cup\{\varphi_{n}\}$ must also be inconsistent. So, for any formula $\psi$, if one has $\Gamma\not\vDash\psi$ and $\Gamma\not\vDash\neg\psi$, $\psi, \neg\psi\notin\Gamma$, and so both $\Gamma\cup\{\psi\}$ and $\Gamma\cup\{\neg\psi\}$ are inconsistent, and therefore $\Gamma\cup\{\psi\vee\neg\psi\}$ is inconsistent, what is absurd given $\Gamma$ is consistent and $\psi\vee\neg\psi$ is an axiom.


问:
Is it possible to find sum of this series?
I am trying to find the sum of the following series asked by my friend.
$$n\cdot\left(\bigl\lfloor\tfrac{n}{2}\bigr\rfloor+
\bigl\lfloor\tfrac{n}{3}\bigr\rfloor+
\bigl\lfloor\tfrac{n}{4}\bigr\rfloor+
\bigl\lfloor\tfrac{n}{5}\bigr\rfloor+
\cdots\right)$$
where the sum is up to $n/2$ terms if $n$ is even, and up to $(n-1)/2$ terms if $n$ is odd.

I am not getting how to solve it. Is this form can be reduced to a formula? Thank you.

Note: $\bigl\lfloor\;\;\bigr\rfloor$ means floor function, like $\bigl\lfloor 4.78\bigr\rfloor=4$.

答:
There is unlikely to be a closed form.

A related function is called the divisor summatory function $$D(n)=\sum_{m\le n} d(m)=\sum_{k=1}^n \left\lfloor\frac{n}{k}\right\rfloor $$  where $d(n)$ is the divisor function, the number of positive integers which divide $n$.  Your function is $nD(n)-n = n\sum\limits_{m\le n} \big(d(m) -1\big)$.

Since $D(n) \approx n\log_e(n) + n(2\gamma-1)$, where $\gamma \approx 0.5772156649$ is the Euler–Mascheroni constant, you might say your function was about $$n^2\big(\log_e(n) + 2\gamma-2\big)$$ though the absolute errors (positive or negative) can tend to grow faster than $n$ even if the relative error can tend to reduce: as an example, when $n=10079$ your function would give $850294677$ while the approximation would be about $850545092$, while for the highly composite $n=10080$ your function would jump up to $851094720$ while the approximation would be about $850723956$.


问:
Expectation of the area of the triangle
A point X is randomly chosen from the interval (0, a) of the x-axis and Y respectively from the
interval (0, b) of the y-axis. Suppose that X and Y are independent. Determine the expectation
of the area of the triangle OXY (O = origin).

My probability density function is ab/2? I have previously used Lebesgue integration to determine the expected value. However I do not understand how to use it when I have two unknown variables.

答:
The area of a triangle with the coordinates $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is
$$
S = \frac{1}{2} \left | x_1y_2 + x_2y_3 + x_3y_1 - y_1 x_2 - y_2x_3 - y_3x_1 \right |
$$

When you sample $X$ and $Y$, you will have a triangle with points $(0,0), (X,0), (0,Y)$, so you are interested in the random variable
$$
S = \frac{1}{2} |XY| = \frac{1}{2}XY
$$
since we are guaranteed that $X,Y$ are positive. Then by the independence, we simply have
$$
\mathbb{E}[S] = \frac{1}{2} \mathbb{E}[XY]=\frac{1}{2} \mathbb{E}[X] \mathbb{E}[Y] = \frac{1}{2} \frac{a}{2} \frac{b}{2} = \frac{ab}{8}.
$$

Title: Expectation of the area of the triangle


答: Define a random variable $A:(0,a)\times (0,b)\to \Bbb R$ by $$A(x,y)=\frac{xy}{2}.$$By definition, the expected value of $A$ is $$\Bbb E[A]=\int_0^a\int_0^b A(x,y)\,{\rm d}\Bbb P(x,y),$$where $\Bbb P$ is the Lebesgue probability measure on $(0,a)\times (0,b)$. But ${\rm d}\Bbb P = (ab)^{-1} {\rm d}x\,{\rm d}y$. Hence $$\Bbb E[A] = \frac{1}{ab}\int_0^a\int_0^b\frac{xy}{2}{\rm d}x\,{\rm d}y =\frac{1}{2ab}\frac{a^2}{2}\frac{b^2}{2} = \frac{ab}{8}.$$

Title: Expectation of the area of the triangle

Body:
A point X is randomly chosen from the interval (0, a) of the x-axis and Y respectively from the
interval (0, b) of the y-axis. Suppose that X and Y are independent. Determine the expectation
of the area of the triangle OXY (O = origin).

My probability density function is ab/2? I have previously used Lebesgue integration to determine the expected value. However I do not understand how to use it when I have two unknown variables.


问:
Locus of the center of a semi-circle tangent to a fixed semicircle, both contained in a circle
Following with a curious result of
the Cut-the-Knot entry "A Property of Semicircles:
What is this about? A Mathematical Droodle",

Let's suppose $AB$ is fixed diameter of one semicircle, but $CD$ (as a diameter) is a dynamic semicircle which always is tangent with semicircle ($AB$ as diameter). What's the track function will it be for point $N$ (middle point of $CD$)?

答:
The shortest possible answer to your question is “the Limaçon of Etienne Pascal”. This also known as the conchoid of a circle and its equation in the polar coordinate system is given by
$$r=a+b\cos\left(\theta\right). \tag{1}$$

However, the midpoint $N$ describes only the part of this quartic curve in the range $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. The curve called cardioid appears as a special case of the Pascal limaçon when $a = b$.

Let $r_0$ be the radius of the semicircle $AQPB$, which is inscribed in the circle $\Gamma$ having radius $R$ and the point $O$ as its center (see $\mathrm{Fig.\space 1}$). Please note that the origin of the Cartesian coordinate system is located at $M$, the center of this semicircle. Let $r_1$ and $G$ be the radius and the center of the semicircle $EQF$, which is also inscribed the circle $\mathit{\Gamma}\space$ such that the two semicircles externally tangential to each other at $Q$.

The auxiliary line $MG$ is drawn to join the centers of the two semicircles and it is obvious that  it goes through $Q$ as it is their common point. Furthermore, the auxiliary lines $OG$ and $OF$ were also drawn. Since $G$ is the midpoint of $FE$, which is a chord of $\Gamma$, $OG$ is its perpendicular bisector. Let $MG$ and $\measuredangle OMG$ be $r$ and $\theta$ respectively, where $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.

We start our search for the locus of $G$ by expressing $r_1$ in terms of $r$ and $r_0$.
$$MG = r_0 + r_1 = r \quad\rightarrow\quad r_1 = r\enspace –\space r_0 \tag{2}$$

Since $OE$ is a radius of $\Gamma$, $OE = R$. Likewise, since $GE$ is a radius of the semicircle $EQF$, $GE = r_1$. The triangle $OGE$ is a right angle triangle. Therefore,
$$OG^2 = R^2 - r_1^2. \tag{3}$$

The triangle $BMO$ is also a right angle triangle, hence, we shall write,
$$OM^2= R^2 – r_0^2. \tag{4}$$

We apply cosine law to $\triangle OMG$ to obtain,
$$OG^2=OM^2+MG^2-2\times OM\times MG\times \cos\left(\theta\right).\tag{5}$$

From (3), (4), and (5), it follows,
$$R^2 – r_1^2 = R^2 – r_0^2 + r^2 – 2r\sqrt{ R^2 – r_0^2}\cos\left(\theta\right).\tag{6}$$

Once we simplify the equation, which we obtained by eliminating $r_1$ from (6) using (2), we have,
$$r = r_0 + \sqrt{ R^2 – r_0^2}\cos\left(\theta\right). \tag{7}$$

This is the equation in polar form of the locus of both $G$ and $N$. This equation resembles (1). Therefore, we can conclude that the sought locus is the part of a Pascal limaçon that traverses from  $A\space\left(\theta =-\frac{\pi}{2}\right)\space$ to $\space B\space\left(\theta=\frac{\pi}{2}\right)$.

The $x-$ and $y-$coordinates of the midpoint $G$ can be expressed as,
$$x_G = r\cos\left(\theta\right) = r_0\cos\left(\theta\right) + \sqrt{ R^2 – r_0^2}\cos^2\left(\theta\right) \quad\quad\mathrm{and}\quad$$
$$y_G = \pm r\sin\left(\theta\right) = \pm \Bigl[ r_0\sin\left(\theta\right) + \sqrt{ R^2 – r_0^2}\cos\left(\theta\right) \sin\left(\theta\right)\Bigr].$$

The equation of the locus in the Cartesian coordinate system is given by
$$\left(x^2+y^2-x \sqrt{ R^2 – r_0^2}\right)^2=r_0^2 \left(x^2+y^2\right). \tag{8}$$

It is evident from (8) that the locus of $N$ and $G$ is a quartic (i.e. fourth order) curve.

We would like to present some additional information for OP’s perusal, which would answer the question “How does one construct the semicircles, such as $FEQ$ or $CDP$ shown in $\mathrm{Fig.\space 1}$, when the circle $\it\Gamma$ and the inscribed semicircle with the fixed radius are given?”

In the first scenario illustrated in $\mathrm{Fig.\space 2}$, we are given $\measuredangle \theta$ as well. Draw a line through $M$, which makes $\measuredangle \theta$ with the $x$-axis, to intersect the given semicircle at $Q$. Draw the line $OT$ through $O$ parallel to $MQ$. Draw a line through $Q$ parallel to $x$-axis to meet $OT$ at $H$. Drop a perpendicular from $H$ to $MQ$ to meet it at $G$. Now, $G$ is the center of the sought semicircle and $GQ$ is its radius. Draw the semicircle $EGF$ to complete the construction.

In the second scenario illustrated in $\mathrm{Fig.\space 3}$, $r_1$, the radius of the sought semicircle, is given. Extend the $y$-axis up to $J$ so that $AJ = r_1$. Draw a circular arc with $M$ as the center and $MJ$ as the radius. Construct a right angle triangle $OKB$ such that $KB = r_1$ and $OB = R$. Draw a second circular arc with $O$ as the center and $OK$ as the radius to intersect the first arc at $G$, which happens to be the center of the sought semicircle having the given radius $r_1$.


问:
MLE of $\frac{1}{\theta}$ Uniform distribution.
Let $X_1,X_2...X_n$ be a random sample from $U(\theta-5,\theta+5)$ where $\theta \in(0,\infty) $ is unknown. Let $T=\max(X_1,X_2...,X_n)$ and $U=\min(X_1,X_2...,X_n)$. Then which of the following statements is TRUE?

$(A) \dfrac{2}{T+U}$ is M LE of $\dfrac{1}{\theta}$

$(B) MLE$ of $\dfrac{1}{\theta}$ doesnt exist

$(C)U+8$ is MLE of $\theta$

$(D)\dfrac{T+U}{2}$ is unique MLE of $\theta$

$\theta-5<x_{(1)} < x_{(2)}....x_{(n)}< \theta+5$

$\theta<x_{(1)}+5$ and $\theta>x_{(n)}-5$

$x_{(n)}-5<\theta<x_{(1)}+5$

Any statistic satisfying above condition is MLE of $\theta$ so there is no unique MLE here option $(D)$ rules out

$x_{(1)}+8<x_{(1)}+5$

Option $(C)$

$8<5$ which is false so option $C$ is false.

I am not sure how to prove $(A)$ and $(B)$ but in the exam, I have ticked $(B)$ so am I right?   If I am wrong can anyone prove how?

答:
Because of the equivariance of MLEs, the MLE of $1/\theta$ is the reciprocal of the MLE of $\theta.$

The joint density is
$$
\begin{cases} 1/10^n & \text{as long as $\theta$ is so located that} \\
& \text{the interval $(\theta\pm5)$ includes} \\
& \text{both $U$ and $T$,}  \\[6pt] 0 & \text{otherwise.} \end{cases}
$$

So $\theta$ must be in the interval $(T-5,U+5)$ in order that the likelihood be positive; otherwise the likelihood is $0.$ Except that we have this complication: If it is known that $\theta>0,$ then one must use instead $(\min\{T-5,0\},\, U+5).$

Thus one can say there is a non-unique MLE: Any number in that interval will serve.


答:
If $\theta\in \mathbb R$, any MLE of $\theta$ can be expressed as a convex combination $X_{(n)}-5$ and $X_{(1)}+5$:

$$\hat\theta_{\alpha}=\alpha(X_{(n)}-5)+(1-\alpha)(X_{(1)}+5)\,,$$

where $\alpha\in (0,1)$ is free of $\theta$.

Under the restriction $\theta>0$, MLE is $\hat\theta_{\alpha}$ provided $\hat\theta_{\alpha}>0$. If $\hat\theta_{\alpha}<0$, then technically MLE of $\theta$ and hence MLE of $1/\theta$ does not exist.

So for $\alpha=\frac12$, one MLE of $\theta$ is  $\hat\theta=\frac{X_{(n)}+X_{(1)}}{2}$ if $X_{(n)}+X_{(1)}>0$. Then $1/\hat\theta$ is guaranteed to be one MLE of $1/\theta$ provided $X_{(n)}+X_{(1)}>0$.

So I think the correct option among (A) and (B) depends on the sample observed.

Title: MLE of $\frac{1}{\theta}$ Uniform distribution.

Body:
Let $X_1,X_2...X_n$ be a random sample from $U(\theta-5,\theta+5)$ where $\theta \in(0,\infty) $ is unknown. Let $T=\max(X_1,X_2...,X_n)$ and $U=\min(X_1,X_2...,X_n)$. Then which of the following statements is TRUE?

$(A) \dfrac{2}{T+U}$ is M LE of $\dfrac{1}{\theta}$

$(B) MLE$ of $\dfrac{1}{\theta}$ doesnt exist

$(C)U+8$ is MLE of $\theta$

$(D)\dfrac{T+U}{2}$ is unique MLE of $\theta$

$\theta-5<x_{(1)} < x_{(2)}....x_{(n)}< \theta+5$

$\theta<x_{(1)}+5$ and $\theta>x_{(n)}-5$

$x_{(n)}-5<\theta<x_{(1)}+5$

Any statistic satisfying above condition is MLE of $\theta$ so there is no unique MLE here option $(D)$ rules out

$x_{(1)}+8<x_{(1)}+5$

Option $(C)$

$8<5$ which is false so option $C$ is false.

I am not sure how to prove $(A)$ and $(B)$ but in the exam, I have ticked $(B)$ so am I right?   If I am wrong can anyone prove how?


问:
Prove that $2^{s(n)} \le d(n) \le 2^{t(n)}$ with given conditions on $d(n)$, $s(n)$ and $t(n)$
Let $d(n)$ denote the number of divisors of $n$, $s(n)$ denote the number of prime divisors of $n$ not counting repetitions, and $t(n)$ denote the number of prime divisors of $n$ counting repetitions. We would like to show that $2^{s(n)} \le d(n) \le 2^{t(n)}$ for all positive integers $n$.

For example, if we consider $n=12$, then $d(12)=6$, $s(12)=2$ and $t(12)=3$. Thus $2^{2} \le 6 \le 2^{3}$ is true. If $n$ is prime, then $2^{1} \le 2 \le 2^{1}$.

I'm struggling to come up with a proof for the composite case. I figure that the Fundamental Theorem of Arithmetic is required here, but not sure how to use it.

答:
In fact $s(n)$ and $t(n)$ are known as prime omega functions $\omega(n)$ and $\Omega(n)$. The "definitions" of the functions, for a given $n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot ... \cdot p_k^{\alpha_k}$, are
$$\omega(n)=k$$
$$\Omega(n)=\alpha_1+\alpha_2+...+\alpha_k$$
and
$$d(n)=(\alpha_1+1)\cdot(\alpha_2+1)\cdot...\cdot(\alpha_k+1)$$
The easy part is $\alpha_i+1\geq2$ for $i=1..k$, thus
$$d(n)\geq 2^k=2^{\omega(n)}$$
The slightly more complicated (but not too) is to show that $$m+1\leq2^m, \forall m\in \mathbb{N}$$ (left as an exercise! e.g. use induction). Then
$$d(n)\leq 2^{\alpha_1}\cdot2^{\alpha_2}\cdot...\cdot2^{\alpha_k}=
2^{\alpha_1+\alpha_2+...+\alpha_k}=2^{\Omega(n)}$$


问:
If I am solving a limit on 2 variables and after changing to polar coordinates the angle dissapears, what does that mean?
For example, I have: $\lim_{(x,y)\to(0,0)}\frac{e^{x^2+y^2}\ln(2(x^2+y^2)+1)-\frac{3}{2}(x^2+y^2)-1}{x^2+y^2}$

After a change to polar coordinates: $\lim_{r\to0}\frac{e^{r^2}\ln(2(r^2)+1)-\frac{3}{2}(r^2)-1}{r^2}$

I think that since the angle disappears the function has some sort of radial symmetry, so it doesn't matter how do I approach to $(0,0)$. After that I solved the limit like I would solve a single variable limit using L'Hopital's and the limit was equal to $\frac{3}{2}$ which is correct. Am I correct in assume that every time the angle disappears I can evaluate the limit as if it were one variable? Or what happened there?
(Sorry if I have some spelling mistakes, I'm not english native speaker).

答: Yes,  every time the variables show up they show up in the form of $r^2$ as your transformation shows, so in polar the angle is indeed not needed and you get a one variable limit


问:
How does one transform a vector from one coordinate system to another?
I know a point can be translated to new position in new coordinate system just by translation with difference of old origin to new origin.

How does this work with Vectors?

I know we need to rotate too in case of vectors if the new coordinate system is oriented differently but a little unclear how the translated vector would look like. For example: Say vector from origin to (3,4), would it result in a vector from new origin to new_position?

答:
I think things are much clearer if you differentiate between a vector (an object) and its coordinates (its representation).

When we talk about a vector, say in $\mathbb{R}^2$, we mean a specific point in it. When we talk intuitively about the coordinates of a vector, this is just actually just shorthand for its "parts" with respect to other vectors (the basis) which we have fixed as "building blocks."

For example, with the usual $\mathbb{R^2}$, the standard basis is $\begin{bmatrix} 1\\0  \end{bmatrix}$ and $\begin{bmatrix} 0\\1  \end{bmatrix}$. Then, a vector, say $\begin{bmatrix} 5\\6  \end{bmatrix}$ is actually $$\begin{bmatrix} 5\\6  \end{bmatrix}=5\begin{bmatrix} 1\\0  \end{bmatrix}+6\begin{bmatrix} 0\\1  \end{bmatrix}.$$ In this case, the vector is no different to its coordinates.

When we change coordinates, we change our reference point (our building blocks; our basis) but not the vector itself. For example, if instead I use the vectors $\begin{bmatrix} 1\\0  \end{bmatrix}$ and $\begin{bmatrix} 1\\1  \end{bmatrix}$ as basis, the vector $\begin{bmatrix} 5\\6  \end{bmatrix}$ is now represented by the coordinates $\begin{bmatrix}-1 \\6  \end{bmatrix}$ since $$\begin{bmatrix}5 \\6  \end{bmatrix} = -1\begin{bmatrix} 1\\0  \end{bmatrix} + 6\begin{bmatrix}1 \\1  \end{bmatrix}.$$

The coordinates tell us how much of each building block we need as these are the components that make up our vector.

If you wanted a slightly more involved explanation: When working with linear spaces, these transformations of coordinates are typically handled by change of basis matrices $P$. These represent the identity map with respect to different bases $B$ and $D$. What these do is that they take the coordinates of a vector $v$ in $B$ and then output the coordinates of $v$ with respect to $D$.

In symbols, with $P=[id]_{B,D}$, $$[id_{B,D}][v]_B = [v]_D.$$

As an example, let $B$ be the standard basis, $D =\lbrace \begin{bmatrix} 1\\0  \end{bmatrix},\begin{bmatrix}1 \\1  \end{bmatrix} \rbrace$ and $v=\begin{bmatrix} 5\\6  \end{bmatrix}.$

We then have $P=[id]_{B,D}=\begin{bmatrix} 1&-1\\0&1  \end{bmatrix}$.

The coordinates of $\begin{bmatrix} 5\\6  \end{bmatrix}$ with respect to $D$ are then given by $$[\begin{bmatrix} 5\\6  \end{bmatrix}]_D = P[v]_B=\begin{bmatrix} 1&-1\\0&1  \end{bmatrix}\begin{bmatrix} 5\\6  \end{bmatrix}=\begin{bmatrix} -1\\6  \end{bmatrix}$$ as we have obtained prior.

If you want to visualize these, think of rotating, changing the angle between or stretching the $x$ and $y$ axis in $\mathbb{R}^2$ and then find the coordinates of a vector with respect to the squished and rotated coordinates. You can also easily add translations, although these are no longer linear in the strict sense (linear operations fix the origin). If you involve translations, the transformations you are looking at for changes of coordinates will be called "affine".

For more basic information, I suggest the introductory book by Jim Heffron on linear algebra, which is readily available for free online: https://hefferon.net/linearalgebra/. It is a friendly read and has many examples and exercises.


问:
Prove that a product set is dense.
Given $z = (z_i)_{i \in I}$ a fixed point of $\displaystyle\prod_{I}X_i$, where each $X_i$ is non-empty and connected.

For any finite subset $K \subset I$, we define $X_K = \{ (x_i) ∈ \displaystyle\prod_{I}X_i \ | \ x_i = z_i \text{, if } i \in I \setminus K \}$

Finally, we define $A = \{X_K \ | \ K \text{ finite subset of } I\} $

How can one prove that $\bigcup A$ is dense in $\displaystyle\prod_{I}X_i$ ?

One needs to show that for any point of $\displaystyle\prod_{I}X_i$ any open set containing it will intersect $\bigcup A$.

I just don't see why it is the case.

答:
If we have a (non-empty) basic open product set $B$ in $\displaystyle \prod_{i \in I} X_i$ it is of the form

$$B=\prod_{i \in I} U_i$$

where there is a finite set $F \subseteq I$ such that $U_i = X_i$ for all $i \in I\setminus F$ and all $U_i$ are (non-empty) open in $X_i$. (These are the finite intersections of sets of the form $p_i^{-1}[U_i]$ which form the standard subbase for the product topology).

Then it's clear that $X_F \cap B \neq \emptyset$ : pick $a_i \in U_i$ for $i \in F$ and define $a_i = z_i$ for all $i \notin F$. Then $(a_i)_i \in X_F \cap B$ and so $B \cap A \neq \emptyset$ as $A \supseteq X_F$ for this $F$ too.

And as all basic open sets intersect $A$, all open sets do too (they are unions of basic open sets after all). So $A$ is dense.

Title: Prove that a product set is dense.


问:
How many solutions are there for $3x_1+3x_2+x_3+x_4=30$?

Find how many solutions there are for $$ 3x_1+3x_2+x_3+x_4=30$$

I know how to solve this: $x_1+x_2+x_3+x_4=30$
and I read this link, but I am still not sure about my answer.
I wrote:
$x_4+x_3=30-3(x_1+x_2) $
then I get this: $0 \le x_1+x_2 \le 10$ (that I know to solve: $66$ combination)
and now I'm not sure what to do with $x_3+x_4$.

答:
If you are allowing $x_i$ to be greater than or equal to $0$ in the equation $\color{red}{3x_1}+\color{blue}{3x_2}+\color{green}{x_3}+\color{brown}{x_4}=30$, then if we consider the expansion $$\displaystyle
\color{red}{\left((x^3)^{0}+(x^3)^1+(x^3)^2+\cdots\right)}\color{blue}{\left((x^3)^{0}+(x^3)^1+(x^3)^2+\cdots\right)}\color{green}{\left(x^0+x^1+x^2+\cdots\right)}\color{brown}{\left(x^0+x^1+x^2+\cdots\right)}$$

Then the number of solutions to the given equation will be the coefficient of $x^{30}$ in the above expansion (Why?).

Now we simplify it a bit, we know that $1+x+x^2+\cdots$ is a geometric progression whose sum equals $\dfrac{1}{1-x}$.

Using this, our expansion becomes, $$\dfrac{1}{\color{red}{(1-x^3)}\color{blue}{(1-x^3)}\color{green}{(1-x)}\color{brown}{(1-x)}}=(1-x^3)^{-2}(1-x)^{-2}$$

Now using binomial theorem, we can rewrite it as $$\left[\sum_{k=0}^{\infty}\binom{k+1}{k}x^{3k} \right]\left[\sum_{r=0}^{\infty}\binom{r+1}{r}x^r\right]$$

So coefficient of $\displaystyle x^{3k+r}=\binom{k+1}{k}\binom{r+1}{r}=(k+1)(r+1)$ and when $r=30-3k$, it is $(k+1)(31-3k)$ which is summed over from $k=0$ to $k=10$ which turns out to be the same summation as written by @Aqua, but note that with this, we can solve it even if there were restrictions on $x_i$, for example, if it were given that $1\le x_1 \le 5$, and rest everything the same, then we would only consider the product $$\displaystyle
\color{red}{\left((x^3)^{1}+(x^3)^2+(x^3)^3+\cdots+(x^3)^5\right)}\color{blue}{\left((x^3)^{0}+(x^3)^1+(x^3)^2+\cdots\right)}\color{green}{\left(x^0+x^1+x^2+\cdots\right)}\color{brown}{\left(x^0+x^1+x^2+\cdots\right)}$$ and proceed accordingly.


问:
Does every element of the weak-star closure of a set belong to the weak-star closure of a bounded subset?
I feel like this must be a monumentally stupid question. Say $X$ is a Banach space, $S\subset X^*$, and $x^*$ is in the weak* closure of $S$. Must $x^*$ lie in the weak* closure of some norm-bounded subset of $S$?

(If $x^*$ is the weak* limit of a sequence of elements of $S$ then this is clear by Banach-Steinhaus. But a convergent net of scalars need not be bounded...)

答:
The reference to Krein-Smulian in one of the answers is actually a little misleading, and is related to the following warning in Conway's Functional Analysis, page 161:

In fact, you can find a Banach space $X$ and $S\subseteq X^\ast$ a subspace (so certainly convex) such that $S$ is weak$^*$-dense in $X^\ast$, but there is a vector $f_0\in X^\ast$ such that no bounded net in $S$ converges to $f_0$; equivalently, $f_0$ is not in the weak$^\ast$-closure of $S$ intersected with any bounded subset of $X^\ast$.

I do not know a canonical reference for a construction.  However, it is fairly easy to adapt this Mathoverflow answer.  I wrote up the details here (github).


问:
Set of functions from $SS = \{ A_{1}, A_{2}, A_{3},...\}$ to $\{ \mathbf{T}, \mathbf{F} \} $ is countable?
Let $SS=\{ A_1,A_2,A_3,\ldots\}$, and let $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ .Is the set  V countable?  Justify your answer.

My instinct is to say that $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ is uncountable, and to prove this using a diagonalization argument, i.e. create a table of values of the functions $v_{1}, v_{2},...$ for natural numbers $n_1, n_2,\ldots$ and define a function $v_{m} : SS \to \{ \mathbf{T}, \mathbf{F} \}$ where it takes all values in the diagonals of the table $T, F$ and flips them, and then show that $v_{m}$ cannot appear anywhere in the list. Is my guess correct, and if so would this be a reasonable approach to the problem?

答: Yes, this is (essentially) the countably infinite case of Cantor's Theorem.</p>


问:
Help understanding Garling's proof that closed and bounded is equivalent to compactness
I'm self-studying real analysis using A course in mathematical analysis by Garling

In the proof that, in $\Bbb{R}$, compactness $\Leftrightarrow$ closed and bounded I don't understand how they deduce that  $V_n = \{x \in \Bbb{R} : |x-a| \gt \frac{1}{n} \}$ covers the arbitrary compact set B if $a \not\in B$

I don't believe any more of the proof is relevant to this but to this point they have proven compactness $\Rightarrow$ boundedness and that closed and bounded $\Rightarrow$ compactness

答:
I'm going to use the definition of compactness that says that every open cover of a compact set has a finite subcover.

Assume the condition doesn't hold. What then? Well, you can come up with at least one sequence of real numbers contained in the compact set that converge monotonically towards a. You know this because $\forall n\in \mathbb{N}$ we can find $x\in B$ such that $|x-a|<1/n$ by assumption.

Well, now that you have this sequence, you can come up with an open cover of B that doesn't have a finite subcover. So, say our sequence approaches a from below. Then take the open set $(-\infty,a_1)$ where $a_1$ is the first value in our sequence. Then, for the rest of the points in B between $(a_1,a)$ you can cover these with the the set containing all intervals $(a_{i-1},a_{i+1})$.

We know that this covers B completely between $a_1$ and a because $a_i$ converges to a and the intervals contain all $a_i$ and all points between them. We also know that there is no finite subcover since that would mean that there is some M for which $a_M$ is not in the subcover, but it is in B. But that's a contradiction because B is compact.


问:
Dropping an assumption on bounded convergence theorem
If $f_n$'s are uniformly bounded complex valued functions converging pointwise to $f$ and $E$ is of finite measure, then we know that $f$ is integrable on $E$ and $\int_Ef_n\rightarrow \int_Ef$. How about we drop the assumption that $E$ is of finite measure provided that we a priori know $|\int_Ef|<\infty$?  Can we still say $\int_Ef_n\rightarrow \int_Ef$?

答:
No. Consider the line with Lebesgue measure.  The functions
$$f_n(x) = \left\{ \begin{array}{ll} \frac 1n & |x| \le n \\ 0 & |x| > n \end{array} \right.$$ converge uniformly to $0$ yet satisfy $\displaystyle \int f_n = 2$ for all $n$.


问:
Show that this infimum is greater than zero
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 $R_i=(c_{i1},d_{i1}) \times \dots \times (c_{id},d_{id})$ is an open rectangle $\mathbb{R}^d$.

For $x \in K$, let $$\delta(x) = \sup \{ r>0 : B(x,r) \subset R_i \text{ for some } i\}$$

where the ball is taken with respect to the maximum norm $||x||_{\infty}=\max\{|x_1|,\dots,|x_d|\}$. Let $\delta= \inf\{\delta(x):x \in K\}$.  I want to show that $\delta>0$ and that for any $x$ there is some $R_i$ such that $B(k,\delta) \subset R_i$.

EDIT: It was suggested in the comments that Lebesgue's number lemma could do it. But is there a more direct way?

Any help is greatly appreciated.

答: You can go directly.   You have a finite length of $K$ and a finite amount of $R_i$'s.   Each possible combination gets you a closed ball (rectangles are balls in the max norm) that lives inside an open ball.   Thus,  there's an $\epsilon>0$  distance between the two,  so you have each supremum involve is bounded below by $\frac \epsilon 2$.   Take the minimum of all of those finite values, and that will be a lower bound on your final set


问:
Specific value of $\zeta(3/2)$?
Is anything known about the value of $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}?$$ It is a classical result that $\displaystyle \zeta(2)= \frac{\pi^2}{6}$ and $\zeta(3)$ has been shown to be irrational by Roger Apéry in 1979.

Do we even know if $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}$$ is an irrational number or not? Is it  true that $$\zeta(3/2)=\sum_{n\geq 1}\frac{1}{n^{3/2}}=\frac{a}{b}\sum_{n\geq 1}\frac{(-1)^{k-1}}{n^{3/2}\binom{2n}{n}}?$$ Not sure if Apéry's proof can be adapted here since I haven't read it.

答:
By no means an answer but an extended comment. I should also preface I am an amateur mathematician with no formal education. Following Wikipedia's entry for the Riemann zeta function we have the following functional equation:
$$
\zeta(s)=2^{s}\pi^{s-1}\sin\left(\pi s/2\right)\Gamma(1-s)\zeta(1-s)\label{a}\tag{1}
$$
Suppose $s=3/2.$ Via direct substitutions I can rewrite LHS side of $\ref{a}$ explicitly as
$$
2^{3/2}\pi^{3/2-1}\sin\left(3\pi/4\right)\Gamma(1-3/2)\zeta(1-3/2);\label{b}\tag{2}
$$
which after regrouping and gathering like terms is equal to
$$
2^{3/2+1/2}\pi^{3/2-1+1/2}\zeta(-1/2);\label{c}\tag{3}
$$
which is reducible to
$$
4\pi\zeta(-1/2).\label{d}\tag{4}
$$
And so
$$
\zeta(3/2)=4\pi\zeta(-1/2), \label{e}\tag{5}
$$
Observe:

Case 1: Note $4\pi$ is an irrational number. If $\zeta(-1/2)$ is rational then RHS side of $\ref{e}$ is an irrational number and so to by equality is $\zeta(3/2).$

Case 2:  Again note $4\pi$ is an irrational number. If $\zeta(3/2)$ is a rational number then the RHS of $\ref{e}$ is a rational number. If $\zeta(-1/2)$ is an rational number then the RHS of $\ref{e}$ is irrational; which is a contradiction. Subsequently $\zeta(-1/2)$ is an irrational number.

From cases 1 and 2 it seems at least one of $\zeta(3/2)$ or $\zeta(-1/2)$ is an irrational number. In particular they cannot both be rational. More strongly it would seem of the infinite number of pairs
$$
\{\zeta(j/2),\zeta(1-j/2)\}_{j\in\mathbb{N}}
$$
at least one number from each of the infinite pairs is irrational.

Title: Specific value of $\zeta(3/2)$?


答: Your approach is correct. Note that $V$ is bijective to the power set of $SS$ by mapping $v \in V$ to $v^{-1}(\mathbf{T}) \in 2^{SS}$, the subset of $SS$ that maps to $\mathbf{T}$.

Title: Set of functions from $SS = \{ A_{1}, A_{2}, A_{3},...\}$ to $\{ \mathbf{T}, \mathbf{F} \} $ is countable?

Body:
Let $SS=\{ A_1,A_2,A_3,\ldots\}$, and let $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ .Is the set  V countable?  Justify your answer.

My instinct is to say that $V = \{ v \mid v: SS \to \{ \mathbf{T}, \mathbf{F} \} \}$ is uncountable, and to prove this using a diagonalization argument, i.e. create a table of values of the functions $v_{1}, v_{2},...$ for natural numbers $n_1, n_2,\ldots$ and define a function $v_{m} : SS \to \{ \mathbf{T}, \mathbf{F} \}$ where it takes all values in the diagonals of the table $T, F$ and flips them, and then show that $v_{m}$ cannot appear anywhere in the list. Is my guess correct, and if so would this be a reasonable approach to the problem?


问:
Maximum of the convex linear combination of continuous functions on a compact domain
I can't figure out a way to prove, or disprove, the following statement: let $(f_i(x))_{i\in I}, I=\{1,...,n\}$ be a collections of real valued, continuous functions defined on a compact domain, i.e. $\forall i \in I, f_i:[0,1]\rightarrow\mathbb{R}$. Moreover, let us assume that each functions has exactly one maximum. Define $g(x):=\sum_{i=1}^n \omega_i \cdot f_i(x)$, with $\omega_i \in [0,1], \sum_{i=1}^n \omega_i=1$. $\forall i \in I$, let $x^*_i= arg max_{x\in[0,1]}f_i(x)$. I suspect that $argmax_{x\in[0,1]}g(x)\subseteq [min\{(x^*_i)_{i \in I}\}, max\{(x^*_i)_{i \in I}\}]$ but I can't find a way to prove it. Can someone help me?

Body:
For $0\leq a < b \leq 1$, let $t_{a,b}$ be the "triangular" function such that
$t_{a,b}(x) = 0$ if $x\leq a$ or $x\geq b$, $t_{a,b}(\frac{a+b}{2}) = 1$.
Let
$$
f_1(x) = 4 t_{0, 1/4}(x) + 2 t_{3/4, 1}(x), \qquad
f_2(x) = - 4 t_{0, 1/4}(x) + t_{1/4, 1/2}(x),
$$
so that $x_1^* = 1/8$, $x_2^* = 3/8$.
Choosing $\omega_1 = \omega_2 = 1/4$ it holds that
$g(x) = t_{3/4, 1}(x) + \frac{1}{2} \, t_{1/4, 1/2}(x)$, hence $\text{argmax} g = \{7/8\}$.


问:
What is an implicit function?
According to the definition of implicit function we cannot determine the value of one variable explicitly from the function. I have gone through many websites, few books and few youtube videos but could not understand implicit function clearly.

Equation of a circle is an implicit function as mentioned in most of the sources because we cannot get the value of y from the function f(x,y)=x^2+y^2=25 explicitly.But why can't we get the value of y explicitly i mean according to this function we can get two values for y at a certain abcissa. We are clearly getting two values of y for a certain value x then why is it implicit function?

I could not get the concept of this type of function. Pardon me for writing anything wrong if I have done any.Looking for an explanation

Again I apologise for asking such a silly question in this forum.

答:
One can find a rigorous definition in [Wikipedia][1] , however I will attempt to provide a practical way of thinking of implicit function's.

An explicit single variable function can be thought of as a 'mapping' from one set to another set of the form $x \to y$, this can be graphed on a cartesian grid by highlighting all the pair $(x,y)$ included in this mapping.

Now, in an implicit function, we no longer have the idea of mapping from the previous definition but we can think of the solution's to the implicit equation which defines the function. If we took a graph and highlighted all the points which solved the equation of the implicit function, then we would get the curve of the implicit function analogous to how there was a curve corresponding to the explicit function.

Some times we can write implicit functions as an 'multipile' explicit function, for example how we see writing the circle's equation as:

$$ y = \begin{cases} \sqrt{1-x^2} \\  - \sqrt{1-x^2}  \end{cases}$$

Notice that we have two possible definitions of 'y' when we isolate the implicit equation and solve for $y$, this two definitions can be thought of as pieces of the original curve. For example, $y=\sqrt{1-x^2}$ defines the upper disc and $ y = - \sqrt{1-x^2}$ defines the lower one.

[1]: https://en.wikipedia.org/wiki/Implicit_function#:~:text=An%20implicit%20function%20is%20a,)2%20%E2%88%92%201%20%3D%200

Title: What is an implicit function?


问:
Faithful semi-finite normal weights

The above screenshot is from Takesaki's book(Vol 2, Chapter VIII)
I wonder how to prove the implication (ii)→(i)?

By Theorem 2.11, we have $\sigma_t^{\psi}(x)=h^{-it}\sigma_t^{\varphi}(x)h^{it}$ for all $x\in M$. We have $\psi\circ\sigma_t^{\psi}(x)=\psi\circ\sigma_t^{\varphi}(h^{it}xh^{-it})$.

How to show that $\psi\circ\sigma_t^{\varphi}(x)=\psi(x)$?

答:
The modular group is defined via the GNS representation for $\psi$. So
\begin{align}
\psi(\sigma^\psi_t(x))&=\langle \sigma_t^\psi(x)\Omega,\Omega\rangle
=\langle \Delta^{-it} x\Delta^{it}\Omega,\Omega\rangle
=\langle x\,\Delta^{it}\Omega,\Delta^{it}\Omega\rangle=\langle x\,\Omega,\Omega\rangle=\psi(x).
\end{align}
The equality $\Delta^{it}\Omega=\Omega$ follows from $\Delta\Omega=\Omega$, and this in turn follows by definition:
$$
\Delta\Omega=FS\,\Omega=F\Omega=\Omega.
$$

Title: Faithful semi-finite normal weights


问:
Finding, writing down and visualizing images of functions
I apologize in advance if this is a duplicate (I would expect to see something like this on here), but I haven't found anything that matches my question.

I have this general difficulty (perhaps, to lack of practical experience) of finding and writing down images of functions given a domain. For example, I have recently seen this problem in a book:

Given $f(x,y) = \left(\begin{array}{c}u(x,y)\\v(x,y)\end{array}\right) = \left(\begin{array}{c}x^2-y^2\\2xy\end{array}\right)$, find image of $f$ when $x>0$, $y>0$, $x^2+y^2<1$

Basically, it's asking for a set of inequalities for what $u$ and $v$ trace out in the $u$-$v$ plane. How do I go about solving such problems for a general case of $f : \mathbb{R}^m \to \mathbb{R}^n$? For example, in the above problem, the domain of $f$ is a quarter of a circle of radius 1 in the first quadrant of the $x$-$y$ plane. How do I find what $f$ maps this to in the $u$-$v$ plane? Where do I start?

Update:
So apparently (by empirically constructing a table of values), $f$ maps the quarter of a circle to a half-circle ($v$>0, $u^2+v^2<1$). Is there a way to see this intuitively? Most importantly, how can I get these two inequalities algebraically?

答: You can try domain coloring. For the squaring map, see this page which includes an animation.</p>