问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
actually numbers from 11 to 16 are quite regular in French (and in Italian) too: they just are a derivation from Latin.

```<code>        | French    | Italian       | Latin
un      | on·ze     | un·dici       | un·decim
deux    | dou·ze    | do·dici       | duo·decim
trois   | trei·ze   | tre·dici      | tre·decim
quatre  | quator·ze | quattor·dici  | quattuor·decim
cinq    | quin·ze   | quin·dici     | quin·decim
six     | sei·ze    | se·dici       | se·decim
sept    | dix-sept  | dici(as)sette | septem·decim
huit    | dix-huit  | dici-otto     | duo·de·viginti
neuf    | dix-neuf  | dici(an)nove  | un·de·viginti
</code>```

(18 and 19 in Latin are computed as 20-2 and 20-1).
Each language has its own way to cope with small numbers.

Title: Why do the French count so strangely?

Body:
actually numbers from 11 to 16 are quite regular in French (and in Italian) too: they just are a derivation from Latin.

        | French    | Italian       | Latin
un      | on·ze     | un·dici       | un·decim
deux    | dou·ze    | do·dici       | duo·decim
trois   | trei·ze   | tre·dici      | tre·decim
quatre  | quator·ze | quattor·dici  | quattuor·decim
cinq    | quin·ze   | quin·dici     | quin·decim
six     | sei·ze    | se·dici       | se·decim
sept    | dix-sept  | dici(as)sette | septem·decim
huit    | dix-huit  | dici-otto     | duo·de·viginti
neuf    | dix-neuf  | dici(an)nove  | un·de·viginti

(18 and 19 in Latin are computed as 20-2 and 20-1).
Each language has its own way to cope with small numbers.


问:
Inequality with mean inequality

If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.

ContentLicense: CC BY-SA 4.0

Title: Inequality with mean inequality

Body:
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$
$$
= \sqrt{
9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
2\sum_{cyc}
{\sqrt{
\left(3 +
\frac{1}{a^{2}}\right)
\left(3 +\frac{1}{b^{2}}\right)}}
}
$$
$$\tag{By C-S}\geqslant \sqrt{9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
2\sum_{cyc}{3 + \frac{1}{ab}}}$$
$$= \sqrt{9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
54}$$
$$\geqslant \sqrt{9 +18 + 54} =\sqrt{81} = 9$$
Which uses $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 18$ and $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geqslant \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.


问:
Evaluate $\int_{-\frac{1}{\sqrt 3}}^{\frac{1}{\sqrt 3}} \frac{x^4}{1-x^4} \cos^{-1} (\frac{2x}{1+x^2}) dx$
Note that the integral can(not) be simplified as
$$
2\int_0^{1/\sqrt 3} \frac{x^4}{1-x^4} \cos^{-1} \left(\frac{2x}{1+x^2}\right)dx
$$
Since $\cos^{-1}$ is not an even function. Let $x=\tan y$
$$
\implies
\int_0^{\pi/6}\frac{x^4}{1-x^4} \left(\frac{\pi}{2} -2y\right) \sec^2(y)\ dy
$$

So how do I solve the original problem?

Body:
Hint:

$$\int_{-a}^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx$$

$$=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx+\int_0^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}dx=I_1+I_2$$

For the first integral set $x=-y,dx=-dy$

$$I_1=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx=-\int_a^0\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$

$$=\int_0^a\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$

Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,

$$I_1=\int_0^a\dfrac{y^4}{1-y^4}\left(\pi-\cos^{-1}\dfrac{2y}{1+y^2}\right)dy=\pi\int_0^a\dfrac{y^4}{1-y^4}dy-I_2$$

Hope you can take it from here!


问:
Evaluate $6\log_8(4)$ without using a calculator
I am to evaluate $6\log_8(4)$ without a calculator. The answer is provided as 4 but I cannot see how to arrive at 4.

Ignoring the 6 at the beginning of the expression, $log_8(4)$ can be written as $8^x=4$.

Without the use of a calculator I cannot simplify that part any further. I know that the root or 3rd root of 8 is not 4.

I'm not sure if the leading 6 helps or if I am to multiply the end result by 6 or some part of the expression by 6 while doing the working out part?

Completely stuck here. How can I arrive at 4? Granular, baby steps appreciated.

[EDIT]
I'd like to add that this is the textbook chapter I am working of. It's the absolute beginning of learning about logarithms. Looking at the comments and answer so far, there's reference to dividing logs which has not been covered in this book so far. Given the content of this chapter, I wonder if it's expected of me that I know how to solve this question?

答:
Let's go step by step.

We start with the fact that $8=2^3$.  That's equivalent to $2=8^{1/3}$ from which we immediately deduce that $$\log _{8} 2=\frac{1}{3}$$

Now, of course, we really wanted $\log_84$.  But, as $4=2^2$ we have $$\log_84=\log_82^2=2\log_82=\frac 23$$  It follows that $$\boxed{6\log_84=6\times \frac 23=4}$$


问:
Cumulative probability of increasing probability for game fail stacking
In regards to Cumulative probability of increasing probability

Hello, sorry for digging up a grave post. I've been trying to wrap my head around this. Could you answer a few questions?

First, is this a Binomial Distribution problem?

Second, If I was doing some equipment upgrading, where every fail gives a % increase ( eg fail stacking) how would this work?

Third, Do you have an excel friendly formula? I've been trying to use BINOM.DIST on google sheet, and I believe it works with a non increasing %.

Thank you kindly and stay safe.

答:
Yes, if you have a chance of 3% to happen, the question 'How likely is it that after 20 trials it has not occurred even once?' is a question where the binomial distribution will give the correct result. In particular, you are looking for $P(X=0)$ for $X\sim B(0.03,20)$.

There are various ways to think about this. First of all, in the video game context designers will often 'cheat' and not actually implement the stated probabilities, because humans are terrible at accepting true randomness as random. Even if you have just a few hundred players in a coin-flipping game one of them will get 10 Heads or Tails in a row, which will make their personal experience different from what the developer might have intended.

What you can do however is just multiply the success chance by a factor or just add something, in both cases with a cutoff of 1 since probabilities cannot be bigger than 1.
So if your starting probability is $p_0$ and you want to increase the probability by $q$ then you might just set $p_1=p_0+q$ or $p_1=p_0(1+q)$ (these approaches will need different values of $q$ of course, they are not at all equivalent). There are smoother ways to increase probabilities without ever going over $1$ but that does not seem to be what you are looking for.
This would mean that you are having a different distribution after every additional failure. You can model this via Markov Chains, for example, they are nice and visual.

I might be wrong but if you want the distribution of your events to depend on the previous events then I think it will be very hard to model this in a single excel cell. If you want to get some simulated data for various implementations I would not use a spreadsheet but whatever programming language you find most approachable.


问:
Find relation between Fourier coefficient of $ f(x) $ and $ \overline{f\left(x\right)} $
Let $ f:\mathbb{R}\to\mathbb{C} $ be integrable and periodic with period $ 2\pi $.
I have to find relation between Fourer coefficients of $ h\left(x\right)=\overline{f\left(x\right)} $ and $ f(x) $

Here's what Ive tried:

$ \tilde{h}\left(k\right)=\frac{1}{2\pi}\intop_{-\pi}^{\pi}\overline{f\left(x\right)}e^{-ikx}dx $

Now what I want to do is this:

$ \tilde{h}\left(k\right)=\frac{1}{2\pi}\intop_{-\pi}^{\pi}\overline{f\left(x\right)}e^{-ikx}dx=\overline{\frac{1}{2\pi}\intop_{-\pi}^{\pi}f\left(x\right)e^{ikx}dx}=\overline{\tilde{f}\left(-k\right)} $

But I cannot justify that for complex function $ g $ the following equallity holds:

$ \intop_{a}^{b}\overline{g\left(x\right)}dx=\overline{\intop_{a}^{b}g\left(x\right)dx} $.

Any ideas how to make it work?

Thanks in advance.

答: By definition, $ \int {g(x) \mathrm{d}x} = \int {a(x) \mathrm{d}x } + i \int {b(x) \mathrm{d}x} $, where $ a(x) = \mathrm{Re}(g(x)) $ and $ b(x) = \mathrm{Im}(g(x)) $.  Apply this definition to $ \bar{g}(x) = a(x) - i b(x) $.


问:
If $S \subset A$ and $S \neq A$, then is $\text{cl}(S) \subset A$?
Synopsis

I was wondering if it were a true statement that the closure of a set $A$ is the smallest set containing $A$ that is not equal to $A$. In other words, if there exists a set $S$ such that $A \subset S$ but $A \neq S$, then is it true that $\text{cl}(A) \subseteq S$, regardless of if $S$ is open or closed? Can I prove it? Can you give hints if I can? If this statement is false, then please give a counterexample.

I know that the closure of a set is the smallest closed set, but for an exercise I'm working on, I don't know if the set containing $S$ is open or closed.

答:
It depends on whether:

You are looking at $A$ and $S$ as subsets of another, bigger, ambient topological space $T$, or

You are looking at the topological space $S$ with its own topology (which it may have inherited from some bigger space $T$ - or not).

In the first case, your statement is invalid. Take $A$ to not be closed in $T$ (say $A=B(0,1)$ in $T=\mathbb R^2$ - the open ball) and $S=A$. Then the closed ball $\text{cl}(A)=B[0,1]\not\subseteq S$.

In the second case, your statement is valid. Namely, $\text{cl}(A)$ is the intersection of all closed sets that contain $A$. Thus, as one of those sets is $S$ itself ($S$ is, in its own topology, both open and closed!), it follows that $\text{cl}(A)\subseteq S$.

The reason the statement is invalid in the first case but valid in the second is that the "closure operator" $\text{cl}$ is in one case taken in $T$ and in the second in $S$ - and it is not the same operator, i.e. closure on $S$ is not a restriction of closure in $T$.


问:
Show that if $ b_1, b_2, …, b_n $ is a superincreasing sequence then $ b_{i+j} \gt 2^{j-1} b_i $ for $ 1 \leq i \leq n $ and $ 1 \leq j $.
Superincreasing sequence definition: https://en.wikipedia.org/wiki/Superincreasing_sequence.

I tried converting $ b_{i+j} $ using property from this question, but didn't get anywhere.

Any help would be greatly appreciated.

答:
So $b_{i+j}>b_{i+j-1}+\cdots +b_{i}+b_{i-1}+\cdots +b_1$ but $b_{i+j-1}>b_{i+j-2}+\cdots b_i+\cdots +b_1,$ hence
$$b_{i+j}>(b_{i+j-2}+\cdots b_i+\cdots +b_1)+b_{i+j-2}+\cdots b_i+\cdots +b_1=2\left (b_{i+j-2}+\cdots b_i+\cdots +b_1\right ),$$
so doing this $j-1$ times...
$$b_{i+j}>2\left (b_{i+j-2}+\cdots +b_i+\cdots +b_1\right )>2\left (2\left (b_{i+j-3}+\cdots +b_i+\cdots +b_1\right )\right )>\cdots >2^{j-1}\left (b_i+\cdots+ b_1\right ),$$
but the $b_j$ are positive reals, so $b_j>0$ hence put everyone to $0$ except $b_i$ then
$$b_{i+j}>2^{j-1}\left (b_i+0+\cdots +0\right)=2^{j-1}b_i.$$


问:
If $\sum\limits_{j=0}^{\infty}a_j (e^t)^j=0$ for all $t$, then $a_j = 0$ for all $j$.
Consider the series:

$$\sum\limits_{j=0}^{\infty}a_j (e^t)^j=0 $$
for $t \in (-t_0,t_0)$ and $t_0>0$. How could I argue for $a_j=0$ for all $j$?

Maybe I consider this as an infinite polynomial in $x:=e^t$, which can only be zero if all coefficient $a_j$ are $0$.

Is this possible?

Alternative:

$$f^{n}(0)= \sum_{j=n}^{\infty} \frac{j!}{(j-n)!} 0^{k-n}=n! a_n$$

答:
Both approaches work.

If $g(x) = \sum a_n x^n$ then the radius of convergence of $g$ is (at least) $e^{t_0}$. Since $g$ vanishes in the nondegenerate interval $e^{-t_0} < x < e^{t_0}$ it must vanish everywhere, so its coefficients vanish.

Proving that vanishing power series have vanishing coefficients can be done using derivatives, like your second approach.


答: With so little information about sets A, B and the space in which those sets exist, it is very easy to come up with an example when the second inclusion does not hold. Take some space X with trivial topology (that is, only the whole space and the empty set are open) and a least 3 elements a, b, c. Then set A = {a}, S = {a, b}. A $\subset$ S, but cl(A)=X $\nsubseteq$ S.

Title: If $S \subset A$ and $S \neq A$, then is $\text{cl}(S) \subset A$?

Body:
Synopsis

I was wondering if it were a true statement that the closure of a set $A$ is the smallest set containing $A$ that is not equal to $A$. In other words, if there exists a set $S$ such that $A \subset S$ but $A \neq S$, then is it true that $\text{cl}(A) \subseteq S$, regardless of if $S$ is open or closed? Can I prove it? Can you give hints if I can? If this statement is false, then please give a counterexample.

I know that the closure of a set is the smallest closed set, but for an exercise I'm working on, I don't know if the set containing $S$ is open or closed.


问:
Integral of $\int^1_0 \frac{dx}{1+e^{2x}}$
I am trying to solve this integral and I need your suggestions.

I think about taking $1+e^{2x}$ and setting it as $t$, but I don't know how to continue now.

$$\int^1_0 \frac{dx}{1+e^{2x}}$$
Thanks!

答:
HINT:

Putting $e^{2x}=v\implies 2x=\ln v\implies 2dx=\frac{dv}v$

$$\int^1_0 \frac{dx}{1+e^{2x}}=\int_1^{ e^2}\frac1{2v(1+v)}dv=\frac12\cdot\int_1^{ e^2}\left(\frac1v-\frac1{v+1}\right)dv $$

ContentLicense: CC BY-SA 3.0


问:
What is the value of $\int\frac{\sin(3x)}{\cos(7x)\cos(4x)}$
The integral is $\int\frac{\sin3x}{\cos7x\cos4x}$

I have tried sum to product rule and $\sin3x$ formula:
$\int\frac{3\sin x-4\sin^3 x}{\frac{1}{2}(\cos3x+\cos11x)}$

How should I proceed further?

答:
$$\frac{\sin 3x}{\cos 7x\cos 4x} = \frac{\sin (7x-4x)}{\cos 7x\cos 4x}=\frac{\sin 7x\cos 4x - \cos 7x \sin 4x}{\cos 7x \cos 4x} = \tan 7x - \tan 4x$$

I think you can take it from here.


问:
Transformations with Inverse Gamma RVs
I have two independent random variables:

$X_{1}$ is governed by an Inverse Gamma(a, b).

$X_{2}$ is governed by an Inverse Gamma(a, b).

Given the transformation $Y = X_{1}/X_{2}$, I want to find the marginal distribution for $Y$, $f_{Y}(y)$; however, I am a little confused how to proceed with the transformation. Here is how I tried thinking of it:

To make things easier, I tried assigning $W = X_{2}.$

$$Y = X_{1}/X_{2} \rightarrow Y = X_{1}/W$$

$$W = X_{2} \rightarrow X_{2} = W$$

I am unsure if this is correct and how to proceed.

答:
To simplify the notation let's set $Z=\frac{X}{Y}$

Let's set

$$\begin{cases}
z=\frac{x}{y}\\
u=x,
\end{cases}\rightarrow\begin{cases}
x=u\\
y=\frac{u}{z},
\end{cases}$$

The jacobian is $|J|=\frac{u}{z^2}$ and thus the joint density $(U,Z)$ is

$$f_{UZ}(u,z)=\frac{z^{a-1}}{\Gamma(a)\Gamma(a)}\cdot \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{u^{2a+1}}$$

This can be rewritten as

$$\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{\Gamma(2a)}\Bigg(\frac{1+z}{u}\Bigg)^{2a+1}\cdot\frac{1}{1+z}$$

To get your $f_Z(z)$ you must integrate

$$\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot\underbrace{\int_0^{\infty} \frac{b^{2a}e^{-\frac{b}{u}(z+1)}}{\Gamma(2a)}\Bigg(\frac{1+z}{u}\Bigg)^{2a+1}d\Bigg(\frac{u}{1+z}\Bigg)}_{=1}$$

Thus your density is a Beta prime as the integral is an integral of an Inverse gamma

$$ \bbox[5px,border:2px solid red]
{
f_Z(z)=\frac{\Gamma(a+a)}{\Gamma(a)\Gamma(a)}z^{a-1}(1+z)^{-a-a}\cdot\mathbb{1}_{[0;+\infty)}(z)
\
}
$$

If you do not like to deal with Inverse Gamma rv's you can rewrite you $Z$ in the following way

$$Z=\frac{X}{Y}=\frac{\frac{1}{Y}}{\frac{1}{X}}$$

and thus you can use Gamma distributions if you are familiar with them.

In fact, if $X\sim Gamma$ then $\frac{1}{X}\sim InverseGamma$

Title: Transformations with Inverse Gamma RVs


问:
Infinite Series of Positive Elements in a $\textrm{C}^*$-algebra

Let $A$ be a $\textrm{C}^*$-algebra and let $(a_n)$ be a sequence of elements of $A$. Is it then true that
$$\sum_{n=1}^\infty a_n^*a_n\ \text{converges} \iff \sum_{n=1}^\infty a_na_n^*\ \text{converges?}$$

I've seen this stated as obvious, albeit in a more general setting in a paper, and I agree that it feels like something that might be true, but I don't seem to be able to get anywhere with it (other than noting its trivial if $A$ is abelian).

My only real attempt went along the lines of if $\sum_{n=1}^\infty a_n^*a_n$ converges, then it converges to some positive element $a^*a\in A$, then the obvious guess of what $\sum_{n=1}^\infty a_na_n^*$ might converge to is $aa^*$. But then I got no further.

It feels like quite an elementary question, but I'm completely stumped! Can anyone help with this?

答:
In the algebra of all bounded operators on $\ell^2$,  consider the operator
$a_n$
defined by
$$
  a_n(\xi ) = \frac 1{\sqrt n}\langle \xi , e_1\rangle e_n, \quad\forall \xi \in  \ell^2,
  $$
where $\{e_n\}_{n\geq 1}$ is the canonical basis.  I'll leave it up to you to verify that
$$\sum_{n=1}^\infty a_na_n^*$$ converges in norm,  but
$$\sum_{n=1}^\infty a_n^*a_n$$ doesn't converge, not even weakly!


问:
Prove the formula of combination number
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{x+k}=\frac{n!}{x\left( x+1 \right) \cdots \left( x+n \right)}}
\\
$$
I recently read an article and found this formula above. I tried to prove it, but only proved the following result. I don't know how to prove this more general combinatorial equation.
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{m+k+1}=\int_0^1{\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \left( -1 \right) ^kx^{m+k}\text{d}x}=\int_0^1{x^m\left( 1-x \right) ^n}\text{d}x}=\frac{m!n!}{\left( m+n+1 \right) !}}
$$

答:
By partial fractions, we know
$$\frac{n!}{x(x+1) \cdots (x+n)} = \frac{a_0}{x} + \frac{a_1}{x+1} + \dots + \frac{a_n}{x+n}$$
for some set of numbers $a_0, a_1, \dots , a_n$. To find $a_k$ we multiply both sides of the equation by $x+k$ and then set $x=-k$ in the resulting equation.  The result is
$$a_k = (-1)^k \binom{n}{k}$$

Title: Prove the formula of combination number


问:
Solve $2^x + 3^{2x + 2} = 5$
Out of curiosity, how would you approach the following exponential equation:
$$2^x + 3^{2x + 2} = 5$$

This resulted from a work on a system of logarithmic equations and I know for a fact that it has a single solution, $x \approx -0.345464$, which I obtained with the help of Wolfram Alpha. Anyhow, I simply can't figure out a way to solve for $x$ in an enclosed form.

My main question would be: is the solution generally expressable as a real number in a closed form? Or is this equation a dead end and maybe I should reconsider my approach to the system of equations?

That also brought me to the following thought. The equation can be rewritten as:
$$2^x + 9\cdot 9^x - 5 = 0$$
which is a general composition of exponential functions. I couldn't find any general approaches to such general exponential equations, but my intuition is that there's not much 'general' stuff to do with them and all depends on the current case.

I hope you could understand me, my terminology is perhaps not very precise and I'm a bit self learned in mathematics, so any clarification is welcome! Thanks in advance!

答:
I don't think one can solve this "nicely". However, the following graph may help! Even if you can't solve it by hand, you can always get an idea of what range of values should the solution fall within.

If you really want to solve it numerically, the Newton-Raphson method is just one of the many methods you could try out to obtain the root of $f(x) = 2^x + 3^{2x+2} - 5$. Note that $f'(x)>0$ so there's only one solution.

Title: Solve $2^x + 3^{2x + 2} = 5$


问:
If $0<x<1$, show that $0<2x-2x^2<1$
If $0<x<1$, show that $0<2x-2x^2<1$

I have so far:
$0<2x<2$ and $-2<-2x^2<0$

So, adding both inequalities, I get:
$-2<2x-2x^2<2$

Any hint on how to make the bounds tighter?

Title: If $0<x<1$, show that $0<2x-2x^2<1$


问:
Curve fitting with an invertible function
I have some data that need to be transformed as shown in the picture given some function.

example of input data and desired result

The function in question can take whatever form but with the following requirements:

It needs to be able to fit the data shown (i.e f(orange data) = blue data as much as possible)

It needs to be invertible

I need to be able to adjust its parameters so that in certain conditions it becomes identity

I've played around with various functions and although I can find various options that give a perfect fit, either I can't invert them or I can't figure out how to adjust the parameters to get an identity.

To give a more concrete example, let's say that a function of the form f(x) = ax + b could work (it doesn't). In this case the inverse function would be finv(x) = (x-b)/a.
Now we could imagine a weight that is provided as an external parameter which goes between 0-1, in which case I could adjust parameters a and be so that

a'= w*a - (1-w) -> so that when the weight is at 1 we get the desired fitting behavior, when the weight is at 0 we get 1

b' = w*b -> so that b becomes 0 when the weight is 0
The above would give me f(x) = x when the weight is at 0 and ax+b when the weight is at 1.

What function could fulfil all 3 criteria given the form of the data I have?

Any ideas would be extremely welcome, I've been looking at this for days and I'm getting nowhere.

Thanks!

Edit to add some examples that give me a good fit:
f(x) = ax/(b-x)^c

good fit, I can get it to become identity, but I don't see how to invert
f(x) = (ax/(b-x))^c

good fit, invertible but can't get it to become identity

Throwing more parameters at it:
f(x) = (ax^b)/(c-x^d)

again good fit, identity if I set a=1, b=1, c=2, d=0 but can't see how to invert

答:
Thaking account of your comment, you have to chose only analytically invertible functions. So your choice is very limited : polynomials of degree lower than five and a few other elementary functions without combination between them. I doubt that with a so limited choice an acceptable fitting to data be possible in general.

May I suggest another approach ?

First, fit to data any convenient function possibly not invertible, say $y=f(x)$ .

Second, fit to "inverted data" any other convenient function possibly not invertible, say $x=g(y)$. "Inverted data" means $(x,y)$ replaced by $(y,x)$.

So, you have a couple of functions which can be used the same way than if you have $f$ and $f^{-1}$. You can compute $g\big(f(x)\big)$ and compare to $x$ in order to evaluate the deviation.  Or you can compute $f\big(g(y)\big)$ and compare to $y$.


答:
Your bounds could be tighter, because $2x$ and $-2x^2$ aren't independent.

Complete the square:  $-2x^2+2x=-2x^2+2x-\frac12+\frac12=-2(x-\frac12)^2+\frac12\le\frac12$.

On the other hand, if $0<x<1$, then $x^2<x$, so $2x-2x^2>0$.

Title: If $0<x<1$, show that $0<2x-2x^2<1$

Body:
If $0<x<1$, show that $0<2x-2x^2<1$

I have so far:
$0<2x<2$ and $-2<-2x^2<0$

So, adding both inequalities, I get:
$-2<2x-2x^2<2$

Any hint on how to make the bounds tighter?


答:
You can take the inequality, split it by the two sides:
$$2x-2x^2>0$$
$$2x(1-x) > 0 \tag{1}$$

Divide both sides of $(1)$ by $2x$  since we know $0<x<1$:
$$1-x > 0$$
Rearranging:
$$\boxed{x<1}$$
Divide both sides of $(1)$  by $1-x$  since we know $0<x<1$ ,  then $1-x$ will be a positive:
$$2x > 0 $$
Dividing by 2:
$$\boxed{x>0}$$

This one could be solved by sketching:
$$-2x^2+2x-1<0\tag{2}$$
$$x\in \mathbb{R}$$

$$$$
$$\therefore \quad 0<x<1 $$

EDIT: Fixed typos and added full solution.

Title: If $0<x<1$, show that $0<2x-2x^2<1$


问:
In how many ways can we distribute $m$ balls to $n$ baskets when $n > m$

Given $m$ balls and $n$ baskets, in how many ways can we distribute the $m$ balls to the $n$ baskets when given that $n > m$ (the number of baskets is greater than the number of balls)?

So I said, that it does not differ if $n>m$ the solution is still ${n+m-1 \choose n-1}$, is that correct? I merely considered it as the number of solutions $x_1 + x_2 + ... + x_n = m$ when $0 \leq x_i$. Is that correct?

Also, as an addition:

In addition to the question, what if there can be only a maximum of one ball in each basket?

So I said, it is still the same $x_1 + x_2 + ... + x_n = m$ but this time $0 \leq x_i \leq 1$ and it is still the same solution. Is that correct?

答:
We assume the usual convention: the balls are indistinguishable, and the baskets are distinguishable.

For your first question, yes, it is precisely the formula you are familiar with.

When we are restricted to at most one ball per basket, then yes, we are counting the solutions of $x_1+x_2+\cdots+x_n=m$ with the restriction that for any $i$ we have $x_i=0$ or $x_i=1$.

But to me the easiest way to think of it is that we are choosing the $m$ baskets that will be lucky and get a ball. Or if you wish choosing the $i$ such that $x_i=1$. Or else think of it as forming an $n$-letter "word" made up of $m$ $1$'s and the rest $0$'s.

Whichever way one views it, this can be done in $\dbinom{n}{m}$ ways.


问:
Trouble with the proof of Proposition 4.3.18 of Pedersen's Analysis Now
I am currently trying to understand the proof of Proposition 4.3.18 in Pedersen's Analysis now, which reads

To each Tychonoff space $X$ there is a Hausdorff compactification $\beta(X)$, with the property that every continuous function $\Phi: X \to Y$, where $Y$ is a compact Hausdorff space, extends to a continuous function $\beta \Phi: \beta(X) \to Y$.

The proof starts by noting that $C_b(X)$ is a commutative unital C$^*$-algebra, and is therefore isometrically isomorphic to a (commutative and unital) C$^*$-algebra of the form $C(\beta(X))$, where $\beta(X)$ is a compact Hausdorff space.

By the Gelfand duality between the category of commutative and unital C$^*$-algebras and the category of compact Hausdorff spaces, we can take $\beta(X) = \Omega(C_b(X))$, the space of characters on $C_b(X)$.

Then we can define a map $\iota: X \to \beta(X)$, where $\iota(x)(\phi) := \phi(x)$ for all $x \in X$ and $\phi \in \beta(X)$.

The particular part of the proof that I am struggling to understand is the proof that $\iota(X)$ is dense in $\beta(X)$.

He argues that if $\iota(X)$ is not dense in $\beta(X)$, then there is a non-zero continuous map $f: \beta(X) \to \mathbb{C}$ vanishing on $\iota(X)$. This I understand. He then says that under the identification $C_b(X) = C(\beta(X))$, this is impossible. This is the sentence I am stuck on. Why is it impossible under this identification?

We have that $C_b(X)$ is isometrically isomorphic to $C(\Omega(C_b(X)))$ via the map $\delta: g \mapsto (\delta_g: \Omega(C_b(X)) \to \mathbb{C}, \phi \mapsto \phi(g))$. I am pretty sure what Pedersen is getting at is that the map $\delta^{-1}(f)$ is zero, but I am not able to show that this is the case. This answer also claims that a similar map is zero.

In summary, my question is:

Can we show that $\iota(X)$ is dense in $\beta(X)$ by showing that $\delta^{-1}(f) = 0$? If so, how do we do this?

答:
Consider a special set of characters of $C_b(X)$, for each $x\in X$ define:

$$\delta_x: C_b(X)\to\Bbb C, \quad g\mapsto g(x)$$

Since the (non-zero) characters of $C_b(X)$ are the points of $\beta X$ this gives you a way of embedding $X$ into $\beta X$. Now if $f$ is some continuous function on $\beta X$ we may identify it also with an element $\tilde f\in C_b(X)$, namely $\tilde f = \delta^{-1}(f)$ using your notation. Remember that
$$f(\delta_x) = \delta(\tilde f)\,(\delta_x) = [\phi \mapsto \phi(\tilde f)]\,(\delta_x)= \delta_x(\tilde f) = \tilde f(x) $$

Asking that $f$ vanishes on $X$ is asking that $f(\delta_x)=0$ for all $x\in X$, in particular looking at $\tilde f$ this becomes:
$$\tilde f(x)=0\quad \forall x\in X$$
the only function in $C_b(X)$ satisfying this property is the zero function.


问:
Show that $f(n)$ tends to $+\infty$ if $f$ is injective
Given that $f$ is an application from $\Bbb N \to  \Bbb N$  :

Show that the sequence $(f(n))_n$ tends to $+\infty$ if $f$ is injective.

Since $f$ is injective I can see that every natural number has at most one antecedent by $f$ and since the set of antecedent is infinite the set of images by $f$ is also infinite. I have some trouble writing this correctly let alone find where to go from there to prove the question. Any help would be appreciated!

答:
We need to show, for all $M$ (without loss of generality, assume $M \in \Bbb{N}$), we can find an $N \in \Bbb{N}$ such that
$$n > N \implies f(n) > M.$$
The set $\{1, 2, \ldots, M\}$ is a finite set. Note that
$$f^{-1}\{1, 2, \ldots, M\} = \bigcup_{n=1}^M f^{-1}\{n\},$$
where the right hand side is a finite union of singleton or empty sets (since $f$ is injective), which means the left hand side finite. Thus, we may define
$$N = \max f^{-1}\{1, 2, \ldots, M\}.$$
Hence,
$$n > N \implies n \notin f^{-1} \{1, 2, \ldots, M\} \implies f(n) \notin \{1, 2, \ldots, M\} \implies f(n) > M,$$
as required.


问:
Help in understanding special algebra expansions
enter image description here

I cannot understand how this formula is called and in what cases it is applied.  Can you help with this and give examples of equations using these formulas?

答:
$$S = 1 + a + ... + a^{n-1}$$
$$aS = a + a^2 + ... + a^{n}$$
So by subtracting $$(a-1)S = a^n - 1$$
for $n \in \mathbb{N}$

So this is basically just geometric progression! You can derive the second result similarly - i.e. divide LHS and RHS by $(a+1)$ and observe that the sum $(1-a+a^2...-a^{2k-1}+a^{2k})$ is a geometric progression with common ratio $=-a$. Doing a manipulation similar to what I did above with $S$ will lead you to the result.

Examples of where these are used? Well, it's always nice to have some factorization tricks up your sleeve.


问:
Understanding the definition of a local ring.
Here is the definition my professor gave to us:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

My question is:

Why the only proper ideal of a field is $(0),$ could anyone explain this to me please?

答: If $0\ne a\in\mathfrak a\subset F$ then for all $c\in F$, we have $c=\frac caa\in \mathfrak a$ and so $\mathfrak a=F$.


答:
If $F$ is a field and $I$ is an ideal, suppose $j \neq 0 \in I$. Then since $I$ is an ideal, for any $x \in F$, we have
$$
x = (xj^{-1})j \in I.
$$
So either $I = (0)$ or $I = F$.

Title: Understanding the definition of a local ring.

Body:
Here is the definition my professor gave to us:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

My question is:

Why the only proper ideal of a field is $(0),$ could anyone explain this to me please?


答:
Suppose $I$ is an ideal of a field $F$. We want to show that either $I = F$ or $I = {0}$. We have two cases to consider:

Case1: Suppose $I$ has no non-zero element, then $I=0$

Case2: Suppose $I$ has a non-zero element, say $x$. Then, of course, $x$ lies in the field $F$. Hence, $x$ has a multiplicative inverse. Since, the ideal is closed under multiplication $x \times x^{-1} = 1$ must in the ideal. This means that, $I = F$, again due to closedness of the ideal with respect to the multiplication.


答:
As you expect, you cannot get a closed form solution.

Consider that ou are looking for the zero of function
$$f(x)=2^x + 3^{2x + 2}-5$$ By inspection, you know that the solution is somewhere between $-1$ and $0$.

Try Taylor expansion
$$f(x)=5+x (\log (2)+18 \log (3))+x^2 \left(\frac{\log ^2(2)}{2}+18 \log ^2(3)\right)+O\left(x^3\right)$$
Using the first term only, you have, as an estimate
$$x=-\frac{5}{\log (2)+18 \log (3)}\approx -0.244282$$ Now, use more terms and use series reversion to get
$$x=t-\frac{ \left(\log ^2(2)+36 \log ^2(3)\right)}{2 (\log (2)+18 \log (3))}t^2 +O\left(t^3\right)\quad \text{with} \quad t=\frac{f(x)-5}{\log (2)+18 \log (3)}$$ Making $f(x)=0$ as we desire, then another estimate
$$x=-\frac{5 \left(7 \log ^2(2)+828 \log ^2(3)+72 \log (2) \log (3)\right)}{2 (\log   (2)+18 \log (3))^3}\approx -0.30832$$

Now, let us polish the root using the so simple Newton method and, since I am lazy, I shall start with the first estimate I wrote. The successive iterates are
$$\left(
\begin{array}{cc}
 0 & -0.2442817505 \\
 1 & -0.3353427850 \\
 2 & -0.3453567313 \\
 3 & -0.3454642504 \\
 4 & -0.3454642627
\end{array}
\right)$$ and we could get as many decimal figures as required. For the fun of it, the solution
$$-0.345464262655186305826853008027166943006071510\cdots$$

Title: Solve $2^x + 3^{2x + 2} = 5$

Body:
Out of curiosity, how would you approach the following exponential equation:
$$2^x + 3^{2x + 2} = 5$$

This resulted from a work on a system of logarithmic equations and I know for a fact that it has a single solution, $x \approx -0.345464$, which I obtained with the help of Wolfram Alpha. Anyhow, I simply can't figure out a way to solve for $x$ in an enclosed form.

My main question would be: is the solution generally expressable as a real number in a closed form? Or is this equation a dead end and maybe I should reconsider my approach to the system of equations?

That also brought me to the following thought. The equation can be rewritten as:
$$2^x + 9\cdot 9^x - 5 = 0$$
which is a general composition of exponential functions. I couldn't find any general approaches to such general exponential equations, but my intuition is that there's not much 'general' stuff to do with them and all depends on the current case.

I hope you could understand me, my terminology is perhaps not very precise and I'm a bit self learned in mathematics, so any clarification is welcome! Thanks in advance!


问:
Logical equivalence proof of $\forall x[\forall y(P(x,y))] \equiv \forall y[\forall x(P(x,y))]$
So I have been asked to prove the following logical equivalence:

$\forall x[\forall y(P(x,y))] \equiv \forall y[\forall x(P(x,y))]$

Of course, I understand intuitively that the equivalence is true but I'm having a hard time to actually prove it. Should it be done through other equivalences? If so, through which ones? Or maybe I should explain in my own words since it so obvious?

答:
Assume $$\forall x[\forall y(P(x,y))]$$
By specialization of $x$
$$\forall y(P(x,y)) $$
and by sepcialization of $y$
$$ P(x,y).$$
Now generalize over $x$
$$\forall x(P(x,y)) $$
and generalize over $y$:
$$\forall y[\forall x(P(x,y))]$$


问:
Fourier coefficients implies absolutely continuity
When I read some paper, it says

$u\in H^{3/2}$ if and only if $u$ is absolutely continuous with $u'\in H^{1/2}$, where
$$H^s=\left\{ u\in L^1(\mathbb{S}^1):\|u\|_{H^{s}} \doteq\left(\sum_{n=-\infty}^{+\infty}|n|^{2 s}\left|a_{n}(u)\right|^{2}\right)^{\frac{1}{2}} \right\}, a_{n}(u)=\frac{1}{2 \pi} \int_{0}^{2 \pi} {u}(e^{i\theta}) e^{-i n \theta} d \theta.$$

"If part" is easy to understand, but how to prove if $u\in H^{3/2}$, $u$ is absolutely continuous?

Any hint will be appreciated!

答:
The absolute continuous part of the assertion is somewhat useless, it is just here to say that u' makes sense for people who do not know distribution theory.

Nevertheless your question boils down to the fact that $H^{3/2}$ is imbedded into absolutely continuous functions. The proof depends on how you define $H^{3/2}$. If for example you  follow the definition with Fourier coefficients, then you can check that
$v=\sum ina_ne^{inx}$ is an $L^2$ function (this is true even in $H^1$), and that actually $u-\int_0^x v$ is a constant. This last point is due to the fact that for any smooth periodic $\phi$, $\int_0^{2\pi}(u-\int_0^x v)\phi'dx=0$. (corrected after the comment)

Now obviously $\int_0^x v$ is absolutely continuous, as it is the integral of an $L^2$ (hence $L^1$) function. Note that this argument works for any $H^s$ function, $s\geq 1$.

There is maybe a simpler argument, but this is not trivial as you can only embed $H^{3/2}$ into $C^{0,\alpha},\ \alpha<1$, and unfortunately these spaces are not comparable to absolutely continuous functions.


答:
Sometimes a change of variable helps:

Let $x=(1+u)/2$ with $-1\lt u\lt1$. Then

$$2x-2x^2=2x(1-x)=2\left(1+u\over2\right)\left(1-u\over2\right)={1-u^2\over2}$$

and we have $-1\lt u\lt1\implies 0\le u^2\lt1$, so that $0\lt(1-u^2)/2\le1/2\lt1$

Title: If $0<x<1$, show that $0<2x-2x^2<1$


问:
Maximal order of magnitude of Prime Omega Function
Let $\omega(n)=\sum_{p|n}1$ be the prime omega function.

Prove that for any $\epsilon > 0$,

There exists $N>0$ such that for all $n > N$,

$$\omega (n)<\frac{(1+\epsilon)\log{n}}{\log{\log{n}}}$$

There exists infinitely many $n$ such that

$$\omega (n)>\frac{(1-\epsilon)\log{n}}{\log{\log{n}}}$$

I don't even know how to begin the question.

I only know that $\displaystyle \left| \frac{\sum_{i=1}^n\omega(i)}{n}-\log{\log{n}}\right|<C$ for some constant $C$ and $n\geq 3$.

For (1), I don't know how to come up with $N$ with $\epsilon$;

For (2), I suppose somehow we can construct a formula for those $n$'s, but I don't know how to deal with the $\log{\log{n}}$.

Thank you for any help.

答:
We'll need the PNT: $\pi(x) \sim \frac{x}{\ln x},\theta(x)\sim x$.

Let $N_x = \prod \limits_{p \leq x} p = e^{\theta(x)} $ and so $\omega(N_x) = \sum \limits_{p|N_x} 1 = \sum \limits_{p\leq x} 1 = \pi(x) \sim \frac{x}{\ln x}$.

$\ln N_x = x$ and $ \ln \ln N_x = \ln x$ and so $ \omega(N_x) \sim \frac{x}{\ln x} \sim \frac{\ln N_x} {\ln \ln N_x}$, this means that for all $\epsilon >0$ there exists $x_0 $ such that for all $x> x_0$ we have that $ \frac{(1-\epsilon) \ln N_x}{\ln \ln N_x} < \omega(N_x) < \frac{(1+\epsilon) \ln N_x}{\ln \ln N_x}$, the lower bound is enough to answer your second question.

For the first you need to notice that when $N$ is a multiplication of $\pi(x)$ different primes then $N \geq N_x$ since the primes in $N_x$ are in ascending order and so $\omega(N) =\omega(N_x) < \frac{(1+\epsilon) \ln N_x}{\ln \ln N_x} \leq  \frac{(1+\epsilon) \ln N}{\ln \ln N}$ because $\frac{\ln t}{\ln \ln t}$ is monotonic function.

Note : you can use lower/upper bound on $\theta(x),\pi(x)$ to make the upper argument an explicit one rather than implicit.


问:
Edge Coverings of Complete Graph With Loops
Let $G_{n,m} = (V,E)$ be a complete graph on $n$ vertices, $n \in \mathbb{Z}^+$, with one additional feature: there are $m$ loops at every vertex, $m \in \mathbb{N}$. In other words, for every vertex $u$ and for every vertex $v$ in $V$, if $u \neq v$, then there is exactly one $\{u,v\}\in E$, and we have $\{v,v\}_i \in E$ for $i=1,2,...,m$.

I am trying to determine the number of edge coverings on such a graph, which I denote as $C(G_{n,m})$. In other words, how many subsets of $E$ exist in which every vertex in $V$ is incident with at least one edge?

I stumbled upon a paper conveying the edge coverings of a complete graph on $n$ vertices, denoted as $K_n$, as

$$C(K_n) = \sum_{j=0}^{n} (-1)^j \binom{n}{j} 2^{\binom{n-j}{2}}$$

It is also given for $k$ edges, $0 \le k \le \binom{n}{2}$,

$$C(K_n,k) = \sum_{j=0}^{n} (-1)^j \binom{n}{j} {\binom{\binom{n-j}{2}}{k}}$$

but I am still having difficulty given the graph I'm examining contains loops as well. My best attempt has consisted of subtracting from the powerset of $E$ all those subsets in which there is at least $1$ isolated vertex. I ended up with a recursive equation as follows:

$$ C(G_{n,m}) = 2^{\binom{n}{2}+nm} - 1 - \sum_{t=1}^{n-1} \binom{n}{t} C(G_{n-t,m})$$

Any help is appreciated. Thanks!

答:
Notice, first, that the first formula is an inclusion-exclusion one, so you can not just take out the elements without including the repetition back. In that way, call
$$A_i = \{\text{collection of edges not covering vertex }i\},$$
so $$|A_i|=2^{\binom{n-1}{2}+n\cdot m-m},$$
and so, if you want to get the edges that do not cover any vertex in a set $X,$ you obtain $$\left | \bigcap _{x\in X} A_x\right |=2^{\binom{n-|X|}{2}+n\cdot m-|X|\cdot m},$$
because you are taking away the edges connected to $x\in X$ and each one has $m$ loops that you have now to take away. Then you are interested in
$$\left |\mathcal{P}(E)\setminus \bigcup _{v\in V}A_v\right |=2^{\binom{n}{2}+n\cdot m}-\left | \bigcup _{v\in V}A_v\right |$$
using PIE and the formula we found for intersection, we get that
$$C(G_{n,m})=\sum_{j=0}^{n} (-1)^j \binom{n}{j} 2^{\binom{n-j}{2}+(n-j)\cdot m}.$$


问:
Find the number of integral ordered pairs $(a, b)$ such that $a^2 + b^2 + ab = 1$
I simplified it to $(|a+b|+1)(|a+b|-1) = ab$ because I thought that i could find the no. of ordered pairs by finding the number of divisors of RHS, but the RHS isn't a constant so now I'm stuck. Please help.

答:
Hint (for an elementary answer) :

$$a^2+b^2+ab=1 \Longleftrightarrow \left( 2a + b\right)^2 + 3b^2=4$$


答: Perhaps first try to find all complex numbers $z=a+\frac{1+i\sqrt 3}2b$ with $a,b\in\Bbb Z$ and $|z|=1$.

Title: Find the number of integral ordered pairs $(a, b)$ such that $a^2 + b^2 + ab = 1$

Body: I simplified it to $(|a+b|+1)(|a+b|-1) = ab$ because I thought that i could find the no. of ordered pairs by finding the number of divisors of RHS, but the RHS isn't a constant so now I'm stuck. Please help.


问:
Using the ZFC axioms, is {1} = 2?
I am looking at the power set of two, $P(2)$ = {$\emptyset$,  {$\emptyset$},  {1}, {$\emptyset$, 1} } = {0, 1, {1} ,2}.

I consider {1} = {{0}}.  By the Empty Set Axiom and the Subset Axiom, this set includes the empty set, ie 0 $\in$ {1}. So, I can write {1} = {{0}} as {0, {0}} = {0, 1} = 2. This would mean the power set of two only has three distinct elements.

I am using ZFC Axioms.  Where am I going wrong?

Thanks for the comments.  I realise that my confusion was in how I was thinking about the Subset Axiom - not distinguishing "includes as an element" from "includes as a subset".

The question is maybe kind of ridiculous.  I am not sure if I should delete it or leave it.  I would be happy if a mod wants to delete it.

答:
Let $0:=\emptyset$, $1:=\{\emptyset\}$, so that $ \{1\}:=\{\{\emptyset\}\}$.

We do have $0\in1$ as well as $0\subset1$ and $0\subset\{1\}$, but $0\notin\{1\}.$

Title: Using the ZFC axioms, is {1} = 2?


问:
Why is it always necessary that a second order differential equation have 2 degrees of freedom?
I'm an AL student , and i have been recently introduced to the concept of Homogeneous Second Order Linear Differential Equations.I could understand how the Characteristic Equation forms and why it was necessary but in the Case of Repeated Real Roots , our teacher explained that the solution - ce^(rx) was not the most general solution because second order differential equations are supposed to always have "2 degrees of freedom".I was trying to figure out Why it is the case that second order differential equations should always have 2 degrees of freedom , but I cant seem to find a reasonable answer.Any answers would be much appreciated</p>

答: You're free to specify two initial conditions: $ y(x_0) $ and $ y'(x_0) $, and then the solution is determined (assuming certain mild hypotheses about the differential equation---look up theorems on existence and uniqueness if you are curious what these are).


答:
A 2nd order DE specifies the second derivative of the solution.  This leaves the value at some time $t_0$ and the first derivative at that same time unspecified.  Those are your two degrees of freedom.  One example might be a ball on a spring.  Wherever and however it starts, the equation of motion tells you that the ball will follow a sinusoid motion.  But it doesn't tell you where the initial location of the ball has to be, or the initial kick you can give it in its motion.

ADDED: This is not a complete explanation, but might give you an idea of how it all works.  Let's say we have a DE

$\displaystyle
\frac{d^2x}{dt^2} = f(t)$

and let's say we find out that it is satisfied by some $x_0(t)$.  Now, what if we randomly pick two numbers $a$ and $b$ and add the following to $x_0(t)$ to make $x_1(t)$:

$\displaystyle
x_1(t) = x_0(t) +a +bt $

Is $x_1(t)$ also a solution of the equation, or not?  Yes, it is, because

$\displaystyle
\frac{d^2x_1}{dt^2} = \frac{d^2x_0}{dt^2} = f(t)$

That is, the second derivative "killed" both the $a$ and $bt$ terms.  Note that this is going to happen regardless of what values you choose for $a$ and $b$.  So you have two "degrees of freedom" in your solution.

Title: Why is it always necessary that a second order differential equation have 2 degrees of freedom?

Body: I'm an AL student , and i have been recently introduced to the concept of Homogeneous Second Order Linear Differential Equations.I could understand how the Characteristic Equation forms and why it was necessary but in the Case of Repeated Real Roots , our teacher explained that the solution - ce^(rx) was not the most general solution because second order differential equations are supposed to always have "2 degrees of freedom".I was trying to figure out Why it is the case that second order differential equations should always have 2 degrees of freedom , but I cant seem to find a reasonable answer.Any answers would be much appreciated</p>


问:
conic sections, ellipse
A particle is travelling clockwise on the elliptical orbit given by
$$\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$$
The particle
leaves the orbit at the point $(-8, 3)$ and travels in a straight line tangent to the ellipse.
At which point will the particle cross the $y$-axis?</p>

答: We have $x^2/100+y^2/25=1$ so if we set $F(x,y)=x^2/100+y^2/25-1=0$ then $$y'=m_{\text{tangant} }=\frac{-F_x}{F_y}=\frac{-x}{4y}$$ and so $$m_{(-8,3)}=\frac{8}{12}$$ and the equation of the tangent line as @lab noted is $$y=\frac{8}{12}(x+8)+3$$

Title: conic sections, ellipse


问:
Does $\vDash t=s$ imply that $t$ and $s$ are notations for the same term?
Let $\mathcal L$ be a (fixed) language and let $t$ and $s$ both denote $\mathcal L$-terms.

My question is:

Is $\vDash t=s$ enough to conclude that $t$ and $s$ are notations for the same term?

Contrapositively in words: if two terms are distinct then is it always possible to find an interpretation (i.e. structure plus assignment) that confirms this fact?

I suspect this is the case but did not meet this (yet) in my efforts to get more familiarity with logics.

Thank you in advance.

答:
Yes, this is true - and this is one of those things which can seem more mysterious than it actually is if we think too hard about it.

Specifically, let $\mathcal{A}$ be any structure in which each function symbol is interpreted as an injective function and different function symbols' interpretations have disjoint ranges, and let $\nu$ be any variable assignment giving distinct variables different values. By induction on complexity of $s$ and $t$, we get $s^{\mathcal{A},\nu}\not=t^{\mathcal{A},\nu}$.

HINT: thinking of $s$ and $t$ as labelled trees growing downwards - e.g. "$f(g(x), y)$" has a top node labelled "$f$" which has two children labelled "$g$" and "$y$," the latter of which is terminal and the former of which has a child labelled "$x$" which is then terminal - look at the "highest" point of difference between them. For instance, if the outermost function symbols are different then disjoint ranges condition on $\mathcal{A}$ automatically gives the desired result.


问:
Is there an infinite product of reals in $(0,1)$ that doesn't converge to $0$?
I seem to recall hearing something like this from one of my professors last year.

Trying to be a little bit more succinct, does there exist a sequence $\{a_n\}_{i=0}^\infty$ such that $(\prod_{n=0}^\infty a_n) \ne 0,$ where $ \forall n,$ $ a_n \in (0,1)$?

It seems very counterintuitive, since if $\forall n, a_n = 1-\epsilon$, with arbitrarily small $\epsilon > 0$, the product still converges to zero, but I suppose if the sequence converges really quickly to $1$, it might be possible to come up with such an example. Is it?

答:
In fact, we can generate such a product that evaluates to any number $x\in(0,1)$ as follows:

Since
$$
\frac{k+x-1}{k}\frac{k+1}{k+x}=1-\frac{1-x}{k^2+kx}
$$
the product telescopes
$$
\prod_{k=1}^n\left(1-\frac{1-x}{k^2+kx}\right)=x\,\frac{n+1}{n+x}
$$
Therefore,
$$
\prod_{k=1}^\infty\left(1-\frac{1-x}{k^2+kx}\right)=x
$$


问:
Simple extension of purely inseparable extension
It is Albert "Modern Higher Algebra", chapter 7, section 9, exercises 5 and 6.

Let $K$ be a field of degree $n$ over $F$ of characteristic $p$ such that every quantity $\alpha$ of $K$ is a root of a corresponding equation $x^{p^t}=a$ in $F$. Prove that $n$ is a power of $p$ and that there exists an integer $e$, called the exponent of $K$ over $F$, such that $k^{p^e}$ is in $F$ for every $k\in K$, and $e$ is the least integer with this property.

Let $Z>K>F$ where $K$ is as in Ex. 5 and $Z=K(\xi)$ is a simple separable extension of degree $n$ over $F$. Prove that $Z=K\times F(\xi_0)$ where $\xi_0=\xi^{p^e}$ and $e$ is the exponent of $K$ over $F$. Prove in fact that $Z=K(\xi)$ and $Z_0=F_0(\xi_0)$ are equivalent (but not necessarily equivalent over $F$) under the correspondence $a\longleftrightarrow a^{p^e}$ for every $a$ of $Z$, where $F_0$ is a certain subfield of $F$.

Question

I do not know how to prove that $Z$ is generated by $K$ and $F(\xi_0)$, or to prove that $\xi\in K(\xi_0)$, and I do not know how to use the isomorphism between $Z$ and $Z_0$ to conclude that $[Z:F]=[K:F][F(\xi_0):F]$.

Notation

If $K$ is finite field extension of $F$ and $A$, $B$ are subfields of $K$ containing $F$, we say $K=A\times B$ if $K$ is the subfield of $K$ generated by $A\cup B$ and $[K:F]=[A:F][B:F]$.

My attempt

Induction on $[K:F]=n$. If $k\in K$, $k\notin F$, then let $u$ be the least such that $k^{p^u}\in F$, so let $a=k^{p^u}$, then the polynomial $x^{p^u}-a$ is irreducible in $F[x]$. Indeed, if $x^{p^u}-a=g(x)h(x)$ where $g(x),h(x)\in F[x]$, then, in $K[x]$: $$g(x)h(x)=x^{p^u}-a=x^{p^u}-k^{p^u}=(x-k)^{p^u},$$ so $g(x)=(x-k)^s$ where $s\leq p^u$, but $g(x)\in F[x]$, so $k^s\in F$, so, if $d=\mathrm{gcd}(s,p^u)$, then $d=p^v$ for some $v\leq u$, and $k^d\in F$, so $k^{p^v}\in F$, so $u\leq v$, so $p^u=d\mid s$, so $p^u\leq s$, so $s=p^u$, and therefore: $$g(x)=(x-k)^{p^u}=x^{p^u}-a.$$ Therefore $[F(k):F]=p^u$. The hypothesis is certainly valid for $K$ over $F(k)$ and $[K:F(k)]=n/p^u$, so, by the induction hypothesis, $n/p^u$ is a power of $p$, so that $n$ is also a power of $p$, and there is an $f$ such that $l^{p^f}\in F(k)$ for every $l\in K$. Also, for every $m\in K$, then there are $a_i\in F$, $i=0,\dots,p^u-1$, such that $m=\sum a_ik^i$, so: $$m^{p^u}=\left(\sum a_ik^i\right)^{p^u}=\sum a_i^{p^u}(k^{p^u})^i\in F.$$ Therefore for every $l\in K$ we have $l^{p^f}\in F(k)$, so $l^{p^{f+u}}=(l^{p^f})^{p^u}\in F$.

Because the characteristic of $F$ is $p$, it is easy to see that: $$F_0=\{k^{p^e}:k\in K\}$$ is a subfield of $F$ and $a\longrightarrow a^{p^e}$ is an injective homomorphism from $Z$ to $Z_0$. Also, for every $b\in Z_0$, there are $b_i\in F_0$ such that $b=\sum b_i\xi_0^i$, so there are $a_i\in K$ such that $b_i=a_i^{p^e}$, so: $$b=\sum b_i\xi_0^i=\sum a_i\xi^{p^e\cdot i}=\left(\sum a_i \xi^i\right)^{p^e}.$$ For $Z=K\times F(\xi_0)$, I have to show that: (a) $Z$ is generated by $K$ and $F(\xi_0)$; (b) $[Z:F]=[K:F][F(\xi_0):F]$. The item (a) is equivalent to prove that $\xi\in K(\xi_0)$, and I do not know how to do this. The item (b) is equivalent to $[Z:K]=[F(\xi_0):F]$, and, by the isomorphism between $Z$ and $Z_0$, this is equivalent to $[Z_0,F_0]=[F(\xi_0):F]$, and this is equivalent to prove that the minimal polynomial $P_0(x)$ of $\xi_0$ in $F_0$ has the same degree as the minimal polynomial $P(x)$ of $\xi_0$ in $F$. Obviously $\mathrm{deg}P(x)\leq\mathrm{deg}P_0(x)$. But I do not know how to prove the other inequality.

答:
$K/F$ is a finite extension. For $\alpha \in K$ we are told that $\alpha^{p^t}\in F$ for some $t$, thus $\alpha$ is a root of $x^{p^t}-\alpha^{p^t}=(x-\alpha)^{p^t}$, its $F$-minimal polynomial must be of the form $h(x)=(x-\alpha)^m$, $(-1)^m h(0) = \alpha^m$ is in $F$ so $\alpha^{\gcd(m,p^t)}=\alpha^{p^r}$ is in $F$, repeating we get that the minimal polynomial is of the form $x^{p^t}-\alpha^{p^E}$ where $E$ is the least integer such that $\alpha^{p^E} \in F$.

Since $K/F$ is finitely generated there is a least $e$ such that the $p^e$-th powers of the generators are in $F$ and since $\beta\to \beta^{p^e}$ is a ring homomorphism then that $e$ is the least $e$ such that $K^{p^e}\subset F$.

Given $Z/K$ finite separable, thus $Z=K(\xi)$. With $g(x) = \sum_{n=0}^N a_n x^n$ the $K$-minimal polynomial of $\xi$ which is separable (coprime with $g'$) we get that $g^{p^e}(x)=\sum_{n=0}^N a_n^{p^e} x^n$ is separable, in $F[x]$, and $g^{p^e}(\xi^{p^e}) = g(\xi)^{p^e}=0$. Thus $F(\xi^{p^e})/F$ is separable.

For each $K$-conjugate $\sigma(\xi)$ distinct from $\xi$ we have that $\sigma(\xi^{p^e})$ is distinct from $\xi^{p^e}$. Thus $\xi$ and $\xi^{p^e}$ have the same number of $K$-conjugates, their $K$-minimal polynomials have the same degree, ie. $[K(\xi):K]=[K(\xi^{p^e}):K]$ and $K(\xi)=K(\xi^{p^e})$.

We also obtained that $[K(\xi^{p^e}):F]=[F(\xi^{p^e}):F][K:F]$ thus $K(\xi^{p^e}) = F(\xi^{p^e})\otimes_F K$.

The extensions $K(\xi)/K$ and $K^{p^e}(\xi^{p^e})/K^{p^e}$ are isomorphic.


问:
$13$ integers from $S = \{150 , 151 , ...... , 173 \}$. Any choice of $13$ integers will have a pair of integers whose difference is $3$.
We have to choose $13$ integers from this set $S = \{150 , 151 , 152 , 153 , .............. 172 , 173 \}$.

I have to show that for any choice of $13$ integers there will be a pair of integers whose difference is $3$.

I was trying to apply Pegion Hole principle. But I am not being able to do so.

Can anyone please help me ?

答:
Partition into sets of size double the difference, $3$.

$$ \{150,151,152,153,154,155\} , \{156,\ldots,161\}, \ldots , \{168\ldots,173\}  $$

These are $4$ sets. In each set we have $3$ pairs differing by $3$.

E.g., ${(150,153), (151,154),(152,155)}$.

Hence we have $12$ pairs of numbers differing by $3$.

Any choice of $13$ integers would include two numbers from same pair. Hence proved.


答:
Consider $3$ sets/rows/vectors of numbers (let $a=150$):

$A = (a, a+3, a+6, a+9, \ldots, a+21)$,

$B = (a+1, a+4, a+7, a+10, \ldots, a+22)$,

$C = (a+2, a+5, a+8, a+11, \ldots, a+23)$.

If we choose $13$ numbers, then one of the sets ($A,B,C$) provides at least $5$ numbers (according to pigeonhole principle).

And that's impossible to choose $5$ numbers out of $8$ to have no neighboring numbers.
These neighboring numbers will have difference $3$.

Title: $13$ integers from $S = \{150 , 151 , ...... , 173 \}$. Any choice of $13$ integers will have a pair of integers whose difference is $3$.

Body:
We have to choose $13$ integers from this set $S = \{150 , 151 , 152 , 153 , .............. 172 , 173 \}$.

I have to show that for any choice of $13$ integers there will be a pair of integers whose difference is $3$.

I was trying to apply Pegion Hole principle. But I am not being able to do so.

Can anyone please help me ?


问:
Covariance of Gamma Transformations
If I have two independent random variables: $X$ is governed by Gamma$(a,b)$ and $Y$ is governed by Gamma$(c,b)$, then how can I find the Cov($Z,X$), where $Z = X/Y$?

I tried using properties of covariance, but the transformation was messing me up.
Cov(Z,X) = E(ZX) - E(X)E(Z)

答: Note that$$\begin{align}\operatorname{Cov}(Z,\,X)&=E[ZX]-E[Z]E[X]\\&=E[X^2/Y]-E[X/Y]E[X]\\&=E[X^2]E[1/Y]-E[X]^2E[1/Y]\\&=\operatorname{Var}(X)E[1/Y].\end{align}$$I leave it to you to write the final expression in terms of $a,\,b,\,c$.

Title: Covariance of Gamma Transformations


问:
How do I prove $\lim_{n \rightarrow \infty} \mu(E_n \Delta E)=0$ iff $1_{E_n} \rightarrow 1_{E}$ in measure $\mu$?
Let $(X,M, \mu)$ be a measure space and let $E_n, E \in M$. Prove that

$\lim_{n \rightarrow \infty} \mu(E_n \Delta E)=0$ iff $1_{E_n} \rightarrow 1_{E}$ in measure $\mu$.

Can anyone suggest me some hints for this question?

答:
Hint: Show that
$$E_n\Delta E =\{x\in X:|1_E-1_{E_n}|=1\}.$$
Recall that convergence in measure means that for every $\epsilon $ $$\lim _{n\rightarrow \infty}\mu (|1_E-1_{E_n}|>\epsilon)=0.$$


问:
Largest possible radius of a circle lying in a parabola

Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$

and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?

答:
First of all, you have a typo, the circle is $ (x + r)^2 + y^2 = r^2 $.

Clearly since $ x \leq 0 $ on the parabola, any intersections of the circle with the parabola will have $ x \leq 0 $.  $ x = 0 $ will always correspond to the vertex.  The question is asking you to set the radius $ r $ as big as possible without having any other intersection points.  Through the calculations you/they have determined that any other intersection points can only occur when $ x = 12 - 2r $ or $ x = 0 $.  But since $ y^2 $ has to be non-negative ($ y $ is real after all) and we have $ y^2 = -12x $, $ x = 12 - 2r $ corresponds to an intersection point if and only if $ x \leq 0 $.  So we get three intersection points if and only if $ 12 - 2r < 0 $, in which case the three intersection points are $ (0,0) $, $ \left(12 - 2r, \sqrt{12(2r-12)} \right) $, and $ \left(12 - 2r, -\sqrt{12(2r-12)} \right)$.  Put differently, the circle is in the parabola if and only if there is (at most) one intersection point, which in turn is true if and only if $ r < 6 $.

Clearly if you have three intersection points, the circle is not "inside" the parabola.


问:
Inequality with mean inequality

If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.

答:
let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$
$$xy+yz+zx=18 \tag{given}$$
$$\color{red}{x+y+z\ge 3\sqrt{6}} \tag1$$

notice that $${(x-\sqrt{6})}^2\ge 0\Rightarrow \sqrt{3+x^2}\ge \frac{\sqrt{6}}{3}x+1$$

Thus $$\sum \sqrt{3+x^2}\ge \sum \frac{\sqrt{6}}{3}x+1\ge\frac{\sqrt{6}}{3}\color{red}{3\sqrt{6}}+3=9$$

Note the inequality marked $(1)$ is left as an exercise

Title: Inequality with mean inequality


问:
Why is $|\lambda u|=|\lambda||u|$
My text book in linear algebra - out of the blue - claims that:

$|\lambda u|=|\lambda||u|$

Where u is a vector and $\lambda$ is a constant.

I would understand if || were used to denote absolute numbers, but in this book, || is used to denote length (so the dot product of the vector v with itself is written as $|v|^2$)

I'm positively dumbfounded. How do I prove this and why would av scalar like $\lambda$ even have a length in the first place?

EDIT:

This is used in a later section to prove that the projection of v on u equals:

$$|\frac{u \cdotp v}{|u|^2}u|=\frac{|u \cdotp v|}{|u|^2}|u|=\frac{|u \cdotp v|}{|u|}$$ so it definitly seems they're referring to lengths, and not absolute numbers. Otherwise I can't really see that working.

答:
Perhaps this makes it clear: $|\lambda u|_V=|\lambda|_{\mathbb R}\,|u|_V$.

It reads: the length of $\lambda u$ in the vector space $V$ is the product to the absolute value of $\lambda$ in $\mathbb R$ by the length of $u$ in $V$.

Using subscripts as above sometimes helps to make sense of what is going on.
For instance, if $T:V\to W$ is a linear transformation and $v_1,v_2 \in V$, then
$$
T(v_1+ v_2) = T(v_1) + T(v_2)
$$
actually means
$$
T(v_1+_V v_2) = T(v_1) +_W T(v_2)
$$

Title: Why is $|\lambda u|=|\lambda||u|$


答:
Also note that $\mathbb{E}\Big[\frac{1}{Y}\Big]$ is known as $\frac{1}{Y}\sim \text{Inverse Gamma}$

Inverse Gamma Distribution

Title: Covariance of Gamma Transformations

Body:
If I have two independent random variables: $X$ is governed by Gamma$(a,b)$ and $Y$ is governed by Gamma$(c,b)$, then how can I find the Cov($Z,X$), where $Z = X/Y$?

I tried using properties of covariance, but the transformation was messing me up.
Cov(Z,X) = E(ZX) - E(X)E(Z)


问:
$\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$

Body:
@Ofir, you are almost right:

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}=\frac{1}{2}\left(\int^1_0\frac{2x+2}{x^2+2x+1}dx+\int^1_0\frac{-2}{x^2+2x+1}dx\right)=\frac{1}{2}(\ln|{x^2+2x+1}|_{0}^{1}+\frac{2}{1+x}|^{1}_{0})=\frac{1}{2}(\ln{4}-\ln{1}+1-2)=\frac{1}{2}(\ln^4-1)$$


问:
What is minimum number of edges should be removed from $K_6$ to get planar graph?
What is minimum number of edges should be removed from $K_6$ to get planar graph?

It is easy to show it just with picture.. but is it possible to prove it analytically?

答:
All the faces must be "triangles" (If a face is not a $3$ cycle then draw in a "diagonal"). So $3f=2e$.

Planar ? ... Euler's formula ... $v-e+f=2$. ... $K_6$ : Also $v=6$.

Solve these $3$ equations ( $3 \times (v-e+f=2)$ etc ...) gives $e=12$.

$K_6$ has $15$ edges ... so we need to remove $\color{red}{3}$ edges.


答:
You may know about the dot product that $(\lambda u)\cdot (\mu v)=\lambda\mu (u\cdot v)$.
Hence
$$ |\lambda u|=\sqrt{\lambda u\cdot \lambda u)}=\sqrt{\lambda ^2 (u\cdot u)}=|\lambda|\sqrt{u\cdot u}=|\lambda|\,|u|.$$

Title: Why is $|\lambda u|=|\lambda||u|$

Body:
My text book in linear algebra - out of the blue - claims that:

$|\lambda u|=|\lambda||u|$

Where u is a vector and $\lambda$ is a constant.

I would understand if || were used to denote absolute numbers, but in this book, || is used to denote length (so the dot product of the vector v with itself is written as $|v|^2$)

I'm positively dumbfounded. How do I prove this and why would av scalar like $\lambda$ even have a length in the first place?

EDIT:

This is used in a later section to prove that the projection of v on u equals:

$$|\frac{u \cdotp v}{|u|^2}u|=\frac{|u \cdotp v|}{|u|^2}|u|=\frac{|u \cdotp v|}{|u|}$$ so it definitly seems they're referring to lengths, and not absolute numbers. Otherwise I can't really see that working.


答: Notice that for x $ \not \in E_n \Delta E$, it breaks into two cases: $x \in E_n \cap E$, in which case $|1_{E_n}(x)- 1_E(x)|= 0$, or $x \not \in E \cup E_n$, in which case $|1_{E_n}(x) - 1_E(x)| = 0$ again. Moreover, $|1_{E_n}(x) - 1_{E}(x)| = 1$ iff $x \in E_n \Delta E$. Thus, $\mu(x : |1_{E_n}(x) - 1_E(x) | \geq \epsilon) = \mu(E_n \Delta E)$ for any $\epsilon \leq 1$ and $\epsilon > 0$. Now just consult the definition of convergence in measure.

Title: How do I prove $\lim_{n \rightarrow \infty} \mu(E_n \Delta E)=0$ iff $1_{E_n} \rightarrow 1_{E}$ in measure $\mu$?

Body:
Let $(X,M, \mu)$ be a measure space and let $E_n, E \in M$. Prove that

$\lim_{n \rightarrow \infty} \mu(E_n \Delta E)=0$ iff $1_{E_n} \rightarrow 1_{E}$ in measure $\mu$.

Can anyone suggest me some hints for this question?


答:
It is part of the definition of an inner product $\langle \;,\; \rangle$ on a complex vector space $V$ that

$\langle \lambda v, w\rangle = \lambda\langle v, w\rangle,$ and

$\langle v, \lambda w\rangle = \bar{\lambda}\langle v, w\rangle,$

for all $\lambda \in \Bbb C$ and $v, w \in V$. (As usual, $\bar{\lambda}$ is the complex conjugate of $\lambda$.)

After this, a norm is defined as
$$\|v\| = \sqrt{\langle v, v\rangle}.$$
This then gives us

\begin{align}
\|\lambda v\| &= \sqrt{\langle \lambda v, \lambda v\rangle}\\
&= \sqrt{\lambda\langle v, \lambda v\rangle}\\
&= \sqrt{\lambda\bar{\lambda}\langle v, v\rangle}\\
&= \sqrt{|\lambda|^2\langle v, v\rangle}\\
&= |\lambda|\sqrt{\langle v, v\rangle}\\
&= |\lambda|\|v\|.
\end{align}

In case you're working over $\Bbb R$, you just have $\lambda$ everywhere instead of $\bar{\lambda}$.

Title: Why is $|\lambda u|=|\lambda||u|$


问:
Euler totient function sum of divisors. Theorem 2.2 Apostol

Prove that :
  If $ n\ge 1 $,
  then
  $ \sum_{d|n}\phi(d)=n $.

Let $S$ denote the set $\{1,2,...,n\}$. We distribute the integers of $S$ into disjoint sets as follows. For each divisor $d$ of $n$, let

$A(d) = \{k \in S :(k,n) = d\}$

That is, $A(d)$ contains the elements of S which have the gcd d with n.  The sets $A(d)$ for a disjoint collection whose union is S. Therefore if $f(d)$ denotes the number of integers in $A(d)$ we have
$\sum_{d|n}f(d)=n$

I don't understand why the sum of $f(d)$ equals $n$. Can someone explain this?

答:
Here is another approach to solve this problem if you are familiar with cyclic groups although it is equivalent to approaches given in other answers. But knowing the interdependency of mathematical disciplines is always beneficial.

Let $G(a)$ be the cyclic group generated by the element $a$ of order $n$.  By fundamental theorem of cyclic groups we know that for each divisor $k$ of $n$ , $G(a)$ has excatly one subgroup of order $k$ - namely $G(a^{\frac{n}{k}})$. Also we know that $G(a^k) = G(a)$ if and if only $gcd(n,k) = 1$.

Now if $d$ divides $n$ then there is exactly one subgroup of order $d$ and let it be $G(b)$. Then every element of order $d$ generates $G(b)$. But $G(b^k) = G(b)$ only if $gcd(k,d)=1$. So number of elements having order $d$ is $\phi(d)$.

Now if the total number of elements in given cyclic group is $n$, then by fundamental theorem of cyclic groups,
$$
\sum_{d|n} \phi(d) = n
$$


问:
Good source to learn about surface singularities?
I am looking for something that treats singularities on algebraic surfaces and curves over $\mathbb{C}$, starting from the very basics but not stopping there.

I checked out Miles Reid his lectures on algebraic surfaces and they seem promising (especially the exercises). Also Harris - algebraic geometry a first course seems nice. However, the first gets very technical quickly while the second book's treatment of singularities is pretty short.

Is there also something that treats loads and loads of examples, furnishing a good intuition about the subject?

Thanks!

答: "Basic Algebraic Geometry 1: Varieties in Projective Space" by Igor R. Shafarevich is a nice text that is quite elementary. Surface singularities are studied on page 264, it may be what you are looking for.</p>


答:
$(x+r)^2+y^2=r^2$ (check the equation)

$y^2 + 12x = 0$

Equating the two,

$x^2 -12x + 2xr = 0$

$x^2 + (2r-12)x = 0$

So two solutions in $x$, $x = 0, (12-2r)$

For any value of $r \gt 6$, $x = 12 - 2r$ is negative and so you have another value of $x$ other than $0$ that satisfies the parabola equation $y^2 = -12x$ which means the circle will intersect the parabola for $r \gt 6$.

Note that for $r \lt 6$, $x = 12-2r$ is positive which cannot be on the parabola $y^2 = -12x$.

Title: Largest possible radius of a circle lying in a parabola

Body:
Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$

and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?


问:
Trouble with the proof of Proposition 4.3.18 of Pedersen's Analysis Now
I am currently trying to understand the proof of Proposition 4.3.18 in Pedersen's Analysis now, which reads

To each Tychonoff space $X$ there is a Hausdorff compactification $\beta(X)$, with the property that every continuous function $\Phi: X \to Y$, where $Y$ is a compact Hausdorff space, extends to a continuous function $\beta \Phi: \beta(X) \to Y$.

The proof starts by noting that $C_b(X)$ is a commutative unital C$^*$-algebra, and is therefore isometrically isomorphic to a (commutative and unital) C$^*$-algebra of the form $C(\beta(X))$, where $\beta(X)$ is a compact Hausdorff space.

By the Gelfand duality between the category of commutative and unital C$^*$-algebras and the category of compact Hausdorff spaces, we can take $\beta(X) = \Omega(C_b(X))$, the space of characters on $C_b(X)$.

Then we can define a map $\iota: X \to \beta(X)$, where $\iota(x)(\phi) := \phi(x)$ for all $x \in X$ and $\phi \in \beta(X)$.

The particular part of the proof that I am struggling to understand is the proof that $\iota(X)$ is dense in $\beta(X)$.

He argues that if $\iota(X)$ is not dense in $\beta(X)$, then there is a non-zero continuous map $f: \beta(X) \to \mathbb{C}$ vanishing on $\iota(X)$. This I understand. He then says that under the identification $C_b(X) = C(\beta(X))$, this is impossible. This is the sentence I am stuck on. Why is it impossible under this identification?

We have that $C_b(X)$ is isometrically isomorphic to $C(\Omega(C_b(X)))$ via the map $\delta: g \mapsto (\delta_g: \Omega(C_b(X)) \to \mathbb{C}, \phi \mapsto \phi(g))$. I am pretty sure what Pedersen is getting at is that the map $\delta^{-1}(f)$ is zero, but I am not able to show that this is the case. This answer also claims that a similar map is zero.

In summary, my question is:

Can we show that $\iota(X)$ is dense in $\beta(X)$ by showing that $\delta^{-1}(f) = 0$? If so, how do we do this?

答:
Recently, I wrote all this out in detail for myself, so here I share my notes with you. Note that the assumption that $X$ is Tychonoff can be ommitted. The construction works for every topological space. The Tychnoff assumption is only there to ensure that the canonical inclusion is injective.

Recall that if $A$ is a commutative $C^*$-algebra, then we can consider the space of characters $\Omega(A)$ . If $A$ is a unital $C^*$-algebra, then this becomes a compact Hausdorff space for the weak$^*$-topology. Note that we have a natural map
$$i_X: X \to \Omega(C_b(X)): x \mapsto \text{ev}_x.$$
Clearly this is a continuous map, as an easy argument with nets shows.

Lemma: The map $i_X$ has dense image.

Proof: Assume to the contrary that $\overline{i_X(X)}\subsetneq \Omega(C_b(X))$. Then Urysohn's lemma applied to the compact Hausdorff space $\Omega(C_b(X))$ gives a non-zero continuous function $f: \Omega(C_b(X))\to \mathbb{C}$ that is zero on $i_X(X)$. Consider the canonical isomorphism
$$\Psi: C_b(X) \to C(\Omega(C_b(X))): \omega \mapsto \text{ev}_\omega.$$
Choose $\omega \in C_b(X)$ with $\text{ev}_\omega = f$. Then for all $x \in X$, we have
$$\omega(x) = \text{ev}_x(\omega) = \text{ev}_\omega(\text{ev}_x) = f(i_X(x)) = 0$$ so $\omega = 0$, which is a contradiction. $\quad \square$

Theorem: If $X$ is a topological space, then $(\Omega(C_b(X)), i_X)$ is a Stone-Čech compactification of $X$.

Proof: Let $K$ be a compact Hausdorff space and let $f: X \to K$ be a continuous map. This induces a $*$-morphism $$C(f): C(K) \to C_b(X): g \mapsto  g \circ f$$ and this then induces a continuous map $$\Omega(C(f)): \Omega(C_b(X)) \to \Omega(C(K)): \chi \mapsto \chi \circ C(f)$$
Consider the homeomorphism $$i_K: K \to \Omega(C(K)): k \mapsto \text{ev}_k.$$

Then we define the continuous map $F:= i_K^{-1}\circ \Omega(C(f)): \Omega(C_b(X)) \to K$. Moreover, we have $F\circ i_X= f$. Indeed, if $x \in X$, then
$$i_K(F \circ i_X(x)) = i_K (F(\text{ev}_x)) = \Omega(C(f))(\text{ev}_x) =  \text{ev}_x \circ C(f)= \text{ev}_{f(x)}= i_K(f(x))$$
so that by injectivity of $i_K$ we obtain $F \circ i_X = f$.

The condition $F \circ i_X = f$ determines $F$ uniquely on $i_X(X)$, which is dense in $\Omega(C_b(X))$ by the preceding lemma. Thus $F$ is unique. $\quad \square$


问:
How to show that $a_{n+2}=\frac{a_{n+1} ^2 -1}{a_n}$ . is bounded?
Let $a_{n+2}=\frac{a_{n+1} ^2  -1}{a_n}$  be a sequence of real numbers where $a_n>0$ for all $n \in \mathbb{Z}_{+}$. It is given that $a_1=1 , a_2=b>0$. It is given that $1<b<2$ .Is it possible to show that $a_n$ is bounded from this given information?

答:
Well, bounded or not, it depends on $b$. I won't go in details but will give the outline in the following.

Firstly, you can observe that $ a_{n+2}a_n -a_{n+1}^2= a_{n+1}a_{n-1} -a_n^2$, and this leads to :
$$ \dfrac{a_{n+2}+a_n}{a_{n+1}}= \dfrac{a_{n+1}+a_{n-1}}{a_{n}}= \dots = \dfrac{a_3+a_1}{a_2} =b$$

So now you have the following recursive relation:
$$ a_{n+2} -ba_{n+1}+a_n=0$$

Solving that, we have some formula of $(a_n)$ as:
$$ a_n= a \phi_1^n+b\phi_2^n$$

And rest is simple, normally, the first necessary condition we would get for the boundedness is $|b| \le 2$.


问:
What is the best way to arrange trapezoidable bricks to form a giant circle?
I have some bricks that form a trapezoid shape. They have the following dimensions:

a = 4.25 in

b = 9.05 in

c = 2.375 in

d = 9.05 in

I want to arrange these such that they form a giant lawn circle, with a radius of 255 inches (circumference of 1602 inches), without any gaps between them.

Gaps are to be avoided as much as possible.

The bricks are to be set on top of a perfectly circular wall that is already constructed. It has a 255-inch radius.

The bricks can overlap the 255-inch circle, in fact that is desired as it would look poor if they are perfectly flush, but they can't overlap more than 1".

They can be arranged so either side $\overline{a}$ or $\overline{c}$ are pointing outwards.

The bricks sit on a wall like this, except the wall forms a perfect circle:

If I simply arrange them within the circle, with side $\overline{a}$ along the circumference, it has many large gaps between the bricks, which is not good:

If I arrange them in an alternating pattern, some with side $\overline{a}$ along the circumference, while others with $\overline{c}$ along the circumference, I can reduce the gaps, but there are still many large gaps visible:

Unfortunately, with this pattern, if no gaps are given, the bricks just form a straight line and do not match the curve of the circle. Note that the goal is a nice fit around the circumference, without any gaps:

If I create some pattern, such that it alternates between sides $\overline{a}$ and $\overline{c}$ being placed outwards, before having two $\overline{a}$'s outwards in a row, this creates a giant regular polygon. This is an ideal solution, as no gaps are present between bricks. A small extra overlap outside the circle is also welcome athletically.

The best solution I can come up with it to create a giant tetracontakaitrigon (43-sided regular polygon) using 43 sets of six $\overline{a}$-out-facing and five $\overline{c}$-out-facing blocks. Each set is rotated by 8 degrees, leaving a gap of ~3 degrees of gap between the sets. I believe I calculated the degrees right, as in there will be an empty gap between sets, and it doesn't require something impossible, like bricks magically overlapping.

Each set would measure 37.375 inches, so 43 sets would create a circumference of 1607.125 inches.

This slightly larger circumference would mean the bricks need to overlap the circumference by .39 inches at the center of each set of bricks.

The solution requires 473 bricks.

Did I make any mistakes in my calculations?

Is there a better solution that reduces the 3-degree gaps between bricks?

答:
Comment:

The angle of brick $2\alpha$ is:

$sin(\alpha)=\frac{4.25-2.375}{2\times9,05}=0.1000..$

$\alpha≈ 5.75^o $ ⇒ $2\alpha=11.5^o$

The radius of the garden is $255"$ we have:

$d=255 sin (11.5) ≈ 50.84" $

Now if you make a set of 5 bricks, you get a trapezoid ( as shown in figure) we have:

Short side of trapezoid : $2\times4.25+3\times2.375=7.125$

Long side of trapezoid: $3\times 4.25+2\times 2.375=17.5$

Now a set out of three of these sets has outer side as:

$2\times 17.5+15.625=50.625 $

This almost equal to $d=50.84$

That means you have to make sets of 5 bricks; total number of brick is:

$\frac{360}{11.5}\times 15=469.5 ≈470$


问:
Basis coefficients in strong convergence of sequence in Hilbert space
Let $\{v_i\}_{i\in I}\subset H$ be a basis of a Hilbert space which is not necessarily orthogonal, and a sequence $x_n\in H$ such that:
$$x_n=\sum_{i\in I}{\alpha_n^iv_i}$$
Now, assume that $x_n\to x$ (strongly convergence) and:
$$x=\sum_{i\in I}{\alpha^iv_i}$$
Is it true that $\alpha_n^i\to \alpha^i$ for every $i\in I$?

答:
Consider the Hilbert space $\ell^2$ and let $(e_n)_{n=1}^\infty$ be the canonical vectors.

Denote $e = \sum_{n=1}^\infty \frac1n e_n$ and let $B$ be an algebraic basis for $\ell^2$ containing the linearly independent set $S = \{e_n : n\in \Bbb{N}\} \cup \{e\}$.

Consider
$$x_n := \sum_{k=1}^n \frac1k e_k = \sum_{b \in B} \alpha_n^b b, \quad \alpha_n^b = \begin{cases}
\frac1k, &\text{if $b = e_k$ with $1 \le k \le n$}, \\
0, &\text{if $b = e_k$ with $k > n$}, \\
0, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} $$
$$x := \sum_{k=1}^\infty \frac1k e_k = e = \sum_{b \in B} \alpha^b b, \quad \alpha^b = \begin{cases}
0, &\text{if $b = e_k$ with $k\in\Bbb{N}$}, \\
1, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} $$

Then $x_n \to x$ strongly in $\ell^2$ but
$$\lim_{n\to\infty} \alpha^b_n = \begin{cases}
\frac1k, &\text{if $b = e_k$ with $k\in\Bbb{N}$}, \\
0, &\text{if $b = e$}, \\
0, &\text{if $b \in B\setminus S$}, \\
\end{cases} \ne \alpha_b.$$


问:
Prove that $\inf(A^{-1}) = \sup(A)^{-1}$
This is very similar to this question, whose answer I do not understand. I want to proved that for $A \subseteq \mathbb{R}$:

$$
\inf(A^{-1}) = \sup(A)^{-1}
$$

where $A^{-1} = \{\frac{1}{a} \mid a \in A\}$ and it is given that $\sup(A) < 0$.

My first attempt was this:

Let $i := \inf(A^{-1})$, then we know:

$$
\begin{align}
&(1) \quad \forall a \in A: \frac{1}{a} \ge i \\
&(2) \quad \forall \epsilon > 0 \in \mathbb{R}: \exists a \in A: \frac{1}{a} < i + \epsilon
\end{align}
$$

from which it follows that:

$$
\begin{align}
&(1) \quad \forall a \in A: a \leq \frac{1}{i} \\
&(2) \quad \forall \epsilon > 0 \in \mathbb{R}: \exists a \in A: a > \frac{1}{i + \epsilon}
\end{align}
$$

Now if I could show that $\frac{1}{i + \epsilon} \geq \frac{1}{i} - \epsilon$ that would show that $\sup(A) = \frac{1}{i}$ and conclude the proof. But I don't know how to show that that's true.

My second attempt was this:

Choose an arbitrary $\epsilon > 0$, then pick an $a \in A$ such that $a > \sup(A) - \epsilon$. Then we have:

$$
\frac{1}{\sup(A) - \epsilon} > \frac{1}{a} \geq \inf(A^{-1})
$$

and if we let $\epsilon$ go to zero we get:

$$
\frac{1}{\sup(A)} \geq \inf(A^{-1})
$$

but I don't know how to show that it also holds that:

$$
\frac{1}{\sup(A)} \leq \inf(A^{-1})
$$

because if repeat the procedure, picking an  $a$, such that $\frac{1}{a} < \inf(A^{-1}) + \epsilon$ then I also get:

$$
\frac{1}{\inf(A^{-1}) + \epsilon} < a \leq \sup{A}
$$

i.e.

$$
\frac{1}{\sup(A)} \geq \inf{A^{-1}}
$$

again.

How can fix one or both of these approaches? And where do I actually need to use $\sup(A) < 0$? Is the equality not true for $\sup(A) > 0$?

答:
Here is a simple approach.

We know that for any $a$ in A, $$ a \leq Sup(A)$$
Or
$$\frac{1}{Sup(A)} \leq \frac{1}{a}$$

So $Sup(A)^{-1}$ is a lower bound to $A^{-1}$.
Since $Inf(A^{-1})$ is greatest lower bound to $A^{-1}$ we must have $Sup(A)^{-1} \leq Inf(A^{-1})$.

Now similarly $\frac{1}{Inf(A^{-1})}$ is an upper bound to the set $A$, $Sup(A)^{-1} \geq Inf(A^{-1})$. So we have your equality.


问:
local extremum of a function

Let $A = \{(x_1,x_2, x_3) \in \mathbb{R}^3 : x_1^2 + x_2^2 + x_3^2 \leq 1\}$ and $f : A \to \mathbb{R}$ be defined by $f(x_1,x_2,x_3) := x_1^2 + x_2^3 - 2x_1x_2 - x_3^2$ for $(x_1,x_2,x_3) \in A.$ By the EVT and continuity of $f, f$ must attain a global maximum. Prove that all points of global maximum are on the unit sphere $S = \{(x_1,x_2,x_3) \in \mathbb{R}^3 : \sum_{i=1}^3 x_i^2 = 1\}$.

I know that points of local maximum are critical points. We have that $\nabla f(x_1,x_2,x_3)$ is $(2x_1 -2x_2, 3x_2^2 - 2x_1, -2x_3).$ This is zero iff $(x_1, x_2, x_3) = (0,0,0)$ or $(0,\frac{2}3, \frac{2}3),$ but neither of these points are global maxima for $f$ (we easily have $f(1,0,0) = f(0,1,0) = 1$, which is greater than the value of $f$ at both of these points).  By the second derivative test, $f$ attains a local maximum at a point where the Hessian of $f$ at $\vec{a} \in A$, whose $(i,j)$ entry is $\partial_i \partial_j f(\vec{a})$, is negative semi-definite. We have that the Hessian of $f$ is given by $Hf(\vec{a}) = \begin{bmatrix}2 & -2 & 0\\
-2 & 6a_2 & 0 \\
0 & 0 & -2\end{bmatrix}\,\forall \vec{a} = (a_1, a_2, a_3).$ For a column vector $\vec{z} = (a,b,c)\in\mathbb{R}^3, z^T Hf(\vec{a})z = (2a-2b)a + (-2a + 6a_2 b) -2c^2.$ I'm not sure how to use this to determine local maxima though.

答:
Let's define $\mathring{A}:=\{(x_1,x_2,x_3)\mid x_1^2+x_2^2+x_3^2<1\}$ and $cl(A):=\{(x_1,x_2,x_3)\mid x_1^2+x_2^2+x_3^2=1\}$, so $A=\mathring{A}\cup cl(A)$.

We can find all possible points $(x_1,x_2,x_3)\in \mathring{A}$ which might attain a global maximum by setting $Df(x_1,x_2,x_3)\overset{!}{=}0$ (first order condidtion aka F.O.C.) and then verifying that those points indeed lie in $\mathring{A}$. This is what you have already done and as you have pointed out those points are no global maxima nor do they lie on the unit sphere. As the function is continuous and defined on a compact set it must attain a global maximum (and a global minimum). We can easily find points $(x,y,z)$ and $(a,b,c)$ on $cl(A)$ which satisfy $f(x,y,z)>f(0,0,0)$ and $f(a,b,c)<f(0,\frac{2}{3},\frac{2}{3})$. So the global extrema must be attained by points in $cl(A)$. Hence, they lie on the unit sphere.

Title: local extremum of a function


问:
Prove a ray inside compact convex set will always hit the boundary
Let $K$ be a compact convex set in Normed linear space.Prove given two distinct point $x,y \in K$,the ray $\{x + t(y-x)\mid t>0\}$ will always hit the boundary point of $K$.

I can formalize this question as follows:

$\sup\{t\mid z = x+t(y-x)\in K,t>0\}<\infty$ and the sup is attainable at some $z\in \partial K$

I have a not very good proof,is there some good proof for this proposition?

I do as follows,if sup is unbounded there exist a sequence of $t\to \infty$,by convexity,which means ray contains in the bounded set,which is impossible.(due to the fact that subset of bounded set is still bounded,and ray is unbounded in normed linear space(I don't know how to show unbounded of ray exactly at topological vector space)).

Now there exist a sequence that approximate the sup,since closeness of K,the limit point also lies in K.(as follows since $t_n \to t^*$ we have $\|t_n-t^*\|\|y-x\| \to 0$ which means $z_n\to x+t^*(y-x)$)

The limit for TVS is proved as follows:$z = x +t^*(y-x)$ since any neiborhood around this point is abosrbing,which means at direction $y-x$ exist a small $\epsilon$ segment lies inside it.that is exist $x + (t^*-\epsilon)(y-x)$ contains inside neiborhood,since $t_n\to t^*$ hence must exist some $z_n$ lies in this neiborhood.

答:
Consider the set
$$T = \{t \ | \ x + t(y-x) \in K \}$$
Since $K$ is bounded, $T$ is a bounded subset of $\mathbb{R}$.  Since $K$ is closed, $T$ is a closed subset of $\mathbb{R}$. Note that $1\in T$. Consider $t^{*} = \sup T$. Clearly $t^{*}\in T$.  Moreover for every $t > t^{*}$, we have $x + (t^{*}+\frac{1}{n}) (y-x) \not \in K$ and $x + (t^{*}+\frac{1}{n}) (y-x) \to x + t^{*} (y-x)$. Therefore $x + t^{*} (y-x) \in \partial K$

Note: We did not use $K$ convex. If $K$ is moreover convex, $T$ is a closed segment containing $[0,1]$. The points on the boundary of $T$ are in $\partial K$.

$\bf{Added:}$ The only condition needed is $K$ bounded.  Since the we have the partition of the full space
$$X = \mathring{K}\cup \partial K \cup \mathring  {X\backslash K}$$

If the ray $\{ x + t (y-x) \ | t>0\}$ did not intersect the boundary, it would be contained in $\mathring{K}$, since $(0, \infty)$ is connected, contradiction with $K$ bounded.


问:
For $a\in (0,1)$,$ \lim_{h\to 0^+} \int_h^{1-h} t^{-a}(1-t)^{a-1} dt$ exists. Let the limit be $g(a)$. Then evaluate $g’(\frac 12)$
$$g(a) = \int_0^1 t^{-a}(1-t)^{a-1}$$

$$g’(a) =0$$

The answer seems straightforward, but it doesn’t feel right, so I would like someone to reaffirm it so that i know this is the way to solve it

Thanks!

答:
I hope you're familiar with Leibniz integral rule (of differentiating under the integral sign).

$$\frac{d}{dx}\int_a^bf(x,t)dt = \int_a^b\frac{\partial}{\partial x} f(x,t)dt $$

Apply this rule on your integral. I believe you will easily find your answer then.

EDIT:
Your hypothesis that $g'(a) = 0$ is unfounded. Please perform the calculation again. If you face the problem again, just comment below. I or somebody will surely post the complete solution.

EDIT 2:
$g'(a) = 0$ does hold true at $a  = \frac{1}{2}$, so your answer might be correct too.

EDIT 3:
Differentiating under the integral sign, we get
$$g'(a) = \int_0^1 -t^{-a}\log{(t)}(1-t)^{a-1}+(1-t)^{a-1}t^{-a}\log{(1-t)} dt$$

Put $a = \frac{1}{2}$ here. You get
$$g'(\frac{1}{2}) = \int_0^1 \frac{\log{(1-t)}-\log{t}}{\sqrt{t}\sqrt{1-t}}dt$$

Let this eqn be e1.

Now, use the $a-x$ property in definite integration and replace $t$ by $1-t$ and you will get

$$g'(\frac{1}{2}) = \int_0^1 \frac{\log{(t)}-\log{(1-t)}}{\sqrt{t}\sqrt{1-t}} dt$$

Add them and I think you will get your answer. Please point out any mistake if you find.


问:
Interval of unique solution of differential equation $y’=x^2+y^2, y(0)=1 .$
Consider the following initial value problem  $$y’=x^2+y^2, y(0)=1$$

Which of the following statements is/ are

$1.$ there exist a unique solution in $[0,\frac{\pi}{4}]$.

$2.$ every solution is bounded on $[0,\frac{\pi}{4}]$.

$3.$ The solution has a singularity at some point in $[0,1]$.

$4.$ The solution because unbounded at some sub interval of $[\frac{\pi}{4}, 1]$.

I tried to find interval for unique solution by using existence and uniqueness theorem as

For $$|x|\leq a$$ $$|y-1|\leq b$$ we have

$$|f(x,y)|=|x^2+y^2|\leq a^2+(1+b^2)$$
$$h=min\{a,\frac{b}{M}\}=min\{a,\frac{b}{1+a^2 +b^2+2ab}\}$$ Now to find maximum of $$g(b)= \frac{b}{1+a^2 +b^2+2ab}$$ we have $$g’(b)= \frac{1+a^2-b^2}{(1+a^2 +b^2+2ab)^2}=0\iff b=\pm\sqrt{1+a^2}$$ So $$h=min\{ a,\frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}\}$$ $$a= \frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}$$ gives $$ 2a(1+a^2)=(1-2a^2)\sqrt{1+a^2}$$ $$a=\frac{1}{2\sqrt{2}}$$ Now it seems no conclusions. If it is unique( that I don’t know how ) in $[0,\frac{\pi}{4}]$ then it will be bounded also in this interval as continuous functions are bounded on compact set and if solution can’t be extended to $[0,1]$ then it is because of unboundedness of solution as I guess . I don’t know how to handle last two portions. Please help. Thanks.

答:
Clearly the equation has a unique solution since $f(x,y)$ is locally Lipschitz. Since $y'\ge0$ for all $x\ge0$, we have $y\ge1$. Note
$$ y'=x^2+y^2\ge y^2 $$
and hence
$$ \frac{y'}{y^2}\ge1. $$
Integrating both sides from $0$ to $x$ gives
$$ -\frac1{y(x)}+1>x $$
or
$$ y(x)>\frac{1}{1-x}, x\in[0,1). \tag1$$
Similarly for $x\in[0,\frac{\pi}{4}]$, we have
$$ y'=x^2+y^2\le (\frac\pi4)^2+y^2 $$
or
$$ \frac{y'}{(\frac\pi4)^2+y^2}\le 1. $$
Integrating both sides from $0$ to $x$ gives
$$ \frac{4}{\pi}\arctan(\frac{4}{\pi}y(x))-\frac{4}{\pi}\arctan(\frac{4}{\pi}y(0))\le x $$
or
$$ y(x)\le\frac{\pi}4\tan(x+\frac{4}{\pi}\arctan(\frac{4}{\pi})). \tag2$$
Note that (2) holds only for $x\in[0,\frac{\pi}{4}]$. We can
Combining (1) and (2) gives, for $x\in[0,\frac{\pi}4]$,
$$ \frac1{1-x}< y(x)\le\frac{\pi}4\tan(x+\frac{4}{\pi}\arctan(\frac{4}{\pi})). $$
Thus (1), (2), and (3), and (4) are true.


问:
Is V a Vectorspace?
I want to show that $V$ is a vectorspace.

Let $K = \mathbb{R}$ and $V= (\mathbb{R}^{2\times1}, \oplus,\odot)$ and $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, $\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} \in \mathbb{R}^{2\times1}$

with $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \oplus\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} := \begin{pmatrix} x_1 + y_1 \\ 0 \end{pmatrix}$
and $ \lambda\odot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} := \begin{pmatrix} \lambda x_1 \\ \lambda x_2 \end{pmatrix}, \lambda \in \mathbb{R}$.

Now we could just check if all vectorspace axioms are fullfilled (that would require alot of writing and doesn't seem smart).

So my take on this is that we just show that $V \subseteq \mathbb{R}^{2 \times 1}$ ($V$ is a vector subspace of $\mathbb{R}^{2 \times 1}$) because if V is a subspace, it is by definition a vectorspace.

Alright, let's start.

$ V \neq \emptyset$

Completion under addition

Completion under scalarmultiplication

(1) Because if $x_1 = 0$ and $y_1 = 0$ we get $\begin{pmatrix} 0 \\ x_2 \end{pmatrix} \oplus \begin{pmatrix} 0 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0+0 \\ 0 \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ We have $0v \in V$ therefore $ V \neq \emptyset$

(2) Suppose $\overrightarrow{x},\overrightarrow{y} \in V$ and $\overrightarrow{x} = \begin{pmatrix}x_1 \\ x_2\end{pmatrix}, \overrightarrow{y} = \begin{pmatrix}y_1 \\ y_2\end{pmatrix}, x_1,x_2,y_1,y_2 \in \mathbb{R}$.

Now $\overrightarrow{x} \oplus \overrightarrow{y} = \begin{pmatrix}x_1 \\ x_2\end{pmatrix}\oplus \begin{pmatrix}y_1 \\ y_2\end{pmatrix} = \begin{pmatrix}x_1+y_1 \\ 0\end{pmatrix} \in \mathbb{R}^{2\times1}$. Therefore $V$ is completed under addition.

(3) Suppose $\overrightarrow{x} \in V $ and $\overrightarrow{x} = \begin{pmatrix}x_1 \\ x_2\end{pmatrix}, x_1,x_2,\lambda \in \mathbb{R}$.

Now $\lambda \odot \overrightarrow{x} = \lambda \odot \begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \begin{pmatrix} \lambda x_1 \\ \lambda x_2\end{pmatrix} \in \mathbb{R}^{2\times1}$. Therefore V is completed under scalarmultiplication.

With (1),(2),(3) we've shown that $V \subseteq \mathbb{R}^{2\times1}$. Because $V$ is a subspace of $\mathbb{R}^{2\times1}$ we've shown that $V$ must be a vectorspace.

Can someone check if this proof is valid? I'm really confused because of the $\odot$ and $\oplus$.

EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT-EDIT

Alright, now I have another kinda similar problem to this one.

Let $K = \mathbb{R}$ and $U= (\mathbb{R}^{3\times1}, \oplus,\odot)$ and $\begin{pmatrix} x_1 \\ x_2  \\ x_3 \end{pmatrix}$, $\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \in \mathbb{R}^{3\times1}$

with $\begin{pmatrix} x_1 \\ x_2 \\ x3 \end{pmatrix} \oplus\begin{pmatrix} y_1 \\ y_2 \\ y3 \end{pmatrix} := \begin{pmatrix} x_1 + y_1 + 3 \\ x_2 + y_2 + 3 \\ x_3 + y_3 + 3 \end{pmatrix}$
and $ \lambda\odot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} := \begin{pmatrix} \lambda x_1 \\ \lambda x_2 \\ \lambda x_3  \end{pmatrix}, \lambda \in \mathbb{R}$.

Now if I understood your answers correctly we can just say that $U$ is not a vectorspace because

Let $a \in U$ then $2 \odot a = (a) \oplus (a)$ must be true, which is not the case because: $2 \odot \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} 2 \cdot a_1 \\ 2 \cdot a_2 \\ 2 \cdot a_3 \end{pmatrix}$ but
$(a) \oplus (a) = \begin{pmatrix} a_1 + a_1 + 3 \\ a_2 + a_2 + 3 \\  a_1 + a_1 + 3\end{pmatrix} = \begin{pmatrix} 2a_1 + 3\\ 2a_2 + 3 \\  2a_3 + 3\end{pmatrix}$

Or am I missing something?

答:
Important point: You can't just use the subspace test with $\mathbb{R^2}$ because, first of all, $V = \mathbb{R}^{2}$ (as a set) and is not a subset. But more importantly, $\mathbb{R^{2}}$ has different operations (in the way you're thinking of it). A subspace must have the same operations, and clearly $+$ doesn't (I'm just gonna call $\oplus$ as +). It is an important subtlety and you should make sure you understood this.

So if given $V \subseteq W$, where $W$ is a known vector space, to check that $V$ is a vector subspace we would need that $V$ inherits the same operations from $W$. Otherwise $V$ must be treated as its own thing, and you must check from scratch.

Now that being said, the space you described is not a vector space.

For one thing there's no additive identity. For example, given $(1, 2)$, no matter what we choose as our identity, say, $(0, 0)$, we don't retain $(1,2)$. In this case, $(1, 2) + (0, 0) = (1, 0) \not = (1,2)$. More generally you will never retain the second entry if it is non zero.

As noted in the comments, associativity isn't satisfied either. This object fails being a vector space pretty badly!

Title: Is V a Vectorspace?


问:
Tao Analysis 2 Lemma 6.6.6 (lemma for inverse function theorem)

Let $B(0,r)$ be a ball in $\mathbb{R}^n$ centered at the origin, and let $g: B(0,r) \to \mathbb{R}^n$ be a map such that $g(0) = 0$ and
$$||g(x) - g(y)|| \le \frac12 ||x-y||$$
for all $x,y \in B(0,r)$ (here $||x||$ denotes the length of $x$ in $\mathbb{R}^n$). Then, the functions $f : B(0,r) \to \mathbb{R}^n$ defined by $f(x) : = x + g(x)$ is one to one, and furthermore the image $f(B(0,r))$ of this map contains ball $B(0,r/2)$.

I understand how to prove one to one. The proof for the second part is on the picture below.

It is not clear to me how the same argument shows that $F$ maps $\overline{B(0, r-\epsilon)}$ to itself. If I use the same argument when $x \in \overline{B(0, r-\epsilon)}$,
$$||F(x)|| \le ||y|| + ||g(x)|| \le \frac{r}2 + ||g(x) - g(0)|| \le \frac{r}2 + \frac{r-\epsilon}2 = r- \frac{\epsilon}2$$
This is not sufficient to show that it maps to $\overline{B(0, r-\epsilon)}$. I think I am missing something here. I would appreciate if you give some help.

答: What happens is that $y$ is in the open ball of radius $r/2$. So, if $\epsilon$ is small enough, you can get $$\|y\|<r-\epsilon/2.$$


问:
Confidence Interval for $2\theta^{-2}x$
For a sample size of one from the population with distribution:

$f(x; \theta)=2\theta^{-2}x$  for $0<x<\theta$

Find a $100(1-\alpha)\%$ Confidence Interval for $\theta$.

Anytime I see $\theta$, I feel like the pivot method is probably the way to go, but any help would be taken.

答:
For any value of $\theta$, the probability must be at most $\alpha$ that the confidence interval for the observed value does not contain $\theta$. That can be achieved in different ways. For instance, we can require that only the lowest $x$ values lead to a confidence interval not containing $\theta$:

$$
\int_0^b2\theta^{-2}x\,\mathrm dx=\theta^{-2}b^2\stackrel!=\alpha\;,
$$

which yields $b=\sqrt\alpha\theta$ as the upper bound of $x$ values for which $\theta$ lies outside the interval, leading to the confidence interval $[x,x/\sqrt\alpha]$. Alternatively, we can require that only the highest values of $x$ lead to a confidence interval not containing $\theta$:

$$
\int_c^\theta2\theta^{-2}x\mathrm dx=1-\theta^{-2}c^2\stackrel!=\alpha
$$

yielding $c=\sqrt{1-\alpha}$ and thus the confidence interval $[x/\sqrt{1-\alpha},\infty]$. Of course we could also use anything in between, e.g. a fraction $\alpha/2$ of the highest and a fraction $\alpha/2$ of the lowest $x$ values, yielding $b=\sqrt{\alpha/2}\theta$ and $c=\sqrt{1-\alpha/2}\theta$ and the confidence interval $[x/\sqrt{1-\alpha/2},x/\sqrt{\alpha/2}]$.

Title: Confidence Interval for $2\theta^{-2}x$


问:
Largest possible radius of a circle lying in a parabola

Circle of radius $r$ touches the parabola $y^2+12x=0$ at its vertex. Centre of circle lies left of the vertex and circle lies entirely within the parabola. What is the largest possible value of $r$?

So my book has given the solution as follows:

The equation of the circle can be taken as: $(x+r)^2+y^2=r^2$

and when we solve the equation of the circle and the parabola, we get $x=0$ or $x=12-2r$.

Then, $12-2r≥0$ and finally, the largest possible value of $r$ is $6$.

This is where I got stuck as I'm not able to understand why that condition must be true. I get that the circle must lie within the parabola...

Can someone please explain this condition to me?

答:
It is helps to learn and remember standard forms of the parabola along with the osculating circle radii at their vertices.
$$ y^2= \pm 2Rx , x^2=\pm 2R y$$
Differentiating $ 2 R y = x^2$ twice show that the minimum touching circle radius is R.. in this case $6$.