问:
$f:\mathbb{R} \mapsto \mathbb{R}$ convex and increasing. Show there is $a \in \mathbb{R}$ such that $\lim\limits_{x\to -\infty}\frac{f(x)}{x}=a$
Let $f:\mathbb{R} \to \mathbb{R}$ be a convex and increasing function such that $\lim\limits_{x\to -\infty}f(x)\to-\infty$ Show that there is an $a \in \mathbb{R}$ such that $\lim\limits_{x\to -\infty}\frac{f(x)}{x}=a$.

First of all, i can't really find an example respecting the hypothesis. Is there a convex function which is increasing with $\lim\limits_{x\to -\infty}f(x)=-\infty$? If $f$ was supposed concave, it appears true to me.

Then, if the statement holds, I don't really see how to prove it. I tried to apply the following property of convex function on the interval $x<a<b$: $$\frac{f(a)-f(x)}{a-x}\le\frac{f(b)-f(x)}{b-x}\le\frac{f(b)-f(a)}{b-a}.$$

But, even if i suceed to "sandwich" ${f(x) \over x}$, i don't really see how it could help.. Thank you in advance for help.

答:
We know that $f(c) \le 0$ for some $c \in \Bbb R$. For $x < y < c$ we have $f(x)  \le 0$ and the convexity condition
$$
 f(y) \le \frac{c-y}{c-x} f(x) + \frac{y-x}{c-x} f(c) \le \frac{c-y}{c-x} f(x) \, .
$$
It follows that
$$
 0 \le \frac{f(x)}{x-c} \le \frac{f(y)}{y-c}
$$
which shows that $f(x)/(x-c)$ decreases with decreasing $x$ and is bounded below by $0$. Therefore
$$
 a = \lim_{x \to -\infty} \frac{f(x)}{x-c}
$$
exists, and then
$$
 \frac{f(x)}{x} = \frac{x-c}{x} \frac{f(x)}{x-c}
$$
has the same limit $a$ for $x \to -\infty$.


问:
A problem from distribution theory.
Let $f$, $g\in C(\Omega)$, and suppose that $f \neq g$ in $C(\Omega)$. How can we prove that $f \neq g$ as distributions?

Here's the idea of my proof.

$f$ and $g$ are continuous functions, so they will be locally integrable.

Now, take any $\phi \neq 0 \in D(\Omega)$. Let us suppose that $\langle T_f,\phi\rangle =\langle T_g,\phi \rangle $ and $f\neq g$

$\langle T_f,\phi \rangle = \langle T_g,\phi \rangle$

$\implies \int_\Omega f(x) \phi(x) dx = \int_\Omega g(x) \phi(x) dx$

$\implies \int_\Omega \phi(x)[f(x)-g(x)] dx =0$

i.e., the area under above function is zero.

We know that $\phi$ is non zero, we just need to prove that this integral will be zero only when $f(x)-g(x)=0$ and it will make contradiction to our supposition that $f$ and $g$ are not equal.

[Please help me prove the last point]

答:
Let me add a little more detail to my comment.

You are correct up to the $\int_{\Omega} \phi(x)[f(x)-g(x)]dx = 0$.

Now, for $\varepsilon > 0$ and $x_0 \in \Omega$, let $B_{\varepsilon}(x_0)$ denote the ball of radius $\varepsilon$ centered at $x_0$. Then, you can find a test function $\psi$ with the following properties:

(i) $\psi(x) > 0 \, \forall x \in B_{\varepsilon}(x_0)$

(ii) $\psi(x) = 0 \, \forall x \notin B_{\varepsilon}(x_0)$

(iii) $\int_{\Omega} \psi = \int_{B_{\varepsilon}(x_0)}\psi = 1.$

Now, choose $\phi = \psi$ and for any $x_0 \in \Omega$
$$
0 = \int_{\Omega} \psi(x)[f(x)-g(x)]dx = \int_{B_{\varepsilon}(x_0)} \psi(x)[f(x)-g(x)]dx \approx f(x_0)-g(x_0)
$$

As $\varepsilon \rightarrow 0$, the equation is exact, i.e. $f(x_0) - g(x_0) = 0$. Since $x_0$ was arbitrary, $f = g$ in $\Omega$, a contradiction.

ContentLicense: CC BY-SA 3.0

Title: A problem from distribution theory.


问:
Are there implicit parentheses around fractions?
I was just looking at a simple math question that came across my social media feed and noticed something I must have always ignored in the past.

In this equation $x = 3 \div \frac{1}{3}$ there seems to be an implicit set of parentheses around $\frac{1}{3}$ which yields a result of $x= 9$.

This seems strange to me because I look at $\frac{1}{3} = 1 \div 3$ and wonder why we can't rewrite it as $x = 3 \div \frac{1}{3} = 3 \div 1 \div 3 = 1$

Is omitting the the parentheses just shorthand that I never realized I was using; $\frac{1}{3} \neq 1 \div 3$, but instead $\frac{1}{3} = (1 \div 3)$?

Should fractions come at the front of (F)PEMDAS/(F)BEDMAS?

Or is it just that the definition of a fraction entails that it is an individual unit and expressing      $\frac{1}{3}$ as $(1 \div 3)$ doesn't really make sense either?

答:
The mnemonics BEDMAS and PEMDAS have some utility, but they're not perfect.
They work relatively well for parsing computer code,
where the operands and operators are all written in one linear string.
Even then, neither mnemonic literally describes the correct order of application of operations in all cases.
Just look at the ambiguity between DM in one mnemonic and MD in the other.

As soon as we start arranging things in two dimensions we have the effect of implicit parentheses. There should be no doubt that $\frac 2{3-1}$ is meant to be evaluated like $2/(3-1)$ and not like $2/3-1.$
When we write $e^{2x}$ it means the multiplication $2x$ is performed before the exponentiation, regardless of the occurrence of E before M in the mnemonics.

The mnemonic really should be something like PE(MD)(AS), and I would say it only applies when we have no two-dimensional clues to say what operation applies first, such as in the expression
$3 - 1 \times 3$ (which is $3 - (1\times 3),$ not $(3-1)\times 3)$)
or in the expression $3x^2$ (which is $3\times(x^2),$ not $(3 \times x)^2$).

The two-dimensional clues I would take from a horizontal fraction bar would be that it always has implicit parentheses like this:
$$ \left(\frac{( \cdots )}{( \cdots )} \right).$$

Even these implicit parentheses can be unclear in some cases, so we simply should not write expressions like
$$ \frac{1}{\frac AB}, $$
which require carefully looking at the relative sizes of typefaces and slightly different lengths of horizontal lines.

ContentLicense: CC BY-SA 4.0


问:
Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.
Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.
Is this solution enough?... I cannot understand the solution provided in the below mentioned link:  https://www.stat.cmu.edu/~larry/=stat705/Lecture4.pdf

答:
Yes, it is enough. And as @noam.szyfer suggested, this does need independence.

I think you are okay with the first equality, since $X_i$-s cannot exceed 1.

For the second, note that $X_{(n)}\leq 1-\epsilon$ means that the maximal one of the first $n$ trials is smaller than $(1-\epsilon)$, therefore every one of the first $n$ trials should be smaller than that value. And they are independent, so the propability of $\{X_{(n)}\leq 1-\epsilon\}$ is the product of all $n$ probabilities of $\{X_k\leq 1-\epsilon\}$.

Hope this helps.


问:
Formulate and Prove by induction a general formula
$1^3=1$

$2^3=3+5$

$3^3=7+9+11$

$4^3=13+15+17+19$

Please help me to find the formula. I have no idea about this

答: Hint: notice that the terms that appear on the RHS are all the sums of odd numbers. Try and use this to come up with a formula.


问:
Proof of the inverse of a matrix multiplication from the relation $\operatorname{inv}(A) =\operatorname{adj}(A)/\det(A)$
I am trying to prove that (1) $(\mathbf{AB})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$ using the relation (2)$$ \mathbf{A}^{-1}=\text{adj}\mathbf{A}/\det \mathbf{A}$$ where $\operatorname{adj}\mathbf{A}$ is the adjugate of $\mathbf{A}$, and $\det\mathbf{A}\neq 0$

For the special case where both matrices are $n$ x $n$, I came down to try to prove that (3)  $$\text{det}(\mathbf{AB})(i/j)=\sum_{k=1}^n\text{det}\mathbf{A}(i/k)\det\mathbf{B}(k/j)$$
where $\det\mathbf{A}(i/j)$ is the determinant of $\mathbf{A}$ minus row $i$ and column $j$.

I have checked the validity of this relation for a $2×2$ matrix, but cannot prove it in general.  On the other hand, is there a more straightforward way of proving (1) using (2) without going through (3)?

答:
There is an easy way: Suppose that $A, B$ are $n \times n$ matrices such that $AB$ is invertible.
This means that $\det(AB) \neq 0$ and hence, by the formula you gave we have
$$
\begin{align}
(AB)^{-1}
&= \frac{\operatorname{adj}(AB)}{\det(AB)} \\
&= \frac{\operatorname{adj}(B) \cdot \operatorname{adj}(A)}{\det(B) \cdot \det(A)} \\
&= \frac{\operatorname{adj}(B)}{\det(B)} \cdot \frac{\operatorname{adj}(A)}{\det(A)} \\
&= B^{-1} A^{-1}
\end{align}
$$
by using $\operatorname{adj}(AB) = \operatorname{adj}(B)\operatorname{adj}(A)$.


答:
For any invertible matrix $\;X$, its inverse is denoted by $\;X^{-1}$. Thus, the inverse of the product $\;AB\;$ of square matrices $\;A,\,B\;$ is $\;(AB)^{-1}$. Yet

$$(AB)\cdot(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$$

using associativity, and thus also $\;B^{-1}A^{-1}\;$ is the inverse of $\;AB$. By uniqueness of the inverse, $\;(AB)^{-1}=B^{-1}A^{-1}\;$ and we're done.

Title: Proof of the inverse of a matrix multiplication from the relation $\operatorname{inv}(A) =\operatorname{adj}(A)/\det(A)$

Body:
I am trying to prove that (1) $(\mathbf{AB})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$ using the relation (2)$$ \mathbf{A}^{-1}=\text{adj}\mathbf{A}/\det \mathbf{A}$$ where $\operatorname{adj}\mathbf{A}$ is the adjugate of $\mathbf{A}$, and $\det\mathbf{A}\neq 0$

For the special case where both matrices are $n$ x $n$, I came down to try to prove that (3)  $$\text{det}(\mathbf{AB})(i/j)=\sum_{k=1}^n\text{det}\mathbf{A}(i/k)\det\mathbf{B}(k/j)$$
where $\det\mathbf{A}(i/j)$ is the determinant of $\mathbf{A}$ minus row $i$ and column $j$.

I have checked the validity of this relation for a $2×2$ matrix, but cannot prove it in general.  On the other hand, is there a more straightforward way of proving (1) using (2) without going through (3)?


问:
What are some examples of proof by contrapositive?
Applying the Modus Tollens argument to Fermat's Little Theorem really helped me to understand logical implication. I never knew that FLT was actually a compositality test.

Theorem (FLT): given integers $a>1$ and $n>1$, if $n$ is prime, then $a^n$ is congruent to $a\  (\bmod\  n)$.

By the contrapositive, if $a^n$ is not congruent to $a\  (\bmod\  n)$, then $n$ is not prime. Thus $n$ is composite.

What are some other simple and instructive examples of proof by contrapositive?

答: Look at Richard Hammack's "Book of Proof" for a detailed discussion of proof forms. There are several "proof technique" and such introductory courses (perhaps under discrete mathematics and similar) with lecture notes on the 'net. </p>


问:
Maximum and minimum involving functional equation.
Let $f$ be continous in $\mathbb{R}$ with $f(x+1)=f(x).$ Prove: $f$ takes on minimum and maximum.

答: A function that satisfies this condition is called a $1$-periodic function. For such functions, their behavior is completely described by their restriction to any interval of length 1. Since $f$ is continuous, take its restriction on $[0,1]$. This restriction is a continuous function over a compact set, so it admits a maximum and a minimum value. Can you see how periodicity forces these values to be the maximum and the minimum of $f$ over the entire $\mathbb{R}$?


问:
Constructing a Random Variable From a Law
Let $X$ be a polish space and let $\mathcal{B}(X)$ be its Borel $\sigma$-algebra.  Let $\mu$ be a probability measure on $(X,\mathcal{B}(X))$.  How can we construct an $\mathbb{R}^n$-valued random element with law $\mu$?

答:
Since I cannot comment, I have to write this as an answer:

the law of a random variable is the measure induced by it on the target space. You can not use it to induce a measure on the original measure space(here is $\mu$). You need to find a probability measure on $\mathbb{R}^n$ to be the candidate.

And I am not sure about the criteria for this to work, waiting for others to solve it :)

One thing I am sure is that it depends on the topology on $X$. If the topology is trivial with all open sets $\{\emptyset,X\}$, then you cannot define a random variable inducing law on $\mathbb{R}^n$ not concentrating on one single value.


问:
Alternative Methods for Trigonometry Proofs (Varying Angles)
Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?

答:
Partial Answer

Firstly,

The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

$$\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)$$

Most of the trigonometry formulas can be derived after establishing the basic ones (especially the sum and difference identities). In the unit circle, by rotating an angle (in the standard position) by the multiples of 90 degrees counterclockwise around the origin (or by using congruent triangles), you can prove the phase shift identities and much more. Sketching graphs often helps too. For example, $\sin(x-180°)=-\sin(x)$ can be seen by sketching $\sin(x-180°)$, noting that it is simply a sine curve with a horizontal shift of $180°$. Complex numbers are especially useful for double/triple angle formulae and power reduction etc.


问:
Do we really know the value of expressions with irrational powers?
The way we evaluate decimal powers such as $a^.75$ is by splitting it into $(a^3)$^(1/4). How then can we evaluate irrational powers? I know that we can approximate, but whenever we graph a^x we assume continuity at the irrational numbers, but how can we be certain of this?</p>

答: The definition is that $a^b = e^{b \ln a}$ for positive $a$. This agrees on the rational points with what you get by roots and powers, and is continuos in $b$, so it is a decent definition.


问:
Cayley graph of an infinite group using GAP
I have a finitely presented group which is infinite, and I wish to create a visual representation of its Cayley graph using GAP.

If possible I would like to do this using YAGS package.

My group is $\langle a, b\mid a^3 = b^3 = (ab)^3 = e\rangle$.

Its easy to define it in GAP, but it is infinite and for a visualization I should somehow choose only finitely many of its elements.

I'm looking for something like this:

答:
You can use Bartholdi's package FR:

```<code>gap> LoadPackage("FR");

gap> f := FreeGroup( "a", "b" );;

gap> g := f / [ f.1^3, f.2^3, (f.1*f.2)^3 ];
<fp group on the generators [ a, b ]>

gap> WordGrowth(g,rec(limit:=4,draw:="Cayley.dot"));
</code>```

this returns a dot file which you can open with e.g. imagemagick or convert to another format. DOT is a widely used format for graph so you'll find tool if the layout doesn't suits you.

<img src="https://i.stack.imgur.com/O7jCo.png" alt="{Cayley graph of the group $F_2/ {a^3 ; b^3 ; (ab)^3}$}" />

Body:
You can use Bartholdi's package FR:

gap> LoadPackage("FR");

gap> f := FreeGroup( "a", "b" );;

gap> g := f / [ f.1^3, f.2^3, (f.1*f.2)^3 ];
<fp group on the generators [ a, b ]>

gap> WordGrowth(g,rec(limit:=4,draw:="Cayley.dot"));

this returns a dot file which you can open with e.g. imagemagick or convert to another format. DOT is a widely used format for graph so you'll find tool if the layout doesn't suits you.

<img src="https://i.stack.imgur.com/O7jCo.png" alt="{Cayley graph of the group $F_2/ {a^3 ; b^3 ; (ab)^3}$}" />


问:
test for convergence in this series $(-1)^n\frac{2^{2n+1}}{n!}$
I'm testing for convergence in this series:

$(-1)^n\frac{2^{2n+1}}{n!}$

I approached it with the ratio-test, although is there a better way to test for convergence in alternating series?

This is what I've done so far:

$$(-1)^{n+1}\frac{2^{2n+1+1}n!}{(-1)^n(n+1)!2^{2n+1}}$$

Then cancelling the terms:
$$-\frac{1}{n+1}\frac{1}{2}$$ The series converges to $-\frac{1}{2}$?

答:
You could notice that the series is a modification of the Taylor series of e^-x,

e^-x = sigma of (-1)^n * (x^n)/n!

First we will simplify the series
sigma of (-1)^n * (2^(1+2n))/n!

= 2 * sigma of (-1)^n * (2^(2n))/n!
-- Here we just take out a multiplication of a constant out of the sum

= 2 * sigma of (-1)^n * (4^n)/n!
-- By using the identity, 2^(2n) = 2^2^n = 4^n

Therefore, 2*e^-4 = 2 * sigma of (-1)^n * (4^n)/n! = sigma of (-1)^n * (2^(1+2n))/n!

Working with Taylor expansions can subtract our need to calculate the limit of some tests (which sometimes can be a tedious work), the series does not converge to -1/2,

Using the ratio test you'd need to calculate the lim of a_(n+1) / a_(n) in absolute value, I mean lim of |a_(n+1) / a_(n)|, if that limit is less than 1 then you know that the series converges but that limit does not indicate to what it converges to, only that it converges.


答:
Yes @noam.szyfer, you need independence. In order for $X_{(n)}$ to converge to 1 in probability, we need to show that $\forall \epsilon>0, P(|X_{(n)}-1|<\epsilon)=1$. This is given via the following:
\begin{equation}
P(|X_{(n)}-1|<\epsilon)\\=P(|1-X_{(n)}|<\epsilon)\\
=P(1-X_{(n)}<\epsilon)\\
=1-P(1-X_{(n)}\geq \epsilon)=1-P(1-\epsilon\geq X_{(n)})\\
\end{equation}
However, note that $X_{(n)}\leq 1-\epsilon$ if and only if $X_i\leq 1-\epsilon$ for all i. This implies that
\begin{equation}
1-P(1-\epsilon\geq X_{(n)})=1-P(1-\epsilon\geq X_1,1-\epsilon\geq X_2,..)\\
=1-\prod_{i=1}^\infty P(1-\epsilon\geq X_i)\,\,(\text{By Independence})\\
=1-(1-\epsilon)^n
\end{equation}
The final term limits to one as $1-\epsilon<1$ for any $\epsilon>0$ and hence $\lim_{n\rightarrow \infty}(1-\epsilon)^n=0$.

Body:
Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.
Is this solution enough?... I cannot understand the solution provided in the below mentioned link:  https://www.stat.cmu.edu/~larry/=stat705/Lecture4.pdf


问:
Complex numbers and $ e^{i\theta} $
I have a complex $ z $ that I want to write as $ z=|z|e^{i\theta} $ where $ \theta $ is its phase.

I tried using 2 "different" ways and got different results, so I assume one of them is wrong.

This is the number: $ z=\frac{a}{b-ic} $

where $ a,b,c \in \mathbb{R} $ and positive.

Now we can write:

$ z=\frac{a}{b-ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(0-\arctan\left(\frac{-c}{b}\right)\right)} $

But also:

$ z=\frac{-a}{-b+ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(\pi-\arctan\frac{c}{-b}\right)} $

So where is the mistake?

Thanks in advance.

答:
I think this was actually a better question when it still had all the variables $\omega$, $R$, $L$, and $C.$
Eliminating those variables eliminates one of the sources of ambiguity inherent in the original problem, which is that we don't know which of the two quantities $\omega^2 RLC$ or $R$ is greater.

The problem is the in the conversion of
$p + iq$ to $\sqrt{p^2+q^2} \, e^{i \arctan(q/p)}.$
That formula is correct only when $p > 0.$
If $p < 0$ a correct formula is
$$p + iq = \sqrt{p^2+q^2} \, e^{i (\pi + \arctan(q/p))}.$$

And of course if $p = 0$ neither formula works at all.

So if you have $b - ic,$ where $b$ and $c$ are positive,
you have the case $p = b > 0$, so
$$b - ic = \sqrt{b^2+c^2} \, e^{i \arctan(-c/b)}.$$

But when you look at $-b + ic,$ you now have the case $p = -b < 0,$ so
$$-b + ic = \sqrt{b^2+c^2} \, e^{i (\pi + \arctan(-c/b))}.$$

You applied the wrong formula to the case $-b + ic,$ and that's where the extra $\pi$ comes from.

In your original exercise, you had the expression
$\left(R - \omega^{2}RLC\right)+i\omega L$
in one case and
$\left(\omega^{2}RLC - R\right)-i\omega L$
in the other.
If $R - \omega^{2}RLC > 0,$ then $\omega^{2}RLC - R < 0,$ and vice versa.
That is, one of these expressions is the $p > 0$ case
and the other is the $p < 0$ case.
The plain arc tangent formula only works in one case;
the other case needs some modification, such as adding $\pi$ to the arc tangent.
(Subtracting $\pi$ works equally well, because that's the same as adding $\pi$ and then subtracting $2\pi$.)

That's why there's a difference of $\pi$ between the angles you found in your results. The correct result is the one where you applied your formula to
$p + iq$ where $p > 0.$
In the other case your angle is shifted by $\pi$ radians.

Title: Complex numbers and $ e^{i\theta} $


答:
As the different commentors here agree we also need the $X_i$ to be independent for this to work. Here is the same proof as in the PDF you linked, with additional explanations.

We need to show that $\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon)$ converges to 0 as $n\to \infty$. (This is the definition of convergence in probability). So let $\varepsilon>0$, then we have:

$$\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon) = \Bbb{P}(1-X_{(n)}\geq\varepsilon) = \Bbb{P}(X_{(n)}\leq1-\varepsilon)$$

For the first equality we used that $X_{(n)}-1$ is non-positive, since $X_{(n)}$ is bounded by 1.

Now comes the crucial step: What is the probability that the maximum of all the random variables $X_i$ is smaller than some value? It's the same as saying: every $X_i$ must be smaller than this value (since then also the maximum of all of them are smaller. If one was bigger, then also the maximum would be bigger!). So in that sense, the following two events are the same:

$$\Bbb{P}(X_{(n)}\leq1-\varepsilon) = \Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon).$$

Finally, we use the independence assumption to get

$$\Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon) = \prod_{i=1}^{n}\Bbb{P}(X_i < 1-\varepsilon) = (1-\varepsilon)^n.$$

We see that the last term tends towards 0 as $n$ tends towards infinity.

Title: Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.


问:
The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

The length of projection of the line segment joining the points $(1, –1, 0)$ and $(–1, 0, 1)$ to the plane $2x + y + 6z = 1$ is equal to?

I am confused if i have to find the length of perpendicular from line segment to plane or the length of projection

答:
If we name the given points as $A(1, -1, 0)$ and $B(-1,0,1)$, then,

$ \vec {AB} = (-2, 1, 1), |\vec {AB}| = \sqrt6$

Equation of plane is $2x+y+6z = 1$. So normal vector to the plane, $\vec{n} = (2,1,6)$ and $|\vec n| = \sqrt{41}$.
So unit normal vector, $\hat{n} = \frac{1}{\sqrt {41}}(2,1,6)$.

Now projection of $\vec{AB}$ in the direction of normal vector to the plane is $|AB| \cos\theta$ and onto the plane is $|AB| \sin\theta$ where $\theta$ is the angle between the normal vector and $\vec{AB}$.

Projection $P_{n}$ of $\vec {AB}$ in the direction of normal vector can be simply found using dot product.
$P_{n} = \vec {AB} \cdot \hat{n} = \frac{3}{\sqrt {41}}$.

Then the projection onto the plane is $P = \sqrt{|AB|^2 - P_n^2} = \sqrt{\frac{237}{41}}$


问:
Random varible defined from another ($Y:=\left | X \right |-1$)
Given a PMF of $X$

$$p(x)=\left\{\begin{matrix}
0.1, & x=-4, \\
0.2 & x=-1\\
0.6 & x=1\\
0.1 & x=2\\
0 & else
\end{matrix}\right.$$
Find $\mathbb{P}(X \geq 0.8)$, $\mathbb{E}X$, var$(X)$ and sd$(X)$. Furthermore, find the mean and pmf for $Y:=\left | X \right |-1$.

I have found the firs 4 solutions $\mathbb{P}(X \geq 0.8)=0.7$, $\mathbb{E}X=1$, The varians to be $17/5$ and thus the sd to be roo square root of that answer.

Now for the last 2: Would this mean that the PMF's entries of $Y$ would just correspond to the aboslute value of the discrete r.v $X$ minus 1? Would you have to compute the mean from the start even though we just calculated the expected value?

答:
Hint: To start make a table of the transformed x-values.

$$\begin{array}{|c|c|c|c|c|} \hline x &-4&-1&1&2  \\  \hline y=|x|-1 &3&0&0&1  \\  \hline \textrm{prob.} &0.1&0.2&0.6&0.1 \\ \hline  \end{array}$$

Now you can write down the pmf of y, $P(Y=y)$. What is $P(Y=0)$ for instance? Then you calculate the expected value and the variance as usual.


问:
$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$
$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$. First of all, it seems to me that the condition that $f''$ bounded is useless... Isn't it?

My attempt is to pass by mean value theorem, so i would like to know if my approach holds, please.

First, consider the interval $[x,x+a]$ with $x>0, a>0$. As $f$ is differentiable on $[x,x+a]$, by mean value theoreme there exists some $c_x \in ]x,x+a[$ such that

$f'(c_x)=\frac{f(x)-f(x+a)}{a}$.

Inserting the limit on RHS and LHS we get the following:

$\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x)-f(x+a)}{a} = 0$.

Therefore, $\lim_{x\to\infty}f'(x)=0$ as wanted.

I know there is a post already on this statement, but they use the fact that $f''$ is bounded in theirs proofs... If this hypothesis is necessary, could someone explain why or give a counter example on my proof?

答:
hint

Apply second order Taylor-Lagrange formula in $ [0,x] $ for an $ x >0 $.

there exists $ c\in (0,x) $ such that

$$f(0)=f(x)-xf'(x)+\frac{x^2}{2}f''(c)=0$$


问:
Winding number of a curve (not complex analysis)
I am asked to calculate the winding number of an ellipse (it's clearly 1 but I need to calculate it)

I tried two different aproaches but none seems to work.

I would like to know why none of them work (I believe it is because these formulas only work if I have a curve parametrized by arc lenght).

Approach 1:

A valid parametrization : $\gamma=(a\cos t,b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\dot{\gamma}(t)=(-a\sin t,b\cos t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\ddot{\gamma}(t)=(-a\cos t,-b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$

$\det(\dot{\gamma}(t)|\ddot{\gamma}(t)) = \renewcommand\arraystretch{1.2}\begin{vmatrix}
-a\sin t & -a\cos t \\ b\cos t & -b\sin t \end{vmatrix}=ab \sin^2 t+ab \cos^2 t=ab$

$||\dot{\gamma}(t)||^3=(\displaystyle\sqrt{(-a\sin t)^2+(b\cos t)^2})^3=(\displaystyle\sqrt{a^2\sin^2 t+b^2\cos^2 t})^3=a^3b^3$

$\kappa(t)=\displaystyle\frac{ab}{a^3b^3}=\displaystyle\frac{1}{a^2b^2}$

$\mathcal{K}_\gamma = \displaystyle\int_{0}^{2\pi} \displaystyle\frac{1}{a^2b^2} \ dt= \displaystyle\frac{2\pi}{a^2b^2}$, $\mathcal{K}_\gamma$ is the total curvature of the curve.

$i_\gamma=\displaystyle\frac{\displaystyle\frac{2\pi}{a^2b^2}}{2\pi}=\displaystyle\frac{1}{a^2b^2}$...which is not necessarily 1.

Approach 2:

Winding # = $\displaystyle\frac{1}{2\pi}\displaystyle\int_{\gamma}\displaystyle\frac{-y}{x^2+y^2}\>dx+\displaystyle\frac{x}{x^2+y^2}\>dy$

That gives us $\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{-b\sin t}{a^2\cos^2 t+b^2\sin^2 t}(-a\sin t)+\displaystyle\frac{a\cos t}{a^2\cos^2 t+b^2\sin^2 t}(b\cos t) \right)\>dt$

$\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{ab}{a^2\cos^2 t+b^2\sin^2 t }\right)\>dt$, which I computed and cannot be calculated.

Clearly the second approach is valid if we are dealing with a circumference of radius 1. We can generalize for the elipsee using Green's Theorem. I would also like if someone could show me this way as well.

Thank you

答:
Green's Theorem will tell you that the winding number (given by approach #2 — your total curvature in #1 is incorrect because you need an arclength integral) of the ellipse is the same as the winding number of any circle (which is easy to compute).

Let $P=-\dfrac y{x^2+y^2}$ and $Q=\dfrac x{x^2+y^2}$. Note that on $\Bbb R^2-\{0,0\}$, we have $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$. Thus, if $R$ is the region between the ellipse $E$ and a circle $C$ centered at the origin lying inside it, we have
$$\int_E P\,dx+Q\,dy - \int_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA = 0.$$
But you can easily calculate that $\int_C P\,dx + Q\,dy = 2\pi$.


问:
Volume of $y=x^2$ bounded by $y=a$ with substraction
I'm trying find the volume of the volume of $y=x^2$ rotated around the $y$ axis and bounded by $y=a$. I'm following MIT 18.01 and professor solves the problem by slicing the shape to shells. To be more exact $$y=x^2, y=a\\One \ slice = dV = (2\pi x)(a-x^2)dx\\V=\frac{\pi }{2} a^{2}$$
But I think there is an easier solution. Why don't we just calculate the volume of the cylinder with height $a$ and radius = $\sqrt{a}$ (as $y=x^2$) and subtract the area under the $y=x^2$ curve rotated around the y axis?$$y=x^2, y=a\\V_{cylinder} = \pi\cdot\sqrt{a}^2\cdot a=\pi\cdot a^2\\Area\ under\ x^2 =A_{curve}=\int ^{\sqrt{a}}_{0} x^{2} dx\\V_{curve}=A_{curve}\cdot2\pi\\V=V_{cylinder}-V_{curve}=\pi(a^2-\frac{2}{3}a^{3/2})$$Why they do not equal the same thing?

答:
The way you calculated $V_{\textrm{curve}}$ is incorrect. There is no such formula $V_{\textrm{curve}}=A_{\textrm{curve}}\cdot 2\pi$,

Instead, if you use the shell method:
$$
V_{\textrm{curve}}=\int_{0}^{\sqrt{a}}2\pi x\cdot x^2\ dx=\frac{1}{2}\pi a^2
$$
so that
$$
V_{\textrm{cylinder}}-V_{\textrm{curve}}=\frac{1}{2}\pi a^2
$$


答:
Hint

Alternative approach that avoids using induction.

Each expression $n^3 = a_1 + a_2 + \cdots a_k$

involves exactly $n$ terms (i.e. $k = n$), whose average value is $n^2.$

Of course, proving that the average value of these terms is $n^2$, may involve induction, or may also be proven directly.

Title: Formulate and Prove by induction a general formula

Body:
$1^3=1$

$2^3=3+5$

$3^3=7+9+11$

$4^3=13+15+17+19$

Please help me to find the formula. I have no idea about this


问:
What is meant by "a curve $r(t)$ of constant length $|r(t)|$"?
I am working on an assignment for Calculus III. This particular one has to do with some vector calculus, but I am caught on some wording by my professor and he is not holding office hours at the moment.

In a few questions he asks us to show certain properties of "a curve $r(t)$ with constant length $|r(t)|$." I have been operating under the assumption that this means any curve with a scalar as its length, such as $\langle 0,0,t\rangle$ over $0\le t\le 1$, but am now questioning that assumption, as he asks us to show a property is true "for all of $t$."

Does anyone have any ideas what he means here? I have been working these by creating examples with small bounds such as the one above for simple integration for the lengths.

答:
First off, a couple of nit-picks:

The function $r$ is a curve in $\mathbb{R}^3$.  That means that it has the form $$ r : \mathbb{R} \to \mathbb{R}^3 : t \mapsto (x(t), y(t), z(t)), $$
where $x(t)$, $y(t)$, and $z(t)$ denote the $x$-, $y$-, and $z$-coordinates of some object at some time $t$.  That is, $r(t)$ is a point in space.  This may be a subtle point, but I think that it is wise to regard the the value of $r$ at a time $t$ to be a point in three-dimensional space, rather than a three-dimensional vector (some books call this a "position vector", but I'm not a huge fan of that terminology, either).  The codomain of $r$ is (presumably) endowed with a smooth structure, but the additional vector space structure which $\mathbb{R}^3$ possesses is largely irrelevant to the material in an introductory calculus class, and is nonexistent in a more general setting (i.e. the study of smooth manifolds).

I am not a fan of the description of $r$, which is "a curve with constant length".  The length of a curve typically refers to the distance that a particle traverses while traveling along that curve.  Instead, what is being described here is something else about the geometry of the curve.  Specifically, there is some fixed number $\varrho>0$ such that for all $t$,
$$ |r(t)| = d(0,r(t)) = \rho. \tag{$\ast$}\label{dist}$$
This is not a curve of constant length, but rather a curve where each point on the curve is some fixed distance from the origin.  I think that the notation in (\ref{dist}) is quite clear, but if one wants a description of this identity in English, I might say "$r$ is a curve on a sphere" or "$r$ is a curve such that each point on $r$ is a fixed distance $\varrho$ from the origin."

The term "curve" is a little overloaded.  We actually typically mean two different things by the word:  first, a curve is the image of an interval with respect to a continuous function, i.e. the "curve" $r$ is, as a set,
$$ \{ r(t) : t \in \mathbb{R} \}. $$
On the other hand, we also thing of the function itself as a curve, i.e. a "parameterized curve"—the curve consists not only of the set described above, but also the parameterization which tells you which point in the set corresponds to which moment in time.  That is, a "curve" can be seen both as a static object and as a dynamic object.  It might be worth noting that there may be many functions which correspond to the same static curve, but each corresponds to a different dynamic.  For example
$$ r_t(t) = \left( \cos(t), \sin(t), 0 \right) \ (t\in [0,2\pi])
\quad\text{and}\quad
r_2(t) = \left( \cos(-t), \sin(-t), 0 \right) \ (t\in [0,2\pi]) $$
are identical as sets, but have different dynamics.  These are distinct parameterizations of the same set.

This is, perhaps, a rather minor point, but I think that it is worth noting when one word is used to simultaneously represent two distinct, though related, ideas.

As I said, I think that the identity in (\ref{dist}) explains what is going on very clearly.  However, if more intuition is required, imagine the following scenario:

Grab a sphere off of your shelf (you have several, right?).  Declare that the center of this sphere is the origin and pick out some preferred coordinates (that is, mark out three coordinate axes).  Next, find an ant or other small insect, then dip this ant in paint (you monster), and drop it on the sphere.

From the moment you place the ant on the surface of the sphere, keep track of the ant's position.  After $t$ seconds have elapsed, the ant's position is denoted by $r(t)$—that is, $r$ is a function which takes a real number as input (a time) and spits out a position in three-dimensional space as an output (the ant's position in space).

Because the ant is on the surface of the sphere, its distance from the center of the sphere (the origin) is constant.  Therefore
$$ |r(t)| = \text{ the distance from center of the sphere to position of the ant at time $t$} = \text{constant}. $$
Now, remember that you dipped the ant in paint (you sadist) before you put it on the sphere, so it has been painting a trail of paint on the sphere which corresponds to the path that it has taken.  This path, which we may regard as the static curve described by the ant's position, is a curve with the property that each point is a fixed, constant distance from the origin.


问:
Pumping Lemma works on language, but language is not regular
So i am given this language:

L = { $c^ma^nb^n $ | $m≥ 1 $ and $n≥ 0$ } U { $a^mb^n$ | $m,n≥ 0$ }

And i have to prove that the pumping lemma property works on L.
Although pumping lemma can work, i then have to prove that L is not a Regular language.

Can anyone help me into proving this thing? And how is it possible that L would not be regular, but pumping lemma property worked?

答:
The easiest way to show that $L$ is not regular is probably to use the Myhill-Nerode theorem, i.e. we have to show that the equivalence relation
$$w \sim w' \text{ iff for all } x \in \Sigma^\ast, wx \in L \Leftrightarrow w'x \in L$$
has infinitely many equivalence classes.

Towards this, consider the words of the form $w_n = ca^n$.
Now for $n \neq m$ we have that $w_nb^n \in L$ but $w_m b^n$ is not, so in particular we have $w_n \not\sim w_m$, i.e. all equivalence classes associated to the $w_n$ are pairwise distinct for all $n$, and hence the language is not regular as the Nerode relation has infinite index.

Another way would be to assume that $L$ is regular and showing that any DFA $\mathcal A$ for $L$ could be modified to be a DFA $\mathcal A'$ for the language
$$ L' = \{c a^n b^n \mid n \in \mathbb N\}$$
which is non-regular, which is (probably) provable using the Pumping lemma.
This is a contradiction and hence $L$ cannot be regular.


问:
What is the name of the function $1/\Gamma(z) \int_0^\infty \frac{x^{z -1}}{e^w-1}dw$?
I recently saw

on the internet, and was curious what the name of the "pineapple" function was?

$$1/\Gamma(z) \int_0^\infty \frac{\color{red}{x}^{z-1}}{e^w-1}dw$$

I recognize the "mango" function as $\Gamma(z)$ the gamma function.  Also, does anyone know what theorem this refers to?

答:
One representation of the Riemann zeta-function
$$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}, \quad \operatorname{Re}s>1.$$

This theorem refers to the Riemann Hypothesis, one of the greatest unsolved problem in mathematics. The RH states:

All non-trivial zeros of the Riemann Zeta-Function lie on the critical line.


问:
Why do I need uniform convergence?
This is from baby Rudin problem 7.9:

Let $f_n$ be a sequence of continuous functions which converges uniformly to a function $f$ on a set $E$. Prove that:

$$\lim_{n\rightarrow \infty} f_n(x_n)= f(x)$$

for every sequence of points $x_n \in E$ such that $x_n \rightarrow x$ for $x \in E$.

What I've tried

We want to show:

$$|f_n(x_n) - f(x)| \rightarrow 0$$

By the hypothesis on $x_n$:

$$|f_n(x_n) - f(x)| = \left|f_n\left(x + o(1)\right) - f(x)\right| = \left| f_n\left(x \right) + o(1) - f(x) \right|$$

Where the second equality comes from the continuity of $f_n$. This clearly goes to $0$ as $n\rightarrow \infty$. Nowhere have I used the uniform convergence of $f_n$ to $f$. What did I miss?

I have seen Why do I need uniform convergence here? but my argument is different.

答: I'm not  a huge fan of the $o(1)$ notation being used here, and this may be hiding where the mistake lies. It is true that for any fixed $n$ you have $f_{n}(x + o(1)) = f_{n}(x) + o(1)$ (using your notation). However, this may not be true for all $n$ sufficiently large. In fact, saying that it is true for all $n$ sufficiently large is virtually the definition of uniform convergence! Without uniform convergence, what can happen is that as $n$ gets larger, the $o(1)$ quantity converges more and more slowly to zero, and in the limit may no longer converge to zero at all!

Title: Why do I need uniform convergence?


问:
Unable to solve definite integral if an implicit function
We were asked to find the area under the given function, as a challenging exercise:

\begin{gather*}
6x^{3} +11x^{2} y+6xy^{2} +y^{3} =x
\end{gather*}
The first thing I tried to do, was probably naive, and I immediately thought of finding the solution to y in terms of x as a cubic equation.

Needless to say, I really made a mess out of it.

Could someone please provide some hints on how to approach this kind of problem?
And it would also be helpful if anyone described a general method to deal with the area under an implicit function (if such a method exists).

答:
$$6 x^3+11 x^2 y+6 x y^2+y^3=x$$
convert in polar coordinates
$$x=r\cos t;\;y=r\sin t$$
we get
$$r^3 \sin ^3 t+6 r^3 \cos ^3 t+11 r^3 \sin  t \cos ^2 t+6 r^3 \sin ^2 t \cos  t=r \cos  t$$
$r$ can be simplified and the equation can be solved wrt $r^2$
$$r^2 \sin ^3 t+6 r^2 \cos ^3 t+11 r^2 \sin  t \cos ^2 t+6 r^2 \sin ^2 t \cos  t- \cos  t=0$$
$$r^2=\frac{\cos  t}{\sin ^3 t+6 \cos ^3 t+11 \sin  t \cos ^2 t+6 \sin ^2 t \cos  t}$$
which can be simplified dividing numerator and denominator by $\cos^3 t$
$$r^2(t)=\frac{1}{\cos^2 t}\cdot \frac{1}{\tan^3 t+6\tan^2 t+11\tan t +6}$$
Any area $S$ we want to calculate can be solved by the integral
$$S=\frac12\int_a^b r^2(t) dt$$
Setting $\tan t=u$ we get
$$\int \frac{du}{u^3+6 u^2+11 u+6}=\int \left(-\frac{1}{u+2}+\frac{1}{2 (u+3)}+\frac{1}{2 (u+1)}\right)\,du=$$
$$=\frac{1}{2} \log (u+1)-\log (u+2)+\frac{1}{2} \log (u+3)+C=$$
$$=\frac{1}{2}\left(\log \left(u^2+4 u+3\right)-2 \log(u+2)\right)+C=$$
$$=\frac{1}{2} \log \left(\frac{u^2+4 u+3}{(u+2)^2}\right)+C=$$
$$=\frac{1}{2} \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)$$
$$S=\left[\frac{1}{4}  \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)\right]_a^b$$
To get the area in the picture below we compute
$$S=\left[\frac{1}{4}  \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)\right]_0^{\pi/2}=\frac{1}{4} \log \left(\frac{4}{3}\right)\approx 0.072 $$


答:
It possible rewrite Gamma function such as
$$\Gamma(s)= \int_{0}^{\infty} e^{-x} x^{s-1} dx = \int_{0}^{\infty} e^{-nu} (nu)^{s-1}ndu = n^s \int_{0}^{\infty} e^{-nu} u^{s-1} du $$
Then
\begin{align*}
\Gamma(s)\zeta(s) = \sum_{n=0}^{\infty} \frac{\Gamma(s)}{n^s} = \sum_{n=0}^{\infty} \int_{0}^{\infty}e^{-nx}x^{s-1} dx &= \int_{0}^{\infty} x^{s-1} \left (\sum_{n=0}^{\infty} e^{-nx} \right)  dx \\
& = \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} dx
\end{align*}
Hence, that is equivalent to Riemann Hypothesis.

Title: What is the name of the function $1/\Gamma(z) \int_0^\infty \frac{x^{z -1}}{e^w-1}dw$?

Body:
I recently saw

on the internet, and was curious what the name of the "pineapple" function was?

$$1/\Gamma(z) \int_0^\infty \frac{\color{red}{x}^{z-1}}{e^w-1}dw$$

I recognize the "mango" function as $\Gamma(z)$ the gamma function.  Also, does anyone know what theorem this refers to?


答:
Counterexample. Let $f(x) = \frac 1x \sin(x^2)$. Then $f(x) \to 0$ as $x \to \infty$, but
$$
f'(x) = - \frac{1}{x^2} \sin(x^2) + \frac 1x \cos(x^2) \cdot 2x
= - \frac{1}{x^2} \sin(x^2) + 2 \cos(x^2),
$$
which is not convergent to $0$ as $x \to \infty$. The problem here lies of course in the second derivative, which is not bounded.

Problem with your reasoning. For each $x > 0$, you find just one point $c_x \in [x,x+a]$ for which $f'(c_x)$ is small. And indeed, if you fix any sequence $x_n \to \infty$, your construction yields a corresponding sequence $c_n \in [x_n,x_n+a]$ that satisfies $c_n \to \infty$ and $f'(c_n) \to 0$. But there is no reason why every sequence $y_n \to \infty$ should be obtained as $c_n$ in this way.

In a way, this may be surprising. Your family of $c_x$ is indexed by all points $x > 0$, and the intuition may tell us that those $c_x$ should cover at least the whole halfline $[a,\infty)$. But this intuition is wrong! It would be true if $x \mapsto c_x$ was a continuous function, while it is not. The best way to see this is by studying some counterexample.

How to solve this. As you noted yourself, this is the easy part. This question was already asked and answered, e.g. here.

Body:
$f$ twice differentiable on $(0,\infty)$, $f''$ is bounded on $(0,\infty)$, $\lim_{x\to\infty}f(x)=0$. Show that $\lim_{x\to\infty}f'(x)=0$. First of all, it seems to me that the condition that $f''$ bounded is useless... Isn't it?

My attempt is to pass by mean value theorem, so i would like to know if my approach holds, please.

First, consider the interval $[x,x+a]$ with $x>0, a>0$. As $f$ is differentiable on $[x,x+a]$, by mean value theoreme there exists some $c_x \in ]x,x+a[$ such that

$f'(c_x)=\frac{f(x)-f(x+a)}{a}$.

Inserting the limit on RHS and LHS we get the following:

$\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x)-f(x+a)}{a} = 0$.

Therefore, $\lim_{x\to\infty}f'(x)=0$ as wanted.

I know there is a post already on this statement, but they use the fact that $f''$ is bounded in theirs proofs... If this hypothesis is necessary, could someone explain why or give a counter example on my proof?


问:
Equality of markov transition kernels implies that the processes have the same distribution
My question is if two continuous Markov processes $(Y(t))$ and $(X(t))$ have the same transition kernel is it then true that they have the same finite-dimensional distribution?

More precisely following the notation from p. 49 and onwards in Brownian Motion by Mörters and Peres we say that $p:[0,\infty)\times \mathbb{R}\times \mathscr{B}\rightarrow \mathbb{R}$, where $\mathscr{B}$ denote the Borel subsets of $\mathbb{R}$, is a Markov transition kernel if

$p(\cdot,\cdot ,A)$ is measurable

$p(t,x,\cdot)$ is a Borel probability measure.

For all $x\in \mathbb{R}$, $t,s>0$ and $A\in \mathscr{B}$
$$p(t+s,x,A) = \int_{\mathbb{R}}p(t,y,A)p(s,x,\mathrm{d}y).$$

We say that $X:[0,\infty)\times \Omega\rightarrow \mathbb{R}$ is a Markov process if there exists a Markov transition kernel $p$ and a filtration $\mathcal{F}(s)$ on $\Omega$ such that
$$\mathbb{P}(X(t)\in A\mid \mathcal{F}(s)) = \mathbb{E}(1_{X(t)\in A}\mid \mathcal{F}(s)) = p(t-s,X(s),A).$$

My question regards when two Markov processes are the same (in distribution). Suppose that $(Y(t))$ and $(X(t))$ are two Markov processes such that $t\mapsto (X(t,\omega), Y(t,\omega))$ is continuous for each $\omega$. Let $p_X$ and $p_Y$ be the Markov transition kernels of $X$ and $Y$ respectively. If $p_X = p_Y$
does it then hold that $X$ and $Y$ have the same finite-dimensional distributions?

To show this we need to show that for each Borel set $A\in \mathbb{R}^n$
$$\mathbb{P}((X(t_1),\dotsc,X(t_n))\in A) = \mathbb{P}((Y(t_1),\dotsc,Y(t_n))\in A).$$

I only see how this follows in the case where $n=1$, otherwise I don't see how to do this, could someone help me with this?

答:
Provided $X$ and $Y$ have the same initial distributions, then Yes: the finite dimensional distributions of $X$ (for example) can be expressed in terms of integrals involving its initial distribution and the transition kernel.

Taking $n=2$, if $X$ has initial distribution $\mu$ then for $t_1<t_2$,
$$
\Bbb P[(X(t_1),X(t_2))\in A] =\int\int\int 1_A(y,z)\mu(dx) p(t_1,x,dy)p(t_2-t_1,y,dz).
$$
(Each of the integrals is over $\Bbb R$.)


问:
How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?

Let $p(x)=x^4+x+1$ be a polynomial in $\mathbb{R}[x]$. How can I prove that $p$ doesn't have real roots?

My attempt:

From calculus, I know that

$$\lim_{x \to \pm\infty} p(x) = \infty\,.$$

Then, if it has real roots, then we have two or four real roots. I know that if $$\gcd \left( p(x), p'(x) \right) = 1\,,$$ then the roots is simple. Is there another hint?

Body:
$f'(x) = 4x^3 +1$.

To find extreme points $f'(x) = 4x^3 + 1=0 \implies x =-\sqrt[3]{\frac 14}$

$f''(x) = 12x^2$ and $f''(-\sqrt[3]{\frac 14})= 12(-\sqrt[3]{\frac 14}) > 0$ so this is a minimum.

So $f(x) \ge f(-\sqrt[3]{\frac 14})= \sqrt[3]{\frac 1{4^4}} -\sqrt[3]{\frac 14} + 1  > -\sqrt[3]{\frac 14}_1 >-1+1 = 0$ for all $x$


问:
Primitive polynomials of finite fields
there are two primitive polynomials which I can use to construct $GF(2^3)=GF(8)$:

$p_1(x) = x^3+x+1$

$p_2(x) = x^3+x^2+1$

$GF(8)$ created with $p_1(x)$:

0

1

$\alpha$

$\alpha^2$

$\alpha^3 = \alpha + 1$

$\alpha^4 = \alpha^3 \cdot \alpha=(\alpha+1) \cdot \alpha=\alpha^2+\alpha$

$\alpha^5 = \alpha^4 \cdot \alpha = (\alpha^2+\alpha) \cdot \alpha=\alpha^3 + \alpha^2 = \alpha^2 + \alpha + 1$

$\alpha^6 = \alpha^5 \cdot \alpha=(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha+1+\alpha^2+\alpha=\alpha^2+1$

$GF(8)$ created with $p_2(x)$:

0

1

$\alpha$

$\alpha^2$

$\alpha^3=\alpha^2+1$

$\alpha^4=\alpha \cdot \alpha^3=\alpha \cdot (\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1$

$\alpha^5=\alpha \cdot \alpha^4=\alpha \cdot(\alpha^2+\alpha+1) \cdot \alpha=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1$

$\alpha^6=\alpha \cdot (\alpha+1)=\alpha^2+\alpha$

So now let's say I want to add $\alpha^2 + \alpha^3$ in both fields. In field 1 I get $\alpha^2 + \alpha + 1$ and in field 2 I get $1$. Multiplication is the same in both fields ($\alpha^i \cdot \alpha^j = \alpha^{i+j\bmod(q-1)}$. So does it work so, that when some $GF(q)$ is constructed with different primitive polynomials then addition tables will vary and multiplication tables will be the same? Or maybe one of presented polynomials ($p_1(x), p_2(x)$) is not valid to construct field (altough both are primitive)?

答: The generator $\alpha$ for your field with the first description cannot be equal to the generator $\beta$ for your field with the second description.  An isomorphism between $\mathbb{F}_2(\alpha)$ and $\mathbb{F}_2(\beta)$ is given by taking $\alpha \mapsto \beta + 1$; you can check that $\beta + 1$ satisfies $p_1$ iff $\beta$ satisfies $p_2$.</p>

Title: Primitive polynomials of finite fields


问:
What are some examples of proof by contrapositive?
Applying the Modus Tollens argument to Fermat's Little Theorem really helped me to understand logical implication. I never knew that FLT was actually a compositality test.

Theorem (FLT): given integers $a>1$ and $n>1$, if $n$ is prime, then $a^n$ is congruent to $a\  (\bmod\  n)$.

By the contrapositive, if $a^n$ is not congruent to $a\  (\bmod\  n)$, then $n$ is not prime. Thus $n$ is composite.

What are some other simple and instructive examples of proof by contrapositive?

答:
When you want to prove "If $p$ then $q$", and $p$ contains the phrase "$n$ is prime" you should use contrapositive or contradiction to work easily, the canonical example is the following:

Prove for $n>2$,

If $n$ is prime then $n$ is odd.

Here $q$ is the phrase "$n$ is odd". Here $p$ is exactly the phrase "$n$ is prime" and is very difficult to work with it.
Because the fact that $n$ is prime means that it is not divisible by other number grater than $1$ and different for $n$, so you must to choose from these $n-2$ true sentences the only one that is useful which is "$n$ is not divisible by 2", but you would not know which is the right choice, unless you read $q$.

But proving contrapositive equivalent form is very easy, and you don't to do any choice.

If $n$ is even then $n$ is not prime.

Which follows from the fact that every even number greater than $2$ is divisible by $2$, hence not prime.

So contrapositive (also contradiction) is used to avoid situations where you have a lot of information and very little of it is actually useful.


问:
A confusion regarding the statement about Tarski's Fixed Point Theorem
As title suggests, I have a thought about Tarski's Fixed Point Theorem.

The statement of the theorem I was given was that:

Suppose $X$ is a set and $f:\wp(X)\to\wp(X)$ is an increasing function (with respect to containment) then $\exists Z\in \wp(X)$ such that $f(Z)=Z.$

I understood that prove that was given. However, I was thinking couldn't we just take $Z=\emptyset$ then we have $f(\emptyset)=f(Z)=\emptyset=Z$? (thus the result becomes trivial?)

Or should we add that $Z$ must be non-empty?

Thank you so much in advance!

答:
Take $X=\Bbb N$, and take $f(Z)=Z\cup\{42\}$. Why is $\varnothing$ a fixed point? It's not. $f(\varnothing)=\{42\}\neq\varnothing$.

Or for example, $f(Z)=Z\cup\{\min\{n\in\Bbb N\mid n\notin Z\}\}$, i.e. always add the smallest element not in $Z$. No finite iteration of $f$ from $\varnothing$ is a fixed point.


问:
Alternative Methods for Trigonometry Proofs (Varying Angles)
Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?

答:
A delightful application of linear algebra:

Let $R(\theta)$ be the linear map $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that rotates points counterclockwise an angle $\theta$ about the origin. It is well-known that $R(\theta)$ has the matrix representation:

$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

It is obvious from inspection that $R(\theta + \psi) = R(\theta)  \circ R(\psi)$. That is to say:

$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} \cos \psi & -\sin \psi \\ \sin \psi & \cos \psi \end{bmatrix} = \begin{bmatrix} \cos (\theta+\psi) & -\sin (\theta+\psi) \\ \sin (\theta+\psi) & \cos (\theta +\psi)\end{bmatrix}$$

Hence matrix multiplication gives you  the angle addition identities:

$$\sin (\theta + \psi) = \sin\theta\cos\psi + \cos\theta\sin\psi$$

for example. Of course, identifying $a+bi \equiv \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ takes a little bit of the shock out of this, but I would argue that this is an elegant way of deriving trigonometric identities without relying on Euler's formula.


问:
ratio test on series with logarithms $\frac{n}{\log_{10}(n+1)}$
How do I correctly use the ratio test on series with logarithms to test for convergence?

For example, I have this series:

$$\frac{n}{\log_{10}(n+1)}$$

When I use this part of the ratio test: $a_{n+1}$

Do I re-arrange the series like so:

$$\frac{n+1}{\log_{10}(n+1+1)}$$
Or is there a different way of doing it. Also, can the logarithm be converted into series form, and how would it look like?

答:
You seem to be confused about the ratio test. What the ratio test says is this:

If $(a_n)$ is a sequence of real numbers, then

If $\limsup_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}<1$, then the series $\sum_na_n$ converges absolutely.

If $\liminf_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}>1$, then the series $\sum_na_n$ diverges.

Note that the inequalities have to be strict.

I assume you want to check whether the series $\sum_n\frac{n}{\log(n+1)}$ converges or diverges. You do not have to use the ratio test here, because $\frac{n}{\log(n+1)}\to\infty$, and it is well known that if $\sum_na_n$ converges then $|a_n|\to0$. So your series diverges for trivial reasons.


问:
A logarithmic Calculation
I have encountered a logarithmic question that I could not solve. Here is the question:

$$\dfrac {\log 64}{\log4}+\dfrac {\log 8}{\log 2}$$

A)1 B)2 C)3 D)5 E)6

Rules of Logarithm:

$$\log_a\left(\frac xy\right)=\log_ax-\log_ay$$

$$\log_ax^n=n\log_ax$$

Here is my calculation:

$$Q=\dfrac {\log 2^6}{\log 2^2}+\dfrac {\log 2^3}{\log 2}$$

$$Q=6\log 2-2\log 2+3\log 2-\log 2$$

$$Q=6\log 2$$
What is the error in my calculation? Thanks in advance.

答: You've confused $\log x/\log y$ with $\log(x/y)$. Use the second rule twice, with $a=4$ ($a=2$) in the first (second) fraction.

Title: A logarithmic Calculation


答:
$\def\a{\alpha} \def\b{\beta} \def\t{\theta} \def\ar{\text{area}}$

Here's an unusual way to derive the addition formula for $\sin(\a+\b)$, based on areas of triangles. First, recall that the area of a triangle with two sides $a$ and $b$ and included angle $\t$ is $A=\frac12ab\sin\t$. The steps are outlined.

Consider the diagram below:

The points $D$ and $E$ are obtained by dropping perpendiculars from $B$ and $A$, respectively, onto $\overline{OC}$.

$\bullet$ Show that $\triangle ACE \sim \triangle BCD$.

$\bullet$  Show that $(AE)(CD) = (BD)(CE)$ and conclude that $\ar\triangle ACD = \ar\triangle BCE$.

$\bullet$  Explain why
\begin{align*}
\sin(\a+\b) &= 2\,\ar\triangle OAB = 2\big(\ar\triangle OAC+\ar\triangle OCB\big) \\
&= 2\big(\ar\triangle OAD + \ar\triangle OEB\big).
\end{align*}

$\bullet$  Find $OD$ (in terms of $\b$) and $OE$ (in terms of $\a$), and deduce the formula for $\sin(\a+\b)$.

Title: Alternative Methods for Trigonometry Proofs (Varying Angles)

Body:
Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?


问:
Prove that $P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$
Let $(Ω,S,P)$ be a probability space. Let $A, B, C ∈ S$ with $P(B)$ and $P(C) > 0$. If $B$ and
$C$ are independent show that
$$P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$

$\textbf{My Solution:}$

[Note: All the arrows used in the solution represent the direction of flow of inference in a graphical model.]

[Note: I have used standard notation which are generally in the literature of probabilistic graphical models. Take a look. https://ermongroup.github.io/cs228-notes/representation/directed/]

It is given that A depends on B. So we can start with the fact that $B \rightarrow A$.

Take RHS
$$P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B,C)}P(C) + \frac{P(A,B,\overline C)}{P(B,\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)P(C)}P(C) + \frac{P(A,B,\overline C)}{P(B)P(\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}\ \ \ \text{(eq. 1)}$$

$\textbf{Case I}: A \perp C$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(A,B)P(C)}{P(B)} + \frac{P(A,B)P(\overline C)}{P(B)}$$
$$= \frac{P(A,B)}{P(B)}$$
$$= P(A|B) = LHS$$

$\textbf{Case II}: C\leftarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(A|B)P(C|A)}{P(B)} + \frac{P(B)P(A|B)P(\overline C|A)}{P(B)}$$
$$= P(A|B)P(C|A) + P(A|B)P(\overline C |A)$$
$$= P(A|B) = LHS$$

$\textbf{Case III}: C\rightarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(C)P(A|B,C)}{P(B)} + \frac{P(B)P(\overline C)P(A|B,\overline C)}{P(B)}$$
$$= P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
How to proceed from here?

答:
Notice you ended up where you started in the end of part III. From the second line of part (III) $$\begin{split}\frac{P(B)P(C)P(A|BC)}{P(B)}+\frac{P(B)P(\bar C)P(A|B\bar C)}{P(B)}&=\frac{P(B|C)P(C)P(A|BC)}{P(B)}+\frac{P(B|\bar C)P(\bar C)P(A|B\bar C)}{P(B)}\\
&=\frac{P(C)P(A,B|C)}{P(B)}+\frac{P(\bar C)P(A,B|\bar C)}{P(B)}\\
&=\frac{P(A,B,C)}{P(B)}+\frac{P(A,B,\bar C)}{P(B)}\\
&=\frac{P(A,B)}{P(B)}\\&=P(A|B)\end{split}$$


问:
How to think about this simple differential calculus expression?
In the course of doing differential equation substitutions, I often come across the expression $\frac{d}{dt}\frac{dy}{dx}$, where $y$ is a function of $x$, and $x$ itself is a function of $t$.  For some reason, I have a hard time figuring out how to expand this.  One possibility is to multiply the inside by $\frac{\frac{dx}{dt}}{\frac{dx}{dt}}$, yielding $\frac{d}{dt}\frac{\frac{dy}{dx}\frac{dx}{dt}}{\frac{dx}{dt}} \implies \frac{(\frac{d}{dt}(\frac{dy}{dx}\frac{dx}{dt}))\frac{dx}{dt} - (\frac{dy}{dx}\frac{dx}{dt})\frac{d}{dt}\frac{dx}{dt}}{(\frac{dx}{dt})^2}$, which looks nice because $\frac{dy}{dx}\frac{dx}{dt} = \frac{dy}{dt}$.  However, to proceed further I would need to use the expanded version, and then the initial expression $\frac{d}{dt}\frac{dy}{dx}$ reappears.  Is there a simple chain rule here I'm not spotting?  How should I think about expanding this expression?

答: Let $y=f(x)$ and let $x(t)$ be a given function. Notice that $\frac{dy}{dx}=f'(x)$, so we need to calculate $\frac{d}{dt}(f'(x(t)).$ Applying the chain rule, we find this is $f''(x(t))\cdot \frac{dx}{dt}$. Translating back to your notation, this means that $\frac{d}{dt} \frac{dy}{dx}=\frac{d^2 y}{dx^2} \frac{dx}{dt}.$


问:
Solving equations involving two sets
Out of curiosity, how would one go about solving an equation involving two sets? For example,

$$ \{1, 2, 3\} = \{a + b + c, a + b - c, a - b + c, a - b - c\} $$

An intuitive solution to this is $ \{a = 2, b = 0.5, c = 0.5\} $, but is there a specific process?

答:
hint

In the right set, two elements are equal. The only possibilities are
$$a+b+c=a-b-c$$
or
$$a+b-c=a-b+c$$
which gives
$$b=\pm c$$

The equation  becomes
$$\{a-2b,a,a+2b\}=\{1,2,3\}$$

The sum gives $ 3a=6 $ and $ a=2$.

Title: Solving equations involving two sets


问:
The Fixed-point theorem predicts the existence of quines.
From Odifreddi's "Classical Recursion Theory", Vol 1., page 165.

The Fixed-Point Theorem can be used directly to find programs exhibiting self-referential features. E.g., by considering $f$ recursive such that $\varphi_{f(e)}(x) = e$ we get an index $e$ such that $\varphi_e(x) = e$ which can be interpreted as the code of a program printing itself.

Here is the statement of the theorem:

Fixed-point theorem

Given a recursive function $f$ there is an $e$ such that $\varphi_e \simeq \varphi_{f(e)}$, i.e. $e$ and $f(e)$ compute the same function.

To me it seems that the outlined procedure here is:

pick a recursive $f$ such that $\varphi_{f(e)}(x) = e$

take the $e'$ of the recursion theorem such that $\varphi_{e'} \simeq \varphi_{f(e')}$

we get $\varphi_{e'}(x) = e'$, i.e. a quine

Why such an $f$ exist? Note that $f$ must be recursive.

Related but with other nuances

Does the recursion theorem give quines?

On the existence of an injective recursive function such that all its values are also its indexes.

答:
Let $\psi$ be a recursive function (no need for partial recursive, because is just the projection function) such that: $\psi(x,y) = x$.

By the parameter theorem (S-m-n theorem), there exist a recursive function $f$ such that:

$$\phi_ {f(x)}(y) = \psi(x,y) = x.$$

As $f$ is recursive, by the fix-point theorem, it has a fix-point $e$, i.e. $\phi_{f(e)} = \phi_e$.

Then:

$$ \phi_e(y) = \psi(e,y) = e.$$

I think he is assuming this reasoning for the existence of $f$. Note that it don't matter witch constant you choose for $x$ before using the fix-point theorem (i.e. making $x=e$ we get $\phi_{f(e)}(x) = e$, the before-using-fix-point-theorem part on the Odifreddi text).


答:
Your mistake was stated in the alternative answer

$$\log_yx=\frac {\log x}{\log y}≠\log \frac xy =\log x-\log y$$

Just use

$$\frac {\log x}{\log y} =\log _yx$$

We have

$$\frac {\log 64}{\log 4}=\log_4{64}=3$$

$$\frac {\log 8}{\log 2}=\log_28=3$$

Title: A logarithmic Calculation

Body:
I have encountered a logarithmic question that I could not solve. Here is the question:

$$\dfrac {\log 64}{\log4}+\dfrac {\log 8}{\log 2}$$

A)1 B)2 C)3 D)5 E)6

Rules of Logarithm:

$$\log_a\left(\frac xy\right)=\log_ax-\log_ay$$

$$\log_ax^n=n\log_ax$$

Here is my calculation:

$$Q=\dfrac {\log 2^6}{\log 2^2}+\dfrac {\log 2^3}{\log 2}$$

$$Q=6\log 2-2\log 2+3\log 2-\log 2$$

$$Q=6\log 2$$
What is the error in my calculation? Thanks in advance.


问:
How to evaluate cumulative distribution function w/ sigma of given SRS?
I have been given a set of 10 positive values that make up an SRS (Simple Random Sample) of a population, which has a normal distribution.

I then am asked to evaluate $a = \sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2}$ and provide the distribution. Since it is a sum, I assumed a would simply be a number.

Finally, I am asked to evaluate the following CDF: P ($\sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2} \le a$).

How should I interpret this? If the values in the SRS are all positive, then how could any subset less than the size of $n$ have the value of $a$? Wouldn't the probability of this just be $1$?

答:
For many years, there have printed CDF tables of the chi-squared distribution in most applications oriented basic statistics texts.

Also, statistical software is now widely used: For example, Suppose $Q \sim\mathsf{Chisq}(\nu = 5),$ and that you want $P(Q \le .05) = 0.0000292$ or you want $c=11.0705$ such that
$P(Q \le c) = .95,$ you can use the chi-squared CDF pchisq or the chi-squared quantile function (inverse CDF) qchisq in R, as follows:

```<code>pchisq(.05, 5)
[1] 2.920954e-05
qchisq(.95, 5)
[1] 11.0705
qchisq(.05, 5)
[1] 1.145476       # P(Q < 1.1455) = 0.05
pchisq(1.1455, 5)
[1] 0.05000219
</code>```

Also, the chi-squared density function is dchisq.

```<code>hdr = "Density of CHISQ(df = 5)"
curve(dchisq(x, 5), 0, 15, lwd=2, col="blue",
  ylab="PDF", xlab="q", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = c(.05, 1.1455, 11.0705), col=c("red","cyan3", "purple"), lwd=2, lty="dotted")
</code>```

Application: Suppose you want a 90% confidence interval
for $\sigma^2$ based on a sample of size $n=10$ from
a normal population with unknown mean and variance.
Then $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu = 9).$ [The degrees of freedom are reduced to $\nu=n-1=9$ because
$\bar X$ estimates $\mu$ in the usual formula for the
sample variance $S^2.$ Easy to remember; a bit messy to prove.]

Then if values $L$ and $U$ cut
5% of the probability from the lower and upper tails, respectively of $\mathsf{Chisq}(9),$ we have
$P\left(L \le \frac{9S^2}{\sigma^2} \le U\right) = 0.9.$ This
leads to a 90% CI for $\sigma^2$ of the form
$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$

In particular, suppose we have a normal sample of size $n=10$ with sample variance $S^2 = 3.21,$ then our 90% CI for $\sigma^2$ is
$(1.71, 8.69).$ [Generally, it takes a lot of data to
get a short CI for a population variance. "Variances are very variable."]

```<code>9*3.21/qchisq(c(.95,.05), 9)
[1] 1.707550 8.688427
</code>```

Body:
For many years, there have printed CDF tables of the chi-squared distribution in most applications oriented basic statistics texts.

Also, statistical software is now widely used: For example, Suppose $Q \sim\mathsf{Chisq}(\nu = 5),$ and that you want $P(Q \le .05) = 0.0000292$ or you want $c=11.0705$ such that
$P(Q \le c) = .95,$ you can use the chi-squared CDF pchisq or the chi-squared quantile function (inverse CDF) qchisq in R, as follows:

pchisq(.05, 5)
[1] 2.920954e-05
qchisq(.95, 5)
[1] 11.0705
qchisq(.05, 5)
[1] 1.145476       # P(Q < 1.1455) = 0.05
pchisq(1.1455, 5)
[1] 0.05000219

Also, the chi-squared density function is dchisq.

hdr = "Density of CHISQ(df = 5)"
curve(dchisq(x, 5), 0, 15, lwd=2, col="blue",
  ylab="PDF", xlab="q", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = c(.05, 1.1455, 11.0705), col=c("red","cyan3", "purple"), lwd=2, lty="dotted")

Application: Suppose you want a 90% confidence interval
for $\sigma^2$ based on a sample of size $n=10$ from
a normal population with unknown mean and variance.
Then $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu = 9).$ [The degrees of freedom are reduced to $\nu=n-1=9$ because
$\bar X$ estimates $\mu$ in the usual formula for the
sample variance $S^2.$ Easy to remember; a bit messy to prove.]

Then if values $L$ and $U$ cut
5% of the probability from the lower and upper tails, respectively of $\mathsf{Chisq}(9),$ we have
$P\left(L \le \frac{9S^2}{\sigma^2} \le U\right) = 0.9.$ This
leads to a 90% CI for $\sigma^2$ of the form
$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$

In particular, suppose we have a normal sample of size $n=10$ with sample variance $S^2 = 3.21,$ then our 90% CI for $\sigma^2$ is
$(1.71, 8.69).$ [Generally, it takes a lot of data to
get a short CI for a population variance. "Variances are very variable."]

9*3.21/qchisq(c(.95,.05), 9)
[1] 1.707550 8.688427


问:
Complex numbers and $ e^{i\theta} $
I have a complex $ z $ that I want to write as $ z=|z|e^{i\theta} $ where $ \theta $ is its phase.

I tried using 2 "different" ways and got different results, so I assume one of them is wrong.

This is the number: $ z=\frac{a}{b-ic} $

where $ a,b,c \in \mathbb{R} $ and positive.

Now we can write:

$ z=\frac{a}{b-ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(0-\arctan\left(\frac{-c}{b}\right)\right)} $

But also:

$ z=\frac{-a}{-b+ic}=\frac{|a|}{\sqrt{b^{2}+c^{2}}}e^{i\left(\pi-\arctan\frac{c}{-b}\right)} $

So where is the mistake?

Thanks in advance.

答:
Somewhat off topic:

Since the OP's question was answered by David K's response, I consider it open season to discuss a more straightforward approach.

Without loss of generality, $(b - ic) \neq [0 - i(0)] \implies (b^2 + c^2) \neq 0.$

$$z = \frac{a}{b-ic} \times \frac{b+ic}{b+ic} =
\frac{a(b + ic)}{b^2 + c^2}.$$

Since $(b^2 + c^2) \in \mathbb{R^+}$, you can let $s\in\mathbb{R}$ represent $\frac{a}{b^2 + c^2}$, and then let $z_0$ represent  $\left(\frac{1}{s} \times z\right).$  Note that since $(a)$ is presumed to be positive, $s$ must be positive.

Therefore, the problem reduces to expressing

$$(b + ic) = z_0 = re^{i\theta}.$$

Once this is done, you will have that $z = (sr)e^{i\theta}.$

Let $r$ denote $\sqrt{b^2 + c^2}$ and

let $\theta$ denote the unique angle, within a modulus of $(2\pi)$

such that $\cos(\theta) = \frac{b}{r}$ and $\sin(\theta) = \frac{c}{r}.$

Then $z_0 = (b + ic) = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$

Title: Complex numbers and $ e^{i\theta} $


问:
Pointwise maximum: translation and scaling
From this question, I am getting that $2\times \max(a,b) = \max(2a,2b)$, and more generally, $c\times \max(a,b) = \max(c\times a, c\times b)$ for $c\geq0$, since we have
$$
\begin{align*}
a + b + |a-b| &= a+b + \max(a-b, b-a) \\
&= \max\big[(a+b) + a-b),\; (a+b) + b-a\big] \\
&= \max(2a,2b))\\
a + b + |a-b| &= \big[\max(a,b) + \min(a,b)\big] + \big[\max(a,b) - \min(a,b)\big] \\
&= 2\max(a,b)
\end{align*}
$$

Is this correct? It feels odd to move $(a+b)$ into the max in the first line. This seems to indicate that for $f:\mathbb{R}^n\to \mathbb{R}$ defined by the pointwise max
$$f(x) := \max(x_1,x_2,\ldots,x_n)$$
satisfies $f(c\times x + d) = cf(x) + d$ for $c\geq0$ and $d$ a (specific?) linear combination of $x$. For $d$ a constant scalar, we have that $f$ some translation and scaling properties (not sure what this is called), which is intuitive, but is this right for $d = d(x)$ some linear combination of $x$?

What operations are allowed for moving terms in and out of $\max(\cdot)$?

答:
You are making things too complicated.

It is simply that $a\ge b$ if and only if $2a\ge 2b$; so

when $a\ge b$,
$$
LHS=2\cdot a=2a=\max(2a,2b)
$$

and when $a<b$,
$$
LHS=2\cdot b=2b=\max(2a,2b)
$$

If you want to apply the  formula that
$$
\max(x,y)=\frac{x+y-|x-y|}{2}
$$
then it is simply
$$
2\max(x,y)=\frac{2x+2y-2|x-y|}{2}=\frac{2x+2y-|2x-2y|}{2}=\max(2x,2y)
$$

Added.

Okay, after fixing the typos, in your approach, you have
$$
\begin{align}
a+b+|a-b|
&=a+b+\max(a-b,b-a)\\
&=\max(a-b+(a+b),b-a+(a+b))\\
&=\max(2a,2b)
\end{align}
$$

In particular, it is correct that
$$
a+b+\max(a-b,b-a)
=\max(a-b+(a+b),b-a+(a+b))
$$
because
$$
\max(x,y)+L=\max(x+L,y+L)
$$
since $x\ge y$ if and only if $x+L\ge y+L$.

In general, if you have $x=(x_1,x_2,\cdots,x_2)$ and $c\ge 0$, you have
$$
c\max(x_1,\cdots,x_n)=\max(cx_1,\cdots,cx_n)
$$

So $f(cx)=cf(x)$.

On the other hand, for any $L\in\mathbb{R}$,
$$
\max(x_1+L,\cdots,x_n+L)=\max(x_1,\cdots,x_n)+L
$$
Combining together,
$$
\max(cx_1+L,\cdots,cx_n+L)=c\max(x_1,\cdots,x_n)+L
$$


问:
Random variable with density proportional to a function and finite in some points
Let $X$ be a random variable on $[-1,3]$ with density $f(x) = k x^2$ (with $k \in \mathbb{R} $ to be determined) on $[-1,3]$ apart from some points s.t. $p(X=-1) = p(X=3) = \dfrac{1}{4} $ and $p(X=0) = \dfrac{1}{3}$.

What is the cumulative distribution function of $X$? How much is $p(-1 \leq X < 3) $?

To find $k$ I would just calculate $$ k = \dfrac{1}{\int_{-1}^3 x^2 \ dx} $$ and then the cumulative distribution should be $$F(x) = \int_{-1}^x f(t) \ dt $$
while
$$p(-1 \leq X < 3) = 1 - p(X = 3) = \dfrac{3}{4}$$
Any suggestion would be appreciated.

答:
You can't calculate $k$ the way you suggest because the total probability due to the continuous part of the PDF isn't 1. There are three possibilities, and
$$\mathrm P(X=-1) + \mathrm P(X=3) + \mathrm P(-1<X<3) = 1$$

Therefore we have

$$\mathrm P(X=-1) = \mathrm P(X=3) = \frac 1 4 \implies \mathrm P(-1<X<3) = 1 - \frac 1 4 - \frac 1 4 = \frac 1 2$$

Thus $$k = \frac {1/2}{\int_{-1}^3 x^2}$$

Then e.g. the cumulative function is zero below -1, then jumps to $\frac 1 4$ as you cross $-1$ and smoothly increases to $\frac 3 4$ at $3$, then jumps to one.

Have another go at the question!


问:
Riemann zeta function is analytic for $\operatorname{Re}(z)>1$
I want to show that, Riemann zeta function is analytic on the domain $\operatorname{Re}(z)>1$. I know that it is absolutely and uniformly convergent on the right half of the line $Re(z)=1$. Now, I am trying to invoke Morera's theorem to prove its analyticity, but I am not quite sure how  to do that. Can anyone help?

In general, if anyone can point me to some other direction then that is also welcome.

答:
Otherwise it is immediate that for $\Re(c)> 1$ and $|s-c|<\Re(c)-1$, by absolute convergence
$$\sum_{n\ge 1}n^{-s}=\sum_{n\ge 1}n^{-c}\sum_{k\ge 0}\frac{((c-s)\log n)^k}{k!}=\sum_{k\ge 0} (c-s)^k \sum_{n\ge 1}n^{-c} \frac{(\log n)^k}{k!} $$

We need the whole analytic continuation of $\zeta(s)$ and the Cauchy integral formula to prove that the latter series in fact converges for $|s-c|<|c-1|$.


问:
Computing $\int_0^1 \left(1+\frac{x}{n}\right)^{-n} \; dx$
I am trying to compute $\int_0^1 \left(1+\frac{x}{n}\right)^{-n}  \; dx$. Using an online calculator, I found the answer to be $\frac{(n+1)^{1-n}}{(n-1)} - \frac{n}{n-1}$. When I tried to verify this, I got:

$$\int_0^1 \left(1+\frac{x}{n}\right)^{-n}  \; dx= n^n\int_0^1(n+x)^{-n}\; dx = \frac{n^n}{1-n} \cdot \left[(n+x)^{1-n}\right|_{x=0}^1 = \frac{n^n}{(1+n)^{n-1}(1-n)}-\frac{n}{1-n}$$

This is not the same as the answer I for from the online calculator. Where have I gone wrong?

答:
$$\int_0^1\left(1+\frac xn\right)^{-n}dx=n\int_0^1\left(1+\frac xn\right)^{-n}\,d\left(1+\frac xn\right)=\left.n\frac{\left(1+\frac xn\right)^{-n+1}}{-n+1}\right|_0^1=$$

$$=\frac n{1-n}\left[\left(1+\frac1n\right)^{-n+1}-1\right]=\frac n{1-n}\left[\frac{n^{n-1}}{(1+n)^{n-1}}-1\right]$$

the above is true for $\;n\neq1\;$, and in this last case we'd get

$$\int_0^1\left(1+\frac xn\right)^{-1}dx=n\int_0^1\frac{dx}{n+x}=n\log|x+n|_0^1=n\log\frac{n+1}n$$


问:
Given that $a_n > 0$, prove: $\liminf \left(\frac{a_{n+1}}{a_n}\right)\leq \liminf\; \sqrt[n]{a_n}\leq\limsup\left(\frac{a_{n+1}}{a_n}\right)$
Given that $a_n > 0$, I need to prove: $\liminf \left(\dfrac{a_{n+1}}{a_n}\right)\;\leq \;\liminf\; \sqrt[\Large n]{a_n}\;\leq\;\limsup\left(\dfrac{a_{n+1}}{a_n}\right)$.

I am really confused about how to use the definitions to prove $\liminf \left(\dfrac{a_{n+1}}{ a_n}\right)\;\leq\;\liminf \;\sqrt[\Large n](a_n)$.

As most of you might have guessed the motivation for this comes from the ratio test and the root test for convergence of series.

Any help is much appreciated.

答:
The proof can go as follows. Set $$\ell =\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$$

Choose $\alpha <\ell$. By the definition of $\liminf$, there must exist an $N$ such that, for each $n\geq N$, we have that $$\alpha <\frac{a_{n+1}}{a_n}$$
That is, for $k\geq 0$, we have $$ a_{N+k}>\alpha\cdot a_{N+k-1}$$

This gives that $$a_{N+k}>\alpha^k \cdot a_{N}$$

In paricular $n-N\geq 0$ when $n=N+1,\dots$; so

$$a_{n}>\alpha^{n-N} \cdot a_{N}$$

Now, taking the $n$-th root gives that for $n\geq N+1 $ $$\root n \of {{a_n}}  > \alpha \cdot{\left( {\frac{{{a_N}}}{{{\alpha ^N}}}} \right)^{1/n}}$$

and taking $\liminf\limits_{n\to\infty}$ gives

$$\liminf\limits_{n\to\infty} \root n \of {{a_n}}  \geq \alpha $$

This means that for each $\alpha <\ell$,

$$\liminf\limits_{n\to\infty} \root n \of {{a_n}}  \geq \alpha $$

which is saying that  $$\ell \leq \liminf\limits_{n\to\infty} \root n \of {{a_n}}$$

The proof for $\limsup$ is completely analogous.


问:
$GL_n\mathbb{C}$ as subgroup of $GL_{2n}\mathbb{R}$
I am confused about the following paragraph in Fulton W., Harris J. "Representation Theory: A First Course":

Of course, the group $GL_n\mathbb{C}$ of complex linear automorphisms
of a complex vector space $V=\mathbb{C}^n$ can be viewed as subgroup
of the general linear group $GL_{2n}\mathbb{R};$ it is, thus, a real
Lie group as well, as is the subgroup $SL_n\mathbb{C}$. Similarly, the
subgroups $SO_n\mathbb{C} \subset SL_n\mathbb{C}$ and
$Sp_{2n}\mathbb{C} \subset SL_{2n}\mathbb{C}$ are real as well as
complex Lie subgroups. Note that since all nondegenerate bilinear
symmetric forms on a complex vector space are isomorphic (in
particular, there is no such thing as a signature), there is one
complex orthogonal subgroup $SO_n\mathbb{C} \subset SL_n\mathbb{C}$ up
to conjugation; there are no analogs of the  groups
$SO_{k,l}\mathbb{R}.$

The things I don't understand:

How do we get the subgroup structure on $GL_n\mathbb{C}$? It seems, that according to the definition of a subgroup (it is also a closed submanifold), we should construct a map $f: GL_n\mathbb{C} \rightarrow GL_{2n}\mathbb{R}$. Should $f$ also be a homomorphism of groups? If it is the case, then I can hardly imagine its coordinate representation.

How do we prove that all nondegenerate bilinear symmetric forms are isomorphic?

I have never encountered such a designation $SO_n\mathbb{C}$, it was always $SO_n$. And now I don't understand why we ignore the field.

Also, I want to emphasize that before this paragraph, the structure of a complex manifold was not defined.

答:
Consider, for instance, the case $n=2$. An element of $GL_2\Bbb C$ is a matrix of the form$$\begin{bmatrix}z_1&z_2\\z_3&z_4\end{bmatrix}$$with $z_1z_4-z_2z_3\ne0$. But you can see it as$$\begin{bmatrix}\operatorname{Re}z_1&-\operatorname{Im}z_1&\operatorname{Re}z_2&-\operatorname{Im}z_2\\\operatorname{Im}z_1&\operatorname{Re}z_1&\operatorname{Im}z_2&\operatorname{Re}z_2\\\operatorname{Re}z_3&-\operatorname{Im}z_3&\operatorname{Re}z_4&-\operatorname{Im}z_4\\\operatorname{Im}z_3&\operatorname{Re}z_3&\operatorname{Im}z_4&\operatorname{Re}z_4\end{bmatrix}\in GL_4\Bbb R.$$This works in any dimension.

You will find an answer here.

The group $SO_n$ is the group of those matrices $A\in GL_n\Bbb R$ such that $A^TA=\operatorname{Id}$ and that $\det A=1$. The condition $A^TA=\operatorname{Id}$ is very inportant over the reals, since it means that $A$ is the matrix of an isometry. But over $\Bbb C$ the same condition has no particular geometric meaning.


问:
Dense subalgebras of the unitization and their intersection with the original $C^*$-algebra
As I was trying to figure out this question, I wondered if the following is true:

If $A$ is a non-unital $C^*$-algebra and $\tilde{A}$ denotes its unitization, then is it true that any $*$-subalgebra $B\subset\tilde{A}$ that is dense in $\tilde{A}$ satisfies $B\cap A\neq\{0\}$?

It seems to me that this is too much to ask and I cannot see any particular reason for this to be true. Of course, if $1_{\tilde{A}}\in B$, then the result is obvious. Also, (maybe this helps), by approximating the unit through $B$ we get a sequence of invertible elements that belong to $B$, but we cannot deduce anything about their inverses (whether they belong to $B$ or not). Note that this argument proves the result if we also assume that $B$ is a (one-sided, in general) ideal in $\tilde{A}$.

On the other hand, I cannot think of any counter-example.

I am particularly interested for the abelian case, because non-unital abelian $C^*$-algebras are continuous functions over LCH spaces, so  we can identify the unitization with the continuous functions over the one point compactification and maybe some topological arguments can prove/ disprove the result.

答: By definition, $A$ is an ideal in $\tilde A$. Consider the quotient map $\pi\colon \tilde A\to \tilde A/A\cong \mathbb{C}$. By definition, $x\in A$ if and only if $\pi(x)=0$. If $A\cap B$ were trivial, then $\pi$ would be injective on $B$, which clearly cannot be the case unless $\dim B\leq 1$.  And it's not hard to see that $\dim B\leq 1$ can only occur under these circumstances if $A=\{0\}$, $\tilde A=\mathbb{C}$ and $B=\tilde A$. Otherwise $A\cap B\neq\{0\}$.


问:
If two complex numbers are close enough then their Arguments too?
Let $z_1=r_1e^{i\theta_1}$ and $z_1=r_2e^{i\theta_2}$ be two complex numbers such that $-\pi<\theta_1,\,\theta_2\leq \pi$. My question is:  If $|z_1-z_2|<\delta$, for some $\delta>0$ sufficiently small then $|\theta_1-\theta_2|<\lambda$ for some $\lambda$ sufficiently small ?

Any suggestion is welcome.

答: In general, no. If $z_1=\frac\delta4$ and $z_2=-\frac\delta4$, then their arguments are $0$ and $\pi$ respectively, but $|z_1-z_2|=\frac\delta2<\delta.$


答:
Assuming that $n\ne 1$, your calculation is correct.

Consider the case when $n=2$.

Your calculation gives $2-\frac43=\frac23$ but the other one gives $\frac13-2<0$, which cannot be right since the integrand is nonnegative.

Title: Computing $\int_0^1 \left(1+\frac{x}{n}\right)^{-n} \; dx$

Body:
I am trying to compute $\int_0^1 \left(1+\frac{x}{n}\right)^{-n}  \; dx$. Using an online calculator, I found the answer to be $\frac{(n+1)^{1-n}}{(n-1)} - \frac{n}{n-1}$. When I tried to verify this, I got:

$$\int_0^1 \left(1+\frac{x}{n}\right)^{-n}  \; dx= n^n\int_0^1(n+x)^{-n}\; dx = \frac{n^n}{1-n} \cdot \left[(n+x)^{1-n}\right|_{x=0}^1 = \frac{n^n}{(1+n)^{n-1}(1-n)}-\frac{n}{1-n}$$

This is not the same as the answer I for from the online calculator. Where have I gone wrong?


问:
What is the name of the function $1/\Gamma(z) \int_0^\infty \frac{x^{z -1}}{e^w-1}dw$?
I recently saw

on the internet, and was curious what the name of the "pineapple" function was?

$$1/\Gamma(z) \int_0^\infty \frac{\color{red}{x}^{z-1}}{e^w-1}dw$$

I recognize the "mango" function as $\Gamma(z)$ the gamma function.  Also, does anyone know what theorem this refers to?

答:
It's a special case of the Bose-Einstein integral $F_-$. In (statistical) physics, we define the Bose-Einstein integral ($F_-$) and Fermi-Dirac integral ($F_+$) as follows:
$$F_{\mp}(s, \alpha)=\frac{1}{\Gamma(s)} \int_0^{+\infty}\frac{x^{s-1}}{e^{x+\alpha}\mp 1}\, \mathrm{d}x$$

So your function is just $z \mapsto F_-(z, 0)$.


问:
multiplier algebra and intersection
Let $M(A)$ be the multiplier algebra of non-unital C* algebra $A$. So $A$ is essential ideal of $M(A)$. Does there exists a dense $\ast$-subalgebra $B$ of $M(A)$, such that  $B\not\subset A$, and $B\cap A=\{0\}$?

答: Not always. For example, in I. Farah's book there are references for $C^*$-algebras that have the property that $M(A)\cong\tilde{A}$, the unitization. As MaoWao proves in this post, if $B\subset\tilde{A}$ is a dense subspace in $\tilde{A}$, then we cannot have $B\cap A=\{0\}$.

Title: multiplier algebra and intersection


问:
Surjective holomorphic map from a domain to $\mathbb{C}$
Let $U\subset \mathbb{C}$ be a domain (i.e. open and connected). Is it always the case that there exists a holomorphic map $f:U\to \mathbb{C}$ such that $f(U)=\mathbb{C}$?

I managed to prove it for the upper half plane ($(z-i)^2$) (and so for every simply connected set by the Riemann mapping theorem)
and for $\mathbb{C}-\{z_0,\dots,z_n\}$ (just take $f(z)=\prod_{i=0}^{n+1} (z-z_i)$, where $z_{n+1}$ is a point in $\mathbb{C}-\{z_0,\dots,z_n\}$) but I do not seem to be able to prove it in general.

答:
If $U=\mathbb C,$ then $f(z)=z$ does the job. If $U\ne \mathbb C,$ then $\partial U$ is nonempty. In this case take any $a\in U.$ Then $d(a,\partial U)$ is a finite positive number $r,$ and there is $b\in \partial U$ such that $|a-b|=r.$ It follows that $D(a,r)\subset U.$

For $z\in \mathbb C\setminus \{b\}\supset U,$ set $f(z) = \dfrac{1}{z-b}.$ Then $f(D(a,r))$ is an open half plane. As you pointed out, this leads to a holomorphic function from $U$ to $\mathbb C$ that is surjective.


问:
Limit of $F [v_j]$ for $F [w] := \int_Ω f (x, w(x)) dx$.
I was trying to understand a part from a book on calculus of variation, but in the following text, I don't understand how the two problems are equivalent :

Assume that we are given a weakly converging sequence $v_j$  in
$L^2(Ω)$, where $Ω ⊂ \mathbb{R}^d$ is a bounded Lipschitz domain, and
an integral functional $$F [w] := \int_Ω f (x, w(x)) dx, w ∈ L^2
> (Ω)$$, with $f : Ω × \mathbb{R} → \mathbb{R}$ continuous and bounded
(for simplicity). Then, $(F [v_j])_j$ is a bounded sequence and up to
a (non-renumbered) subsequence we may assume that $F [v_j]$ converges
to some limit as $j → ∞$. The question then arises: how can we compute
this limit for every integrand $f$ as above? Equivalently, we could
ask for the weak* limit in $L^∞ (Ω)$ of the sequence of compound
functions $F_j (x) := f (x, v_j (x))$.

Why do we have

Equivalently, we could ask for the weak* limit in $L^∞ (Ω)$ of the
sequence of compound functions $F_j (x) := f (x, v_j (x))$.

?

答:
Let's first unpack what it means for $F[v_j]$ to converge. Suppose $$F[v_j]\rightarrow F^*$$ and note that since f and $\Omega$ are bounded, we must have that
$$\lim_{j\rightarrow \infty}F[v_j]=\lim_{j\rightarrow \infty}\int_\Omega f(x, v_j(x))dx=\int_\Omega \lim_{j\rightarrow \infty} f(x, v_j(x))dx \,\,\,(\text{By Dominated Convergence}).$$ Now note that f is assumed continuous therefore, $\lim_{j\rightarrow\infty} f(x, v_j(x))$ converges pointwise to some measurable function- call it $F(x)$, which yields that $$F^*=\int_\Omega F(x)dx.$$

Now let's look at the definition of weak* convergence in $L^\infty(\Omega)$. We say that a sequence$F_j(x)=f(x, v_j(x))$ converges in the weak* topology to a limit $F(x)$ if and only if $$\int_\Omega F_j(x)g(x)dx\rightarrow \int_\Omega F(x) g(x)dx \,\, , \forall g\in L^1(\Omega).$$

Therefore, the two notions will be equivalent if the following equivalence holds: $$\int_\Omega F_j(x)g(x)dx\rightarrow \int_\Omega F(x) g(x)dx \,\, , \forall g\in L^1(\Omega) \iff \int_\Omega F_j(x)dx \rightarrow \int_\Omega F(x)dx.$$

The $\Rightarrow$ direction is obvious since if the convergence holds for any $g\in L^1(\Omega)$, than just take $g=1$. Conversely, the other direction holds because convergence of the integrals implies we can extract a pointwise convergent sub-sequence such that $F_{jk}[x]\rightarrow F[x]$ pointwise, and then we are free to pull the limit inside of $\int_\Omega F_{jk}(x)g(x)dx$ and the result follows.

Note that this would fail if we got rid of either the continuity or boundedness assumptions on f.


问:
If $P_W$ is the projection onto a linear subspace $span(W)$, is this onto the row or column space of $W$?
Let $W \in \mathbb{R}^{n \times m}$ and $P_W$ be the projection onto a linear subspace $span(W) \subset \mathbb{R}^n$. In this case, is $P_W$ the projection onto the row space or the column space of $W$?

When I see $span(W)$, I interpret it as the span of the vectors in $W$. However, is this referring to the span of the row vectors (row space) or the column vectors (column space)?

答:
Note that the rows of $W$ live in $\mathbb R^m$ and the columns live in $\mathbb R^n$, so $P_W$ must be projecting onto the column space of $W$.

(IMO, it's best to use notation $\text{Span}(S)$ only when $S$ is a set of vectors.)


问:
Show that if $f$ is an entire function with $Re\,f(z) > M\in\mathbb{R}$ for all $z\in\mathbb{C}$, then $f$ is constant
I have the following proof but I don't know if I am right:

$$ \left|\frac{1}{e^{f(z)}}\right| = \left|\frac{1}{e^{Re\,f(z)}}\right| < \left|\frac{1}{e^{M}}\right|$$

Since $1/e^{f(z)}$ is entire, it follows from Liouville's theorem that $1/e^{f(z)}$ is constant.

$\Rightarrow e^{f(z)}$ is constant

$\Rightarrow f'(z)e^{f(z)} = 0 $

$\Rightarrow f'(z)=0$

$\Rightarrow f$ is constant

答:
Your proof is correct. A key step is that for any $z$, one has:
$$
|e^z|=|e^{Re(z)}|
$$

Also, $e^z$ is non-zero for any $z$. You applied the chain rule correctly to $e^{f(z)}$.


问:
Number of roots of $\lfloor\frac x3\rfloor=\frac x2$
How many roots does the equation $$\left\lfloor\frac x3\right\rfloor=\frac x2$$ have?

$1$

$2$

$3$

infinitely many

I checked that $x=0$ and $x=-2$ are the answers, so I think the answer is $(2)$. but I don't know how to solve the problem in general.

答:
hint

Put $ x=3X$.
the equation becomes
$$\lfloor X\rfloor =\frac 32 X$$

which gives

$$X-\lfloor X\rfloor =-\frac X2\in[0,1)$$

So, there are three cases
$1) X\in(-2,-1) \implies$

$$ X=\frac{-4}{3}\implies x=-4$$

$2) X\in[-1,0)\implies $

$$X=\frac{-2}{3}\implies x=-2$$

$3) X=0=x$.

In conclusion, it has three roots.


问:
Proof that $\bigcup_{i \epsilon I} A_{i}$ is an elementary substructure of $\mathfrak{U}$
I'm currently studying model theory with the book "A Course in Model Theory" by Katrin Tent and Martin Ziegler. Currently I have written a proof for Exercise 2.1.1, and I would like to know if it's correct. The problem states:

Exercise 2.1.1. Let $\mathfrak{A}$ be an L-structure and ( $\mathfrak{A}_{i} )_{i \epsilon I}$ a chain of elementary substructures of $\mathfrak{A}$. Show that $\bigcup_{i \epsilon I} A_{i}$ is an elementary substructure of $\mathfrak{A}$.

My solution goes like this:

$\forall i, \forall a, (a \in A_{i} \Rightarrow a \in \bigcup_{i \epsilon I} A_{i})$

$\forall i, \mathfrak{A}_{i} \prec \mathfrak{A}$

$\forall i, \forall \phi, \forall a ,   ((\mathfrak{A}_{i} \prec \mathfrak{A}) \Leftrightarrow  (\mathfrak{A}_{i} \models \phi (a) \Leftrightarrow \mathfrak{A} \models \phi (a) ))$

This first three are basically just definitions, the actual inferences are the following:

The hypothesis: $ \exists i, \exists a \in \bigcup_{i \epsilon I} A_{i},  ( \mathfrak{A} \not\models \phi(a) \Rightarrow \mathfrak{A}_{i} \nprec \mathfrak{A}) $

By Modus Ponens: $\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}$

$(\forall i, \mathfrak{A}_{i} \prec \mathfrak{A}) \wedge (\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}) \Rightarrow \bot$ (by 2 and 5)

And finally, by Modus Tollens on the contradictory hypothesis (4):

$\forall i, \forall a \in \bigcup_{i \epsilon I} A_{i}, \mathfrak{A} \models \phi(a)$

Which should satisfy Tarski's Test which according to the book states:

Let $\mathfrak{B}$ be an $L$-structure and $A$ a subset of $B$. Then $A$ is the universe of an elementary substructure if and only if every $L(A)$-forumla $\phi (x)$ which is satisfiable in $\mathfrak{B}$ can be satisfied by an element of $A$.

So again, does the proof seems correct? are there not any inferential jumps or plain just downright wrong inferences?

答:
I don't understand the proof you've written. (5) does not follow from (4) by Modus Ponens, and (7) is not the negation of the hypothesis (4). Further, your proof concludes that $\phi(x)$ is satisfied by every element of $\bigcup_{i\in I}A_i$, and it should be clear that this is too strong. To apply Tarski's test, you need to consider a formula $\phi(x)$ which is satisfied by some element of $\mathfrak{A}$. But it's quite possible that it's satisfied by only one element!

Anyway, you have the right idea to apply Tarski's test. Let $\varphi(x,\bar{a})$ be an $L$-formula with parameters $\bar{a}$ from $\bigcup_{i\in I}A_i$, such that $\varphi(x,\bar{a})$ is satisfied in $\mathfrak{A}$, i.e., $\mathfrak{A}\models \exists x\, \varphi(x,\bar{a})$. You want to find some $b\in \bigcup_{i\in I} A_i$ such that $\mathfrak{A}\models \varphi(b,\bar{a})$. Then you're done by Tarski's test.

Hint: Use the assumption that $\mathfrak{A}_i\preceq \mathfrak{A}$ to pull satisfaction of $\exists x\, \varphi(x,\bar{a})$ down to $\mathfrak{A}_i$. Careful: not any $i\in I$ works. Why not? How can you pick the right $i$?


问:
Optimization problem solution (water-filling)
Let $x\in\mathbb{R}^n$ be an optimization variable and $\alpha\in\mathbb{R}^n$ be an n-dimensional vector. The standard water-filling problem is formulated as

\begin{equation}
\begin{array}{ll}
\operatorname{minimize} & -\displaystyle\sum_{i=1}^{n} \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1
\end{array}
\end{equation}

and has a well known solution (see Boyd & Vandenberghe pag. 245).

My question is about a similar problem:

\begin{equation}
\begin{array}{ll}
\operatorname{minimize} & \max_{i}- \log \left(\alpha_{i}+x_{i}\right) \\
\text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=1
\end{array}
\end{equation}

The problem is still convex since the maximum of convex function is convex. How the solution would chance for this slightly modified optimization problem? It would be interesting to understand, also in terms of a wireless communication interpretation. Thanks!

答:
I shall prove that the problems are in fact equivalent, in the sense that they both have the same optimal solution $x^*$. Let $P_1$ be the first optimization problem, and $P_2$ the second, and let $\alpha\succ 0$.

If $\tilde{x}$ is the optimal solution for $P_1$, then we know from the book that each $\tilde{x}_i$ satisfies the relation
$$
\tilde{x}_i = \max\{0,1/\tilde{\nu} - \alpha_i\},
$$
where $\tilde{\nu}$ is the optimal dual variable associated with the equality constraint.

Now, write $P_2$ equivalently as
$$\begin{split}
 \text{minimize}  &\quad t\\
 P_2:\quad\text{subject to}&\quad t\geq -\log(\alpha_i + x_i),\quad i=1 ,\ldots,n\\
 &\quad x\succeq0,\quad \mathbf{1}^\top x=1.
\end{split}.$$

Then, the Lagrangian is
$$L(t,x,\mu,\lambda,\nu) = t - \sum_{i=1}^n \mu_i(\log(\alpha_i + x_i) + t) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1) =\\t(1 - \mathbf{1}^\top\mu) - \sum_{i=1}^n\mu_i\log(\alpha_i + x_i) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1).$$

Using this, you can write the KKT conditions of $P_2$ for the primal and dual optimal variables $t^*,x^*,\mu^*,\lambda^*$ and $\nu^*$ as
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad \lambda^*\succeq0,~\mu^*\succeq 0,\quad \lambda_i^*x_i^*=0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n,$$
and
$$\partial/\partial x_i \left[ L(t^*,x^*,\mu^*,\lambda^*,\nu^*)\right]= -\mu_i^*/(\alpha_i^* + x_i^*) - \lambda_i^* + \nu ^* = 0,~i=1,\ldots,n,\quad \mathbf{1}^\top\mu^* = 1.$$
The last equality is necessary, otherwise the dual becomes unbounded in $t^*$. Eliminating $\lambda^*$ yields:
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad x_i^*(\nu^* - \mu_i^*/(\alpha_i + x_i^*))=0,\quad \nu^*\geq \mu_i^*/(\alpha_i + x_i^*),\quad i=1\ldots,n,$$
and
$$\mu^*\succeq 0,\quad \mathbf{1}^\top\mu^* = 1,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n.$$

These are in fact very similar conditions to those of $P_1$. Now, let $M = \{i\in \{1,\ldots,n\} ~|~ \mu_i^* > 0\}$. This is a nonempty set, because $\mathbf{1}^\top \mu^* = 1$. For all $i\in M$, you can thus deduce (in a similar way to the book) that in fact
$$x_i^* = \max\{0, \mu_i^*/\nu^* - \alpha_i\}.$$
Furthermore, we have that $t^* = -\log(\alpha_i + x_i^*)$.

Now, let $M' = \{1,\ldots,n\}\setminus M$. In this case, we have that for all $i\in M'$:
$$
x_i^*\nu^* = 0,\quad \nu^* \geq 0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0.
$$
We can't have $\nu^* = 0$, otherwise it would imply that $0 < \mu_j^*/(a_j + x_j^*) \leq 0$, for all $j\in M$, which is impossible. So, the only option is that $x_i^* = 0$ for all $i\in M'$.

Thus, the problem reduces to solving the equation:
$$
\sum_{i\in M} x_i^* = \sum_{i\in M}\max\{0,\mu_i^*/\nu^* - \alpha_i\} = 1.
$$

However, the choice of $\mu$ can be arbitrary, as long as it satisfies the KKT conditions. Let $K = \{i\in\{1,\ldots,n\}~|~ \tilde{x}_i > 0\}$, and suppose we set the multipliers to
$$
\mu_i^* = \begin{cases}
 1/\lvert K \rvert & i \in K\\
 0     &\text{otherwise,}
\end{cases}
$$
then $\mathbf{1}^\top \mu^* = 1, \mu^*\succeq 0$. Clearly, we also have $M = K$, which implies $x_i^* = \tilde{x}_i=0$ for all $i\in M'$. Now, setting $\nu^* = \tilde{\nu}/\lvert M \rvert > 0$, then this implies $x_i^* = \tilde{x}_i$ for all $i\in M$, and we get exactly the same solution as $P_1$. And since $\tilde{x}_i + \alpha_i = \tilde{x}_j + \alpha_j$ for all $i,j\in M, i\neq j$, then this implies
$$
\log(x_i^* + \alpha_i) = \log(x_j^* + \alpha_j),
$$
thus $t^* = -\log(x_i^* + \alpha_i)$ for all $i\in M$, so all of the KKT conditions are satisfied.

Therefore, solving $P_2$ is equivalent to solving $P_1$. The optimal value of $P_2$ is $-\log(\alpha_i + \tilde{x}_i)$, for any $\tilde{x}_i > 0$.