问:
Find $E[(x+2)^2]$ Given $E[x]$
Find
$
E[(x+2)^2]
$
given that $
E[x] = 5, Var(X) = 2
$.

I'm not sure if I'm making this more complicated than it should be. However, this is what I did.

$$
E[(x+2)^2] =
E[x^2+2x+4] =
E[x^2]+E[2x]+4 = E[x^2]+2E[x]+4 = 25+10+4 = 39.
$$
I'm unsure is if $E[x^2]$ is 25. And furthermore, if there's an easier method to solving this problem if $x+2$ was raised to a higher power

答:
Without any other information, it is not possible to compute the desired expectation.  This is because $\operatorname{E}[X] = 5$ does not uniquely determine $\operatorname{E}[X^2]$.  In fact, if $\operatorname{E}[X] = \mu$, we can write

$$\begin{align}
\operatorname{E}[(X+2)^2]
&= \operatorname{E}[(X - \mu + \mu + 2)^2] \\
&= \operatorname{E}[(X - \mu)^2 + 2(\mu + 2)(X - \mu) + (\mu + 2)^2] \\
&= \operatorname{E}[(X - \mu)^2] + 2(\mu + 2)\operatorname{E}[X - \mu] + (\mu+2)^2 \\
&= \operatorname{Var}[X] + 2(\mu + 2)(\mu - \mu) + (\mu + 2)^2 \\
&= \operatorname{Var}[X] + (\mu + 2)^2.
\end{align}$$

This shows that the desired expectation equals the variance plus $(\mu + 2)^2 = 49$.  So for instance, if $X$ is normally distributed with mean $5$, we can pick any nonnegative variance we please.  All that we can say is that $\operatorname{E}[(X+2)^2] \ge 49$.

Title: Find $E[(x+2)^2]$ Given $E[x]$


问:
Prove complements
can someone help me with this problem.

Notation: $A$' is the complement of A

Prove that $A$ $\cap$ $B$' $\subseteq$ ($A$ $\cap$ $B$)'

My work:

$A$ $\cap$ $B$' = {x $|$ x $\in$ A, x $\notin$ $B$}

$A$ $\cap$ $B$ = {x $|$ x $\in$ A, x $\in$ $B$}

($A$ $\cap$ $B$)' = {x $|$ x $\notin$ A, x $\notin$ B} = {x $|$ x $\notin$ ($A$ $\cap$ $B$)}

I am stuck here. Please help if you can. Thank you!

答: @DeepSea is correct. I just want to touch on something in your solution which may give you insight with future problems you try to solve. There is a problem in your understanding with the last line - $(A \bigcap B)'$ means that $x$ is not in the compliment of the intersection. So $x$ could very well be in $A$ or $B$ - just not both $A$ and $B$.

ContentLicense: CC BY-SA 4.0

Title: Prove complements


答:
A basic fact about the $'$ operator is that it is decreasing - that is, if $X \subseteq Y$ then $Y' \subseteq X'$.

A basic fact about the $\cap$ operator is that $X \cap Y \subseteq Y$.

We note by the $\cap$ basic fact that we have $A \cap B \subseteq B$. Therefore, by the basic fact about $'$, we have $B' \subseteq (A \cap B)'$. Using the $\cap$ basic fact again, we have $A \cap B' \subseteq B'$. Therefore, we have $A \cap B' \subseteq B' \subseteq (A \cap B)'$. Then $A \cap B' \subseteq (A \cap B)'$.

Title: Prove complements

Body:
can someone help me with this problem.

Notation: $A$' is the complement of A

Prove that $A$ $\cap$ $B$' $\subseteq$ ($A$ $\cap$ $B$)'

My work:

$A$ $\cap$ $B$' = {x $|$ x $\in$ A, x $\notin$ $B$}

$A$ $\cap$ $B$ = {x $|$ x $\in$ A, x $\in$ $B$}

($A$ $\cap$ $B$)' = {x $|$ x $\notin$ A, x $\notin$ B} = {x $|$ x $\notin$ ($A$ $\cap$ $B$)}

I am stuck here. Please help if you can. Thank you!


问:
On Sixth Powers $x_1^6+x_2^6+\dots+x_6^6 = z^6$
Fourteen years ago, in 1999 (has it been that long?) Merignac started a search for,

$$x_1^6+x_2^6+\dots+x_6^6 = \color{red}z^6$$

using the five congruence classes,

$$\begin{aligned}
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(1x_6)^6+(42x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(2x_6)^6+(21x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(3x_6)^6+(14x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(6x_6)^6+(7x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_4^6)+(21x_5)^6+(14x_6)^6+(6x_7)^6 = z^6
\end{aligned}$$

where $b =42$. Primitive integer solutions are known for all five classes, but none has one $x_i = 0$ with $\color{red}z<730000$. (See Further work section of The Smallest Solutions to the Diophantine Equation $a^6+b^6+c^6+d^6+e^6=x^6+y^6$.)

To compare to 4th powers, one can primitively solve,

$$a^4(x_1^4+x_2^4+x_3^4)+x_4^4 = z^4$$

for $a=10$ or $a=20$ with the smallest being,

$$10^4(24^4+34^4+43^4)+599^4 = 651^4$$
$$20^4(19^4+83^4+94^4)+4907^4 = 4949^4$$

Question: Is $b=42$ overly restrictive? Can it be reduced to just $b = 21$? (As the fourth power example shows, $a = 10$ has a smaller solution.  If so, then maybe they overshot one in the $z<730000$ range.)

答:
The restriction to $b = 42$ is correct because any solution of the Diophantine equation

$$x_1^6+x_2^6+\dots+x_6^6 = z^6$$

would have to meet constraints associated with each of the prime factors of 42:

a) If $a \not\equiv 0 \pmod{2}$ then $a^6 \equiv 1 \pmod{8}$;

b) If $a \not\equiv 0 \pmod{3}$ then $a^6 \equiv 1 \pmod{9}$;

c) If $a \not\equiv 0 \pmod{7}$ then $a^6 \equiv 1 \pmod{7}$.

(These can be found on this page of Meyrignac's website, albeit applied to a different 6th power equation.)  To meet these constraints, a (primitive) solution must have $x_i \equiv 0 \pmod{2}$ for 5 of the 6 left hand terms, and similarly for mod 3 and mod 7.  Hence a solution must have $x_i^6 \equiv 0 \pmod{42}$ for at least 3 of the 6 left hand terms.

There is also a restriction associated with 13 arising from the fact that:

d) If $a \not\equiv 0 \pmod{13}$ then $a^6 \equiv 1$ or $-1 \pmod{13}$.

This implies that a solution must have an odd number of terms (which could include $z$) congruent to 0 mod 13 and, among the remaining terms, an appropriate balance of sixth powers congruent to 1 and to -1 mod 13.  To combine this restriction with those relating to 2, 3 and 7 would be quite tedious but it can be said that the five congruence classes are not restrictive enough to exclude all cases which are impossible for reasons of congruence.

ContentLicense: CC BY-SA 3.0


问:
How to interpret symbols ∇ Del?
I have the following context

Let $w = [w_1, w_2]^T$

Consider $g(w) = 2w_1w_2 + w_2\cos(w_1)$

Q: Compute $\nabla_w g(w)$ and verify $\nabla_w g([\pi,1]) = [2, \pi - 1]^T$

I'm learning machine learning, but I don't have a lot of advanced mathematics knowledge, how to interpret the symbols $\nabla_w$ in the context?

答: Basically $\nabla_w g(w)$ is to be understood as $(\frac{\partial g}{\partial w_1}, \frac{\partial g}{\partial w_2})^T$, so you get a vector. Now, use your knowledge about how to take partial derivatives of a function of multi-variables, calculate $\nabla_w g(w)$ and evaluate the resulting expression at the prescribed point.

Title: How to interpret symbols ∇ Del?


问:
Fibonacci Sequence & another pattern
While playing around with Python, I saw that if you take any non-zero integers p & q and perform the following operation, the values tend towards phi.

```<code>Flip the numerator and denominator and complete the process by adding the value of the denominator to the numerator.
</code>```

$\dfrac{3}{2},\dfrac{5}{3}, ... \text{after 38 repetitions}... \dfrac{267914296}{165580141} $

Can someone explain how this works?

答: You're sending $\frac{a}{b}$ to $\frac{a + b}{a} = 1 + \frac{b}{a}$. In other words, you're sending $x$ to $1 + 1/x$. The fixed point of this operation is $\phi$.

Title: Fibonacci Sequence & another pattern


答:
Kepler observed the fact you recite:

$$  \lim_{n \rightarrow \infty} \frac{F_{n+1}}{F_n} = \varphi  \text{,}  $$
where $F_n$ is the $n^\text{th}$ Fibonacci number and $\varphi$ is the golden ratio.

The usual way to show this is with Binet's formula.
$$  F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi} = \frac{\varphi^n - \psi^n}{\sqrt{5}}  \text{,}  $$
where $\psi = -1/\varphi$ is the conjugate golden ratio -- the other root of a quadratic equation constructed from the recurrence for the Fibonacci numbers.  Using Binet's formula,
\begin{align*}
\frac{F_{n+1}}{F_n} &= \frac{\varphi^{n+1} - \psi^{n+1}}{\varphi^{n} - \psi^{n}}  \\
    &= \frac{\varphi^{n+1}}{\varphi^n} \cdot \frac{1 - \psi^{n+1}/\varphi^{n+1}}{1 - \psi^{n}/\varphi^{n}}  \\
    &= \varphi \frac{1 - (-1)^{n+1}/\varphi^{2(n+1)}}{1 - (-1)^{n}/\varphi^{2n}}
\end{align*}

As $n$ increases, since $\varphi > 1$, the fractions $(-1)^{n+1}/\varphi^{2(n+1)}$ and $(-1)^{n}/\varphi^{2n}$ approach $0$.  So, in the limit as $n \rightarrow \infty$, this is
$$  \varphi \frac{1 - 0}{1 - 0} = \varphi  \text{.}  $$

Title: Fibonacci Sequence & another pattern

Body:
While playing around with Python, I saw that if you take any non-zero integers p & q and perform the following operation, the values tend towards phi.

```<code>Flip the numerator and denominator and complete the process by adding the value of the denominator to the numerator.
</code>```

$\dfrac{3}{2},\dfrac{5}{3}, ... \text{after 38 repetitions}... \dfrac{267914296}{165580141} $

Can someone explain how this works?


问:
One-point compactification of a linear space
Is it possible to take the one-point compactification of a linear space and get again a linear space? Like if $X$ is a normed space add $*$ to it and define $|x-*| = 1$?</p>

答:
This can't be achieved if you want the normed space structure to be compatible with the topology in the sense that the norm is continuous. Since otherwise, the norm function, now having compact domain, must achieve a maximum, which is impossible due to homogeneity.

In case what you are after is just 'compactifying' vector spaces (with no underlying topology), then you'll have to be more precise about what compact should mean. In any case, you can't take a vector space (in the presence of enough scalars), add a point to it, and extend the vector space structure to a vector space. The reason is that for every scalar $\alpha\ne 0 $ in the ground field, $x\mapsto \alpha x$ is injective. So if you add one new element, you must add at least $|\mathbb F|-1 $ other elements, where $\mathbb F $ is the ground field.


问:
Quick Area of Cross Section Question
"The base of a solid is the region in the first quadrant between the graph of $y=2x$ and the $x$-axis for $0\leq1\leq2$. For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the $x$-axis and one radius in the $xy$-plane. What is the volume of the solid?"

I guess I'm a bit confused about both the wording and where I'd have to go in this question, I understand how to find the area of a cross-section when these cross-sections might be a square or a rectangle, but I'm a bit lost when it comes to circles as well. Any help would be appreciated!

答:
The hardest part of this problem is understanding the wording. Trying to describe a $3$-d shape using words is sometimes difficult, but what it is trying to describe is a pretty simple shape. The base is a right triangle with a length of two units and a height of $4$ units. To find the total volume we need to find how this shape looks in the $z$-direction. If you take a cross-section perpendicular to the $x$-axis you should see a quarter circle, with radius $2x$, because that is the distance from the $x$-axis to the given line. from this we can tell that one cross section has volume $(\pi/4) \cdot (2x)^2 \, dx$. This can be simplified to the $\pi \cdot x^2 \, dy$. Now we simply take the integral from $0$ to $2$ of this volume of a single cross section to add up all the cross sections from $x=0$ to $x=2$. This equation looks like the following:

$$ \pi \, \int_{0}^{2} x^2 \, dx = \frac{8 \, \pi}{3}.$$

Title: Quick Area of Cross Section Question


问:
$\int \frac{1}{x^2\sqrt{x-1}}\; dx$
I am trying to do the following integral
$$
\int \frac{1}{x^2\sqrt{x-1}}\; dx
$$
My first thought is to do a substitution $u = x - 1$ so that I get
$$
\int \frac{1}{(u + 1)^2\sqrt{u}}\; du
$$
The maybe another substitution $t = \sqrt{u}$, so that I get
$$
\int \frac{1}{(t^2 + 1)^2} 2\;dt
$$
This might then be a trig. sub., but I don't think it is supposed to be this hard. Help please.

答:
A different approach to calculate

$$\int\frac{1}{(1+t^2)^2}dt$$

is the following:

$$\int\frac{1}{(1+t^2)}\cdot\frac{1}{(1+t^2)}dt$$

$$\int\frac{1}{(1+t^2)}\cdot\frac{d}{dt}\arctan(t)dt$$

$$\int\frac{1}{(1+t^2)}\cdot d[\arctan(t)]$$

$$\int \cos^2(\arctan(t))d[\arctan(t)]=\int \cos^2(y)dy$$

Title: $\int \frac{1}{x^2\sqrt{x-1}}\; dx$


问:
Why does $e^{t+4}(t-1)=0$ only yield one solution?
I understand that $e^{t+4}(t-1)=0$ only yields one solution because $e^x >0$ what I am wondering is why you can't $\ln$ to cancel out the $e$ to give $(t+4)(t-1)=0$

This is clearly invalid from looking at online resources and there's also no need to do it but Maths doesn't lie after all and I'm wondering why this wouldn't be possible. The only thing I can possibly think of is that you have to $\ln 0$ on the right hand side despite it being $0$

答:
If you tried to use $\ln$ on both sides you get

$\ln [e^{t+4}(t-1)] = \ln 0$

$\ln (e^{t+4}) + \ln (t-1) = \ln 0$

$t+4 + \ln(t-1) = \ln 0$.

And from here we could not the RHS is undefined $\lim_{x\to 0^+} x=-\infty$.  So we must have the LEFT also undefined.  So $t-1 \le 0$.  But if $t-1 < 0$ then we don't have any value for $\lim_{x\to t-1} \ln x$ not even $-\infty$. As $x$ can't be near $t-1$ if $t-1 < 0$ then no limit makes any sense.

So we must have $t-1 = 0$ and $t =1$.

So... you can almost do that.

.......

What you seem to want to do is selectively apply $\ln e^{t+4}$ but NOT apply it to $(t-1)$ and .... you just can't do that.

You certainly can NOT say $[e^{t+4} ](t-1) = 0$ so $[\ln e^{t+4}](t-1) = 0$.  That's just nutty.  $A\ne \ln A$ so it make no sense to say $AB = (\ln A)B$ but that's exactly what you are trying to say.


问:
If $f(x)$ irreducible degree $p$, over $\mathbb{Q}$ with exactly two nonreal roots in $\mathbb{C}$ Then the Galois group of $f(x)$ is $S_p.$
Let $f(x)$ be an irreducible polynomial of degree $p$, $p$ prime, over $\mathbb{Q}$. Suppose that $f(x)$ has exactly two nonreal roots in $\mathbb{C}$. Then the Galois group of $f(x)$ over $\mathbb{Q}$ is the symmetric group $S_p.$

I have the following ideas (written in a non-rigorous way):

Let $K$ the splitting field of $f(x)$ over $\mathbb{Q}$. By hypotesis, exists $\alpha\in\mathbb{C}\setminus\mathbb{R}$ root of $f(x)$. Further, $\beta=\overline{\alpha}$ is the other root of $f(x$) in $\mathbb{C}\setminus\mathbb{R}$.

Let $\sigma\in Gal(K/\mathbb{Q})$ given by $\sigma(\alpha)=\beta$, $\sigma(\beta)=\alpha$ and $\sigma(k)=k$ for any $k\in K$, $k\neq \alpha,\beta.$

Then $\sigma$ "behaves like a permutation" in $Gal(K/\mathbb{Q})$.

By other hand, $p\mid |Gal(K/\mathbb{Q})|$. By Cauchy's theorem, exists element of order $p$, i.e, a permutation of order $p$.

Therefore, $Gal(K/\mathbb{Q})$ contains an permutation order 2 and an permutation of order $p$.

an permutation of order $2$ and an permutation of order $p$ generates $S_p$. Therefore, $Gal(K/\mathbb{Q})=S_p$.

But, there is something that is wrong or I do not understand.
By hypotesis, $p=[K:\mathbb{Q}]=|Gal(K/\mathbb{Q})|$ but if $Gal(K/\mathbb{Q})=S_p$ then $|Gal(K/\mathbb{Q})|=p!\neq p$...

Actualization 1. Now I have better understood the situation. I have the general idea but the proof is not written rigorously.

Let $K$ the splitting field of $f(x)$ over $\mathbb{Q}$ i.e. $K=\mathbb{Q}(r_1,\ldots, r_p)$ with $r_i$ the roots of $f(x)$.

Let $Gal(K/\mathbb{Q})\times \left\{r_1,\ldots, r_p\right\}\to \left\{r_1,\ldots, r_p\right\}$ an action. Then exists $\psi:Gal(K/\mathbb{Q})\to Aut(\left\{r_1,\ldots, r_p\right\})\simeq S_p$ homomorphism.

Therefore $\psi(Gal(K/\mathbb{Q}))$ subgroup of $S_p$.

$f(x)$ is the minimal polynomial of $r_i$ over $\mathbb{Q}$ for any $i=1,\ldots, p$ then $[\mathbb{Q}(r_i):\mathbb{Q}]=p$. Because $[K:\mathbb{Q}]=[K:\mathbb{Q}(r_i)][\mathbb{Q}(r_i):\mathbb{Q}]$ then $p\mid [K:\mathbb{Q}]=|Gal(K/\mathbb{Q})|$.

Therefore exists $g\in Gal(K/\mathbb{Q})$ of order $p$. Then $\psi(g)$ has order $p$ in $S_p?$

Therefore $\psi(g)$ is a $p$-cycle in $S_p$.

By the other hand, Let $\sigma\in Gal(K/\mathbb{Q})$ given by $\sigma(\alpha)=\beta$, $\sigma(\beta)=\alpha$

Then $\sigma$ "behaves like a permutation" in $Gal(K/\mathbb{Q})$.
Therefore $\psi(\sigma)$ is a permutation in $S_p$

We also know that $S_p$ is generated by a permutation and $p$-cycle.

Therefore $Gal(K/\mathbb{Q})\simeq S_p$.

答:
Two non-real roots imply that complex conjugation is an element of order...

As you build the splitting field, the very first step gives you an extension of degree...

Now you know that the Galois group has elements of two special orders. Both of them must have a very specific cycle structure, when you consider them as elements of the symmetric group acting on the $p$ roots. What can you say about a subgroup of $S_p$ containing two such elements?


问:
$n$ people form $k$ clubs. Each club has at least 3 people. every two club has exactly 1 member in common. Maximum number of $k$?
$n$ people form $k$ clubs. Each club has at least 3 people. every two club has $1$ and exactly $1$ member in common. What's the maximum number of $k$?

My thoughts: Similar to oddtown problem, I can think of this as a linear algebra problem. Let $A$ be a matrix of $n$ by $k$ where $A_{ij} = 1$ when member $i$ is in club $j$ and $A_{ij} = 0$ otherwise. $A^TA $ has diagonal greater or equal to 3 and non diagonal entries equal to $1$. What's the biggest dimension of $k$? At the very least $A^TA$ needs to be positive semidefinite. But I can't go any further...

答:
Each person $p$ is a member of at most $\lfloor {n-1\over2}\rfloor$ clubs, because removing $p$ from those clubs leaves several disjoint sets of size at least $2$. Therefore, the number of pairs $(p,C)$ such that $p$ is a person, $C$ is a club, and $p$ is a member of $C$, is at most $n\cdot \lfloor \frac{n-1}2\rfloor$. On the other hand, the number of such membership pairs is at least $3k$, since each $C$ contains at least $3$ members. Therefore, we get that
$$
n\cdot \left\lfloor \frac{n-1}2\right\rfloor\ge 3k\implies k\le \left\lfloor\frac{n}3 \left\lfloor\frac{n-1}2 \right\rfloor\right\rfloor
$$
When when $n\equiv 1$ or $3\pmod6$, the floors disappear, and this becomes $n(n-1)/6$. For these $n$, the bound is exactly attainable using a Steiner triple system of order $n$. See [Skolem] for the details of how to construct Steiner triple systems for these $n$.

Edit 2: Okay, I found this problem in the CRC Handbook of Combinatorial Designs. It is known as a packing design. It turns out that the upper bound $\left\lfloor\frac{n}3 \left\lfloor\frac{n-1}2 \right\rfloor\right\rfloor$ is attainable, except for when $n\equiv 5\pmod 6$, in which cases you can only get that upper bound minus one. The construction for $n\equiv 0,2\pmod 6$ is easy, just take a Steiner triple system of order $n+1$ and delete one person, and all clubs containing them. For $n\equiv 4,5$, you will have to consult the handbook for the construction, in the Packings chapter (chapter 40 page 550 in the second edition).

Charles J. Colbourn and Jeffrey H. Dinitz. 2006. Handbook of Combinatorial Designs, Second Edition (Discrete Mathematics and Its Applications). Chapman & Hall/CRC.

Skolem, T. (1958). Some Remarks on the Triple Systems of Steiner. MATHEMATICA SCANDINAVICA, 6, 273-280. https://doi.org/10.7146/math.scand.a-10551


问:
A sequence of continuous functions on $[0,1]$ which converge pointwise a.e. but does not converge uniformly on any interval
How to construct a sequence of functions that are defined and continuous on $[0,1]$ and it converges to zero a.e. but on any interval it does not converge uniformly?</p>

答:
See example 7.4, page 77, in Counterexamples in Analysis, Gelbaum and Olmsted (the nice diagram on page 79 gives the essential idea of the construction).

There, a sequence of continuous functions is constructed that converges to the function $f(x)=\cases{1/q,&$x=p/q$ in lowest terms, $p$ and $q$ integers with $q>0$\cr 0,&$x$ irrational}$.

That  the convergence is not uniform on any interval follows from the fact that $f$ is discontinuous at every rational $x$, and the fact that a uniform limit of continuous functions is continuous.


问:
Skewed Projector is Nonexpansive
I read the book "Finite-Dimensional Variational Inequalities and Complementarity Problems I", and try to solve Exercise 1.8.16.

1.8.16. Let K be a closed convex set in $\Re^n$ and $A$ be a symmetric positive definite matrix of order $n$. Use (1.5.11) to show that the skewed projector is nonexpansive under the $A$-norm; i.e.,
$\|P_{K,A}(u)-P_{K,A}(v)\|_A \leq \|u-v\|_A, \forall u,v\in\Re^n$,
where $\|\cdot\|_A$ is a $A$-norm, e.g., $\|x\|_A=\sqrt{x^T Ax}$.

The inequality (1.5.11) is
$(P_{K,A}(v)-P_{K,A}(u))^T A(v-u) \geq (P_{K,A}(v)-P_{K,A}(u))^T A (P_{K,A}(v)-P_{K,A}(u))$

It does not seem difficult to show but I cannot find any way to solve this.

If someone already knows how to solve it, could you answer this question? Thank you.

答:
For vectors $x, y \in \Bbb{R}^n$, let
$$\langle x, y \rangle = y^\top A x.$$
As you may or may not know, this forms an inner product on $\Bbb{R}^n$. Linearity in the first coordinate follows from basic properties of matrix multiplication. Similarly, symmetry follows from $A$ being symmetric, and basic properties of the transpose operator. Finally, positive-definiteness of the inner product corresponds precisely to the positive-definiteness of $A$.

Further, the canonical norm for this inner product, defined by $\|x\| = \sqrt{\langle x, x \rangle}$, is precisely the $A$-norm.

So, inequality 1.5.11 says
$$\langle v - u, P_{K, A}(v) - P_{K, A}(u)\rangle \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A^2.$$
Cauchy-Schwarz inequality states,
$$\|v - u\|_A \cdot \|P_{K, A}(v) - P_{K, A}(u)\|_A \ge \langle v - u, P_{K, A}(v) - P_{K, A}(u)\rangle.$$
Therefore,
$$\|v - u\|_A \cdot \|P_{K, A}(v) - P_{K, A}(u)\|_A \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A^2.$$
If $\|P_{K, A}(v) - P_{K, A}(u)\|_A = 0$, then non-expansiveness holds trivially. If not, then $\|P_{K, A}(v) - P_{K, A}(u)\|_A > 0$, so we can divide both sides by it, to get
$$\|v - u\|_A \ge \|P_{K, A}(v) - P_{K, A}(u)\|_A,$$
as required.

Title: Skewed Projector is Nonexpansive


答:
Find $E[(x+2)^2]$ given that $E[x]=5$ ,$Var(X)=2$.

@Jones, observe that:

When you refer to rv you have to use always a Capital letter

your binomial expansion is wrong.

In the post you stated $\mathbb{V}(X)=2$, in your comment you stated it as 5. I assume correct what you stated in your post as it is the more recent edit.

remember that $\mathbb{E}(X^2)=\mathbb{V}(X)+\mathbb{E}^2(X)=2+5^2=27$

Given this, you get

$$\mathbb{E}[(X+2)^2]=\mathbb{E}(X^2)+4\mathbb{E}(X)+4=27+4\times 5+4=51$$

Title: Find $E[(x+2)^2]$ Given $E[x]$

Body:
Find
$
E[(x+2)^2]
$
given that $
E[x] = 5, Var(X) = 2
$.

I'm not sure if I'm making this more complicated than it should be. However, this is what I did.

$$
E[(x+2)^2] =
E[x^2+2x+4] =
E[x^2]+E[2x]+4 = E[x^2]+2E[x]+4 = 25+10+4 = 39.
$$
I'm unsure is if $E[x^2]$ is 25. And furthermore, if there's an easier method to solving this problem if $x+2$ was raised to a higher power


问:
please help me finish the proof of this equation
problem: prove that $ 2(\sqrt{2}+\sqrt{6})\sqrt[4]{3}\int_{0}^{1}\frac{1}{\sqrt{x}\sqrt{1-x}\sqrt{(\sqrt{2}+\sqrt{6})^{2}-x}} dx = \beta (\frac{1}{6},\frac{1}{3}) $

My attempt:
use the relationship between hypergeometric series and elliptic integral, we have

$F_{1}(\alpha,\beta,\gamma,x) = \frac{1}{\beta (\beta,\gamma-\beta)}\int_{0}^{1}t^{\beta-1}(1-t)^{\gamma -\beta -1}(1-tx)^{-\alpha} dt $
(here $ \beta() $ refers to the Beta function)

then

$ LHS = 2\sqrt[4]{3}\int_{0}^{1}x^{\frac{1}{2}-1}(1-x)^{1-\frac{1}{2}-1}(1-(\sqrt{2}+\sqrt{6})^{-2}x)^{-\frac{1}{2}} dx $

$ = 2\sqrt[4]{3}\beta (\frac{1}{2}, \frac{1}{2}) F_{1}(\frac{1}{2}, \frac{1}{2},1,(\sqrt{2}+\sqrt{6})^{-2}) $

$ = \pi \cdot 2\cdot \sqrt[4]{3} \cdot \frac{2}{\pi} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}}) $

$ = 4 \sqrt[4]{3} \kappa ((\frac{1}{\sqrt{2}+\sqrt{6}})$

(here we use $ \kappa (k) = \frac{\pi}{2}F_{1}(\frac{1}{2}, \frac{1}{2},1, k^{2}$)

Thus, we only need to prove the following equation:

$ \kappa (\frac{1}{\sqrt{2}+\sqrt{6}})=\frac{1}{4\sqrt[4]{3}}\beta (\frac{1}{6},\frac{1}{3}) $

I got stuck here, can anyone help me finish this?

note: $ \kappa () $ is the complete elliptic integral of the first kind

答:
$\frac1{\sqrt{2} + \sqrt{6}} = \lambda^*(3),$ where $\lambda^*$ is defined here.

Now, this seems to imply that
$$K(\frac1{\sqrt{2} + \sqrt{6}}) = \frac{3^{\frac14} \Gamma^3(1/3)}{2^{7/3}\pi},$$ as per this page.


问:
Expand $5(x + \Delta x)^2$
During class the tutor expanded $5(x+\Delta x)^2$ as follows:

$$5x+10x\Delta x+5\Delta x^2.$$

However, when I expand it the way I know how (and check with Wolfram Alpha), it expands as follows:

$$ 5 x^2 + 10 x^2 \Delta  + 5 x^2\Delta^2.$$

Which is correct? If both are correct, how? How was the first expansion performed exactly?

答: Your tutor intends $\Delta x$ to be a single value, not the product of $\Delta$ and $x$.  $\Delta x$ is usually some change to $x$, often rather small.  It may be a measurement error or a change in conditions.  With this reading, your tutor's expansion would be the same as $(a+b)^2=a^2+2ab+b^2$ if the first term were $5x^2$

Title: Expand $5(x + \Delta x)^2$


问:
Second Derivative of a matrix
Pardon me for not knowing LateX representation,
I have following function, where $\mu$ and $\Sigma$ are both Matrices.
$$ h = \mu^T \Sigma \mu $$
which is a function of $\alpha$, such that its derivative can be written as
$$ \def\p#1#2{\frac{\partial #1}{\partial #2}}
 \p h\alpha = \left(\p \mu\alpha\right)^T\Sigma^{-1}\mu - \mu^T\Sigma^{-1}\p\Sigma\alpha \Sigma^{-1}\mu  + \mu^T \Sigma \p \mu\alpha
$$
I need to find out the Second derivative wrt $\alpha$. Please help, I searched through net, but couldn't fond any such property.</p>

答:
Assuming that $h = \mu^T \Sigma^{-1}\mu$, as suggested by the derivative, we have: Taking another derivative we apply the product rule as we did for the first and use the chain rule for $\Sigma^{-1}$, in using $\def\pd#1#2{\frac{\partial #1}{\partial #2}}\def\pdd#1#2{\frac{\partial^2 #1}{\partial #2^2}}$
$$ \pd{\Sigma^{-1}}{\alpha} = -\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1} $$
So for the first term of $\pd h\alpha$ we have
$$ \pd{}\alpha\left(\pd{\mu^T}\alpha \Sigma^{-1}\mu \right)
   = \pdd{\mu^T}\alpha\Sigma^{-1}\mu - \pd{\mu^T}\alpha \Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu + \pd{\mu^T}\alpha \Sigma^{-1}\pd\mu\alpha
$$
For the second term
\begin{align*}
&\!\!\!\!\!\!\!\pd{}\alpha\left( \mu^T\Sigma^{-1}\pd\Sigma\alpha \Sigma^{-1}\mu\right)\\
  &= \pd{\mu^T}\alpha \Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu
  - \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu
  + \mu^T\Sigma^{-1}\pdd\Sigma\alpha\Sigma^{-1}\mu\\& \ \ {} - \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu + \mu^T\Sigma^{-1}\pd\Sigma\alpha \Sigma^{-1}\pd \mu\alpha
\end{align*}
And for the third term
$$ \pd{}\alpha\left(\mu^T \Sigma^{-1}\pd\mu\alpha \right)
   = \pd{\mu^T}\alpha\Sigma^{-1}\pd\mu\alpha - {\mu^T}\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\mu\alpha + {\mu^T} \Sigma^{-1}\pdd\mu\alpha
$$
Adding everything, we should have
\begin{align*} \pdd h \alpha &=
  \pdd{\mu^T}\alpha \Sigma^{-1}\mu - \mu^T\Sigma^{-1}\pdd\Sigma\alpha\Sigma^{-1}\mu + \mu^T\Sigma^{-1}\pdd\mu\alpha\\
  &{}+ 2\left(\pd{\mu^T}\alpha \Sigma^{-1}\pd\mu\alpha - \pd{\mu^T}\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu-\mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\mu\alpha + \mu^T\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\pd\Sigma\alpha\Sigma^{-1}\mu\right)
\end{align*}</p>

Title: Second Derivative of a matrix


问:
Decidability of whether a set of matrices generate a free group.
Is there an algorithm to determine whether a specified set of $k$ invertible $n \times n$ matrices generate a free group $F_k$?

We have to make some assumption on the matrix entries: it seems reasonable to go with algebraic entries in $\mathbb{C}$, but I'd be interested in any results in this vein, such as integer entries, entries in other fields, etc.

Partial results for low $n$ and $k$ are of course also interesting: even $n=2, k=2$ seems challenging!

答: The case $n = 2, k=2$ integer coefficients is decidable (the magic words are "Poincare polygon theorem" - the group is a subgroup of $SL(2, \mathbb{Z})$ so a fuchsian group, so the geometric machinery works. For higher $n$ it is believed that the question is undecidable, but this is open, to the best of my knowledge. Other algebraic number fields reduce to the integer case.


问:
Prove that the union is the maximum of the indicator function
Here is the problem:
Prove that $I_{A_1\cup A_2 \cup\dots\cup A_n}(\omega)=\text{max}\{I_{A_2}(\omega),I_{A_1}(\omega),\dots,I_{A_n}(\omega)\}$ for any sets $A_1,A_2,\dots,A_n$

Here is what I know:
$$I_{A_1\cup A_2 \cup\dots\cup A_n}(\omega)\begin{gathered}=\begin{cases} 1\quad\text{if}\quad\omega\in A_1\cup A_2 \cup\dots\cup A_n
\\ 0\quad\text{if}\quad\omega\notin A_1\cup A_2 \cup\dots\cup A_n \end{cases}\\=\begin{cases} 1\quad\text{if}\quad\omega~\text{belongs in one of the}~A_n's
\\ 0\quad\text{if}\quad\omega~\text{belongs in none of the}~A_n's\end{cases}\\=\begin{cases} 1\quad\text{if}\quad I_{A_n}(\omega)=1~\text{for at least one}~A_n
\\ 0\quad\text{if}\quad I_{A_n}(\omega)=0~\text{for all}~A_n's\end{cases}\end{gathered}$$

I don't know where to go from here.

答: You can proceed by formulating a similar casewise definition of the max function on the RHS, i.e.$$\max_i\{I_{A_i}(\omega)\}=\begin{cases}1,&\exists A_j|I_{A_j}(\omega)=1\\0,&I_{A_k}(\omega)=0\forall k\end{cases}$$which is identical to the function definition you wrote.


问:
How do I know if two events are independent or not?
Let's say we flip a coin twice:

The sample space (set of all possible outcomes) is:

$$\Omega = \{HH, HT, TH, TT\}$$

Now let's say we have the following two events (sets of outcomes):

$$A= \{HH, HT, TH\}, B = \{HT, TH, TT\}$$
where $A$ is the event that at least one coin is heads and $B$ is the event that at least one coin is tails.

Now the intersection between A and B ($A\cap B$) is the set of outcomes that are shared between A and B,  the event (C) where at least one coin is heads AND at least one coin is tails:

$$C = A\cap B = AB = \{HT, TH\}$$

In this trivial example, I can calculate the probability of C by taking the size of the event over the size of the sample space: $$P(C) = \frac{size(C)}{size(\Omega)} = \frac{2}{4} = 0.5$$

But I am struggling to understand how I would find this answer if the sets were too large or complicated to count out.

The calculation $P(A)*P(B) = \frac{9}{16}$ clearly gives a different answer. I think this means that events A and B are not independent. Two events are dependent if the outcome of the first event affects the outcome of the second event, so that the probability is changed. So conceptually, how/why is A affecting B (or vice-versa)?

And more generally, if it's not conceptually clear that one event is influencing another, does that mean the only way to determine if they are independent is by collecting data from observations?

答:
I have a potential answer to my question.
I think the issue I am having is conceptualizing the definition of "dependent events."
I think this is a more useful definition: Events are dependent if the outcome of one affects the probability that the conditions of the other event will be satisfied (AKA, that it would occur).

In the example in the question, we can see that the outcomes of event A will indeed affect the probability that event B will occur. If the outcomes of A are either HT or TH, then the probability of B occurring is 100%. That is because the conditions of B, at least one Tail, is satisfied by both those outcomes. If the outcome of A is HH, then the probability of B occuring is 0%. Again, because the conditions of B, at least one Tail, is not met by that outcome.


问:
Finding the $\sigma$-algebra generated by $\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}$.
In my study of $\sigma$-algebra, I am having difficulty understanding the concept of the smallest $\sigma$-algebra.

Here, the set $\mathcal{F}\in P(\Omega)$ (where $P(\Omega)$ denotes the power set of $\Omega$) is a $\sigma$-algebra if $\mathcal{F}$ has the following properties:
\begin{align}
(1):& \ \ \emptyset\in\mathcal{F} \\
(2):& \ \ A\in\mathcal{F}\implies A^c\in\mathcal{F} \\
(3):& \ \ A_1,A_2,...\in\mathcal{F}\implies \bigcup_{i=1}^\infty A_i\in\mathcal{F}
\end{align}

Now, given a family $\mathcal{U}$ of subsets of $\Omega$, there is a smallest $\sigma$-algebra $\sigma_{\mathcal{U}}$ containing $\mathcal{U}$. This $\sigma$-algebra is the intersection of all $\sigma$-algebra containing you.

To help improve my understanding, I considered the following example. Let $\Omega=\{1,2,3,4,5\}$ and let $\mathcal{U}=\{\{1,2,3\},\{3,4,5\}\}=\{A,B\}$. Then,

\begin{align}
\sigma_{\mathcal{U}}&=\{\underbrace{\emptyset}_{\text{by $(1)$}},\underbrace{\Omega,A,B,A^c,B^c}_{\text{by $(2)$}},\underbrace{A\cup B,A^c\cup B, A\cup B^c, A^c\cup B^c}_\text{by $(3)$},\underbrace{(A\cup B)^c,(A^c\cup B)^c,(A\cup B^c)^c,(A^c\cup B^c)^c}_\text{by $(2)$}\} \\
&=\{\emptyset,\{1,2,3,4,5\},\{1,2,3\},\{3,4,5\},\{4,5\},\{1,2\}\}
\end{align}

Is this correct?

答:
Assuming that $A$ equals $\{1,2,3\}$, and not $\{1,2\}$, the sigma-algebra generated by $\{A, B\}$ is the power set of the three sets $\{1,2\}$, $\{3\}$, $\{4,5\}$, hence it has eight elements. (You are missing the set $\{3\}$ and its complement $\{1,2,4,5\}$.)

This follows from the result: If the universe $\Omega$ is countable, then every sigma-algebra on $\Omega$ can be generated by a partition of $\Omega$.

So to determine $\sigma({\mathcal U})$ for your example, your first step is to find the partition that corresponds to $\mathcal U$: that is, for every $\omega\in\Omega$ find the smallest member of $\sigma({\mathcal U})$ that contains $\omega$. Check that there is no way to separate $1$ and $2$ using unions and/or intersections from $\mathcal U$, and similarly you cannot find a way to separate $4$ from $5$. On the other hand, the element $3$ is covered by the singleton set $\{3\} = \{1,2,3\}\cap\{3,4,5\}$. The three sets $\{1,2\}$, $\{3\}$, $\{4,5\}$ are disjoint and their union is $\Omega$, so that's your partition.


答:
If $|x - *| = 1$ for all $x \in X$, then
$$
|x| \le |x - *| + |*| = 2
$$
for all $x$. If there exists $y \in X$ with $|y| > 0$, then there exists an integer $n$ with $n|y| = |ny| > 2$. Therefore, all $x \in X$ must satisfy $|x| = 0$.</p>

Title: One-point compactification of a linear space

Body: Is it possible to take the one-point compactification of a linear space and get again a linear space? Like if $X$ is a normed space add $*$ to it and define $|x-*| = 1$?</p>


问:
Using a binomial model, find an expression for N that guarantees that the probability of turning away a customer is less than 1%
Two movie theaters compete for the business of 1, 000 customers. Assume that each customer chooses between the movie theaters completely at random. If they find the chosen theater is full, then they will return home, instead of going to the other theater. Let N denote the number of seats in each theater.

For one of the movie theater A, using a binomial model, find an expression for N that will guarantee that the probability of turning away a customer (because of a full house) is less than 1%.

My Attempt
Let X be a random variable representing the number of people that choose a certain movie theater. So, $X\sim\text{Bin}(1000,\frac{1}{2})$.

The probability that we turn away a customer is $\mathbb{P}(X>N)$ and we know that this is the same as $\mathbb{P}(X=N+1)+\mathbb{P}(X=N+2)+...$. So we have
$$\sum^{1000}_{X=N+1}\binom{1000}{X}\left(\frac{1}{2}\right)^X\left(1-\frac{1}{2}\right)^{1000-X}<0.01$$.

My question is, how would I go about finding the value of N for this. I tried guessing but the expression is too much for my calculator and I get an overflow error. (I could cheat and use a normal approximation to get my N but I want to know how to solve it without that). Thank you

Body:
Assuming customers are equally likely to choose either theater, and choices are independent, then we want the least $N$ such that $\Pr[X > N] < 0.01$, where $X \sim \operatorname{Binomial}(n = 1000, p = 1/2)$.  Because the mean is $np = 500$ and the standard deviation is $\sqrt{np(1-p)} = 5\sqrt{10} \approx 15.8$, in order for the right-tailed probability to be less than $0.01$, we'd have to be somewhere between $2$ and $3$ standard deviations above the mean. This estimate uses the $68$-$95$-$99.7$ rule.  The actual $z$-score should be $$z_{0.01}^* = \Phi^{-1}(0.99) \approx 2.32635.$$  So using a continuity-corrected approximation,
$$\Pr[X > N] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{N - 500 + 1/2}{5 \sqrt{10}}\right] \approx \Pr\left[Z > \frac{N - 499.5}{5 \sqrt{10}}\right],$$ hence we require $$\frac{N - 499.5}{5 \sqrt{10}} \ge z_{0.01}^* = 2.232635,$$ or $N \ge 537$.  This happens to be the exact bound, mainly because $n = 1000$ is so large and $p = 1/2$ is optimal for the approximation.

How then do we calculate the exact probability?  It requires computing $$\Pr[N > 537] = 2^{-1000} \sum_{x=538}^{1000} \binom{1000}{x}.$$  This is highly problematic without a computer, since the exact value is

$$\frac{23656541122627242261775011991151552914411782707957602657343671920897378833066433409453904020724128142312409463116182645799035563749820974026705212163870470815028374811940015472552709729473060463950492383604650435801919238682815367875510082809740313834734023280158062894359172404052828968053317369907}{2678771517965668302371062622650004526403512029263834018609375970925877627812340306232995947039239645318986682293882867062967863214230785108996144393674643700983641943706057746355268651265592785469488545538261618745895485316849691889791385986519265728642799119421635541915107457913156096709301417017344}.$$

And you have to show that adding $\binom{1000}{537}2^{-1000}$ makes this value greater than $0.01$.  I see no way to do this analytically, by hand, without a computer.  Perhaps some calculators have a built-in function to compute the binomial CDF, but this is essentially the same thing as using a computer.


问:
Calculate this integral
I am having trouble calculating this integral.

Let $1<p<q<\infty$ and $a,b\geq 0$. Calculate:

$$ \int_{0}^{x} \frac{a}{(1+bt^{\frac{q}{p-1}})^\frac{q}{q-p}}dt $$

It would take 'a' out of the integral. But she wouldn't know how to go on. Could you help me?

答:
For the time being, forget the $a$ and let $$\alpha=\frac{q}{p-1}\qquad \text{and}\qquad \beta=\frac{q}{q-p}$$ and consider the problem of
$$I=\int \frac{dt}{(1+b~ t^\alpha)^\beta}$$
$$b~ t^\alpha=u \implies t=b^{-\frac{1}{\alpha } }\,u^{\frac{1}{\alpha }}\implies dt=\frac{b^{-\frac{1}{\alpha } }}{\alpha } \,u^{\frac{1}{\alpha }-1}\,du$$
$$I=\frac{b^{-\frac{1}{\alpha } }}{\alpha } \int u^{\frac{1}{\alpha }-1} (u+1)^{-\beta }\,du=\left(\frac{u}{b}\right)^{\frac{1}{\alpha }} \, _2F_1\left(\frac{1}{\alpha },\beta ;1+\frac{1}{\alpha
   };-u\right)$$

Title: Calculate this integral


问:
Spectrum of an invertible operator
Let $H$ be a Hilbert space.

Let $A, T \in B(H)$ with $A$ invertible.

Show $\sigma(A^{-1}T A) = \sigma (T)$.

Attempt:

$\subseteq$:

We will show the contrapositive.
Let $\lambda \in  \sigma(T)$ such that $(\lambda I -T)$ is  invertible.

So $(\lambda I -T)^{-1}$ exists...

$\supseteq:$

Let $\lambda \in \sigma(A^{-1}TA )$. We want to show that $\lambda\in \sigma(T)$.

We know that $(\lambda I - A^{-1}TA) $ is not invertible. And I want to show $(\lambda I -T)$ is not invertible either...

So basically for both directions I tried contrapositive and direct proof, but not sure how to show the other part is invertible or not invertible...

Thank you for your help!

答: Hint: $\lambda I - A^{-1} T A = A^{-1}(\lambda I - T) A.$

Title: Spectrum of an invertible operator


问:
Determine if the following function is continuous on $ (0,0) $
Determine if the following function is continuous on $ (0,0) $
$$f(x,y)=\begin{cases}
 (1+x^2y^2)^{-1/(x^2+y^2)},& \text{ if } (x,y)\neq (0,0) \\
 1,& \text{ if } (x,y)=(0,0)
\end{cases}$$

Conjecturing that $ \lim _ {(x, y) \to (0,0)} f (x, y) = 1 $.

My attempt is,
$$\begin{align*}
0 &\leq | f (x) -L|\\
 &=|(1+x^2y^2)^{-1/(x^2+y^2)}-1|\\ &\leq |(1+x^2y^2)^{-1/(x^2+y^2)}|+1\end{align*}$$

Hence I do not know what else to do because in another limit like this but with the euler function, you can limit that using that the function is strictly increasing, but here you cannot do that because $ (1 + x ^ 2y ^ 2) $ is not strictly increasing , so I don't know how to limit that exponent with the function $ (1 + x ^ 2y ^ 2) $.

Body: $0 \leq \frac 1 {x^{2}+y^{2}} \ln (1+x^{2}y^{2}) \leq \frac 1 {x^{2}+y^{2}} x^{2}y^{2}\leq \frac 1  2|x||y| \to 0$. So $ \frac 1 {x^{2}+y^{2}} \ln (1+x^{2}y^{2}) \to 0$. Multiply by $-1$ and  take exponential.


问:
Self-adjoint and Unitary Operator in a Hilbert Space
Let $H$ be a Hilbert space and $U$ be a unitary operator.

Suppose $T$ is self-adjoint.

Show that $T' = U^*TU$ is self-adjoint, and if $f: \sigma(T) \rightarrow \mathbb{C}$ is a continuous function, then $f(T') = U^*f(T)U$.

My attempt:

Since $T$ is self-adjoint, $T= T^*$. Also, since $U$ is unitary $U^*U=I = UU^*$.

So $\langle  x ,U^*TU y\rangle =  \langle Ux , TU y\rangle = \langle Ux , TU y\rangle = \langle TUx , U y\rangle =\langle U^*TUx ,  y\rangle$.

Hence $T'$ is self-adjoint.

Now suppose $f$ is continuous. Then we have $f(T') = f (U^*TU)$. From here, I'm not sure what properties of the continuity of $f$ I can use.

Thank you in advance.

答: The spectral theorem shows that the result is true for polynomials. Stone-Weierstrass tells you that polynomials are dense in the continuous functions..</p>


答:
$$(T')^*=(U^*TU)^*=U^*T^*U^{**}=U^*TU=T'$$

For any holomorphic function $f$ on $\sigma(T')=\sigma(T)$,
\begin{align}f(T')&=\frac{1}{2\pi i}\oint f(z)(z-T')^{-1}\,dz\\
&=\frac{1}{2\pi i}\oint f(z)(z-U^*TU)^{-1}\,dz\\
&=\frac{1}{2\pi i}\oint f(z)U^*(z-T)^{-1}U \,dz\\
&=U^* f(T)U\end{align} since $(z-U^*TU)^{-1}=(U^*(z-T)U)^{-1}=U^*(z-T)^{-1}U$ and the fixed $U$ operators can be taken outside the integral.

Hence, since $C^\omega(\Omega)$ is dense in $C(\Omega)$, it follows that $f(T')=U^*f(T)U$ for continuous functions $f$.

Title: Self-adjoint and Unitary Operator in a Hilbert Space

Body:
Let $H$ be a Hilbert space and $U$ be a unitary operator.

Suppose $T$ is self-adjoint.

Show that $T' = U^*TU$ is self-adjoint, and if $f: \sigma(T) \rightarrow \mathbb{C}$ is a continuous function, then $f(T') = U^*f(T)U$.

My attempt:

Since $T$ is self-adjoint, $T= T^*$. Also, since $U$ is unitary $U^*U=I = UU^*$.

So $\langle  x ,U^*TU y\rangle =  \langle Ux , TU y\rangle = \langle Ux , TU y\rangle = \langle TUx , U y\rangle =\langle U^*TUx ,  y\rangle$.

Hence $T'$ is self-adjoint.

Now suppose $f$ is continuous. Then we have $f(T') = f (U^*TU)$. From here, I'm not sure what properties of the continuity of $f$ I can use.

Thank you in advance.


问:
Show that $e^x\ge x^e$ for all $x\ge 0$
Show that $e^x\ge x^e$ for all $x\ge0.$

I was playing around with splitting a number into parts (that add to the original number) and then multiplying the parts to get the result, and I found that the product is highest when the parts tended towards $e$. This inequality is a consequence of that. Is there a proof or intuitive explanation for this?

答: Start with $e^x \geq 1 + x$. Substitute $x \leftarrow x - 1$ to find $e^x \geq e x$. Substitute $x \leftarrow x/e$ to find $e^{x/e} \geq x$.

Title: Show that $e^x\ge x^e$ for all $x\ge 0$


问:
Freeing the trapped knight
The trapped knight problem is as follows. Place all natural numbers (starting with 1) in a spiral on an infinite square grid. A knight begins in the cell labeled 1. Each turn it jumps (in an L-shape) to an unvisited cell with the lowest number. It turns out that the knight eventually gets trapped - has nowhere to move. Neil Sloane made a great video about it.

Maarten Mortier simulated this game for all knights with movements $(x,y)$ such that $1 \leq x,y \leq 80$ and found that they all get trapped eventually. The knight $(8,71)$ takes the longest to get trapped at $2400005$ steps.

Now I have the following questions:

Is there a knight (x,y) that never gets trapped? Assume that $1 \leq x \leq y$.

Can we number the infinite grid in a different pattern, such that the standard (1,2) knight does not get trapped?

答: For 2, the spreadsheet below gives one approach.  I am making a straight path for the knight to follow.  I put 1,2,3 to start the path, then continued with odd numbers.  I put the even numbers starting with 4 clockwise to fill the cells the knight could move to but I don't want it to.

Title: Freeing the trapped knight


问:
Will $3i$ be a positive number?
I came across this question while reading quadratic equations ?
$3i$ which can be expressed as $\sqrt{-9}$ will be a positive or a negative number ?

答: $3i$ is a complex number that is not a real number, so it's neither positive nor negative. Positive numbers have argument $0$ and negative numbers have argument $\pi$, whereas $\operatorname{Arg}(3i)=\frac\pi2$.

Title: Will $3i$ be a positive number?


问:
Let $\mu$ be finite measure and $\lVert f\rVert_\infty>0$ and $a_n=\int |f|^n d\mu$. Prove $\lim {a_{n+1}\over a_n}=\lVert f\rVert_\infty$
Here $\lVert f\rVert_\infty=\text{ess sup } |f|=\inf\{c\in\mathbb{R}|\ \mu(|f|\ge c)=0\}$ i.e. $\lVert f\rVert_\infty$ denotes least essential upper bound of $|f|$.

Now $\lVert f\rVert_\infty>0\implies \exists k>0 $ such that $f\ge k$ almost everywhere $\mu$. So $|f|^n\ge k^n$ almost everywhere, this implies $a_n\ge k^n\mu(\Omega)$ for all $n$ where $\Omega $ is the measure space.

Now $\left|\frac{a_{n+1}}{a_n}-\lVert f\rVert_\infty\right|\\\le\frac{1}{|a_n|}\int |f|^n[|f|-\lVert f\rVert_\infty]\ d\mu\\\le \frac{2\lVert f\rVert_\infty}{a_n}\int |f|^n\ d\mu\\=2\lVert f\rVert _\infty$

But I need to have $\left|\frac{a_{n+1}}{a_n}-\lVert f\rVert_\infty\right|<\epsilon$ for all but finitely many $n$.

Can anyone help me in this regard? Thanks for your help in advance.

答: Hint: By Holder's inequality, $\int |f|^{n}d\mu \leq (\int |f|^{n+1}d\mu)^{\frac n {n+1}} C^{\frac 1  {n+1}}$ where $C=\mu (X)$. Now use the fact that $\int |f|^{n+1}d\mu \geq \int_E |f|^{n+1}d\mu$ where $E=\{x: |f(x)| >\|f\|_{\infty} -\epsilon\}$. Can you finish?


问:
Question regarding complex conjugates on the Argand plane
The following diagram appears in my textbook:

They follow with

I am unsure of this visualization as I've learnt that the conjugate of a complex number in the form
$$z=x+yi,x,y,\in\mathbb{R}$$ is simply
$$z^*=x-yi.$$
This would mean, in polar form,

$$\DeclareMathOperator\cis{cis} z^*=r\cis(-\theta).$$

Is there something I am missing here? I'm genuinely confused as it seems that sources online disprove what is claimed in my textbook, yet my textbook continues to use this definition for later questions.

答:
Among

$r\,\mathrm{cis}(\pi+\theta),$

$r\,\mathrm{cis}(\pi-\theta),$ and

$r\,\mathrm{cis}(2\pi-\theta),$

only the third one refers to the complex conjugate of $r\,\mathrm{cis}(\theta),$ according to the standard definition.

Assuming that in this textbook $\mathrm{Re}(z)$ is indeed along the $x$-axis (i.e., $z=x+iy$), we can infer that the author

uses $(0,2\pi]$ as the principal argument, and

either writes $z$'s conjugate as $(-z^*),$

or writes $z$'s conjugate as $(z^*)$ but defines it as $(-x+iy).$

(1.) is fine (both $(-\pi,\pi]$ and $(0,2\pi]$ are conventional), but (2.) is strange and highly nonstandard.


答:
I'll perhaps write a longer explanation than Igor Rivin.

If $A,B,C$ are invertible bounded operators, then $A\circ B \circ C$ is also invertible.

To show $\lambda \in \sigma(T)$ iff $\lambda\in\sigma(A^{-1}T A)$, you can show instead that $T-\lambda I$ is invertible iff  $A^{-1}TA-\lambda I$ is invertible.

You can then use the identity Igor Rivin wrote:

$$A^{-1}(T-\lambda I)A= A^{-1}TA-\lambda I $$

for both directions. Where you should remember that

$$ \big(A^{-1} \big)^{-1} \Big( A^{-1}TA-\lambda I   \Big)A^{-1}=T-\lambda I $$

for one direction.

Title: Spectrum of an invertible operator

Body:
Let $H$ be a Hilbert space.

Let $A, T \in B(H)$ with $A$ invertible.

Show $\sigma(A^{-1}T A) = \sigma (T)$.

Attempt:

$\subseteq$:

We will show the contrapositive.
Let $\lambda \in  \sigma(T)$ such that $(\lambda I -T)$ is  invertible.

So $(\lambda I -T)^{-1}$ exists...

$\supseteq:$

Let $\lambda \in \sigma(A^{-1}TA )$. We want to show that $\lambda\in \sigma(T)$.

We know that $(\lambda I - A^{-1}TA) $ is not invertible. And I want to show $(\lambda I -T)$ is not invertible either...

So basically for both directions I tried contrapositive and direct proof, but not sure how to show the other part is invertible or not invertible...

Thank you for your help!


问:
Closed form of power series with Gamma function as coefficient
Is it possible to get the closed form of the below series?
$$\sum_{k=0}^{\infty}(-1)^k\frac{x^{p(k+1)}}{\Gamma(p(k+1)+1)}$$
with $p \in (0, 1)$, and $\Gamma(x)$ is the Gamma function.

I know that without $p$, the formula is just $1-e^{-x}$.

And I also tried substitution $y = x^p$.

In addition, searching stackexchange and google, I think this may relate to generalized binomial theorem.

答:
$$f_p(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{p(k+1)}}{\Gamma[p(k+1)+1]}$$

The very first ones are "simple"
$$f_1(x)=1-e^{-x} \qquad f_2(x)=1-\cos (x)$$ $$f_3(x)=-\frac{1}{3} e^{-x} \left(1-3 e^x+2 e^{3 x/2} \cos \left(\frac{\sqrt{3}
   x}{2}\right)\right)$$
$$f_4(x)=1-\cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right)$$ For $p>4$, the only expressions are generalized hypergeometric functions. I give you the very first in order you see the patterns
$$f_5(x)=1-\,
   _0F_4\left(;\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};-\left(\frac{x}{5}\right)^5
   \right)$$
$$f_6(x)=1-\,
   _0F_5\left(;\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};-\left(\frac{x}{6}\right)^6\right)$$
$$f_7(x)=1-\,
   _0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac
   {6}{7};-\left(\frac{x}{7}\right)^7\right)$$ making the general form to be
$$f_p(x)=1-\,
   _0F_{p-1}\left(;\frac{1}{p},\frac{2}{p},\cdots,\frac{p-1}{p};-\left(\frac{x}{p}\right)^p\right)$$


问:
What is MLE of $U[-k,7k]$?
I need to find maximum likelihood estimator of uniform distribution $X : U[-k,7k]$.
I think that the answer is $max\{x_1, ..., x_n \}$ . However, I think I have a mistake because if $X : U[0,k]$ the answer would be the same. So, it seems strange to me why MLE does not depend on the length of the interval.

答:
Suppose $k = 1$ and your sample is $(-0.9, 0.2, 2.5, 3.7, 4.2)$.  Based on the sample maximum $4.2$, your estimate of $k$ would be $(4.2)/7 = 0.6$.  But based on the sample minimum, $-0.9$, your estimate would be $0.9$.  Which one is better?  Can you think of a way to write a mathematical expression to reflect this?

Here is the sufficient statistic calculation.

Let the sample be $\boldsymbol x = (x_1, \ldots, x_n)$, and as usual, define the minimum and maximum order statistics $x_{(1)} = \min \boldsymbol x$, $x_{(n)} = \max \boldsymbol x$, respectively.  Then we have $$f(\boldsymbol x \mid k) = \prod_{i=1}^n \frac{1}{8k} \mathbb 1 (-k \le x_i \le 7k) = (8k)^{-n} \mathbb 1 (-k \le x_{(1)} \le x_{(n)} \le 7k).$$  So by the Factorization theorem, a sufficient statistic is $T(\boldsymbol x) = (x_{(1)}, x_{(n)})$.  This happens to be minimal, but it is not necessary to prove this.  The point of finding $T$ is to show that any MLE of $k$ will be expressible as a function of $T$.

As for the MLE, we have the likelihood
$$\mathcal L(k \mid x_{(1)}, x_{(n)}) \propto k^{-n} \mathbb 1(-k \le x_{(1)} \le x_{(n)} \le 7k).$$  Because $k^{-n}$ is a monotone decreasing function of $k$, $\mathcal L$ is maximized for the smallest possible positive choice of $k$ that keeps the indicator function equal to $1$; that is to say, we require that $k \ge x_{(n)}/7$ and $k \ge -x_{(1)}$.  Equivalently, $$k \ge \max\left(\frac{x_{(n)}}{7}, -x_{(1)}\right).$$  Therefore, the desired MLE is $$\hat k = \max\left(\frac{x_{(n)}}{7}, -x_{(1)}\right).$$

Title: What is MLE of $U[-k,7k]$?


问:
Determinant change with column multiplication
Given an n x n matrix A, what happens to its determinant when each of its columns are multiplied by its column number?

I am teaching myself linear algebra with Elementary Linear Algebra by Anton and Rorres. I don't understand the new set of operations that are permissible for solving determinants. This problem in particular stumped me.

答:
If a row or column of a n x n matrix $ A$ is multiplied by k, then the determinant of the new matrix $B$ is

$$ det(B)=k \text{ } det(A) $$

The proof of the above property comes from the definition of determinant. You can look it up Change in determinant when multiplying row of a matrix

By this property, if each of the columns are multiplied by its column number, then

$$ det(B)= (1 * 2 * 3 *........* n) det(A) $$


问:
A probability game
Motivation: A friend asked me this question.

The Problem: Suppose you start off with a dollar. You flip a fair coin, if it lands on heads you win $50$ cents otherwise you lose $50$ cents. If after $n$ flips you have a nonzero amount of money, you win. What's the probability you win? What about the limiting case as $n$ tends to infinity?

edit: In this game you are not allowed to have negative money. Thanks, Jonathan Fischoff, the linked helped greatly.

答:
Here is a completely probabilistic proof inspired by Moron in Finding a clever solution to a game of chance.

Let p be the probability that I end up with a net loss of 50 cents, then $p = \frac{1}{2} + \frac{1}{2} p^3$. That is to say, I either lose 50 cents, or gain a dollar and lose a dollar and 50 cents eventually. Solving for p I get 1, $(-1-\sqrt{5})/2$, and $(1-\sqrt{5})/2$ and it is not hard to see that we can dismiss the first two. Now the probability I end up with a net loss of one dollar is $p^2 = (3-\sqrt{5})/2$.

Title: A probability game

ContentLicense: CC BY-SA 2.5


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答:
The roots of a continuous function is always a closed subset of $\mathbb{R}$ : $\{0\}$ is closed, thus $f^{-1}(\{0\})$ is closed too.

If you have a closed set $S$, you can define a function $f : x \mapsto d(x,S)$, which is continuous and whose set of roots is exactly $S$ : you can make a continuous function have any closed set as its set of roots.

Therefore you only have to look for closed sets that are uncountable.


答:
Alternatively, you can use Bernoulli's inequality to obtain the desired result. We have:
$\left(1+x^2y^2\right)^{\frac{1}{x^2+y^2}}\ge 1+\dfrac{x^2y^2}{x^2+y^2}=\dfrac{x^2+y^2+x^2y^2}{x^2+y^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\le\dfrac{x^2+y^2}{x^2+y^2+x^2y^2}= h(x,y)$. Also: $x^2+y^2 \ge x^2 \implies \dfrac{1}{x^2+y^2} \le \dfrac{1}{x^2}\implies -\dfrac{1}{x^2+y^2} \ge -\dfrac{1}{x^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\ge \left(1+x^2y^2\right)^{-\frac{1}{x^2}} = \left(\left(1+x^2y^2\right)^{-\frac{1}{x^2y^2}}\right)^{y^2}=g(x,y)$. So: $g(x,y) \le f(x,y) \le h(x,y)$. But both $h,g \to 1$ when $(x,y) \to (0,0)$ because $0 \le \dfrac{x^2y^2}{x^2+y^2} \le x^2$, and thus $\dfrac{x^2y^2}{x^2+y^2} \to 0$. Thus $f(x,y) \to 1 = f(0,0)$, proving continuity of $f$ at $(0,0)$. Note that $g(x,y) \to e^{-0} = 1$ when $(x,y) \to (0,0)$.

Body:
Determine if the following function is continuous on $ (0,0) $
$$f(x,y)=\begin{cases}
 (1+x^2y^2)^{-1/(x^2+y^2)},& \text{ if } (x,y)\neq (0,0) \\
 1,& \text{ if } (x,y)=(0,0)
\end{cases}$$

Conjecturing that $ \lim _ {(x, y) \to (0,0)} f (x, y) = 1 $.

My attempt is,
$$\begin{align*}
0 &\leq | f (x) -L|\\
 &=|(1+x^2y^2)^{-1/(x^2+y^2)}-1|\\ &\leq |(1+x^2y^2)^{-1/(x^2+y^2)}|+1\end{align*}$$

Hence I do not know what else to do because in another limit like this but with the euler function, you can limit that using that the function is strictly increasing, but here you cannot do that because $ (1 + x ^ 2y ^ 2) $ is not strictly increasing , so I don't know how to limit that exponent with the function $ (1 + x ^ 2y ^ 2) $.


问:
Arithmetic mean of subsets of prime numbers. A method that counts the convergence to the golden number Phi.
Based on my research I have my own hypothesis about primes.

The arithmetic mean of each subset of primes that fall between consecutive terms in the Fibonacci sequence, divided by the upper bound of the subset that is the greater Fibonacci number, gives approximately half the golden number Phi.

I have created two applications to find this relationship. One work in graphical mode and shows distribution of primes on Fibonacci spiral, and second works only in text mode.

Screenshots from graphic mode app.

Primes on Fibonacci Spiral 1

Primes on Fibonacci Spiral 2

Results from text mode app.

Some results from my console application: (Half of Phi was doubled for better comparision)

THAT APP IS CALCULATING THE PRIME NUMBERS SUBSETS
BETWEEN FIBONACCI AND FINDING APPROXIMATION____
TO THE GOLDEN NUMBER_PHI._______________________
Author: Sylwester B aka Sylvi91_______________

...

Fib(13) = 233 - Fib(14) = 377
Primes list:
239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373
Primes: Sum = 6989 Qty = 23 Avg = 303
Approximation to the The Golden Number Phi = 1.607427

Fib(14) = 377 - Fib(15) = 610
Primes list:
379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607
Primes: Sum = 18165 Qty = 37 Avg = 490
Approximation to the The Golden Number Phi = 1.606557

Fib(15) = 610 - Fib(16) = 987
Primes: Sum = 43635 Qty = 55 Avg = 793
Approximation to the The Golden Number Phi = 1.606890

Fib(16) = 987 - Fib(17) = 1597
Primes: Sum = 109567 Qty = 85 Avg = 1289
Approximation to the The Golden Number Phi = 1.614277

Fib(19) = 4181 - Fib(20) = 6765
Primes: Sum = 1624217 Qty = 297 Avg = 5468
Approximation to the The Golden Number Phi = 1.616556

Fib(27) = 196418 - Fib(28) = 31781
Primes: Sum = 2499980948 Qty = 9738 Avg = 256724
Approximation to the The Golden Number Phi = 1.615577

and without listing of primes.

Fib(30) = 832040 - Fib(31) = 1346269 SUM = 40227359343 QTY = 36981 AVG = 1087784 Approx. to Phi = 1.6159979915
Fib(31) = 1346269 - Fib(32) = 2178309 SUM = 101931466309 QTY = 57909 AVG = 1760200 Approx. to Phi = 1.6161159872
Fib(32) = 2178309 - Fib(33) = 3524578 SUM = 257870996074 QTY = 90550 AVG = 2847829 Approx. to Phi = 1.6159829631
Fib(33) = 3524578 - Fib(34) = 5702887 SUM = 654441400687 QTY = 142033 AVG = 4607671 Approx. to Phi = 1.6159082233
Fib(34) = 5702887 - Fib(35) = 9227465 SUM = 1661677489343 QTY = 222855 AVG = 7456316 Approx. to Phi = 1.6161136347
Fib(35) = 9227465 - Fib(36) = 14930352 SUM = 4221069024488 QTY = 349862 AVG = 12064954 Approx. to Phi = 1.6161647093
Fib(36) = 14930352 - Fib(37) = 24157817 SUM = 10735109882717 QTY = 549903 AVG = 19521824 Approx. to Phi = 1.6161910656
Fib(37) = 24157817 - Fib(38) = 39088169 SUM = 27324559743219 QTY = 865019 AVG = 31588392 Approx. to Phi = 1.6162635809
Fib(38) = 39088169 - Fib(39) = 63245986 SUM = 69592739215201 QTY = 1361581 AVG = 51111714 Approx. to Phi = 1.6162832531
Fib(39) = 63245986 - Fib(40) = 102334155 SUM = 177416424882449 QTY = 2145191 AVG = 82704255 Approx. to Phi = 1.6163568263
Fib(40) = 102334155 - Fib(41) = 165580141 SUM = 452491851513992 QTY = 3381318 AVG = 133821146 Approx. to Phi = 1.6163912555
Fib(41) = 165580141 - Fib(42) = 267914296 SUM = 1155101203883449 QTY = 5334509 AVG = 216533743 Approx. to Phi = 1.6164403784
Fib(42) = 267914296 - Fib(43) = 433494437 SUM = 2949939781199270 QTY = 8419528 AVG = 350368783 Approx. to Phi = 1.6164857174
Fib(43) = 433494437 - Fib(44) = 701408733 SUM = 7539299522749130 QTY = 13298630 AVG = 566923023 Approx. to Phi = 1.6165268447
Fib(44) = 701408733 - Fib(45) = 1134903170 SUM = 19277443065477372 QTY = 21014892 AVG = 917322966 Approx. to Phi = 1.6165660477
Fib(45) = 1134903170 - Fib(46) = 1836311903 SUM = 49319944945730044 QTY = 33227992 AVG = 1484289058 Approx. to Phi = 1.6165979816
Fib(46) = 1836311903 - Fib(47) = 2971215073 SUM = 126244882484469729 QTY = 52565409 AVG = 2401672219 Approx. to Phi = 1.6166263027
Fib(47) = 2971215073 - Fib(48) = 4807526976 SUM = 323315442948007805 QTY = 83198799 AVG = 3886059015 Approx. to Phi = 1.6166561454
Fib(48) = 4807526976 - Fib(49) = 7778742049 SUM = 828390738449096336 QTY = 131744274 AVG = 6287869015 Approx. to Phi = 1.6166801715
Fib(49) = 7778742049 - Fib(50) = 12586269025 SUM = 2123542722014291259 QTY = 208718785 AVG = 10174181121 Approx. to Phi = 1.6167112114
Fib(50) = 12586269025 - Fib(51) = 20365011074 SUM = 5445740631692717113 QTY = 330797447 AVG = 16462462697 Approx. to Phi = 1.6167398718
Fib(51) = 20365011074 - Fib(52) = 32951280099 SUM = 13971588279376661576 QTY = 524513152 AVG = 26637250612 Approx. to Phi = 1.6167657543
Fib(52) = 32951280099 - Fib(53) = 53316291173 SUM = 35859516892843327174 QTY = 831993816 AVG = 43100701235 Approx. to Phi = 1.6167929271
Fib(53) = 53316291173 - Fib(54) = 86267571272 SUM = 92072211900316816991 QTY = 1320232935 AVG = 69739369060 Approx. to Phi = 1.6168154043
Your calculations took 918.00 seconds to run.
64.5 % from 8GB memory usage.

In both apps I have implemented Sieve of Eratosthenes to find primes and iterative function to calculate Fibonacci numbers.
Using this method let me calculate subset of primes up to the value Fib(54) = 86 267 571 272. But now I facing two problems. First I have reached limits in memory of my home PC [8GB], because Eratosthenes Sieve is memory consumable. Second I have reached limits of 64 bits variables in standard C language. When the variable limit is possible to get around by using, a external library for large numbers, the memory limit does not allow you to move forward too much in the calculations.

I have two questions now.

1.) Is any other way to continue proving my findings? I'd like to check this for primes in between really large numbers, for example Fib (250) = 7896325826131730509282738943634332893686268675876375 and Fib(251) = 12776523572924732586037033894655031898659556447352249 . How to do this?

2.) Is this correct and easy to understand in form of LaTex?

$$
\displaystyle{ \frac12 \Phi   \approx \frac{ \frac{p_{1} + p_{2} + ... + p_{x}}{x}}{   F{n} }}
$$

Where:

$$
\displaystyle{ F{n-1} < (p_{1}, p_{2}, ... , p_{x}) \le F{n}}
$$

$p_{1},p_{2},p_{x}$ - Prime numbers

$(p_{1}+p_{2}+p_{x})$  - sum of Prime numbers in a subset

$Fn-1$ - lesser value of Fibonacci sequence (lower border of primes subset)

$Fn$ - greater value of Fibonacci sequence (higher border of primes subset)

$x$ - qty of prime numbers in a subset

$\Phi$- Golden number Phi = 1.618033988749895...

Thanks in advance. Sorry for mistakes, but english is my second language.

答:
Hint:
The average of the primes between two consecutive Fibonacci numbers $\approx$ the average of the two Fibonacci numbers:

$(987+610)/2/987\cdot 2=(987+610)/987.0= 1.61803444782$


问:
Solving $z^4 + 4i\bar{z} = 0$ in detail
I asked this question here Solving $z^4 + 4i\bar{z} = 0$

Though I accepted the answer, later I found out that I and WA don't agree on the result.

What I did:

Find the absolute value of $z$ by solving $|z^4| = |-4i\bar{z}|$

Getting $|z|^4 = 4|z|$ and hence $|z| = 0$ or $|z| = 4^{\frac{1}{3}}$

Leaving the trivial solution $z=0$ aside, we have $|z| = 4^{\frac{1}{3}}$

Now back to the original equation, multiplying it by $z$ we get:

$z^5 = -4i|z|^2$ and hence $z^5 = -4i4^{\frac{2}{3}}$ hence $z^5 = 4^{\frac{5}{3}}\operatorname{cis}(-\frac{\pi}{2})$

Getting the roots $z_k = 4^{\frac{1}{3}}\operatorname{cis}(\frac{-\frac{\pi}{2} + 2\pi k}{5})$ where $k=0,1,2,3,4$

So we get the multiplication of the complex roots to be:

$z_0\cdot z_1\cdot z_2\cdot z_3\cdot z_4 = 4^{\frac{5}{3}}\operatorname{cis}(\frac{-\pi}{10} + \frac{3\pi}{10} + \frac{7\pi}{10} + \frac{11\pi}{10} + \frac{15\pi}{10}) = 4^{\frac{5}{3}}\operatorname{cis}(\frac{35\pi}{10}) = -4^{\frac{5}{3}}i=-2^{\frac{10}{3}}i$

But when calculating with WA I get $-2^{\frac{5}{2}}i$.

What is the correct result?

答:
Wolframalpha returns the product of the roots of $\displaystyle\frac{z^4 + 4 i z^*}z = 0$ as
$$-10.0794 i,$$ which indeed actually equals the $$-2^{\frac{10}{3}}i$$ that you obtained.

(First dividing the equation by $z$ was to discard the trivial root $z=0,$ as desired.)

Title: Solving $z^4 + 4i\bar{z} = 0$ in detail


问:
What does "express this equation in terms of $x$" mean (eg, for $x+y=2$)?
So I just had this controversial question on my maths exam and it said "express this equation in terms of $x$."

To make it simple, let’s assume that the equation was $x+y=2$.

If we are expressing the equation in terms of $x$, then would it be $y=2-x$ or $x=2-y$?

答:
Expressing an equation in terms $x$ means to express the quantity you're finding in terms of $x$, the variable.

For instance, if you're given an equation: $x + 4 = y - 1$, and asked to express it in terms of $x$;
Then $y = x + 5$, would be the required expression.

Thus for your question, $y = 2 - x$, would be the correct expression.


问:
Probability Pigeonhole Principle

Choose any different $38$ natural numbers less than $1000$. Prove by using the Pigeonhole Principle that among the selected numbers there exists at least two whose difference is at most $26$.

I proved an answer by increasing $26$ to $27$ and multiplying it by $38$. This gives an answer of $1026$, but I can't use the pigeon principle to show it.

答:
RTP: there are at least two pairs whose difference is a maximum of $26$

The difference is maximized when $\frac{1000}{38} \approx 26.315$ so let the set of possible differences be $S_{\text{differences}} = {\{1,2,3,...,26\}}$ .

There are $38$ numbers and only $26$ differences when maximised, so the ceiling function of $\frac{38}{26} =2$

Therefore by PHP, there exists at least two numbers whose difference is at most $26$

Title: Probability Pigeonhole Principle


问:
linear algebra over a division ring vs. over a field
When I was studying linear algebra in the first year, from what I remember, vector spaces were always defined over a field, which was in every single concrete example equal to either $\mathbb{R}$ or $\mathbb{C}$.

In Associative Algebra course, we sometimes mentioned (when talking about $R$-modules) that if $R$ is a division ring, everything becomes trivial and known from linear algebra.

During the summer, I'm planning to revisit my notes of linear algebra, write them in tex, and try to prove as much as possible in a general setting.

Are there any theorems in linear algebra, that hold for vector spaces over a field and not over a division ring? How much linear algebra can be done over a division ring?

Also, what are some examples of division rings, that aren't fields? $\mathbb{H}$ is the only one that comes to mind. I know there aren't any finite ones (Wedderburn). Of course, I'm looking for reasonably nice ones...

答: Anything that involves general homothecies (scalar multiples of the identity viewed as linear operators), or multi-linearity (in particular determinants) will not go through over division rings (or maybe only in some some much weakened form; for instance homothecies by central scalars can still be considered). Anything involving eigenvalues and eigenspaces falls into this category (an eigenspace is a subspace where a linear operator acts as a homothecy). As an example of a theorem that won't be available, the Cayley-Hamilton theorem (indeed the characteristic polynomial of a finite-dimensional linear operator over a division ring is not even defined).


问:
Finding conditional distribution in matching ordering situation
Suppose we draw two values $x_1,x_2$ according to a CDF $F$. Independently, we draw another two values $y_1,y_2$ according to another CDF $G$. Both $F$ and $G$ has support $[0,1]$.

Among those four values, I first observe $x_1$ only. And then, I get to observe one of $y_1$ and $y_2$, depending on whether $x_1\leq x_2$ or $x_1>x_2$ following the rule below:

--- If $x_1\leq x_2$, then I observe $y_1$ or $y_2$, whichever is smaller (or equal to) than the other.

--- If $x_1>x_2$, I observe $y_1$ or $y_2$, whichever is greater than the other.

So, if my $x_1$ is smaller than the other $x$, my observation of $y$ equals the smaller value of $y_1$ and $y_2$. If my $x_1$ is larger than the other $x$, the $y$ I observe is the larger one between $y_1$ and $y_2$.

In this case, if I observe $x_1$ and some value $y$, what is the conditional distribution of the other value of $y$? (so, if $y=y_1$, then what is the distribution of $Y_2$?)

答:
Let $Y$ be whichever of $Y_1, Y_2$ you observe, and let $Z$ be the one that you don't observe.

You want
$$\mathbb{P}(Z \leq z \mid X_1 = x, Y = y).$$

The event $Z \leq z$ either happens with $X_1 \leq X_2$ or with $X_1 > X_2$, so the probability we want is equal to

$$\mathbb{P}(Z \leq z \ \wedge X_1 \leq X_2 \mid X_1 = x, Y = y) + \mathbb{P}(Z \leq z \ \wedge X_1 > X_2 \mid X_1 = x, Y = y).$$

In each case we know whether $Y, Z$ are the maximum or minimum of $Y_1, Y_2$, so the probability equals

$$\mathbb{P}(\max(Y_1, Y_2) \leq z \ \wedge X_1 \leq X_2 \mid X_1 = x, \min(Y_1, Y_2) = y) + \\ \mathbb{P}(\min(Y_1, Y_2) \leq z \ \wedge X_1 > X_2 \mid X_1 = x, \max(Y_1, Y_2) = y).$$

The $X_1, X_2$ variables are independent of the $Y_1, Y_2$ random variables so we can split up the conditional probabilities as

$$\mathbb{P}(X_1 \leq X_2 \mid X_1 = x)\mathbb{P}(\max(Y_1, Y_2) \leq z \mid \min(Y_1, Y_2) = y) + \\ \mathbb{P}(X_1 > X_2 \mid X_1 = x)\mathbb{P}(\min(Y_1, Y_2) \leq z \mid \max(Y_1, Y_2) = y).$$

Calculating the $X_1, X_2$ terms is straightforward assuming that there is no specific point $c$ with $\mathbb{P}(X_2 = c) > 0$:

$$\mathbb{P}(X_1 \leq X_2 \mid X_1 = x) = \mathbb{P}(x \leq X_2) = 1 - F(x), \textrm{ and } \ \mathbb{P}(X_1 > X_2 \mid X_1 = x) = F(x).$$

By symmetry (see note at the end for more detail),
\begin{align} \mathbb{P}(\max(Y_1, Y_2) \leq z \mid \min(Y_1, Y_2) = y) &= \mathbb{P}(Y_2 \leq z \mid Y_1 = y, Y_1 = \min(Y_1, Y_2)) \\\\ &= \mathbb{P}(Y_2 \leq z \mid Y_2 \geq y) \\\\ &= \frac{\mathbb{P}(y \leq Y_2 \leq z)}{\mathbb{P}(Y_2 \geq y)} \\\\ &= \max\left(0, \frac{G(z) - G(y)}{1 - G(y)}\right). \end{align}

The other term can be calculated as:
\begin{align} \mathbb{P}(\min(Y_1, Y_2) \leq z \mid \max(Y_1, Y_2) = y) &= \mathbb{P}(Y_2 \leq z \mid Y_1 = y, Y_1 = \max(Y_1, Y_2)) \\\\ &= \mathbb{P}(Y_2 \leq z \mid Y_2 \leq y) \\\\ &= \min\left(1, \frac{\mathbb{P}(Y_2 \leq z)}{\mathbb{P}(Y_2 \leq y)}\right) \\\\ &= \min\left(1, \frac{G(z)}{G(y)}\right). \end{align}

Putting all this together we have

$$\mathbb{P}(Z \leq z \mid X_1 = x, Y = y) = (1 - F(x))\max\left(0, \frac{G(z) - G(y)}{1 - G(y)}\right) + F(x)\min\left(1, \frac{G(z)}{G(y)}\right).$$

The $Y_1, Y_2$ terms can be solved by using
\begin{align} \mathbb{P}(\max(Y_1, Y_2) \leq z \mid \min(Y_1, Y_2) = y) &= \mathbb{P}(Y_1 \leq z \ \wedge \ Y_2 \leq z \mid \min(Y_1, Y_2) = y) \\\\ &= \mathbb{P}(Y_1 = \min(Y_1, Y_2))\mathbb{P}(Y_2 \leq z \mid Y_1 = y, Y_1 = \min(Y_1, Y_2)) \ + \\\\ &\mathbb{P}(Y_2 = \min(Y_1, Y_2))\mathbb{P}(Y_1 \leq z \mid Y_2 = y, Y_2 = \min(Y_1, Y_2)). \end{align}

Since $\mathbb{P}(Y_1 = \min(Y_1, Y_2)) = \mathbb{P}(Y_2 = \min(Y_1, Y_2)) = 1/2$, and, by symmetry, $\mathbb{P}(Y_2 \leq z \mid Y_1 = y, Y_1 = \min(Y_1, Y_2))$ equals $\mathbb{P}(Y_1 \leq z \mid Y_2 = y, Y_2 = \min(Y_1, Y_2))$, we have

$$\mathbb{P}(\max(Y_1, Y_2) \leq z \mid \min(Y_1, Y_2) = y) = \mathbb{P}(Y_2 \leq z \mid Y_1 = y, Y_1 = \min(Y_1, Y_2)).$$


问:
Formal statement that an analytic function injective on the boundary is injective on the closure
I have been told that such a result guaranteeing injectivity on the closure of the domain exists, however I have not been able to find a formal statement of this result. If someone could provide I would be very grateful.</p>

答:
The claims seems to be wrong (or at least needs a more careful formulation than what I read from the OP).

Start with the unit disk $\Bbb D$ and apply the map $f\colon z\mapsto (z+\sqrt 2)^4$, which turns $\Bbb D$ into a domain $\Omega$ that is disk-ish, but its boundary is eight-ish, i.e., the two points $-\frac{\sqrt 2}2\pm \frac{i\sqrt 2}2$ of $\partial \Bbb D$ both end up as $-1\in\partial \Omega$.

Now on $\overline{\Bbb D}$, consider $g\colon z\mapsto (z+\frac{\sqrt2}2)^2$. This is not injective in $\Bbb D$, but it is injective in $\partial \Bbb D$ with the sole exception that $g(-\frac{\sqrt 2}2+ \frac{i\sqrt 2}2)=g(-\frac{\sqrt 2}2- \frac{i\sqrt 2}2)$ (the only points where $S^1+\frac{\sqrt 2}2$ intersects $-(S^1+\frac{\sqrt 2}2)=S^1-\frac{\sqrt 2}2$). Now on $\overline\Omega$, the map $g\circ f^{-1}$ is well-defined, is continuous, is analytic in $\Omega$, is injective on $\partial \Omega$, but is not injective in $\Omega$.


问:
Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$

Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$

$$$$
Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$
I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$
I would be grateful for any help. Many thanks!

Body:
I decided to do @mathlove's case $2$.

CASE $2$

We see that $n \gt 0$.

$$\dfrac{1}{n} + \dfrac{1}{2n + 1}= \delta + \frac 13$$

$$\dfrac{3n+1}{2n^2+n} - \dfrac 13  = \delta$$

$$ \delta = \frac{-2n^2 + 8n + 3}{6n^2 + 3n}$$

\begin{align}
   0 \le \frac{-2n^2 + 8n + 3}{6n^2 + 3n} \lt 1 \\
   0 \le -2n^2 + 8n + 3 \lt 6n^2 + 3n \\
\end{align}

\begin{align}
   0 &\le -2n^2 + 8n + 3 \\
   2n^2 - 8n - 3 &\le 0 \\
   n^2 - 4n &\le \frac 32 \\
   (n - 2)^2 &\le \frac{11}{2} \\
   (n - 2)^2 &\le 4 \\
   n &\in \{1,2,3,4\}
\end{align}

\begin{align}
   -2n^2 + 8n + 3 &\lt 6n^2 + 3n \\
   8n^2 - 5n - 3 &\gt 0 \\
   256n^2 - 160n &\gt 96 \\
   (16n - 5)^2 &\gt 96 \\
   16n - 5 &\gt 10 \\
   16n &\gt 15 \\
   n &\ge 1
\end{align}

$n= -1 \implies \delta = -\frac 73$

$n=1 \implies \delta = 1$

$n=2 \implies \delta = \frac{11}{30} \implies x = 2\frac{11}{30}$

$n=3 \implies \delta = \frac 17 \implies x = 3\frac 17$

$n=4 \implies \delta = \frac{1}{36} \implies x = 1\frac{1}{36}$

$n=5 \implies \delta = - \frac{7}{165}$

So $n \in \{ 2\frac{11}{30}, 3\frac 17,  1\frac{1}{36}\}$


问:
Why is it enough for a Tangent Plane to only agree with the slope of an $m$-input function in $m$ directions?
Thanks for reading.

Short Version

Why is it enough for the tangent plane to only agree with the slope of a differentiable 2-input function in the $x$ and $y$ directions at a certain point for it to agree in every direction at that point?

Isn’t it still possible for the function to cross the tangent plane, even if their slopes agree in the $x$ an $y$ directions, when we move in some other direction?

Long Version

Say we have some two-input differentiable function $f(x,y)$ defining some surface in 3D space. The input into the function is $(x,y)$ coordinates, and the output of the function is the height of the surface (or hill) at each input point.

Consider some point $(x_0,y_0)$ on this surface. At this point, the surface will have an infinite number of slopes, since there are an infinite number of directions we can move in. However, for two of those directions, the slopes are easy to find:

The slope is $\frac{\partial f}{\partial x}$ in the $x$ direction.

The slope is $\frac{\partial f}{\partial y}$ in the $y$ direction.

For a plane to be tangent to a function at $(x_0,y_0)$, it means it only touches the surface at that point, and doesn't cross the surface anywhere.

When we find a tangent plane to the surface at $(x_0,y_0)$, its enough to make the slope of the plane agree with the slope of the surface in 2 directions. Usually the $x$ and $y$ directions.

I'm aware that two slopes, one in the $x$ and one in the $y$, uniquely define a single plane at that point.

However, there are an infinite number of functions that may have those slopes at that point.

Tangency implies that the slope of the tangent plane agrees with the slope of the function at that point in every direction, not just the $x$ and $y$ directions.

If it didn't agree with the slope of the function at that point in some arbitrary direction, then were we to draw lines in that direction on both the "tangent" (not actually tangent) plane and the function, those lines would intersect, and thus the plane wouldn't be brushing the surface at only a single point, and it wouldn't be a tangent plane.

Why is it enough for the tangent plane to only agree with the slope of a differentiable 2-input function in the $x$ and $y$ directions at a certain point for it to agree in every direction at that point?

Isn’t it still possible for the function to cross the tangent plane, even if their slopes agree in the $x$ an $y$ directions, when we move in some other direction?

I'd like to be able to see the answer intuitively...but I can't, and I need help.

Thank you!

Edit:

I added this as a comment to David's answer below, but I'll add it as part of the question too, as I think it may help communicate what I'm trying to "see".

Take the function $f(x)=x^2$. That's a parabola in 2D space.

If a tangent line agrees with the slope $2x$ at a certain point...well, there's only one slope to agree with, so I KNOW that the line will be tangent.

But, now take the surface $f(x,y)=x^2$. That's a surface in 3D space.

A tangent plane at a certain point must have a slope of $2x$ in the $x$ direction and $0$ in the $y$ direction

Lines drawn on a certain "tangent" plane in those two directions will be tangent to the surface at that point, since the slopes agree in those two directions. We used those directions (the $x$ and $y$ directions) to define the plane.

But, how can I intuitively "see" that lines drawn on the plane in ANY direction will be tangent to the function at that point?

Edit 2

As someone pointed out in the comments, a tangent line doesn't not crossing the function only applies if the function is convex (or concave) like $x^2$. Here, let me draw out the function, and draw a tangent line to $f(x)=x^2$.

The tangent line (blue) agrees with the slope of the function (red) in the one direction it has to agree with the slope in (the $x$ direction).

Therefor, I know that the tangent line won't cross the parabola. If it didn't agree with the slope, then it would cross it either when we moved a little bit towards the right from the point of tangency, or a little bit towards the left.

However, lets say we draw a tangent plane the graph of $f(x,y)=x^2$, with a tangent plane drawn to it.

Although I know lines on the plane won't cross the surface when moving in the $x$ direction or in the $y$ direction (the two white tangent lines won't cross the $x^2$ surface) since the plane agrees with the slope of the surface in those two directions, how can I know that if I draw lines on the plane in some other arbitrary direction, those lines also won't cross the surface?

Why was the slope agreeing in the $x$ and $y$ directions enough to define the tangent plane?

Thanks!

Final Thoughts:

After watching some of Ted's (in the comments) lectures on YouTube, I've realized that when I asked this question, although I had an intuitive "feel" for differentiability in one dimension, I hadn't really thought enough about what it meant in higher dimensions.

Differentiability (for a two dimensional surface) means that the function is locally flat.

Say that $T(x ⃗ )$ is the tangent plane to a function $f(x ⃗ )$ at a point $a ⃗$.

Therefor, it must agree with the slopes of $f(x ⃗ )$ in every direction at $a ⃗$, not just the $m$ Cartesian directions, for which we have the partial derivatives of $f(x ⃗ )$.

However, when we define $T(x ⃗ )$, we only make it agree with the slopes of $f(x ⃗ )$ in the $m$ Cartesian directions.

$$T(x ⃗ )=\frac{∂f}{∂x_0} (x_0-a_0 )+\frac{∂f}{∂x_1}(x_1-a_1 )+⋯+\frac{∂f}{∂x_m}(x_m-a_m )$$

(Each of the $x_i$'s are orthogonal Cartesian coordinates, they're the components of the input vector $\vec{x}$, and each of the partial derivatives above are evaluated at $\vec{a}$).

This is because given the $m$ partial derivatives of $f$ at $\vec{a}$, there’s only one unique hyper-plane approximation for $f(x ⃗ )$ which can be defined!!

That is, the $m$ partial derivatives of $f(x ⃗ )$ at $a ⃗$ define a unique hyper-plane. There is no other hyper-plane that has $m$ slopes and passes through $(\vec{a}, f(\vec{a}))$ in $(m+1)$ dimensional space ($m+1$ because we're adding one more dimension for the function's output itself).

My original question was why this implied that the tangent plane agreed with the slopes in every direction.

The answer?

It doesn't!

It's entirely possible for this "tangent" (in quotations because in this case, it's not actually tangent) to agree with the slopes in the $m$ Cartesian directions, but not in every direction (Ted has some good examples in the lectures referenced in the comments, and here's a picture from "Math-Insight").

However, if that unique “tangent” plane doesn’t agree with the function’s slopes in every direction (not only the Cartesian directions) that just means that the function isn’t differentiable at $a ⃗$!!! In other words, at $\vec{a}$ the function isn't locally flat.

To prove that a function is differentiable at $\vec{a}$, we attempt to build a "tangent" plane at that point by making it agree with the slope of our function in the $m$ Cartesian directions, and then show that the "tangent plane" indeed is a tangent-plane by showing that...

$$\mathrm{lim_{(|dx ⃗ |→0)}}(\frac{[(f(a ⃗+dx ⃗ )-f(a ⃗ )]-[T(a ⃗+dx ⃗ )-T(a ⃗ )])}{|dx ⃗ |})=0$$

...regardless of the direction in which we move from $\vec{a}$ (regardless of the direction of $\vec{dx}$).

In his videos, Ted shows an example of this (with the difference that in his example, $T(\vec{x})$ is not a tangent plane but a linear approximation passing through the origin, although the idea is the same.)

In a nutshell, to answer my original question, if the function is differentiable, then it’s enough for the tangent plane to agree with the slope of the function in the $m$ Cartesian directions for it to agree in every direction, because $m$ slopes define a unique tangent-plane in $m+1$ dimensional space, and that the tangent plane agrees with the slopes of the function in every direction is just the definition of differentiability - that the function is locally flat.

And to show that a function indeed is differentiable, we have to show that a potential unique "tangent" (in quotes because it may not be a tangent plane if the function isn't differentiable) plane to the function at a certain point that agrees with the slope of the function in the $m$ Cartesian directions agrees with the slope of the function in every direction, so that the function indeed is locally flat.

Thank you!

答:
I have the same doubts as you. I don't know what kind of books to get the experience of its development.Intuitively, how to judge a tangent has only one intersection point with the function. Then how to judge whether the plane is also applicable. If the surface is regarded as the deformation of the tangent plane from a certain point, it can be confirmed that there is only one point in contact with the surface.In other words, there is only one intersection point between any straight line at this point on the plane and the curve in a certain direction of the surface.Can I state that my minimum acceptance is a tangent plane (a plane)?

I know that the definition of directional derivative exists in the total differential of a multivariate function, but the total differential is like a tangent plane.But the question is why the total differential can use a tangent plane to express small changes in all directions.I can't answer your doubts, I defaulted to the tangent plane as an approximation.

The total differential of a multivariate function$dz=f_x\cdot dx+f_y\cdot dy$,I parameterize the function$f[x(t),y(t)]$,Where $\vec{u}=<u_1,u_2>,|\vec{u}|=1;x(t)=x_0+t \vec{u_1},y(t)=y_0+t \vec{u_2}$

$\Delta z=f_x\cdot x'(t) \Delta t+f_y\cdot y'(t) \Delta t$ $lim_{\Delta t \to 0}=\frac{f_x\cdot x'(t) \Delta t+f_y\cdot y'(t) \Delta t}{\Delta t}=f_x\cdot x'(t) +f_y\cdot y'(t) $ Note that $x'(t)=\vec{u_1},y'(t)=\vec{u_2}$

Sorry, I don’t know how to zoom in the formula. It looks crowded


问:
Continuous function iff $p \not\in fr(A)$
Let $M$ be a metric space, $A \subset M$ and let $\chi_A: M \to \mathbb{R}$ be a map defined by:
$$\chi_A(x)=
\begin{cases}
1 & \text{if $x \in A$} \\
0 & \text{if $x \not\in A$}
\end{cases}$$
For a $p \in M$, show that $\chi_A$ is continuous map in $p$ if and only if $p \not\in fr(A)$.

My idea was supposing that $p \in fr(A)$ and try to get a contradiction, but I couldn't reach a solid result. Any leads?

答:
$M$ splits into $3$ disjoint sets: $M=\operatorname{int}(A) \cup \operatorname{int}(A^\complement) \cup \operatorname{Fr}(A)$, based on whether $x$ has a ball inside $A$, a ball inside $A^\complement$, or whether every ball around $x$ intersects both $A$ and $A^\complement$.

In the former two cases there is one $\delta$ that works for all $\varepsilon>0$ (the function is locally constant). In the latter no $\delta>0$ can work for $\varepsilon = \frac12$.

So $\chi_A$ is exactly non-continuous at $x \in \operatorname{Fr}(A)$.


问:
Dimension of $\operatorname{Hom}(V, W)$
What is the dimension of $\operatorname{Hom}(V, W)$ if at least one of the two vector spaces $V, W$ is infinite dimensional? In the sense of cardinal numbers.

Thanks

答:
It depends ;-). If either of $V$ or $W$ is zero-dimensional, we will have $\def\Hom{\mathop{\mathrm{Hom}}}\Hom(V,W) = 0$, regardless of the dimension of the other space. If both of $V$ and $W$ are at least one-dimensional, and one of them is infinite dimensional, $\Hom(V,W)$ is infinite dimensional also:

So suppose $\dim V = \infty$, and $\dim W  \ge 1$ first, let $\{v_n \mid n \in \mathbb N\}$ be a linear independent set in $V$, $w_0 \in W \setminus \{0\}$. Then there are linear maps $f_n \in \Hom(V,W)$ such that
$$ f_n(v_m) = \delta_{nm}w_0, \quad n,m \in \mathbb N $$
Then $\{f_n \mid n \in \mathbb N\}$ is linear independent, as: Suppose $\sum\alpha_n f_n  = 0$ with only finitely many $ \alpha_n\ne 0$. Applying this to $v_m$ gives
$$ \alpha_m = \sum \alpha_n f_n(v_m) = 0(v_m) = 0. $$
So $\dim\Hom(V,W) = \infty$.

Suppose now $\dim V \ge 1$, $\dim W = \infty$, let $v_0 \in V \setminus \{0\}$, and $\{w_n \mid n \in \mathbb N\}$ be linear independent in $W$. There are linear maps $g_n \in \Hom(V,W)$ such that
$$ g_n(v_0) = w_n, \quad n \in \mathbb N $$
Then $\{g_n \mid n \in \mathbb N\}$ is linear independent: Suppose $\sum\alpha_n g_n = 0$. Applying this to $v_0$ gives
$$ 0 = 0(v_0) = \sum\alpha_n g_n(v_0) = \sum \alpha_n w_n $$
as the $w_n$ are independent, we have $\alpha_n = 0$ for all $n$. So $\dim\Hom(V,W) = \infty$.

Title: Dimension of $\operatorname{Hom}(V, W)$


答:
Consider that for any $X_i$ you have that

$$-k\leq X_i\leq 7k$$

thus

$$\begin{cases}
-k\leq X_{(1)} \\
7k\geq X_{(n)}
\end{cases}$$

Thus

$$k\geq \max\left(-X_{(1)};\frac{X_{(n)}}{7}\right)$$

Given that the likelihood ($1/(8k)^n$) is strictly decreasing in $k$

$$T=\max\left(-X_{(1)};\frac{X_{(n)}}{7}\right)$$

is MLE

Title: What is MLE of $U[-k,7k]$?

Body:
I need to find maximum likelihood estimator of uniform distribution $X : U[-k,7k]$.
I think that the answer is $max\{x_1, ..., x_n \}$ . However, I think I have a mistake because if $X : U[0,k]$ the answer would be the same. So, it seems strange to me why MLE does not depend on the length of the interval.


问:
A reference for Nielsen's First Generalized Polynomial Expansion
In Henry Gould's Combinatorical identities the following appears under the name Nielsen's First Generalized Polynomial Expansion: Let $f(x)=\sum_{k=0}^na_k x^k$ be a polynomial of degree $n$ and let $n\leq m$. Then

$$f(x+y)=(-1)^{m}\sum_{k=0}^{m+1}\binom{x+k-1}{m}\sum_{j=0}^{k}(-1)^{j}\binom{m+1}{j}f(j-k+y).$$ It took me some time to understand that Nielsen refers to Niels Nielsen (and not to Jakob Nielsen).

Can someone please state the exact reference where this identity is taken from?

答:
For future reference, I have found everything I wanted and even more in the book
Combinatorial Identities for Stirling Numbers by Jocelyn Quaintance and H. W. Gould.

The identity may be derived from results by Nielsen that goes back at least to
Trait´e ´el´ementaire des nombres de Bernoulli by N. Nielsen, 1923.


问:
Let S be the set of all triangles in the xy plane, each having one vertex at the origin and the other two vertices lie on the coordinate..

Let S be the set of all triangles in the xy plane, each having one vertex at the origin and the other two vertices lie on the coordinate axes with integral coordinates. If each triangle in S has area 50sq unit, then number of elements in set S is?

Let the coordinates of the other two vertices be $(a,0)$ and $(0,b)$

Clearly
$$\frac {|a||b|}{2}=50$$
$$|a||b|=100$$

Let’s consider both $a,b$ as positive for now

$$100=2^2\times 5^2$$

And consider an order pair $(a,b)$

So number of such order pairs will be
$$\frac{4!}{2!2!}$$
$$=6$$

This was first quadrant. So for 4 quadrants, answer will be $24$, but given is 36. Which combinations am I missing?

Note: manually counting cases gives 9 possibilities, which gives the right answer, but I want to use a formula to solve

答:
$100 = 2^2 \cdot 5^2$.

So possible integer values of $a$ comes from three choices for factor $2$ - either factor $2$ is not present, multiplied one time or multiplied two times $(2^0, 2^1$ or $ 2^2)$. We have same three choices for $5$. We also observe as we choose $a$, $b$ is automatically decided and so we obtain ordered pair $(a, b)$.

That gives us total of $3\cdot 3 = 9$ combinations in one quadrant. We multiply by $4$ to obtain $36$ possible combinations in total across all $4$ quadrants.


问:
Log notation convention
I am studying number theory for cryptography and I've got confused by the log notation that they used.

In one video, prof Jonathan Kat refers to the length of a number as the number of bits that are needed to represent that number. To denote the length of a number two vertical bars surrounding the number are used and that is considered to be equal to the logarithm of the number magnitude. See below the expression:

$$\|a\|=O(\log a); a=2^{\|a\|} $$

My question is regarding the notation used for that logarithm. I have always thought that when a log appears without subscript was referring to $\log_{10}$, however in the above example is referring to $\log_{2}$.  Then I googled and I found the following:

$x = \log y$  often means  $x = \log_e y$  in mathematics texts.

$x = \log y$  often means  $x = \log_{10} y$  in science and engineering
texts.

$x = \log y$  often means  $x = \log_2 y$  in computer science texts.

How often the above statement are met and why the use of different notation between fields?

答:
Computer scientists, information theorists etc. have three choices: just use $\log$, use $\log_2$, and use any of several binary-logarithm symbols ($\operatorname{lb},\,\operatorname{ld},\,\operatorname{lg}$) and hope the reader knows theirs. For them, logarithms are often only important to within a $O(1)$ factor due to the prevalence of big-$O$ notation, and then the base isn't important enough to motivate the second or third option. In contexts where the specific asymptotic behaviour $\log_2n$ is worth mentioning and not obvious (and it would pretty much always be obvious), how often to mention that logarithms are base $2$ is up to the author, but they're likely to do it once rather than in every equation.

For mathematicians, unless you're dealing with things like "when $\log_ab$ is transcendental", or proving how different bases' logarithms are related in the first place, a logarithm will be natural. Any other base is too arbitrary and unhelpful. It's important, however, to understand the usage differences between $\ln$, $\log$ and $\operatorname{Log}$ in complex analysis.

Even in science and engineering, natural logarithms will be common because of all the calculus needed. I'm not sure base-$10$ logarithms would be used all that often: even when orders of magnitude are discussed, $x=10^y$ will probably come up rather than $y=\log_{10}x$. They might be used on log-axes, but hopefully they will either read $\log_{10}$, or be stated as base-$10$ at least once in the text, e.g. with the convention "$\ln$ for base $e$, $\log$ for base $10$".

Title: Log notation convention


问:
Solve system of equations involving cos and sin
I have come up with the following system, I want to solve it for $a$ and $c$:

$
a \sin (x_0) - c \sin(x_0 - L) = 0\\
c \cos(x_0 - L) - a \cos(x_0) = 1
$

In this system $x_0$ and $L$ are arbitrary.

p.s. why can't $L$ be $n\pi$?

答:
I assume you are solving for $a$ and $c$.  Then the solution may be written in matrix form as

$$\begin{align}\left ( \begin{array}\\a\\c\end{array}\right) &= \frac{1}{\sin{x_0} \cos{(x_0-L)} - \cos{x_0} \sin{(x_0-L)}} \left ( \begin{array}\\\cos{(x_0-L)} & \sin{(x_0-L)}\\\cos{x_0} & \sin{x_0}\end{array}\right)\left ( \begin{array}\\0\\1\end{array}\right) \\ &= \frac{1}{\sin{L}}\left ( \begin{array}\\\sin{(x_0-L)}\\\sin{x_0}\end{array}\right)\end{align}$$

Note that I used the sine addition formula to simplify the denominator term.  This explains why $L \ne n \pi$: the denominator would be zero.

Then

$$a = \frac{\sin{(x_0-L)}}{\sin{L}}$$

$$c = \frac{\sin{x_0}}{\sin{L}}$$


问:
Upper bound of rational numbers in Thomae's function
I'm attempting to prove the Riemann integrability of Thomae's function, defined by:

$$
f(x) =\begin{cases}
1 ,& x=0\\
1/n,& x = m/n \in\mathbb{Q} \cap (0, 1], \text{written in lowest terms with $n > 0$}\\
0,& x\in \mathbb{Q}^c \cap [0, 1]
\end{cases}
$$

at the current step, I'm asked to find the upper bound for the number of elements in $E_n = \{x \in[0, 1] : f(x) \geq 1/n\}$ for natural number $n$

By Euler's Totient function (and intuition i suppose) that there are at most $n-1$ numbers relatively prime (and less than) to $n$. Since $|E_1|=1, |E_2|=2,\ldots$ I suppose $|E_{n+1}|\leq |E_n|+n$, or that $|E_n|\leq \frac{1}{2}(n-1)n-1$. This is correct?

答: There is no such thing as the upper bound of $E_n$; each bounded set has infinitely many upper bounds. But it is clear that an upper bound of $E_n$ is $\frac{(n+1)(n+2)}2-1$. This follows from the fact that$$E_n=\bigcup_{m=1}^n\left\{\frac km\,\middle|\,k\in\{0,1,\ldots,m\}\right\}.$$


问:
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{e^{-1/(x^2+y^2)}}{x^4+y^4}=0$.
Study the continuity of the following function at the point $ (0,0) $.
$$f(x,y)=\begin{cases}
\dfrac{e^{-1/(x^2+y^2)}}{x^4+y^4} & \text{ if } (x,y)\neq (0,0) \\
 0& \text{ if } (x,y)=(0,0)
\end{cases}$$

If I take trajectories, it can be conjectured that the limit of said function is $ 0 $. But when limiting the function $ e ^ {- 1 / (x ^ 2 + y ^ 2)} $ I don't know how to do it, any help to limit that function in such a way that the limit is $ 0 $.

Body:
Note that
$$ x^4 + y^4 = \frac12(x^2-y^2)^2 + \frac12(x^2+y^2)^2 \ge \frac12(x^2+y^2)^2$$
so
$$ \frac{e^{-1/(x^2+y^2)}}{x^4 + y^4} \le 2 \frac{e^{-1/(x^2+y^2)}}{(x^2 + y^2)^2} $$
Since $\lim_{(x,y)\to(0,0)}(x^2+y^2) = 0$, $x^2+y^2>0$ it is enough to show that
$$ \lim_{u\to 0_+} \frac{e^{-1/u}}{u^2} = 0$$
or
$$ \lim_{z\to +\infty} z^2 e^{-z} = 0$$
Do you know how to show this?


问:
Continuity of function $f(x)=2^{-\frac{1}{x^{2}}}$
I need help with this task, if anyone had a similar problem it would help me !

The task is: Determine the type of interruption at the point x = 0 for the function

$$f(x)=2^{-\frac{1}{x^{2}}}$$

I did:

$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$L=R=x=0$$

And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?

Thanks in advance !

答: The real function $x\mapsto 2^{-x^{-2}}$ has a removable discontinuity at $0$, there is no doubt about it.


答: You have a removable discontinuity at $x = 0$.

Title: Continuity of function $f(x)=2^{-\frac{1}{x^{2}}}$

Body:
I need help with this task, if anyone had a similar problem it would help me !

The task is: Determine the type of interruption at the point x = 0 for the function

$$f(x)=2^{-\frac{1}{x^{2}}}$$

I did:

$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$
$$L=R=x=0$$

And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?

Thanks in advance !


问:
System of equations in x and y
Solve for $x, y \in \mathbb{R} $

$$ 5x \left(1+\frac{1}{x^2+y^2}\right) =12$$
$$ 5y \left(1-\frac{1}{x^2+y^2}\right) =4$$

I need a Different Approach apart from what i posted..Thank You

答:
with $x,y>0$,we have the equivalevent system
$$\begin{cases}
\dfrac{1}{x^2+y^2}=\dfrac{6}{5x}-\dfrac{2}{5y}\\
1=\dfrac{6}{5x}+\dfrac{2}{5y}
\end{cases}$$

so $$\dfrac{1}{x^2+y^2}=\dfrac{36}{25x^2}-\dfrac{4}{25y^2}$$
$$\Longrightarrow x=2y$$

Title: System of equations in x and y


问:
Multiplication by a smooth compactly supported function in homogeneous Sobolev spaces
Let $f\in C_c^\infty$ and $g\in \dot{H}^s$, $s>1/2$. Is it true that $\|fg\|_{\dot{H}^s}\le c\|g\|_{\dot{H}^s}$? I think this is true for inhomogeneous Sobolev spaces but I wonder whether the same holds for homogeneous Sobolev spaces. I did not find a reference to this result.

答:
This is also true for homogeneous $H^s$ space. For simplicity look at the case $s=1$, you have, denoting $Q$ the support of function $f$ :

$$||fg||^2_{H^1(\mathbb{R}^d)} = \int_Q |\nabla f|^2 g^2 + |\nabla g|^2 f^2 + f^2 g^2 \ \mathrm{d}x$$
then you have

\begin{align*}
||fg||^2_{H^1} & \leq (C_2 +C_1) \int_{Q} g^2 \ \mathrm{d}x  + C_2 \int_Q |\nabla g|^2 \ \mathrm{d}x\\
& \leq (C_2+C_1) (\int_{\mathbb{R^d}} g^2 \ \mathrm{d}x  + \int_{\mathbb{R^d}} |\nabla g|^2 \ \mathrm{d}x)\\
& \leq C ||g||_{H^1(\mathbb{R}^d)}
\end{align*}
where I denoted $C_1$ the max of $|\nabla f|^2$ on $Q$ and $C_2$ the max of $f^2$ on $Q$.