问:
Positive bit count of the $S(n):=\sum_{i=1}^{n} 2^{(i-1)}$ always $n$?
From a discussion on a stack about the programing, I got the following mathematical Conjectures:

Conjectures:

Let n,m be positive integer, $S(n)$ and $CountPosbit(m)$ be as follows.

$S(n):=\sum_{i=1}^{n} 2^{(i-1)}$

$CountPosbit(m):=$Number of positive bits when m is converted to binary.

Then, the following holds;

(1)$CountPosbit(S(n))=n$  for all positive integer, $n$,

(2)For all positive integer, $m$, following holds;

$CountPosbit(m)=n$  ⇒ $m \ge S(n)$

The following example is to clarify the definition of the $CountPosbit (m)$;

If $m=10$ in the decimal notation, then

the binary notation of m is 1010 therefore,

$CountPosbit(10)=2$

If $m=15$ in the decimal notation. then

the binary notation of m is 1111 therefore,

$CountPosbit(15)=4$

For the more detail of the definition of $CountPosbit$ function, refer to the Excel VBA function in Box2 below.

My questions are as follows:

My questions:

(Q1)Are the above conjectures (1) and (2) correct?

(Q2)If these are correct, please prove them.

The results of trying with the Excel VBA functions of Box 1 and 2 below are as follows. As far as I tried, no counterexample was found.
The worksheet can be downloaded from here.

Box1: Excel VBA function for calculating S (n)

```<code>Function S_n(n)
S_n = 0
For i = 1 To n
 S_n = S_n + (2 ^ (i-1))
Next i

End Function
</code>```

Box2: Excel VBA function for CountPosbit(m)

```<code>Function CountPosbit(m)
'Substituting a decimal number, this function counts the number of positive bits when the decimal number is converted to a binary number.

tmp = Convert10to2(m)
tot = Len(tmp)
cnt = 0

For i = 1 To tot
If Mid(tmp, tot - i + 1, 1) = 1 Then cnt = cnt + 1
Next i

CountPosbit = cnt
End Function

Function Convert10to2(Value) As String
'This function is for converting decimal numbers to binary numbers.
'This function was quoted from the following website.
'https://www.umayadia.com/main/Samples/Sample055Convert10and2.htm

    Dim lngBit As Long
    Dim strData As String
    Do Until (Value < 2 ^ lngBit)
        If (Value And 2 ^ lngBit) <> 0 Then
            strData = "1" & strData
        Else
            strData = "0" & strData
        End If
        lngBit = lngBit + 1
    Loop
    Convert10to2 = strData
End Function
</code>```

答: Yes. $S(n)$ is the sum of a geometric progression.  $$S(n)=2^n-1$$  The binary representation of $S(n)$ is $n$ $1$-bits, since the binary representation of $2^n$ is a $1$ followed by $n$ $0$'s, and subtracting $1$ gives $n$ $1$-bits by the usual algorithm.


问:
Complexity of Code Snippet Without Knowing A Function?
I have the code snippet:

```<code>int const n = 300;
int nArr[n];
for(int i = 0; i<n; i++) {
   if(i >x)
     copyPrevious(nArr,i);
}
</code>```

I need to find the complexity for the cases when x = 300 and x = 5.

I have already reasoned that the x = 300 case will give O(n) complexity, based on n + 2 assignments, since it never enters the if statement.

However, for 5, I am a little confused. This time, the if statement is relevant, and I don't know how to handle it, since I don't know the particulars of the copyPrevious function. The only thing I know about it is that it itself has complexity O(n).

My usual approach is to make a table of the different parts of the code and the assignments that occur inside and outside the loops. Obviously, the for loop assigns i++ a total of n times, but what of the copyPrevious function? How do I treat it in this manner when I know only a particular x and its complexity?

Any help would be greatly appreciated!

答:
Assume that copyPrevious(arr,i) has a running time of $O(i)=Ci$ for some constant $C>0$. Basically, you are trying to compute:
$$ \begin{align*}
\sum_{i=0}^{n-1} 1+\sum_{i=x+1}^{n-1}Ci
&= n +C[(x+1)+(x+2)+\cdots+(n-1)] \\
&= n+C \left[\dfrac{n-x-1}{2}(x+n) \right]  \\
&= n+C \left[\dfrac{n-5-1}{2}(5+n) \right]  \\
&= n+\dfrac{C}{2} (n-6)(n+5)  \\
&= n+\dfrac{C}{2} (n^2-n-30)  \\
&= \dfrac{C}{2}n^2 + \left(1-\dfrac{C}{2} \right)n -\dfrac{C}{2}30  \\
&= O(n^2)\\
\end{align*}$$

ContentLicense: CC BY-SA 3.0


问:
if P is sufficient for Q: (P implies Q) or (Q implies P) ??
for the given statement:

" A sufficient condition for network failure is that the central switch
goes down."

let Q be the network failure, and P be the central switch shutting down. Then symbolically, the statement would be: P is sufficient for Q

can I say that: "if P then Q" so that the antecedent is P and the consequent is Q? or is it the opposite??

答:
The following statements are logically equivalent:

A implies B

If A, then B

A only if B

A is sufficient of B

B if A

B is necessary for A

My favourite example: A = Raining, B = Cloudy

Raining implies cloudy

If raining, then cloudy

Raining only if cloudy

Raining is sufficient for cloudy

Cloudy if raining

Cloudy is necessary for raining

Note that rain does NOT cause cloudiness. "Raining implies cloudy" means only that, at the moment, it is not both raining and not cloudy (an example material implication in classical logic).

ContentLicense: CC BY-SA 4.0


问:
Proving that there is a sequence of letters in all words of a formal language
I have a formal language that contains all words that have twice as many b's as a's, for example aabbbb. Now I'm trying to prove that there always must be atleast one sequence of 3 letters inside given word that are either bab, bba or abb. It's a part of a bigger proof that I'm making but it's crucial for it to work and I struggle with it. It looks quite obvious that it must be the case, but how can I prove it? Any hints?</p>

答:
Assume first that the word has at least 2 a's. Then the word must contain two consecutive b's, for otherwise each b would immediately be followed by at least one a, and there could not be twice as much b's as a's in the word. Thus, consider such a sequence of at least 2 consecutive b's. Either at the beginning or at the end of this sequence, there must be an a. Hence, either abb or bba is contained.
If the word has only one a, then there need not be two consecutive b's, but this is possible only for the word bab.</p>


问:
Closed convex hull = closure of convex hull?
If the "closed convex hull" of A is the intersection of all closed convex sets containing A, is this the same as the closure of the convex hull of A?

Many have asked whether the closure of the convex hull is the same as the convex hull of the closure (answer: no), but I think this is a bit different.

I feel like this should have a simple answer, either based on set logic (if it's true) or a simple counterexample (if it's false).

答:
While user147263's answer is correct, the second part of his proof is not fully detailed. Let me elaborate on it a little bit.

First, it can easily be seen that the closure of a convex set is too a convex set.

Let $B$ be the convex hull of A. In symbols, $B = \cap \{M,\, M \text{ is convex, } A \subset M\}$.

Let $C$ be the closed convex hull of $A$. Again in symbols, $C = \cap \{M,\, M \text{ is convex and closed, } A \subset M\}$.

Then clearly, $\overline{B}$ is closed, convex and contains A. It follows from here that $\overline{B}$ is one of the intersected sets in the definition of C, hence $C \subset \overline{B}$.

Conversely, $B \subset C$. Then also $ \overline{B} \subset \overline{C}$. Because $C$ is closed, then $C = \overline{C}$ and $\overline{B} \subset C$.


问:
$ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ is the same as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$
I want to show that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$ Perhaps they are isometrically isomorphic.

What is causing me trouble is the fact that we have an infinite-dimensional space here. Perhaps I could take an orthonormal Schauder basis for $ L^2 $, but the fact that these spaces are the same feel like it should be obvious.

My question is, is it correct that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$? If so, how would one prove it?

答:
You can see this in many ways, but one is through properties of the tensor product:

For modules $M,N_1,$ and $N_2$ over a commutative ring $R$
$$M\otimes (N_1\oplus N_2)=(M\otimes N_1)\oplus (M\otimes N_2)$$
and
$$M\otimes R=M$$
see Lemma 8.12 here or any text covering tensor products.

In your example, $M$ is $L^2(\mathbb{R},\mathbb{C})$ and $N_1$, $N_2$ , and $R$ are all $\mathbb{C}$. From these two general properties you can deduce the desired isomorphism of $\mathbb{C}$-modules, that is vector spaces over $\mathbb{C}$.

This is a little abstract so if it helps you can think of it this way: each element of $L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2$ is a sum of elements of the form $f\otimes(z_1,z_2)$, which using the properties of the tensor product may be rewritten as $z_1f\otimes(1,0)+z_2f\otimes(0,1)$. Thus, if $\{f_i\}$ is a basis of $L^2(\mathbb{C},\mathbb{R})$, $\{f_1\otimes(1,0)\}\cup \{f_i\otimes(0,1)\}$ is a basis of $L^2(\mathbb{R},\mathbb{C})\otimes \mathbb{C}^2$.


问:
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand
Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).

By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.

答:
1875 book by Reuschle. The technique was initiated by Gauss, some 30 years before Galois theory.
Indeed, on page 249 in Cox, Galois Theory, we find
$$ \sum_{w=1}^{16} \; (w|17) \; \zeta_{17}^w \; = \sqrt{17}.  $$
Letting $\eta_0$ refer to the quantity of interest, calling the sum of the others $\eta_1$ as in Reuschle below, we see $\eta_0 + \eta_1 +1=0, $ while $\eta_0 - \eta_1 = \sqrt{17}.$  So  $2\eta_0 +1 = \sqrt{17}.$

For any prime $p \equiv 1 \pmod 4$ we have
$$ \sum_{w=1}^{p-1} \; (w|p) \; \zeta_{p}^w = \sqrt{p}.  $$
while prime $q \equiv 3 \pmod 4$ gives
$$ \sum_{w=1}^{q-1} \; (w|q) \; \zeta_{q}^w = i\sqrt{q}.  $$
Cox refers to Ireland and Rosen for quadratic Gauss sums, chapter 6. The bits just preceding are I+R Theorem 1, on page 75.

Anyway, double your displayed quantity ( that is, drop the $1/2$), call it $\eta.$  Write out $\eta^2$ in terms of powers of $a,$ taking every exponent from $-8$ to $8$ as you began.   Add up $\eta^2 + \eta$ and reduce.

$$  \eta =   a + a^2 + a^4 + a^8 + a^9 + a^{13}   + a^{15}  + a^{16}   $$

If you write it with exponents from $0$ to $16,$  the result of $\eta^2 + \eta$   comes out to ( exponents $\pmod {17}$)

$$ \eta^2 + \eta = 4a^{16} + 4a^{15} + 4a^{14} + 4a^{13} + 4a^{12} + 4a^{11} + 4a^{10} + 4a^9 + 4a^8 + 4a^7 + 4a^6 + 4a^5 + 4a^4 + 4a^3 + 4a^2 + 4a + 8
$$
But this is ( we split the final $8$ into $4+4$)
$$ 4 \left(a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 \right) +4 = 4
$$
because
$$ a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 = 0$$

All together, as Reuschle tells us, $$  \eta^2 + \eta - 4 =0 $$


问:
Minimum and maximum values of a base
Given the number 328662 in base 10. Find the minimum and maximum values of b where the base b representation of the given number has four digits. Find these two four digit representations. If the base is greater than 10, use A = 10, B = 11, C = 12, etc. for the necessary digits.

If the four digit number ends with 0, then I can use the fact that b is a multiple of a factor of the given number to find b. What would you do if the four digit number does not end with 0? Thank you.

答: The number of possible $4$ digit numbers in base $b$ is $b^4$, including $0000$.  So you need $b^4 \geq 328662$, or $b \geq 23.94$.  Meanwhile you need $b$ to be small enough that the number $328662$ manages to make $4$ digits, so you need $b^3 \leq 328662$, which gives $b \leq 69.01$.  There are many representations in between, so presumably they are asking for the base $24$ and base $69$ representations.  Use a division algorithm to find the representations.

Title: Minimum and maximum values of a base


问:
Finding the Closed Form of a Recursion
I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.

So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$

Now, if we let $b_n = \frac{n(n+1)}{2} a_n$, the recursion becomes: $$b_n = \frac{n}{2} + b_{n-1}.$$

How do I continue on from here? I feel like I have to figure out a closed form for $b_n$ and use that to compute $a_n$, but I can't seem to figure out a way to do so.

Any help would be greatly appreciated.

Title: Finding the Closed Form of a Recursion

Body:
It is simple afte OP's step:
$$B_n-B_{n-1}=\frac{n}{2}$$ Do telescoping
$$B_1-B_0=\frac{1}{2}$$
$$B_2-B_1=\frac{2}{2}$$
$$B_3-B_2=\frac{3}{2}$$
$$....................$$
$$B_{n-1}-B_{n-2}=\frac{n-1}{2}$$
$$B_{n}-B_{n-1}=\frac{n}{2}$$
$$\implies B_n=\frac{n}{2}+B_0$$
$$\implies B_n=\frac{n(n+1)}{4} \implies A_n=\frac{1}{2}$$


问:
Find the value of the sum of only 3 digit numbers such that dividing by it leaves REM = 11?
I had asked a Q similar to this.

Find the value of this 3 digit number such that dividing by it leaves REM = 11

But I want to know also if instead we take sum of all those numbers. Is there a way to solve this too in a similar way to that Q?

The problem happening is how to put limits in the sum such that the numbers are in between 100 and 999.

答:
To sum up all the $3$-digit numbers that give remainder $11$ upon dividing by $13$.

Note that they are of the form of $13k+11$.

$$100 \le 13k+11 \le 999$$

$$89 \le 13k \le 988$$

$$\lceil \frac{89}{13}\rceil  \le k \le \lfloor\frac{988}{13} \rfloor$$

$$7 \le k \le 76$$

I will leave the task of evaluating

$$\sum_{k=7}^{76}(13k+11)$$

as an exercise to you.


问:
How to find the closed form of this recurrence relation?
The relation is :

$x(n) = x(n-1)*c + z $,

where $c$ and $z$ are constants , $x[1]$ and $x[2]$ are known .

$c = 2*p - 1$ ;

$z = 1 - p $

$p$ is a constant .

答: A constant solution is $x(n)=\frac z{1-c}$, so we expect the substitution $x(n)=\frac z{1-c}+y(n)$ to simplify things. Indeed, this yields $y(n)=cy(n-1)$, so $y(n)=c^ny(0)$, so $x(n)=\frac z{1-c}+c^n\left(x(0)-\frac z{1-c}\right)$.


问:
How to prove that if $n > 1$ is a natural number, then $n-1$ is also a natural number?
Here is what we have got to work with:

A subset $E$ of the set $\mathbb{R}$ of real numbers is said to be inductive if $1 \in E$ and if for every $x \in E$ the number $x+1 \in E$ also.

The intersection of all the inductive sets is the set $\mathbb{N}$ of natural numbers, which is also inductive.

Moreover, since the sets $\{ x \in \mathbb{R} | x > 0 \}$ and $\{ x \in \mathbb{R} | x \geq 1 \}$ are inductive, the set $\mathbb{N}$ consists of only positive real numbers $\geq 1$.

Last but not least, we of course have the field, the order, and the completeness axioms of the real number system at our disposal.

Using the above machinery, how can we prove (using induction) the following facts?

If $n > 1$ is a natural number, then $n-1$ is also a natural number.

If $n$ and $m$ are natural numbers with $n > m$, then $n-m$ is also a natural number.

答: For each $n\in\Bbb N$, let $P(n)$ be the assertion “$n=1$ or $n-1\in\Bbb N$”. Clearly, $P(1)$ holds. Now let $n\in\Bbb N$ and suppose that $n=1$ or that $n-1\in\Bbb N$. Then $(n+1)-1=n\in\Bbb N$, and therefore $P(n+1)$ holds.


答:
If $n=328662$ has four digits in base $b$, this means that it is between $(1000)_b=b^3$ and $(10000)_b=b^4$. In other words, you are solving this:

$$b^3\le n\lt b^4$$

which is the same as $\sqrt[4]{n}\lt b\le\sqrt[3]{n}$. In our case:

$$23.9434\ldots\lt b\le 69.0107\ldots$$

so, because $b$ is an integer, we have $b\in\{24, 25,\ldots,69\}$.

So, in (smallest) base $24$ we have $n=(NIE6)_{24}$ and in (largest) base $69$ we have $n=(102F)_{69}$

Title: Minimum and maximum values of a base

Body:
Given the number 328662 in base 10. Find the minimum and maximum values of b where the base b representation of the given number has four digits. Find these two four digit representations. If the base is greater than 10, use A = 10, B = 11, C = 12, etc. for the necessary digits.

If the four digit number ends with 0, then I can use the fact that b is a multiple of a factor of the given number to find b. What would you do if the four digit number does not end with 0? Thank you.


问:
two-dimensional probability density
This is how a probability density of X and Y is defined in my worksheet:

$$ f(x,y) = \begin{cases}
\frac{x^2}{192},  & \text{for}\ 0 \le x \le 4, x\le y \le 2x \\
0, & \text{otherwise}  \\
\end{cases}$$

To make sure it is a density I calculated the double integral $$ \int_0^4\int_x^{2x}\frac{x^2}{192} dy\ dx$$

however I had $\frac13$ as a result... does this mean that my worksheet is wrong, since the probability density must equal 1? Or did something go wrong in my integration?

答: I also get 1/3, both using manual integration and Mathematica. So $f(x,y)$ indeed isn't a valid probability density in $x$ and $y$.</p>

Title: two-dimensional probability density


问:
Probability that XOR of an arbitrary number of random bits is 1
How can we say than the XOR between $n$ (uniform, independent) random bits is $1$ with probability $1/2$?
For example if we have 4 random bits, we know that the XOR of them will be $1$ with half probability. The same happens if we have 3, or 4 random bits. But if we have $n$ random bits?

Body:
Let $D = \bigoplus_{k = 1}^{n-1} b_i$. Then,

\begin{align}P(\bigoplus_{k = 1}^n b_i = 1) &= P(\bigoplus_{k = 1}^n b_i = 1 | D=1)P(D=1) + P(\bigoplus_{k = 1}^n b_i = 1 | D=0)P(D=0)
\\ &= \frac12P(D=1)  +  \frac12P(D=0)
\\ &= \frac12 \end{align}


问:
Prove that if $n \geq 2$, then $n! < \{{2n \atop n}\} < (2n)!$

Prove that if $n \ge 2$, then $$n! < \left\{{2n \atop n}\right\} < (2n)!.$$

Here, $\{{2n \atop n}\}$ is a Stirling number of the second kind.

I have absolutely no idea how to start this question - could someone please help? Thanks!

答:
By definition, $S(2n, n) = \left\{ \begin{array}{c} 2n \\ n \end{array} \right\}$ counts the number of ways to partition a $2n$-element set into $n$ non-empty sets. So let's try to compare this number with permutations.

So first, let's look at a permutation of $n$ elements. For instance the permutation

$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 5 & 1 & 4 & 2 \end{pmatrix}$$

and oh hey we can get a set partition out of that:

$$\{1,8\}, \{2, 10\}, \{3, 6\}, \{4,9\}, \{5,7\}.$$

So evidently $n! \le S(2n, n)$ and except when $n = 1$ this inequality should be strict.

Next, let's take a set partition and attempt to get a unique permutation of $2n$ elements out of it. Let's say

$$\{1,2,3\}, \{4,9\}, \{5\}, \{6,8,10\}, \{7\}.$$

Do we get a unique permutation out of it? I only see one reasonable thing to do so let's try it:

$$\{\{1,2,3\}, \{4,9\}, \{5\}, \{6,8,10\}, \{7\}\} \mapsto (1,2,3)(4,9)(5)(6,8,10)(7)$$

Is this well-defined/injective? Well it looks like we need to fix an ordering on the $2n$ elements ahead of time so that $\{1,2,3\} \mapsto (1,2,3)$ is well-defined. But after that, yeah, that looks like it should work.

You should write these maps in a general form (rather than $n = 5$) and check the details (well-defined/injective). Also check if/when you get strict inequalities.


问:
Finding Null space of matrix when m<n?
I know how to find null space of a matrix, by following certain procedure like this.

But how do I find null space for this matrix?

```<code>[ 1 0 0
  0 1 0 ]
</code>```

Here, the number of rows < number of columns (m<n). I'm not able to represent pivot variables in terms of free variables.

答:
You want to solve the linear system
$$
\begin{array}{l l l}
1 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 & = & 0 \\
0 \cdot x_1 + 1 \cdot x_2 + 0 \cdot x_3 & = & 0.
\end{array}
$$
Therefore $x_2 = x_1 = 0$. Those are the pivot variables. The free variable $x_3$ has no influence whatsoever on the left hand side, so $x_3$ is arbitrary. The solutions space is thus
$$
S := \left \lbrace (0, 0, x_3)^\top \in \mathbb{R}^3: x_3 \in \mathbb{R} \right \rbrace.
$$

Title: Finding Null space of matrix when m<n?


问:
Is it true that for every prime number $p >3$, the next one will be less than $p+p/2$?
I know that for every prime $p$, the next prime is less than $2p$. Can we improve this statement? Can it be less than $(3/2)p$? What is the best function of $p$ for which this is true (for every prime, not just infinitively many)? (assume $p>3$, or $p>$ than a fixed $p$ if necessary).</p>

答: Doesn't work for $p=3$. I seem to recall that $p+\ln p$ was a good upper bound on the prime successor of $p$, in general (or at least for $p$ large enough) but I'm not finding any references at the moment.</p>


问:
intersection of two von Nuemann algebras
Let $M_1$ and $M_2$ be two von Neumann algebras. $p=p^*=p^2$ and $p\in M_1\cap M_2$. Can we conclude that $pM_1p\cap pM_2p=p(M_1\cap M_2)p$?

It is trivial that $p(M_1\cap M_2)p\subset pM_1p\cap pM_2p$.

If $x\in pM_1p\cap pM_2p$, we have $x=pyp=pzp$, where $y\in M_1,z\in M_2$.

Can we prove that $y\in M_1$ or $z\in M_1$?

答:
If $x\in pM_1p\cap pM_2p$, then $x\in M_1\cap M_2$, so
$$
  x = pxp \in  p(M_1\cap M_2)p.
  $$


问:
Calculate $e^A$ where $A \in M_{2\times 2}$
Let  $A=\begin{pmatrix}
0 & 1 \\
-1& 0
\end{pmatrix}$ calculate $e^A$.

i find the jordan form of $A$ and i said that $A \sim \begin{pmatrix}
i & 0 \\
0& -i
\end{pmatrix}=B$

so $e^A\sim e^B$

then from the algebra of that diagonal matrix $B$ we can say that $e^B= \begin{pmatrix}
 \sum_{n=0}^{\infty }\frac{i^n}{n!}& 0 \\
0&\sum_{n=0}^{\infty }\frac{(-i)^n}{n!}
\end{pmatrix}$

Can I find where $ \sum_{n=0}^{\infty }\frac{i^n}{n!}$ converges? Or do I leave it such as ?

答:
$$\sum_{n=0}^\infty \frac{i^n}{n!}=e^i =\cos(1) + i \sin (1)$$

where the second equality is due to $\exp(\theta i ) = \cos(\theta)+i\sin\theta$. Similarly,

$$\sum_{n=0}^\infty \frac{(-i)^n}{n!}=e^{-i} =\cos(1) - i \sin (1)$$

To compute $e^A$, consider the eigenvectors too.


问:
An infinite group isomorphic to its own automorphism group, but not complete, part 1
An example of a non-complete group isomorphic to its own automorphism group is the dihedral group of order $8$.

But, this question is about infinite groups (specifically, non-cohopfian groups).

Question: Do there exist (necessarily infinite) groups $G$ for which $\mathrm{Aut}(G) \cong G$ satisfying the following property?

The center of $G$ is trivial, but $G$ has some outer automorphisms (so the conjugation action $G \to \mathrm{Aut}(G)$ is injective but not surjective).

Clearly, if $G$ satisfies this property, then it cannot be cohopfian. In particular, $G$ must be infinite.

Update: The other case has been moved to Part 2 per user1729's comment below.

答: The infinite dihedral group is an example with property 1.</p>


问:
Semisimple complex Lie Algebra and decomposition into weight spaces.
So I was wondering why a semisimple complex Lie Algebra $L$ is a direct sum of its weight spaces.

Given a Cartan Subalgebra of $H$ of $L$ then since $L$ is a semisimple complex Lie Algebra, then every element of ad(H) is semisimple (diagonalizable) . And since H is abelian, then all elements of H can be simultaneously diagonalised w.r.t some basis $x_1, \ldots , x_n$

But then consider $\alpha _i: H \rightarrow \mathbb{K}$ where $\alpha _i(h)$ specifies the $i^{\text{th}}$ diagonal element of $ad_h$.

Then for $L_{\alpha _i} := \{x\in L\space | \space ad_h(x) = \alpha _i(h)x \space \forall h \in H\}$ we know that $x_i \in L_{\alpha _i}$.

So if L is indeed a direct sum of it's weight spaces $L_{\alpha _i}$, then it must be the direct sum of these. Implying each weight space has dimension $1$. But this implies inturn that $H$ has dimension $1$ as $H$ = $L_{0}$. Which surely is false!?

What exactly have I done wrong?

答:
There are a few places where I think things could have gone wrong.

The definition of the weight space (of $L$ as an $H$-module) is
$$
L_{\lambda}=\lbrace y\in L \mid ((\operatorname{ad}(x)-\lambda(x))^ny=0\text{ for all }x\in H\rbrace
$$
where $\lambda:H\rightarrow\mathbb{C}$ is a $1$-dimensional representation, which isn't exactly what you've written: you need that some power of $\operatorname{ad}(x)-\lambda(x)$ annihilates $y\in L$, for all $x\in H$, and so it isn't necessarily true that $\operatorname{ad}(x)=\lambda(x)y$.

You haven't necessarily shown that $L_0$ is $1$-dimensional. You've taken some $\alpha_i:H\rightarrow\mathbb{C}$, but none of these are the $0$ weight. Whilst it's true that every non-zero weight gives rise to a $1$-dimensional weight space, the same is not true of the Cartan subalgebra.

Why is $x_i\in L_{\alpha_i}$?

As to the reason why: every module over a nilpotent Lie algebra can be decomposed into the direct sum of its weight spaces.

Let $V$ be a module over a nilpotent Lie algebra $L$, so $V=V_1\oplus\cdots\oplus V_k$, with $V_j$ an indecomposable submodule inducing a weight $\lambda_j$. Let $W_\lambda=\bigoplus_{\lambda_j=\lambda}V_j$. Then $V=\bigoplus_\lambda W_\lambda$. Clearly $W_\lambda\subset V_\lambda$, so we try to show the reverse inclusion.

Let $v\in V_\lambda\subset V=\bigoplus_\mu W_\mu$. So $v=\sum_\mu v_\mu$. Then $v(x-\lambda(x))^n=0$, and so $\sum_\mu v_\mu(x-\lambda(x))^n=0$. So each $v_\mu(x-\lambda(x))^n=0$. Suppose $\lambda\ne\mu$, so $\lambda(x)\ne\mu(x)$ for some $x\in L$. Then $v\mapsto vx$ has matrix
$$
\left(\begin{matrix}
\lambda(x) & & * \\
& \ddots & \\
0 & & \lambda(x)
\end{matrix}\right)
$$
on $W_\mu$, which is non-singular. So $v\mapsto v(x-\lambda(x))$ is non-singular on $W_\mu$, and $v\mapsto v(x-\lambda(x))^n$ is non-singular on $W_\mu$. So $v_\mu(x-\lambda(x))^n=0\Rightarrow v_\mu=0$. So $v=\sum_\mu v_\mu=v_\lambda\in W_\lambda$.


答:
Using that $A^2=-I$, you can write
$\exp(A)=\sum_{k\in\Bbb N}\frac{A^k}{k!}
 =\sum_{l\in\Bbb N}\frac{(-1)^l}{(2l)!}I+\sum_{l\in\Bbb N}\frac{(-1)^l}{(2l+1)!}A
$.
You should be able to find nice expressions for those two numerical summations. Then you can write out the precise matrix, which you should be able to recognise.

Title: Calculate $e^A$ where $A \in M_{2\times 2}$

Body:
Let  $A=\begin{pmatrix}
0 & 1 \\
-1& 0
\end{pmatrix}$ calculate $e^A$.

i find the jordan form of $A$ and i said that $A \sim \begin{pmatrix}
i & 0 \\
0& -i
\end{pmatrix}=B$

so $e^A\sim e^B$

then from the algebra of that diagonal matrix $B$ we can say that $e^B= \begin{pmatrix}
 \sum_{n=0}^{\infty }\frac{i^n}{n!}& 0 \\
0&\sum_{n=0}^{\infty }\frac{(-i)^n}{n!}
\end{pmatrix}$

Can I find where $ \sum_{n=0}^{\infty }\frac{i^n}{n!}$ converges? Or do I leave it such as ?


问:
Prove that for all $x>0$, $f(x) < xf'(x)$
Let $f(x)$ be continuous on $[0,\infty)$ and twice differentiable on $(0,\infty)$. If $f(0)=0$ and $f''(x)>0$ on $(0,\infty)$ prove that for all $x>0$, $f(x) < xf'(x)$.

Hello, can anyone help me with this question? Thank you!

答:
Let $x > 0$. Since $f(0) = 0$, by the mean value theorem there exists $c\in (0,x)$ such that $\frac{f(x)}{x} = f'(c)$.
Since $f''> 0$ on $(0,\infty)$, then $f'$ is strictly increasing in $(0,\infty)$, hence $f'(c) < f'(x)$.


问:
Relative homology of sublevels is the homology of the attached $k$-cell
Let $M$ be a smooth manifold of dimension $N$, $f: M \to \mathbb R$ smooth and let $p$ be a nondegenerate critical point, $f(p) = c$. Suppose that it is the only nondegenerate critical point in $f^{-1}([c - \varepsilon, c + \varepsilon])$ for some $\varepsilon > 0$. If the Morse index of $p$ is $k$, the Handle Body Decomposition tells us that
$$
M^{c + \varepsilon} \text{ has the homotopy type of } M^{c - \varepsilon} \text{ with a } k \text{- cell attached}
$$

Now, Corollary 12.12 in Nonlinear Analysis and Semilinear Elliptic Problems, from Ambrosetti and Malchiodi, states that
$$
H_q(M^{c + \varepsilon}, M^{c - \varepsilon}) \simeq
\begin{cases}
\mathbb Z \quad \text{ if } q = k \\
0 \quad \text{ if } q \neq k
\end{cases}
\qquad(*)
$$

How to prove this?

It is a bit intuitive. Knowing that the homology of a set relative to a deformation retract is $0$, I have tried decomposing the pair of sets $(M^{c + \varepsilon}, M^{c - \varepsilon})$ into disjoint unions involving the $k$-cell, in order to be able to apply that the homology of the union be the sum of the homologies and that $(*)$ is precisely $H_q(\overline{B_k}, S^{k - 1})$, but to no end.

Thanks in advance.

答:
Knowing that the pair $(M^{c+\epsilon}, M^{c-\epsilon})$ is equivalent to $(M^{c-\epsilon} \cup_f D^k,M^{c-\epsilon})$, the rest is almost completely formal. Namely, $\textit{if}$ the latter is a good pair, then its relative homology is the same as the reduced homology of the quotient $M^{c-\epsilon} \cup_f D^k /M^{c-\epsilon}$ which is homeomorphic to $S^k$ from which the result follows.

So the only question is if this pair is a good pair. In fact, what we have here is called a relative CW complex (the result of attaching cells to some space), and it is a known homotopical result that these are good pairs.


问:
Mathematically Rotating and Scaling Squares
I need help understanding what's happening mathematically when we rotate and scale a square and fit it within itself and repeat the process. For example: I can start with a square that has a dimension of $250 \times 250$ (purple square, shown in the reference image below). Then rotate it $45$ degrees and then scale it down so that it fits perfectly within the original $250 \times 250$ square, thus scaling the square down to the size of $176.78 \times 176.78$ (red square). Then when I do it again, I get the green square, $125 \times 125$.

I want to understand what's happening here mathematically and see if there's a mathematical equation I can use to be able to accurately calculate the size of a square after it's been rotated and scaled-down based on its initial value/state. That if I start with a square that's $250 \times 250$, how can I mathematically rotate and scale the square so that it would output $176.78$? Then I take that number, $176.78$, and do the same thing done on $250$, to arrive at $125$.

I figured the blue square would be useful in some way, still not sure, but it doesn't appear to accurately "lead" me into $125$, but it's a cheat to $125$ after getting $176.78$. But I need to get to $176.78$ first.

I thought I found a solution by doing: $250$ / $70.7%$ / $2$ but that returns $176.8033946$. Why $70.7%$? Idk, was literally just trying anything that would result in something consistent and that was the closest thing I could find, but it's inaccurate and as you repeat the process, it just becomes more and more inaccurate.

Any help would be appreciated, and hopefully, this makes sense.

Reference image

答:
If you draw the original square at $250 \times 250$ and draw a diagonal from one side to the other, you can use the Pythagorean theorem $\sqrt {a^2 + b^2}$ to find the length of the diagonal.

Notice that when you draw the diagonal, you get a right triangle with $45$ degree angles on each side.  If you know trigonometry, $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac {\sqrt 2}{2},$ hence a formula for rotating the square by $45$ degrees and fitting it into new sides  would be $$\text {side length }_{\text {rotated square}} = \dfrac {x \sqrt {2}}{2}$$

Each time you take the new side and and rotate by $45$ degrees, the square will reduce by $70.71\%$  (as  $\dfrac {\sqrt {2}}{2} \approx .7071$).

Using the information you supplied, let $x = 250$, then the first rotated square will be $125 \sqrt{2} \approx 176.78.$  Rotating again with these new sides gives us $125 \sqrt 2 \cdot \dfrac {\sqrt 2}{2} = \dfrac {250}{2} = 125$.


答:
Its true for every prime greater then or equal to $11$

Much stronger results are also known also,
 for example there exists primes between

$n$ and $(1+\frac{1}{5})n $ $ $ $\forall n\ge25$

$n$ and $(1+\frac{1}{16597})n $ $ $  $\forall n\ge 2010760$

$n$ and $(1+\frac{1}{\ln(n)^2})n$ $ $  $\forall n\ge463$

$n$ and $(1+\frac{1}{2\ln(n)^2})n$ $ $  $\forall n\ge3275$

$n$ and $(1+\frac{1}{25\ln(n)^2})n$ $ $  $\forall n\ge 396738 $

It is important to note that results of this sort should not be taken as good estimates of the distribution of the primes

These are useful in that they assert the existence of primes over intervals for relatively small numbers, in general if you order the primes from decreasing to increasing value, then the $n$th prime will be asymptotic to $n\ln(n)$, this suggests that the primes are distributed roughly logarithmically,

That is if your given a prime number $m$ your on average going to see the next prime at around $m+\ln(m)$

Title: Is it true that for every prime number $p >3$, the next one will be less than $p+p/2$?

Body: I know that for every prime $p$, the next prime is less than $2p$. Can we improve this statement? Can it be less than $(3/2)p$? What is the best function of $p$ for which this is true (for every prime, not just infinitively many)? (assume $p>3$, or $p>$ than a fixed $p$ if necessary).</p>


问:
Norm of functional
I'm having little trouble with understanding one proof I saw on mathstack.

Let's consider functional

$$f\colon (C[0,1], \Vert\cdot \Vert_\infty) \ni g\rightarrow g(0) - 7g(\frac12)$$

Calculation $\Vert f\Vert_\infty$ was the following:

The norm of $f$ is equal to $8$ as for the map $h(x)=\vert 4x - 2 \vert -1$ we have $\Vert h \Vert_\infty = 1$ and $f(h)=8$.

And I'm not exactly sure how come it ends the justification. Could you please tell me why with referring to theorems on which this way of thinking is based on ?

答: I'm going to call the operator $T$ for clarity (instead of $f$). For any $g$ which is continuous on $[0,1]$, we have that $$|Tg|\leq |g(0)|+7|g(1/2)|\leq 8\|g\|_\infty.$$ Thus, $$\sup_{\|g\|_\infty=1}|Tg|=\|T\|\leq 8.$$ The particular function in your post shows that this upper bound can be achieved, and since the operator norm is defined as the least upper bound, the above is actually an equality.

Title: Norm of functional


问:
Expansion of Confluent Hypergeometric Function in terms of $\operatorname{erfi}(x)$
I have the following confluent hypergeometric function: $_1F_1\left(2(m+1),\frac{1}{2},-x^2\right)$.

By using Mathematica, I know that for values of $m=0,1,2,...$ this function expands into a power series involving $\operatorname{erfi}(x)$, that can be further simplified by introducing the Dawson function.

For example, for $m=0$ $$_1F_1\biggl(2,\frac{1}{2},-x^2\biggr)=1-\frac{3}{2}xe^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)-x^2+x^3e^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)$$ $$= 1-3xD_+(x)-x^2+2x^3D_+(x)$$

Here are two more examples, for $m=1$:

and $m=2$:

What is the general term of this expansion, and how can I obtain it by hand?

EDIT

As Aaron Hendrickson suggested in the comments, Mathematica expands the function by this formula, so I've found the general term of the sum. But how can I get this expansion starting from the definition of the particular confluent hypergeometric function $_1F_1\left(n,\frac{1}{2},z\right)$?

答:
So as mentioned in the comments this identity provides a "closed form" for your hypergeometric function in terms of Laguerre polynomials. Here is another option that may be considered a more elementary form:

Using DLMF 13.3.17 for $n=2m+1$ and $a=1$ you may write
$$
\begin{aligned}
{_1F}_1\left({2m+2\atop 1/2};z\right)%
&=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} {_1F}_1\left({1\atop 1/2};z\right)\\
&=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} (1+\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z))\\
&=\frac{1}{(2m+1)!}\left((2m+1)!+\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\right)\\
&=1+\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\\
\end{aligned}
$$
You may then apply the General Leibniz rule to evaluate the derivative. Once the derivatives are evaluated you may then make the substitution $z\mapsto-x^2$ to obtain the final result.

There are many other different forms you can put your hypergeometric function into using similar methods and variants of this method. What its going to come down to is what form is useful for your application.


问:
Binomial theorem from first principles?
I suppose I'll give a little context to this...

I was at first really excited by the prospect that both Bernoulli and Taylor's versions of $e^x$ actually ammount to the same thing (where, when you expand Bernoulli's definition $\lim_{n\to\infty}(1+\frac{x}{n})^n$ it simplifies to the Taylor series $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ - unlike, say, trigonometric functions like $\sin(x)$ which look a bit more unsatisfying). However, upon further reflection, to say that one identity 'simplifies' to the other seems almost circular given it presupposes binomial theorem.

So, I decided to do a little scouting online, and found that binomial theorem could be proven using proof by induction. But once again, I felt unsatisfied as it feels as though you've just divined up some formula which works. How did you come up with it in the first place?

Well, that's when I thought of probability theory. We use a binomial expression $^nC_{r}$ when we are looking for the number of combinations we have for arranging things. We can very intuitively derive this formula, however, as I don't want this question to be unecessarily large, I'm not going to go into detail about this here. What counts, is that we can figure out this formula from first principles - we're not getting it from nowhere.

As I'm sure you already know, $^nC_{r}$ is key to figuring out binomial distributions. Once again I'm not going to explaim this, but an example probability calculation becomes $^{10}C_{4}(\frac{1}{6})^4(\frac{1}{6})^{10-4}$ for rolling 4 sixes in 10 total rolls.

According to (Kolmogorov's) axioms of probability, the probability of all total outcomes $P(\Omega)$ is 1. Let's consider the context of total binomial outcomes of 3 trials (I pick this number for simple and illustrative purposes), where we designate the probablity of success, $\frac{1}{5}$, and that of failure, $\frac{4}{5}$, (where the sum of all different outcomes, here success and failure = 1). Well, we know that becuase we have 3 trials, logically, we'll have 4 possibilities: 0 successes, 1 success, 2 successess, or 3 success; we can write out our probability as: $^{3}C_{0}(\frac{1}{5})^3(\frac{4}{5})^0$ + $^{3}C_{1}(\frac{1}{5})^2(\frac{4}{5})^1$ + $^{3}C_{2}(\frac{1}{5})^1(\frac{4}{5})^2$ + $^{3}C_{3}(\frac{1}{5})^0(\frac{4}{5})^3$ = $(\frac{1}{5})^3$ + $3(\frac{1}{5})^2(\frac{4}{5})$ + $3(\frac{1}{5})(\frac{4}{5})^2$ + $(\frac{4}{5})^3$ $= 1$.

This can of course be generalised to $\sum_{k=0}^{n}$$^{n}C_{k}{p}^k{q}^{n-k}$ where $n$ is the total number of trials $k$ the number of successes, $p$ the probability of success, $q$ that of failure.

Now let's consider the binomial expansion of $(a + b)^3$. Well, supposing we know nothing about binomial theory and do it by hand, our algebra nonetheless simplifies to $a^3 + 3a^2b + 3b^2a + b^3$. Hmmm... this looks suspicously like our total probability forumla expanded above. It almost looks as though $a$ could be representing our successes, and $b$ our failures.

So my reasoning is such: it just so happens that in the context of probability, the total probability of your outcomes add up to 1. If $\Omega = \left\{ a, b \right\}$, then, by Kolmogorov's axioms, $a + b = 1$. What we're effectively saying is that for any binomial expansion, we can treat them as 'probability outcomes' where $a + b \ne 1$.

Is this a reasonable proof?

答:
$a$ and $b$ certainly need not be probabilities.

Consider

\begin{align*}
(a+b)^n = (a+b)\cdot(a+b)\cdots(a+b) & \text{ to n terms}
\end{align*}

Any of the terms in the binomial expansion of $(a+b)^n$ such as $aa\cdots a$, $ab\cdots a$, $ba\cdots a$, $bb\cdots a$, $bb\cdots b$ etcetera require you to make $n$ independent selections, one term - either $a$ or $b$ from each of the $n$ factors. Hence, the general term $a^k b^{n-k}$ occurs ${n \choose k}$ times.

Title: Binomial theorem from first principles?


问:
Finding angle in a triangle with one median and an isosceles triangle in it

In the following figure $AF=BD=DC$ and $AE=EF$. Find the angle $\alpha$.

I tried so many different things (constructing triangles, drawing parallel lines, using Ceva theorem in triangles $ABD$ and $CBE$, ...). Any help for solving this will be much appreciated.

A geomtrical solution (without trigonometry) would be very nice.

答:
In isosceles $\triangle AEF$, drop $EG \perp AF$ with $G$ on $AF$. Let $AG=GF=x$. So $BD=2x$. Let $\angle BAD = \theta$. So $AE=x\sec \theta$.

Applying Menelaus' for transversal $CE$ to $\triangle ABD$,
$$\frac{BC}{DC}\cdot\frac{DF}{FA}\cdot\frac{AE}{EB}=1$$
$$\Rightarrow DF=EB\cos \theta$$
$$\Rightarrow AD-2x=(AB-x\sec\theta)\cos \theta$$
$$\Rightarrow AD=AB\cos \theta+x$$

Next drop $BH \perp AD$ with $H$ on $AD$. In right $ABH$, $AH=AB \cos \theta$. Then $AD=AH+DH$ implies $DH=x$!

Hence in right triangle $BHD$, $BD=2x$, $HD=x=BD/2$. It turns out $\triangle BHD$ is $30^{\circ}-60^{\circ}-90^{\circ}$ and we conclude
$$\alpha = 60^{\circ}$$


问:
Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand
Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).

By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.

Body:
To show that $$ \cos{\frac{2\pi}{17}} + \cos{\frac{4\pi}{17}}+ \cos{\frac{8\pi}{17}}+ \cos{\frac{16\pi}{17}} = \frac{ -1+\sqrt{17}}{4} $$
This is related to
$$ \cos{x} +  \cos{2x}+  \cos{4x} +  \cos{8x}$$
Since we already know the values of the multiple angle
$$ \cos{2x} = 2\cos{x}^2-1$$
$$ \cos{x} +  \cos{2x}+  \cos{4x} +  \cos{8x} = n $$
$$ \cos{x} + 2\cos^2{x}-1+ 2( 2\cos^2{x}-1)^2-1 + 2(  2( 2\cos^2{x}-1)^2-1 )^2 -1 )^2 -1 = n $$
$$ 128\cos^8{x}-256\cos^6{x}+168\cos^4{x}-38\cos^2{x}+cos{x}+1 = n $$
Say $\cos{x} = y $
$$ 128y^8 -256y^6+168y^4-38y^2+y+1 = n $$
What is the equation for $\cos{\frac{2\pi}{17}} $
$$(2y)^8+(2y)^7-7(2y)^6-6(2y)^5+15(2y)^4+10(2y)^3-10(2y)^2-4(2y)+1 = 0$$
$ y = \cos{\frac{2\pi}{17}} $, now you can put the value of $y$ into the other polynomial to get an equation involving $n$

This is done by resultant, in your case use Sylvester matrix,

resultant of $ 128y^8 -256y^6+168y^4-38y^2+y+1 - n$ and  $ 256y^8+128y^7-448y^6-192y^5+240y^4+80y^3-40y^2-8y+1$ wrt $y$

Then you get a equation describing $n$

$$1152921504606846976(16n^8+32n^7-40n^6-88n^5+49n^4+88n^3-40n^2-32n+16)$$

$$1152921504606846976(2n^2+n-2)^4$$


问:
Determine function using given sequence
It is given that f(x) is a continous function and f(0)=2

$f(x)-f(x/23)=x/23$

How can I find f(484)

This is how far I got Using replacement

$f(x)-f(x/23^n)=x/23 + x/23^2 .......x/23^n$

I guess this would result in some series but I'm not able to get to the answer

答:
Let's start from the relation
$$
f(x) - f(\alpha x) = \alpha x\,,
\qquad \text{with}\ \alpha\in (0,1)
$$
(In your case $\alpha = 1/23$.)

For every $n\in\mathbb{N}$, $n\geq 1$, it holds that
$$
f(x) - f(\alpha^n x) = \sum_{k=1}^n [f(\alpha^{k-1} x) - f(\alpha^k x)]
= \sum_{k=1}^n \alpha^k x = \frac{\alpha}{1-\alpha} (1-\alpha^n) \, x.
$$
Passing to the limit as $n\to \infty$, and taking into account that $f$ is continuous (only continuity at the origin is needed), we get
$$
f(x) - f(0) = \frac{\alpha}{1-\alpha}\, x\,.
$$

Title: Determine function using given sequence


问:
Roots of a certain sixth order polynomial
I am looking for the roots (or basically any information regarding them) of the sixth order polynomial
$$p(x):=ax^6+(a+1)x^4+2bx^3-b^2$$
for positive, real constants $a,b$.
Since $p(0)=-b^2<0$ and $\lim_{x\to\pm\infty}p(x)=+\infty$ we have at least two real roots $x_0,x_1$. Moreover, plotting the function for several parameters suggests, that the remaining four roots are non-real (i.e. two pairs of complex conjugate solutions).

Unfortunately this is as far as I got. I am especially interested in the complex solutions and their behavior for large parameters $a,b$.
Any idea on how to tackle this problem is very welcome!

答:
One thing to note is that there is no closed form solution in terms of radicals (as hinted in a comment).

Notice that if $p_{a,b}(x)$ admits an expression of solutions by radicals, then specialising $a$ and $b$ yields a radical expression for the roots over $\mathbb{Q}$. In particular, consider
$$p_{1,2}(x) = x^6+ 2 x^4 + 4 x^3 - 4 \in \mathbb{Q}[x]$$

Recall from galois theory that there is a radical expression for the roots of $p_{1,2}(x)$, then the galois group of $p_{1,2}(x)$ must be solvable.

We may verify (by Eisenstein at $2$) that $p_{1,2}$ is irreducible, and furthermore checking the reduction mod $p$ for a few $p$ can eventually see that $\operatorname{Gal}(p_{1,2}(x)) = S_6$ - alternatively use computer algebra (e.g., GaloisGroup in Magma). Indeed $S_6$ is nonsolvable.

As a remark, if you are wondering about a radical expression over $\mathbb{C}$, a similar argument works while considering $p_{a,2}(x) \in \mathbb{C}(a)$, showing this galois group is not solvable.


问:
Why do I get different answers from this matrix equation?
the equation given is $A=(AX+B)^{-1}$, I first multiply by $(AX+B)$ and this yields that $A(AX+B)=I$ I distribute A, $A^2X+AB=I \implies A^2X=I-AB \implies X=(A^2)^{-1}(I-AB)$ If I know try and solve what $(I-AB)$ is, lets say it is C and then compute $A^{-1}$ and compute $(A^{-1})^2$ and then finally compute $(A^{-1})^2 \cdot C$ I get the wrong answer. If I however continue to simplify from this point:$ X=(A^2)^{-1}(I-AB) \implies X=A^{-1}(A^{-1}-B)$ from here I now get the correct answer, how come?

$A=\big(\begin{smallmatrix}
  3 & 2\\
  1 & 1
\end{smallmatrix}\big)$ and $B=\big(\begin{smallmatrix}
  2 & 0\\
  1 & 3
\end{smallmatrix}\big)$ If I compute $(A^2)^{-1}(I-AB)$ I get $X=\big(\begin{smallmatrix}
  0 & -30\\
  0 & 41
\end{smallmatrix}\big)$ and$ A^{-1}(A^{-1}-B) $ gives $ X=\big(\begin{smallmatrix}
  3 & -2\\
  -5 & 2
\end{smallmatrix}\big)$ which is the correct answer.

答:
For $A^2$ you should have $$
    \begin{bmatrix}
    11 & 8 \\
    4 & 3 \\
    \end{bmatrix}
$$
and for $(A^2)^{-1}$ you should have $$
\begin{bmatrix}
3 & -8 \\
-4 & 11 \\
\end{bmatrix}
$$
And you should have the correct answer either way.


问:
Are there any famous/notable mathematicians who have their own YouTube channel?
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.

答:
I appreciate the central lecture course from

The WE-Heraeus International Winter School on Gravity and Light by Frederic P. Schuller.

He formidably presents in $28$ lectures, each roughly one and a half hour long the mathematical foundations necessary to understand general relativity. The lectures are built around the definition of the term spacetime, defined as

Spacetime is a four-dimensional topological manifold with a smooth atlas carrying a torsion-free connection compatible with a Lorentzian metric and a time orientation satisfying the Einstein equations.


问:
Is it true that for every prime number $p >3$, the next one will be less than $p+p/2$?
I know that for every prime $p$, the next prime is less than $2p$. Can we improve this statement? Can it be less than $(3/2)p$? What is the best function of $p$ for which this is true (for every prime, not just infinitively many)? (assume $p>3$, or $p>$ than a fixed $p$ if necessary).</p>

答: There is no best function $f(p)$ for which this is true, since $g(p)$ with $g(p_n)=(f(p_n)+p_{n+1})/2$ would be better. If you replace "less than" by "less than or equal to", there is a unique best function $f(p)$, namely $f(p_n)=p_{n+1}$.</p>


问:
Question about order of a sum of entire functions.
Let $f_1$ and $f_2$ be two entire funcions of finite orders $\lambda_1$ and $\lambda_2$ respectively. Show that $f=f_1+f_2$ has finite order $\lambda= \max\{\lambda_1,\lambda_2\}$ if $\lambda_1\neq \lambda_2$.

I managed to show that $f$ has order $\leq \max\{\lambda_1,\lambda_2\}$ showing that for every $\epsilon>0$, $|f(z)|<\exp(|z|^{\max\{\lambda_1,\lambda_2\}+\epsilon})$ for sufficiently large, but I'm stuck in the rest of the question. I was thinking in prove that $\lambda$ must be a lower bound to the set $\{a;|f(z)|<\exp(|z|^a) \text{ for }|f| \text{ sufficiently large}\}$, so this will say that $\lambda$ it's the infimun, i.e. the order of $f$.

Thank you.

答:
Without loss of generality, $\lambda_1 < \lambda_2$. You already showed that the order $\lambda $ of  $f$ is at most $ \max (\lambda_1,\lambda_2) = \lambda_2$.

By the same argument, applied to $f_2 = f + (-f_1)$, is
$$
 \lambda_2 \le \max(\lambda, \lambda_1) = \lambda
$$
since the maximum can not be $\lambda_1$.

Therefore $\lambda = \lambda_2 = \max (\lambda_1,\lambda_2)$.


问:
Finding a faulty value in a dataset
I have prices from wholesalers for a product. But sometimes the wholesalers doing mistakes and list the product to a unrealistic price. I want to filter the unrealistic prices out. This is an example:

```<code>[0] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_0
        [price] => 37.00
    )

[1] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_1
        [price] => 204.33
    )

[2] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_2
        [price] => 209.32
    )

[3] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_3
        [price] => 211.12
    )

[4] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_4
        [price] => 243.89
    )
</code>```

As you can see wholesaler_0 is offering the product to an unrealistic price (37 compared to >200). Which is the best way to identify the faulty price?

What I have thought:

To check a price I would calculate the average of all other prices. Which would be for the first price:
 37 compared to 217.17, but what now?

I could say everything that is 50% of the average price is invalid, so (217.17/2) = 108.59 > 37
and that would make the price invalid, as every price below 108.59$ would be invalid.

BUT

Lets say we have a 1 dollar price compared to 3 dollar average price, which would mean that the price is one third of the average price. But this is a realistic price, because there is only a difference of 2 dollar. It would mean that the percentage range of accepted prices should increase if the price gets smaller.

What is the correct way to calculate this? Could someone give me an advice?

Thanks in advance!

EDIT: Here is my solution:
I have created a exponential formula which gives me the right percentage which a value can differ to be a faulty price. To build this I have used the help of a coordinate system:

80 * 0.96x +2.0

答: There's no one "correct" way to detect outliers, and heuristics that incorporate your domain-specific knowledge could work better than a general-purpose approach. But I'd consider calculating a z-score for each entry based on the mean and standard deviation of the item prices.</p>

Title: Finding a faulty value in a dataset


Title: Determine function using given sequence

Body:
It is given that f(x) is a continous function and f(0)=2

$f(x)-f(x/23)=x/23$

How can I find f(484)

This is how far I got Using replacement

$f(x)-f(x/23^n)=x/23 + x/23^2 .......x/23^n$

I guess this would result in some series but I'm not able to get to the answer

Body:
As suggested by @lulu

$F(x)=f(x/23^n)+x/23 +x/23^2+x/23^3.......$

For n tending to infinity

$F(x)=f(0)+x/23(1-x/23)$


问:
Inverse of a sixth order polynomial function
I have the following polynomial which has the following form:

$F_X \left ({x} \right)= Ax^2-Bx^4 + Cx^6$

Where $A$, $B$ and $C$ are real positive and non-zero. The roots are $\pm 161.634$, $\pm 125.993$ and $0$ . I am now in need of finding (or approximate) its inverse.

I have found this similar post from a couple of years ago where the Lagrange inversion method is recommended, but I do not think it can be applied to my situation. I found this nice paper where a method was outlined based on finding some coefficients of a pre-defined polynomial, but it yields poor results when far from zero.

This question is part of a problem that I am solving. Any help/guidance would be highly appreciated.

答:
Note that $F_X(x) = P(g(x))$, where $P(x) = Cx^3 - Bx^2 + Ax$ and $g(x) = x^2$. Over domains where they are invertible, $$F_X^{-1}(y) = g^{-1}(P^{-1}(y))$$

Since you are interested in $x > 0$, $g$ is invertible with $g^{-1}(y) = \sqrt y$. And $P$ is a cubic, for which we have well-known means of inverting. For calculations, I recommend the trigonometric method (applied to the equation $P(t) - y = 0$ to solve for $t$). Once you find this $t$, set $x = \sqrt t$ to find $x = F_X^{-1}(y)$.


答:
The side of the red square is the hypotenuse of an isosceles right triangle whose legs are half the side of the purple square:
$$r^2=\left(\frac p2\right)^2+\left(\frac p2\right)^2$$ or $$r=\frac p{\sqrt2}$$

Title: Mathematically Rotating and Scaling Squares

Body:
I need help understanding what's happening mathematically when we rotate and scale a square and fit it within itself and repeat the process. For example: I can start with a square that has a dimension of $250 \times 250$ (purple square, shown in the reference image below). Then rotate it $45$ degrees and then scale it down so that it fits perfectly within the original $250 \times 250$ square, thus scaling the square down to the size of $176.78 \times 176.78$ (red square). Then when I do it again, I get the green square, $125 \times 125$.

I want to understand what's happening here mathematically and see if there's a mathematical equation I can use to be able to accurately calculate the size of a square after it's been rotated and scaled-down based on its initial value/state. That if I start with a square that's $250 \times 250$, how can I mathematically rotate and scale the square so that it would output $176.78$? Then I take that number, $176.78$, and do the same thing done on $250$, to arrive at $125$.

I figured the blue square would be useful in some way, still not sure, but it doesn't appear to accurately "lead" me into $125$, but it's a cheat to $125$ after getting $176.78$. But I need to get to $176.78$ first.

I thought I found a solution by doing: $250$ / $70.7%$ / $2$ but that returns $176.8033946$. Why $70.7%$? Idk, was literally just trying anything that would result in something consistent and that was the closest thing I could find, but it's inaccurate and as you repeat the process, it just becomes more and more inaccurate.

Any help would be appreciated, and hopefully, this makes sense.

Reference image


问:
The Language of the Set Theory (with ZF) and their ability to express all mathematics
Accordingly, the Language of Set Theory (in this case using $ZF$ axioms) is built up with the aim to express all mathematics. Now, I know that, for example, the construction of the numbers ($\mathbf{ \mathbb{N},\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}}$) is got right from the axioms and definitions from the Language of Set Theory, so everything is inside the same theory. But, when we are dealing with especial structures like for example Euclidian Geometry, we need to state new axioms, and, apparently, it doesn't make any sense to add this axioms to our original set theory because it might be possible to find a different model, maybe completly artificial, in which some axiom of Euclidian geometry migth not hold or result in a contradiction. But Euclidian Geometry and all the different theories that one might think of are part of mathematics. So my point is that it is impossible to develop all of mathematics just from the axioms of set theory. Does it make sense? Please tell me if I'm wrong and if I'm missing something. Also I have these questions:

In building these new theories (like euclidean geometrie, etc), each theory uses the language of set theory. But also we add new axioms. So, are we talking of a different language? Are we talking of a different theory? Is the language of set theory a metalanguage in this new language? I'm totally confused.

答:
The way one can develop both Euclidean and non-Euclidean geometry in set theory is by considering the objects of geometry, such as points and lines, as kinds of sets.  For example, it is natural to consider a line as a set of points.  It is not so natural, but still possible, to consider a point as a set.  It doesn't matter much how we do this, only that it is possible via various formalisms. For example, a point $(x,y)$ in $\mathbb{R}^2$ can be considered as the pair $\{\{x\},\{x,y\}\}$ according to Kuratowski's definition of the ordered pair in set theory, and then the real numbers $x$ and $y$ can be considered as sets of rationals according to the definition of $\mathbb{R}$ in terms of Dedekind cuts, and so on.

With a bit more work, one can describe models of non-Euclidean geometry such as the pseudosphere in terms of sets.  One way to do this is to first embed these models into a higher-dimensional ambient Euclidean space.  A more abstract way is to use the notion of a manifold, which can still be described purely in terms of sets, but this involves so many steps now that most people would not want to do this.  Still, experience with such formalizations shows that it can be done in a straightforward (if tedious) manner.

In any case, one does not have to add axioms to $\mathsf{ZF}$ to study the Euclidean or non-Euclidean case.  The axioms of Euclidean or non-Euclidean geometry simply pick out subclasses of the class of manifolds.  Once you pick a formalization of "manifold" in terms of sets, it is a consequence of $\mathsf{ZF}$ that there are many examples of both types of manifold.

It is true that some things are not decided by $\mathsf{ZF}$, such as the Continuum Hypothesis.  However, the statements "there are Euclidean manifolds" and "there are non-Euclidean manifolds" are both simply theorems of $\mathsf{ZF}$.  Note that in the context of $\mathsf{ZF}$ it does not make sense to ask whether the universe itself is Euclidean, because the universe of sets is not a manifold at all.


问:
Is the following sufficient to determine (not prove) that $a_n=\frac{sin(n^2\pi)}{n^2+4}$ converges?
For the following sequences, determine whether they converge, diverge or oscillate
using the relevant convergence tests. If they converge, state their limit. DO NOT try
to prove these limits.

The question was:

$$a_n=\frac{sin(n^2\pi)}{n^2+4}$$

The way I went about solving it was saying $-1\le sin(n^2\pi)\le 1$ and that $n^2+4>1$ and therefore as n $\rightarrow \infty\ ,a_n\rightarrow 0,$ as $n^2+4>sin(n^2\pi)$. The sequence converges

Is this sufficient?

答:
What you did only leads to the conclusion that $|a_n| < 1$. However, this is not enough to say the limit exists. Consider the example $$a_n = \frac{1}{2}\sin n \ \ (\forall n\in \mathbb N)$$ where we again have $a_n < 1$ but the limit does not exist.

For your question, note that
$$-\frac{1}{n^2+4} \le \frac{\sin(\pi n^2)}{n^2+4} \le \frac{1}{n^2+4}$$
Now, remember the Sandwich (or Squeeze or Two Policemen) theorem.


Title: Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand

Body:
Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).

By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.

Body:
Let

$$s_1=\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}
$$
$$s_2=\cos\frac{6\pi}{17} + \cos\frac{10\pi}{17} + \cos\frac{12\pi}{17} + \cos\frac{14\pi}{17}
$$
and note

$$s_1s_2=2(s_1+s_2) =2\cdot (-\frac12)=-1
$$
Thus, $s_1$ and $s_2$ satisfy $s^2+\frac12 s-1=0$, which yields $s_1= \frac{-1+\sqrt{17}}4$ and, as a by-product, $s_2= \frac{-1-\sqrt{17}}4$.


答:
You can use the Squeeze Theorem.

In fact, you noticed that
$$ \frac{-1}{n²+4} ≤ a_n ≤ \frac{1}{n²+4} . $$
Moreover, we know that
$$ \lim_{n → \infty} \frac{-1}{n²+4} = 0 ,$$
$$ \lim_{n → \infty} \frac{1}{n²+4} = 0 .$$

So we can apply the theorem and say that
$$ \lim_{n → \infty} a_n = 0 .$$

Title: Is the following sufficient to determine (not prove) that $a_n=\frac{sin(n^2\pi)}{n^2+4}$ converges?

Body:
For the following sequences, determine whether they converge, diverge or oscillate
using the relevant convergence tests. If they converge, state their limit. DO NOT try
to prove these limits.

The question was:

$$a_n=\frac{sin(n^2\pi)}{n^2+4}$$

The way I went about solving it was saying $-1\le sin(n^2\pi)\le 1$ and that $n^2+4>1$ and therefore as n $\rightarrow \infty\ ,a_n\rightarrow 0,$ as $n^2+4>sin(n^2\pi)$. The sequence converges

Is this sufficient?


答:
(Caveat: I am not a professional statistician, I hope more people will chip in to answer this.)

There isn't one single correct way to calculate this. One common way is to try to:

Figure out what sort of distribution those prices have. Draw the bar chart and see if the prices resemble some known distribution. For example, the "bell" curve (which would imply Gaussian distribution). (NB there are methods to check if the resemblance is just superficial, e.g. Pearson's $\chi^2$ test - but let's not get distracted.) For prices, in particular, it is worth checking if the logarithm of the price fits Gaussian distribution better than the price itself.

Try to estimate the parameters of this distribution. For example, for Gaussian distribution you could try to calculate the mean and the standard deviation, by supplying the full sample or some subset of the sample which you trust. (If the sample is big enough and the mistakes are rare, there won't be too much difference between those two methods.) Excel, for example, has built-in functions to calculate means and standard deviations of samples.

Now for each wholesaler try to estimate the probability that their price would be that far from the mean as it happens to be.

For example, if the distribution is Gaussian, the mean is $m$ and the standard deviation is $\sigma$, as per the Wikipedia table,  about $68\%$ of all data will fall between $m-\sigma$ and $m+\sigma$ and around $95\%$ of data will fall between $m-2\sigma$ and $m+2\sigma$. In other words, it is only $5\%$ probable that a price will fall outside of this interval by chance. With each additional $\sigma$, the probability falls rapidly.

Still, it is your decision what percentage to count as a "cut-off": if, say, you are happy to say you want to declare a mistake if a wholesaler is providing a price that seems so far from $m$ that the chance of this happening is only $1\%$, as per same table you would use the interval $(m-2.576\sigma, m+2.576\sigma)$. So ultimately the maths can only go that far, but it is still your decision what you will count as a probable mistake. Adding to that:

We only model the actual distribution of prices as one of known distributions (e.g. Gaussian). The distribution may be nothing like Gaussian. If it is common to have, say, $50\%$ discounts, then the prices will probably be clustered around two different prices - one half of the other.

The sample may be too small. One common way to have a better estimate of the mean and standard deviation is to up-front "remove the outliers". One common way to do that is to fix a percentage (say $5\%$), and then only take the $5$-percentile to $95$-percentile of the samples (i.e. order/sort the prices and then dismiss the lowest and highest $5\%$). This may be impossible with less than, say, $20$ wholesalers. Going further down, and the very estimates of the mean and standard deviation will also become inaccurate.

If a price is outside the range you have calculated, it doesn't mean it is necessarily a mistake. After all, there is that small probability (which you previously chose) that it did happen by chance. On the other hand, if a price is within the range, it does not mean it is not a mistake.

Altogether, this is not really exact science. Still it is certainly way better than poring over figures and saying "this one looks odd" - at least it lets you quantify how much odd it is.

The whole area where this fits in maths is "statistical hypothesis testing" - try to find some references and see if you can make sense of the bigger theory behind it.

Title: Finding a faulty value in a dataset

Body:
I have prices from wholesalers for a product. But sometimes the wholesalers doing mistakes and list the product to a unrealistic price. I want to filter the unrealistic prices out. This is an example:

```<code>[0] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_0
        [price] => 37.00
    )

[1] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_1
        [price] => 204.33
    )

[2] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_2
        [price] => 209.32
    )

[3] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_3
        [price] => 211.12
    )

[4] => Array
    (
        [articlenumber] => 32158021n
        [wholesaler] => wholesaler_4
        [price] => 243.89
    )
</code>```

As you can see wholesaler_0 is offering the product to an unrealistic price (37 compared to >200). Which is the best way to identify the faulty price?

What I have thought:

To check a price I would calculate the average of all other prices. Which would be for the first price:
 37 compared to 217.17, but what now?

I could say everything that is 50% of the average price is invalid, so (217.17/2) = 108.59 > 37
and that would make the price invalid, as every price below 108.59$ would be invalid.

BUT

Lets say we have a 1 dollar price compared to 3 dollar average price, which would mean that the price is one third of the average price. But this is a realistic price, because there is only a difference of 2 dollar. It would mean that the percentage range of accepted prices should increase if the price gets smaller.

What is the correct way to calculate this? Could someone give me an advice?

Thanks in advance!

EDIT: Here is my solution:
I have created a exponential formula which gives me the right percentage which a value can differ to be a faulty price. To build this I have used the help of a coordinate system:

80 * 0.96x +2.0


问:
Expectation of Geometric random variable .
For a discrete random variable a negative island exists-

I need to use it, to calculate the Expectation of Geometric random variable.

Now, I don't have any data about number of trials. so, how I do it? I know $x$ can be $1,2,3\ldots$
and I know that expectation of $x$ is $1/p$. so, how can I show it?

答:
We know that $\mathbb P(X = n) = p(1-p)^{n-1}$ and from that we conclude that
$$\mathbb P(X \leq n) = p \sum_{i=1}^n (1-p)^{i} = p \frac{1-(1-p)^n}{1-(1-p)} = 1-(1-p)^n$$ from the geometric series formula, namely
$$\sum_{i=0}^{n-1} q^i = \frac{1-q^n}{1-q}.$$

Now we can conclude that $\mathbb P(X >n) = 1- \mathbb P(X \leq n) = (1-p)^n$. Now we use the given formula to calculate the expected value, using $(1-p) < 1$ (because if $p=0$ we don't have an interesting distribution anyway, therefore we have $p>0$) and the formula for the geometric series (where the $q^n$-part goes to zero because $|q|<1$ in our case):
$$\mathbb E[X] = \sum_{x=0}^{\infty} \mathbb P(X >x) = \sum_{x=0}^{\infty} (1-p)^x = \frac{1}{1-(1-p)} = \frac{1}{p}$$


问:
Conditions for which $P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z')$
Given two independent draws $(X,Z,Y)$ and $(X',Z',Y')$ from some distribution $P$, I would like to understand under which conditions does the following hold:

$$
P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z').
$$

This clearly holds when $X \perp Y | Z$ for example, but I want to be able to say that the above statement holds if and only if some condition is met.

答:
Let's rewrite the two expressions using conditional expectation
$$P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z') \tag{1}$$
\begin{align}
& \iff E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |X,Z,X',Z'\right) = E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right)  \\
& \iff E\left(E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right)|X,X'\right) = E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right) \tag{2} \\
\end{align}

$(2)$ holds true if and only if the variable $E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right)$ is dependant on ($X$ and $X'$). In other words, $E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right)\in \sigma(X,X')$ or there exists a $g$ mesurable such that $E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right) = g(X,X')$  .

Hence, we conclude that the necessary and sufficient condition for $(1)$ is   $E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right)$ is dependant on ($X$ and $X'$) or $\exists g:E\left(\mathbb{I}_{ \{ Y \le Y'  \}}  |Z,Z'\right) = g(X,X')$.


问:
One-sided derivative in $a$ of a continuous monotone function whose derivative on $(a,b]$ is bounded
Consider a function $f : [a,b] \to \mathbb{R}$ such that

$f$ is continuous and monotone on $[a,b]$,

$f'$ is bounded on $(a,b]$.

Does $f$ have a right derivative in $a$, i.e. does the limit $\lim_{x \downarrow a} \frac{f(x)-f(a)}{x-a}$ exist?

What if assume in addition that $f'$ is continuous on $(a,b]$?

Counterexample if $f'$ is not bounded: $f(x) = \sqrt{x}$ on $[0,1]$.

答:
Hint: Construct a monotone function with the required properties which satisfies $f(0)=0$ and $f(1/2^n)=1/2^n$ if $n$ is even and $f(1/2^n)=2/2^n$ if $n$ is odd,
that is, $f(1/2^{2n})=f(1/2^{2n+1})=1/2^{2n}$ for all natural numbers $n$.


问:
Solution verification for problem dealing with complex eigenvalues and linear maps

Let $A \in \mathbb{R}^{n\times n}$ be a square matrix and $\lambda \in \mathbb{C}$ be eigenvalue for $A$. Let $z=x+iy \in \mathbb{C}^{n \times 1}$ be the eigenvector corresponding to the eigenvalue $\lambda$ where $x,y \in \mathbb{R}^{n\times 1}$. Show that for the linear map $f:\mathbb{R}^{n\times 1} \to \mathbb{R}^{n\times 1}, x\mapsto Ax$ the following holds $$f(E)=E$$ where $E=\operatorname{span}(x,y)$.

The proposed solution for this was the following. Denote the real part of $z$ by $\Re(z)=x$ and the imaginary part as $\Im(z)=y$. Now $$Ax=\Re(Az) = \Re(\lambda z) = \Re(\lambda)\Re(z) \color{red}{\textrm{$ -\Im(\lambda)\Im(z)$}} = \Re(\lambda)x -\Re(\lambda)y$$

Similarly $$Ay=\Im(Az)=\Im(\lambda z) =\color{red}{\textrm{$\Im(\lambda)\Re(z) + \Re(\lambda)\Im(z)$}} = \Im(\lambda)x+ \Im(\lambda)y$$

thus $f(x), f(y) \in E=\operatorname{span}(x,y)$. So $f(E) \subset E$.

Now let $v=v_1x+v_2y \in P$. If we can find $w=ax+by$ for which $\color{red}{\textrm{$Aw=v$}}$ then $E \subset f(E)$.

Now from above $Aw = \color{red}{\textrm{$(a\Re(\lambda) +b\Im(\lambda))x + (b\Re(\lambda) -a\Im(\lambda))y = v_1x+v_2y$}}$ and this can be represented as a system of equations which can then be represented by the matrix $$\begin{bmatrix} \Re(\lambda) & \Im(\lambda) \\-\Im(\lambda) & \Re(\lambda) \end{bmatrix} \begin{bmatrix} a \\b \end{bmatrix} =  \begin{bmatrix} v_1 \\v_2\end{bmatrix}$$ which is invertible since $$\det\begin{bmatrix} \Re(\lambda) & \Im(\lambda) \\-\Im(\lambda) & \Re(\lambda) \end{bmatrix} = \lambda^2$$ and $\lambda \ne 0$. And thus there is a solution for the system. So $E \subset f(E) \implies f(E) = E$.

I cannot seem to understand where the parts marked red are coming from? Any help with this would be very much appreciated.

答: I think if you just write $\lambda =a+bi,$ with $a,b$ real, and write out the calculations, then this will be clear.  For example, $$\Re(\lambda z)=\Re((a+bi)(x+iy))=ax-by=\Re(\lambda)\Re(z)-\Im(\lambda)\Im(z)$$


问:
Stuck on proof and need help to solve it

This is not homework

I do it for fun.

I am using the SD system.

Derive $L\supset T$

$L\supset\left ( C\vee T \right ).$ Assume

$\left ( \sim L\vee B \right )\wedge\left ( \sim B\vee\sim C \right ).$ Assume

As we see from the image it is valid in SL truth tree form

There does not seem to be an easy way to derive it. I figure the the easiest method is to derive a contradiction, but how to do it after the initial set up ?

$\sim\left ( L\supset T \right ).$ Assume

For "$Q$" and "$\sim Q$" from #2, we could use $B$ and $\sim B,$ but how to separate them so they could be used ?

Any help would greatly appreaciated.

答:
A proof by contradiction for this one will be more work than doing a conditional proof, i.e. start a subproof, assume $L$, try to get to $T$, end the subproof, and perform a Conditional Introduction to get $L \supset T$

The rest of this proof is doing the Disjunctive Syllogism pattern over and over. So, if you have the Disjunctive Syllogism rule from SD+ this should actually be quite straightforward:

\begin{array}{lll}
1. & L \supset (C \lor T) & Assumption\\
2. & (\sim L \lor B) \& (\sim B \lor \sim C) & Assumption\\
3. & \quad L & Assumption\\
4. & \quad \sim L \lor B & \& \ Elimination \ Premise \ 2\\
5. & \quad \sim \sim L & Double \ Negation \ 3\\
6. & \quad B & Disjunctive \ Syllogism \ 4, 5\\
13. & \quad ...\\
? & \quad T\\
?+1 & L \supset T & \supset \ Introduction \ 3-?\\
\end{array}

In SD this will be a lot more tedious., but I'll show you how to do one of these. In particular, once you have assumed $L$ as part of your conditional proof, you should be able to get to $B$ using $\sim L \lor B$:

\begin{array}{lll}
1. & L \supset (C \lor T) & Assumption\\
2. & (\sim L \lor B) \& (\sim B \lor \sim C) & Assumption\\
3. & \quad L & Assumption\\
4. & \quad \sim L \lor B & \& \ Elimination \ Premise \ 2\\
5. & \quad \quad \sim L & Assumption\\
6. & \quad \quad \quad \sim B & Assumption\\
7. & \quad \quad \quad L & Reiteration \ 3\\
8. & \quad \quad \quad \sim L & Reiteration \ 5\\
9. & \quad \quad B & \sim \ Elimination \ 6-8\\
10. & \quad \quad B \ & Assumption\\
11. & \quad \quad B \ & Reiteration \ 10\\
12. & \quad B & \lor \ Elimination \ 4, 5-9,10-11\\
13. & \quad ...\\
? & \quad T\\
?+1 & L \supset T & \supset \ Introduction \ 3-?\\
\end{array}


答: I believe the answer you are looking for is known as the Devil's Staircase. The function is monotone and continuous on $[0,1]$, but is not differentiable at any point in the Cantor set (including failing to have a right derivative at zero). Suprisingly the function is differentiable everywhere outside the Cantor set, hence, we say the function has "derivative zero almost everywhere".

Title: One-sided derivative in $a$ of a continuous monotone function whose derivative on $(a,b]$ is bounded

Body:
Consider a function $f : [a,b] \to \mathbb{R}$ such that

$f$ is continuous and monotone on $[a,b]$,

$f'$ is bounded on $(a,b]$.

Does $f$ have a right derivative in $a$, i.e. does the limit $\lim_{x \downarrow a} \frac{f(x)-f(a)}{x-a}$ exist?

What if assume in addition that $f'$ is continuous on $(a,b]$?

Counterexample if $f'$ is not bounded: $f(x) = \sqrt{x}$ on $[0,1]$.


问:
How do I express the following limit as a definite integral?
I have difficulty understanding the limit since it doesn't include the summation notation and I don't quite understand how should I express it as an integral. Also, if I just evaluate the limit, it seems that it just gives zero, so I'm not sure if there's something that I'm not getting.

$$\lim _{n \rightarrow \infty}\left[\left(\frac{2}{n}\right)\left(\frac{2}{n}+3\right) \sqrt{5+\frac{2}{n}}+\left(\frac{2}{n}\right)\left(\frac{4}{n}+3\right) \sqrt{5+\frac{4}{n}}+\left(\frac{n}{n}\right)\left(\frac{6}{n}+3\right) \sqrt{5+\frac{6}{n}}+\cdots\left(\frac{1}{n}\right)(5 \sqrt{7})\right]$$

答:
Assuming your  limit is $$\lim_{n\to \infty} \sum_{i=0}^n {\left(\frac{2}{n}\right)\left(\frac{2i}{n} + 3\right) \sqrt{5 + \frac{2i}{n} }}$$
the integral is going to be $$\int_0^1{2\left(2 x + 3\right) \sqrt{5 + 2 x} \, dx}$$
In general, the limit $$\lim_{n\to \infty} \sum_{i=0}^n \frac{1}{n}f\left( \frac{i}{n}\right)$$
is equal to the integral $$\int_0^1{f(x)\, dx}$$


问:
A die is throw until 5 followed by 3 appears
A die is thrown until 5 followed by 3 appears. Like 156234753

What is the expected number of throws?

I've been trying to solve this problem but can't find anything similar.

Any help?

My solution:

Let $\displaystyle X$ be the random variable of the qustion (number of throws until 53).

Let the event $\displaystyle A=$after the first $\displaystyle 5$, you get a $\displaystyle 3$.

From the total law of expectation:

$\displaystyle E[ X] =E[ X|A] \cdotp P( A) +E[ W|\overline{A}] \cdotp P(\overline{A}) =E[ X|A] \cdotp \frac{1}{6} +E[ W|\overline{A}] \cdotp \frac{5}{6}$.

$\displaystyle E[ X|A] :$

We have gotten a $\displaystyle 5$ and then a $\displaystyle 3$. So we stop here. The expected number is the expected until $\displaystyle 5$ plus $\displaystyle 1$ for the $\displaystyle 3$: $\displaystyle 6+1=7$.

$\displaystyle E[ X|\overline{A}]$:

We have gotten a $\displaystyle 5$, but now not a $\displaystyle 3$ (assume we got some $\displaystyle x$).

We don't care about all the outputs before $\displaystyle x$, since they're not relevant to question anymore.

We can "pretend" then that we have just started with the output of $\displaystyle x$, and act as if that was our first result. I will count $\displaystyle 6$ for the results until the $\displaystyle 5$, and continue with same expectation.

$\displaystyle E[ X] =7\cdotp \frac{1}{6} +( 6+E[ X]) \cdotp \frac{5}{6}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E[ X] =\frac{7}{6} +5+\frac{5}{6} E[ X]\\
\frac{1}{6} E[ X] =\frac{37}{6}
\end{array}$

$\displaystyle E[ X] =37$.

I do see something wrong with my answer, the thing is the $\displaystyle x$ could be a $\displaystyle 5$ itself. This means I don't think it's very correct to say $\displaystyle E[ X|\overline{A}] =6+E[ X]$.

Because if the $\displaystyle x$ was a $\displaystyle 5$, then the possibility of ending the game next round is $\displaystyle \frac{1}{6}$ (with a $\displaystyle 3$), which of course doens't happen in $\displaystyle E[ X]$.

I believe that's my mistake, agreed?

Is the fact that my result is close to the actual result a pure casuality? Or does this mean my answer is close? I have given this answer on a test, trying to understand if it's completely wrong, or instead slighly wrong.

答:
Let $x$ be the expected additional waiting time if we have not just tossed a $5$. At the beginning, we certainly have not just tossed a $5$, so $x$ is the required expectation. Let $y$ be the expected additional waiting time if we have just tossed a $3$.

If we have not just tossed a $5$, then with probability $5/6$ we toss a non-$5$ (takes us $1$ toss) and our expected additional waiting time is still $x$. With probability $1/6$ we toss a $5$ (takes us $1$ toss) and our expected additional waiting time is $y$. Thus
$$x=1+\frac{5}{6}x+\frac{1}{6}y$$
If we have just tossed a $5$, then with probability $4/6$ we toss a non-$3$ and a non-$5$, and then our expected additional waiting time is $x$, with probability $1/6$, we toss a $5$ again and the additional waiting time is $y$ and with probability $1/6$ the game is over. Thus
$$y=1+\frac{4}{6}x+\frac{1}{6}y$$
We get $x=36$ and $y=30$


答:
For every positive $\epsilon$, there is an $N=N(\epsilon)$ such that if $p\gt N$ there is always a prime between $p$ and $(1+\epsilon)p$. An $N(\epsilon)$ that works can be specified explicitly.

For additional information, please see this article.


问:
Octahedra with four equilateral faces
Let $A_1, A_2, A_3, A'_1, A'_2, A'_3$ be the vertices of a (not-necessarily convex) octahedron; here $X'$ is the vertex not on an edge with $X$. Suppose that the four non-adjacent triangular faces  $A_1 A_2 A_3$, $A_1 A'_2 A'_3$,$A'_1 A_2 A'_3$,$A'_1 A'_2 A_3$ are equilateral with sides of lengths $x_0, x_1, x_2, x_3$.

I am interested in knowing the conditions on $x_0,x_1,x_2,x_3$ that must be satisfied for such a configuration to exist; and how unique the configuration is.  This is a preliminary step to understanding this Question on Octahedra
and the answer by @Blue.

Clearly it is necessary that $x_i>0$ for each $i$; and by using the triangle inequality on the other four faces, that $x_i+x_j>x_k$ for every 3-set $\{i,j,k\}\subset \{0,1,2,3\}$.

Question 1: are these conditions sufficient for existence? I think so, but have only a hand-waving argument and would like something better.

It seems to me that if such a configuration exists for some $x_0,x_1,x_2,x_3$ then in general if we fix  $A_2, A_3, A'_1, A'_2, A'_3$ there are two points $A_1$ which satisfy the right conditions, giving a convex octahedron and a non-convex one. Again I can only wave my hands, and my intuition may be wrong. So,

Question 2: what can  be proved about uniqueness?

Lastly, to get some feeling for which $(x_0,x_1, x_2,x_3)$ yield an octahedron, I would like to understand their geometry.

Question 3: Describe geometrically  the (subset of) the solid bounded by

\begin{align*}
x_0+x_1+x_2+x_3&=1;&\\
x_i&=0, &i=0,\dots,3; \\
x_i+x_j&=x_k &\textrm{  for every 3-set  }  \{i,j,k\}\subset \{0,1,2,3\}
\end{align*}

which yield solutions.

答:
Without loss of generality, let me assume the lengths are scaled so the shortest is 1. I'll label the four lengths $(1, a, b, c)$ withere $1 \leq a \leq b \leq c$.
Then the condition $c < a + 1$ is sufficient to ensure all the triangle inequalities are met.

Approaching it a different way from Intelligenti pauca, I came to a similar obstruction for $(1, 1, 1.8, 1.8)$. If you consider the wedge formed by two equilateral triangles of side length 1.8 meeting at a point, joined by two isosceles triangles with side lengths 1.8, 1.8, and 1, the base is a 1 by 1.8 rectangle; the two equilateral triangles of side length 1 cannot reach across it (since the height of the triangle is $\sqrt{3}/2 \approx 0.87$).

This also fails Intelligenti pauca's condition, which we can rephrase by saying that when we unfold the three non-equilateral triangles adjacent to the equilateral triangle of side length 1, the angles in the gaps must be greater than $60^\circ$.
Thus the sums of the angles in the two non-equilateral triangles at a vertex must be less than $240^\circ$.

Let me crudely edit Intelligenti pauca's picture:

The marked angle must be greater than $60^\circ$, so $\beta+ \gamma$ must be less than $240^\circ$.
If $a = 1$ and $b = c = 1.8$ as in my example, then the angles $\beta$ and $\gamma$ are about $128.3^\circ$, so we see the criterion fails.

To apply this more generally, the law of cosines tells us the angle $\beta$ is $\cos^{-1}(\frac{1+a^2-b^2}{2a})$ and $\gamma$ is $\cos^{-1}(\frac{1+a^2-c^2}{2a})$.
Since $\gamma \geq \beta$, if $\beta$ is more than $120^\circ$ we already know the situation is impossible. Hence
$$
\frac{1+a^2-b^2}{2a} > \frac{-1}{2}, \text{ or } \  a^2 + a - b^2 > -1.
$$
This fails for $1^2 + 1 - 1.8^2 = -1.24$ in my example, but not for
$1.1^2 + 1.1 - 1.6^2 = -0.25$ in Intelligenti pauca's example.

If $\gamma$ is more than $120^\circ$, we can't automatically say if the total angle is too big without knowing $\beta$.
Using the idenity $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}(AB - \sqrt{(1 - A^2)(1-B^2)})$,
we get the condition $AB - \sqrt{(1 - A^2)(1-B^2)} < \frac{-1}{2}$ where $A = \frac{1+a^2-b^2}{2a}$ and $B = \frac{1+a^2-c^2}{2a}$.

We can rearrange that condition to
$$
A^2 + AB + B^2 < \frac{3}{4}.
$$
In terms of the triangle side lengths $a$, $b$, and $c$, that expands to the truly horrendous expression
$$
3 a^4 + 3a^2 + b^4 + b^2c^2 +c^4 + 3 < 3a^2b^2 + 3a^2c^2 + 3b^4 + 3c^2.
$$
When combined with $a + 1 > c \geq b \geq a \geq 1$, this is a necessary and sufficient condition for the four lengths to appear as sides of equilateral triangles on an octahedron.

Title: Octahedra with four equilateral faces


问:
How do we find the derivative of a multivariate expression wrt another multivariate expression?
I have $f(\theta,b) = 1 - \theta x - b$.  I want to find $$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]}$$

i.e. the change in $f$ wrt $\frac{1-b}{\theta}$.  In the language of limits,
$$
\lim_{h \to 0, a \to 0} \frac{f(\theta + h, b + a) - f(\theta, b)}{\frac{1 - b - a}{\theta + h} - \frac{1 - b}{\theta}}
$$

I saw this How to find the derivative of one expression with respect to another expression

which made me think I could try
$$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]} = \frac{\frac{df(\theta,b)}{d\theta}}{\frac{d[\frac{1-b}{\theta}]}{d\theta}} = \frac{\theta^2 x}{1 - b}$$

But if I do the same strategy wrt the other variable, I get
$$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]} = \frac{\frac{df(\theta,b)}{db}}{\frac{d[\frac{1-b}{\theta}]}{db}} = \theta$$ which is a contradiction.

答:
The thing about decomposing $\frac{df(\theta,b)}{dg(\theta,b)}$ into two separate expressions, and then calculate the product is a fundamentally flawed way to calculate multivariate derivatives. It does often work, but not all the time. This article goes into more detail about it.

To respond to your question, with

$$g(\theta,b)=\frac{1-b}{\theta}$$
we know that
$$\theta=\frac{1-b}{g(\theta,b)}$$

Therefore, we plug it into $f$ and get

$$f(\theta,b)=1-\frac{1-b}{g(\theta,b)} x-b$$

And so,

$$\frac{df(\theta,b)}{dg(\theta,b)} = \frac{(1-b)x}{g(\theta,b)^2}$$

which when plugging in for $g(\theta,b)$, gets us the answer you found first,
$$\frac{\theta^2 x}{1-b}$$


问:
Prove that for all $x>0$, $f(x) < xf'(x)$
Let $f(x)$ be continuous on $[0,\infty)$ and twice differentiable on $(0,\infty)$. If $f(0)=0$ and $f''(x)>0$ on $(0,\infty)$ prove that for all $x>0$, $f(x) < xf'(x)$.

Hello, can anyone help me with this question? Thank you!

答: Basically the MVT applies to $f$ on $(0,x)$ to get: for some $c \in (0,x): f(x) - f(0) = f'(c)(x-0) \implies f(x) = xf'(c) < xf'(x)$ since $f'$ is increasing. So $f(x) < xf'(x)$ as claimed.


问:
homogeneous nonlinear functions
Give an example of degree one positively homogeneous function, (i.e. a function $f$, such that $\forall \alpha\ge0, f(\alpha x) =\alpha f(x)$)  that is not linear, and $f: \mathbb{R} \to \mathbb{R}$.</p>

答:
A simple example would be $f(x)=\left|x\right|$.

About the hint in my comment: For $x\in\mathbb{R}$, $\left|x\right| = \sqrt{x^2}$.

Actually the most general function with the desired property would be $f(x)=\alpha x+\beta\left|x\right|$ with $\alpha, \beta \in\mathbb{R}$ arbitrary real constants.

Title: homogeneous nonlinear functions


问:
$ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ is the same as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$
I want to show that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$ Perhaps they are isometrically isomorphic.

What is causing me trouble is the fact that we have an infinite-dimensional space here. Perhaps I could take an orthonormal Schauder basis for $ L^2 $, but the fact that these spaces are the same feel like it should be obvious.

My question is, is it correct that $ L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2 $ can be thought of as the same space as $L^2(\mathbb{R}, \mathbb{C}) \times L^2(\mathbb{R}, \mathbb{C})$? If so, how would one prove it?

答:
Define $V:L^2 (\mathbb{R}, \mathbb{C}) \otimes \mathbb{C} ^2\to L^2 (\mathbb{R}, \mathbb{C}) \oplus_2 L^2 (\mathbb{R}, \mathbb{C}) $ in the following way. Start with
$$
V(f\otimes (a,b))=af\oplus bf.
$$
Extend by linearity to the linear span of a basis. Now, if $f_1,\ldots,f_n$ are orthonormal, then
\begin{align}
\Big\|V\,\Big[\sum_{k=1}^nc_kf_k\otimes (1,0)+\sum_{j=1}^nd_jf_j\otimes(0,1)\Big]\Big\|^2
&=\Big\|\,\sum_{k,j=1}^nc_k f_k\oplus  d_jf_j\Big\|^2\\[0.3cm]
&=\sum_{k,j=1}^n|c_k|^2+|d_j|^2\\[0.3cm]
&=\Big\|\sum_{k=1}^nc_kf_k\otimes (1,0)+\sum_{j=1}^nd_jf_j\otimes(0,1)\Big\|^2.
\end{align}
So $V$ is isometric, which allows us to extend it to all of $L^2(\mathbb R,\mathbb C)\otimes\mathbb C^2$. Its range is dense and closed, so $V$ is surjective. That is, $V$ is an isometric isomorphism.


答:
As already suggested in a comment the system has two states: either we just have thrown a $5$ (and waiting for a 3) or any other number ($X$). The expected number of throws till the consecutive '53' is therefore:
$$\begin{align}
t_x&=1+\frac56t_x+\frac16t_5\\
t_5&=1+\frac46t_x+\frac16t_5
\end{align}$$
Solving the system we find $t_x=36$, $t_5=30$. Since at first throw we can obtain  both states the final result reads:
$$
t=1+\frac56t_x+\frac16t_5=36.
$$

Title: A die is throw until 5 followed by 3 appears

Body:
A die is thrown until 5 followed by 3 appears. Like 156234753

What is the expected number of throws?

I've been trying to solve this problem but can't find anything similar.

Any help?

My solution:

Let $\displaystyle X$ be the random variable of the qustion (number of throws until 53).

Let the event $\displaystyle A=$after the first $\displaystyle 5$, you get a $\displaystyle 3$.

From the total law of expectation:

$\displaystyle E[ X] =E[ X|A] \cdotp P( A) +E[ W|\overline{A}] \cdotp P(\overline{A}) =E[ X|A] \cdotp \frac{1}{6} +E[ W|\overline{A}] \cdotp \frac{5}{6}$.

$\displaystyle E[ X|A] :$

We have gotten a $\displaystyle 5$ and then a $\displaystyle 3$. So we stop here. The expected number is the expected until $\displaystyle 5$ plus $\displaystyle 1$ for the $\displaystyle 3$: $\displaystyle 6+1=7$.

$\displaystyle E[ X|\overline{A}]$:

We have gotten a $\displaystyle 5$, but now not a $\displaystyle 3$ (assume we got some $\displaystyle x$).

We don't care about all the outputs before $\displaystyle x$, since they're not relevant to question anymore.

We can "pretend" then that we have just started with the output of $\displaystyle x$, and act as if that was our first result. I will count $\displaystyle 6$ for the results until the $\displaystyle 5$, and continue with same expectation.

$\displaystyle E[ X] =7\cdotp \frac{1}{6} +( 6+E[ X]) \cdotp \frac{5}{6}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E[ X] =\frac{7}{6} +5+\frac{5}{6} E[ X]\\
\frac{1}{6} E[ X] =\frac{37}{6}
\end{array}$

$\displaystyle E[ X] =37$.

I do see something wrong with my answer, the thing is the $\displaystyle x$ could be a $\displaystyle 5$ itself. This means I don't think it's very correct to say $\displaystyle E[ X|\overline{A}] =6+E[ X]$.

Because if the $\displaystyle x$ was a $\displaystyle 5$, then the possibility of ending the game next round is $\displaystyle \frac{1}{6}$ (with a $\displaystyle 3$), which of course doens't happen in $\displaystyle E[ X]$.

I believe that's my mistake, agreed?

Is the fact that my result is close to the actual result a pure casuality? Or does this mean my answer is close? I have given this answer on a test, trying to understand if it's completely wrong, or instead slighly wrong.


问:
Be $f: [0, 1] \to \mathbb{R}$ differentiable, such that there is no $x \in [0,1]$ with $f(x) = f'(x) = 0$.
Question : Be $f: [0, 1] \to \mathbb{R}$ differentiable, such that there is no $x \in [0,1]$ with $f(x) = f'(x) = 0$.. Proof that set $Z = \{x \in [0,1] : f(x) = 0\}$ is finite.

My answer:

How f is a derivative function, we have that is continuous in your domain [0,1]. Let's supposse that $f(x) = f'(x)\neq 0$ and that $Z \subset [0,1]$ is infinity. How Z is infinity and bounded, exists a subsequence of elements of Z that converge for some $x_0$. Knowing this, if $f'(0) < 0 < f'(1)$ then exists $x \in (0,1)$ such that $f'(x) = 0$. But We had suppose that $f(x) = f'(x) \neq 0$ which is contradition.

That's correct?

答:
Your contradiction follows from three assumptions, so no, it's not correct.

The weird part is you have a part of a valid solution in the middle of your solution when you let $x_0$ denote a limit point of $Z$ (which you've proven exists after assuming $Z$ is infinite).  To finish, you need to show $$f(x_0)=0=f'(x_0)$$


问:
Logic behind base to number conversion
I am trying to understand and visualize the logic behind the base conversion method.

By "base" I mean how many numbers in a number system:

The decimal number system we use every day has 10 digits
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and so it is Base 10

A binary digit can only be 0 or 1, so is Base 2

A hexadecimal digit can be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C,
D,    E, F},  so is Base 16

And we can use letters {A, B, C, ..., X, Y, Z}, and we get Base 26

Like in base 26 for AAA its  1* 26*26 + 1*26 + 1

Why formula is:-  digit * ( base )^position index

答:
Say your number system has 10 digits. This means that you have numbers 0 through 9 to represent quantities 0 through 9. Once you run out of digits, but still want to represent a quantity numberically, it is conventional (at least in the West) to introduce a new digit to your number to the left of it.

What digit should that be? Well, if we have gone through digits 0-9 once, we can say that we have exhausted 1 of these quantities, so lets place a 1 and start the count from zero again, i.e. 10.

We can keep on going until we say that we have exhausted this quantity of 0-9 + 1 (one unit of ten), 9 + 1 times. Well, we've run out of digits again, so lets add a new digit to the left: 100. What does it mean, well, that we've gone through a unit of 10, 10 times. 10*10 = 100.

We can apply this reasoning for any base. 10 in hexadecimal means you've gone through 0-F one time. 100: you've gone through 10 (i.e. 16), 16 times. So, to figure out the value of 1000? You've gone through 16 * 16, 16 times, i.e. 16 * 16 * 16 = $16^3$

So, for any number base, to find out the value of a position index, you can say that you've gone up to the base at that position index a base number of times. Effectively, we can see that the forumla becomes $base^{position}$$ ^{index}$.

And for your forumla: $digit * base^{position}$$ ^{index}$. Well, say that you want to know what 20 represents in hexadecimal. Well, it simply means that you've gone from 0 to F not once, but twice. So that'll be 2 * 16 (as there are sixteen possible values from 0 to F).

And, we can of course apply this to other position indices. 200 in hex? Well, you've used up 16, 16 times, which is $16^2$, and you've done that twice, so $16^2 * 2$.

An extremely similar logic can be used when thinking about what values to apply after a floating point (we can't say decimal point as the decimal specifically refers to base 10). I am happy to elaborate on how we can find this too if need be.

Title: Logic behind base to number conversion


问:
Time complexity - Prove $\binom{n}{3} = \theta(n^3)$
$\binom{n}{3} = \frac{n!}{6(n-3)!}= \frac{n (n-2)(n-1)}{6}=t(n)$

Proving upper bound $O(n^3)= t(n) \leq c*n^3$

$\frac{n (n-2)(n-1)}{6} = \frac{n^3-2n^2+2n}{6}\leq n^3+2n \leq n^3+2n^3\leq 3n^3$

then $c=3, n=1$

My question is why did they made the expression more bigger and based on what?

答:
Typically, in time complexity in comp sci classes, you can just drop the lower ordered terms.

As for the proof: the goal is to have a function that's $cn^h$. The resulting expression from your calculation is $\frac{n^3-2n^2+ 2n}{6}$ and not in the format we want.  While it may be obvious that the numerator is a cubic polynomial and hence you can just say it's $\leq cn^3$, they show the rigorous process at arriving at that conclusion.

So let's start manipulating it with inequalities. Since it's divided by 6, it's $\leq$ than just the numerator by itself. $n^3-2n^2 + 2n \leq n^3+2n$ since $n$ is positive. $n^3 +2n \leq n^3 + 2n^3 = 3n^3 \leq cn^3$


问:
Find the equation of the tangent to the graph of the function at the given value of $x$:

Find the equation of the tangent to the graph of the function at the given value of $x$,
$$f(x) = x/(x+3), x = -5.$$

My problem is that I do not know how to find the derivative using the power rule when there is a fraction in the equation.

答:
The general formula for a tangent line at a point $a$ is given by
$$t(x)=f'(a)\cdot(x-a)+f(a).$$

We have $f(x)=x/({x-3}).$ By the quotient rule we get
$$f'(x)=\frac{3}{(x+3)^2}.$$

Now we just have to plug $-5$ in. $f(-5)=5/2$ and $f'(-5)=-3/4$. That means that the tangent line is
$$t(x)=3/4(x+5)+5/2=3/4x+25/4.$$