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Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following three equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the three equations has only zero real solution
$$
(u,v,w)= (0,0,0).
$$
How can I prove this claim?
I notice that from the equation,
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+v^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example $5$ and $13$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, we could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products of these three cases are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we may get different sum representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is, the product
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
For given $n>1$ I'm looking for the maximum number $k$ of colors such that there exists a $k$-coloring of the edges of the complete graph on $2n$ vertices with the property that a copy of every possible $k$-colored perfect matching is present in the colored complete graph.
For example, we know that there are $(n+1)$ 2-colored perfect matchings. On the other hand, we know that the complete graph can be decomposed into $(2n-1)$ perfect matchings (cf. Theorem 3.2 in 1-factorizations by Wallis). This gives $k>1$. On the other hand, for $(2n-1)$ colors we need to give each of the disjoint perfect matchings a dedicated color to accomodate the monochromatic perfect matchings. But then a perfect matching with $(n-1)$ edges of color $1$ and one edge of color $2$ cannot be present. So we have $k<2n-1$.
Can we improve upon these bounds?
**EDIT**: The problem seems to be harder than I thought, so I narrow it down.
Prove or disprove that $k=\omega(1)$.
|
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following three equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the three equations has only zero real solution
$$
(u,v,w)= (0,0,0).
$$
How can I prove this claim?
I notice that these equations are equivalent to
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+v^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
|
From the answer for this [question][1], I understand that *affine subspace* is the same as *affine subset*, however (despite the somewhat misleading question's title), it doesn't say that *affine space* is the same as *quotient space*.
Also, I found the following definition of *affine space* on [Wikipedia][2]:
> An affine space is a set A together with a vector space $\overrightarrow{A}$ and a transitive and free action of the additive group of $\overrightarrow{A}$ on the set $A$.
Which uses some terminologies, probably in Group theory, that I'm not familiar with. Some other "intuitive" explanations suggest to me that *affine space* is actually the same as *quotient space*.
Is that true that *affine space* is a synonym of *quotient space*?
[1]: https://math.stackexchange.com/questions/929945/quotient-space-and-affine-space
[2]: https://en.wikipedia.org/wiki/Affine_space
|
Is *affine space* the same as *quotient space*?
|
Find the maximum length $m$ of the sequence $a_1 a_2 \dots a_m$ such that
(1) Each $1\le a_i\le n(\in\mathbb{N})$
(2) No $1\le i \le m-1$ such that $a_{i}=a_{i+1}$
(3) Call $(x,y)$ good if $a_i=x, a_j=y, a_k=x$ for some $1\le i<j<k\le m$.
If $(x,y)$ is good, $(y,x)$ is not good.
I found a proof using induction on $n$.
I want a proof with non-inductive argument. (Probably global?)
And if the problem is famous, can you give a link to it?
|
Do you know modular arithmetic? This problem is much clearer to understand in that setting (it becomes obvious why the exponent $m$ is being taken as odd), while it seems totally opaque and strange as a plain induction problem.
Update: since you know modular arithmetic, then in order to show
$1^m + 2^m + \cdots + n^m \equiv 0 \bmod n(n+1)/2$, take cases if $n$ is even or odd.
Suppose $n$ is even, so $n(n+1)/2 = (n/2)(n+1)$ where $n/2$ and $n+1$ are relatively prime. Mod $n$, we have
$$
1^m + 2^m + \cdots + n^m \equiv 1^m + 2^m + \cdots + (n-1)^m \bmod n
$$
and $j^m + (n-j)^m \equiv j^m + (-j)^m \equiv j^m - j^m \equiv 0 \bmod n$ because $m$ is odd. Thus the sums $1^m + (n-1)^m$, $2^m + (n-2)^m$, and so on all vanish mod $n$. The only term that survives is the solitary term $(n/2)^m$, so
$$
1^m + 2^m + \cdots + n^m \equiv (n/2)^m \bmod n.
$$
Now reduce modulo $n/2$ to kill off the right side, giving us
$1^m + 2^m + \cdots + n^m \equiv 0 \bmod n/2$ when $n$ is even.
Next, show $1^m + 2^m + \cdots + n^m \equiv 0 \bmod n+1$ because $j^m + (n+1-j)^m \equiv 0 \bmod n+1$, and these pairings account for everything on the left side since there is no middle term this time: $n+1$ is odd when $n$ is even.
Thus when $n$ is even, $1^m + 2^m + \cdots + n^m$ is divisible by $n/2$ and by $n+1$, which are relatively prime, so that sum is divisble by $n(n+1)/2$.
The case of odd $n$ is similar and left to you.
|
For example, consider the sequence $2,7,3,5$. The sums of the segments of this sequence are as follows, and they are all unique:
$$2, 2+7, 2+7+3, 2+7+3+5, 7, 7+3, 7+3+5, 3, 3+5, 5$$
Can we generate such a sequence using the first $n$ prime numbers for any given $n$?
Here's my attempt:
Initially, consider the case of natural numbers instead of prime numbers. It can be shown that we cannot construct such a sequence when $n\geq 4$. This is because the sum of natural numbers up to $n$ is $\frac{n(n+1)}{2}$, which equals the number of segments. Therefore, there exist segments whose sum is $\frac{n(n+1)}{2}-1$ and $\frac{n(n+1)}{2}-2$, and we need both $1$ and $2$ at the ends. Considering the neighbors of $1$, since the sum of the number and $1$ cannot be $1$ to $n$, the neighbor must be $n$. Similarly, the neighbor of $2$ must be $n-1$, but $2+(n-1)=n+1$, which coincides with the neighbor of $1$.
I also searched for the case of odd numbers using a computer, and found that we can construct a sequence for odd numbers up to $29$, but we cannot construct a set of odd numbers from $1$ to $31$.
Additionally, I conducted a computer search for the case of prime numbers and found that we can construct a sequence for prime numbers up to $n\leq40$. Here are the results, showcasing the construction of the smallest sequence in lexicographical order. However, upon examining this outcome, it appears that the sequence tends to grow larger in lexicographical order as $n$ increases, suggesting the possibility that we may not be able to construct such a sequence at some point.
```
2
2 3
2 5 3
2 7 3 5
2 7 3 5 11
2 7 3 5 11 13
2 7 3 11 5 13 17
2 7 3 11 17 5 13 19
2 7 3 17 5 13 23 19 11
2 7 3 23 5 17 19 11 13 29
2 7 3 23 11 13 19 29 31 5 17
2 7 3 23 13 5 29 11 17 37 31 19
2 7 3 23 13 11 19 37 31 29 5 17 41
2 7 3 23 13 11 31 41 29 5 17 43 19 37
2 7 3 23 13 17 37 5 19 47 11 41 29 31 43
2 7 3 23 13 19 31 11 41 53 29 43 47 37 17 5
2 7 3 23 13 19 31 11 29 53 17 5 59 47 41 43 37
2 7 3 23 13 19 31 29 11 5 61 53 59 41 47 17 37 43
2 7 3 23 13 29 31 19 5 47 41 53 61 43 67 11 59 17 37
2 7 3 23 13 29 31 19 5 59 41 61 17 53 37 43 67 71 11 47
2 7 3 23 13 29 37 17 41 43 31 61 67 19 71 59 47 73 5 11 53
2 7 3 23 13 29 41 11 5 53 31 19 73 67 79 47 43 61 17 37 59 71
2 7 3 23 13 29 43 17 37 41 83 67 79 31 19 71 61 73 11 59 47 53 5
2 7 3 23 13 29 43 19 31 47 17 59 89 83 67 71 53 5 61 37 73 79 11 41
2 7 3 23 13 29 43 19 31 53 11 79 73 67 89 71 41 61 5 83 59 47 97 37 17
2 7 3 23 13 29 43 19 31 53 11 5 97 83 73 101 71 41 47 89 59 67 61 37 17 79
2 7 3 23 13 29 43 19 31 59 5 53 97 103 101 73 89 17 83 71 11 41 61 67 79 47 37
2 7 3 23 13 29 43 19 31 59 5 53 47 79 97 83 71 107 101 89 17 37 61 67 73 11 41 103
2 7 3 23 13 29 43 19 31 67 17 11 41 107 101 97 83 89 103 59 37 109 5 73 79 61 71 53 47
2 7 3 23 13 29 43 19 31 67 17 11 59 89 107 73 71 53 47 109 37 113 79 103 97 41 101 61 5 83
2 7 3 23 13 29 43 47 5 11 53 17 59 67 97 103 113 109 73 83 101 79 89 61 41 37 107 71 127 19 31
2 7 3 23 13 29 43 47 5 11 53 17 59 71 109 103 19 31 67 97 131 127 79 83 113 41 101 107 37 89 61 73
2 7 3 23 13 29 43 47 31 61 37 17 79 73 97 5 71 83 131 109 113 67 89 59 103 19 107 127 101 137 53 41 11
2 7 3 23 13 29 43 47 31 67 17 11 53 5 137 71 101 79 103 19 127 109 139 83 107 37 59 97 131 89 113 41 61 73
2 7 3 23 13 29 43 47 31 67 17 11 53 5 137 113 107 97 71 89 101 109 131 83 73 103 37 19 127 149 59 41 61 139 79
2 7 3 23 13 29 43 47 31 67 17 11 53 61 79 73 137 113 71 59 109 103 139 151 101 127 149 89 131 5 83 107 37 97 19 41
2 7 3 23 13 29 43 47 31 67 17 11 53 5 83 137 149 89 71 59 131 97 107 73 61 79 139 109 113 101 151 103 157 37 127 19 41
2 7 3 23 13 29 43 47 53 5 11 71 41 61 113 17 89 73 139 157 163 107 59 83 109 37 101 137 31 149 97 103 151 19 67 131 127 79
2 7 3 23 13 29 43 53 37 79 5 11 47 41 89 137 101 139 167 83 113 31 163 71 151 107 109 97 127 73 157 61 131 59 17 149 103 67 19
2 7 3 23 13 29 43 61 19 37 59 47 41 11 131 73 97 89 107 101 31 137 167 139 163 109 149 173 151 127 79 103 157 17 67 113 53 83 71 5
```
|
Is it possible to construct a sequence using the first $n$ prime numbers such that each segment has a unique sum?
|
I read in some lecture notes the following definition of contraction rate:
Definition (Posterior rate of contraction) The posterior distribution $\Pi_n\left(\cdot \mid X^{(n)}\right)$ is said to contract at rate $\epsilon_n \rightarrow 0$ at $\theta_0 \in \Theta$ if $\Pi_n\left(\theta: d\left(\theta, \theta_0\right)>M \epsilon_n \mid X^{(n)}\right) \rightarrow 0$ in $P_{\theta_0}^{(n)}$ probability, for a sufficiently large constant $M$ as $n \rightarrow \infty$.
**Q**: Does this formulation imply that under the posterior: $\epsilon_n^{-1} d\left(\theta, \theta_0\right) \rightarrow 0$?
|
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following three equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Surprisingly, computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the three equations has only zero real solution
$$
(u,v,w)= (0,0,0).
$$
How can I prove this claim?
I notice that these equations are equivalent to
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+v^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
|
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following three equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Surprisingly, computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the three equations has only zero real solution
$$
(u,v,w)= (0,0,0).
$$
How can I prove this claim?
I notice that these equations are equivalent to
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+w^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
|
I am looking at the Tor functor for $R$-modules $M,N$. If, for example, $M$ or $N$ is flat then $\text{Tor}_i^R(M,N) = 0$ for all $i > 0$ so this functor can be seen as a "measure" of flatness (and it makes sense that a few criteria for flatness can be stated in terms of Tor, as in Eisenbud)
But I was wondering, does it precisely measure failure of exactness under tensor producting at the $i$-th resolution? That is to say, if $\text{Tor}_i^R(M,N) = 0$ for the first time at $i = 3$, then should it be understood that $ -\otimes M $ preserves exactness only until the third resolution? Since if $\text{Tor}_i^R(M,N) = 0$ at $i = 2$ lets say, then we know $\text{ker}(\varphi_2) = \text{Im}(\varphi_3)$ in our free (projective) resolution so exactness at the 2nd spot was preserved. And, if $\text{Tor}_i^R(M,N) = 0$ then is $\text{Tor}_j^R(M,N) = 0$ for $j > i$ as well?
Some examples suggest that, as
$$\text{Tor}_i^R(R/(x), M) = \begin{cases}
M/xM & i = 0, \\
\text{ann}(M) & i = 1, \\
0 & i > 1
\end{cases}$$
And also $$ \text{Tor}_i^R(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z}/m\mathbb{Z}) = \begin{cases}
\mathbb{Z}/(m,n)\mathbb{Z} & i = 0,1 \\
0 & i > 1
\end{cases} $$
But I don't see how to prove it, apriori I don't see any reason for it to be true besides these examples.
|
Does $\text{Tor}_i^R(M,N) = 0$ imply $\text{Tor}_j^R(M,N) = 0$ for $j > i$?
|
Let $V$ be an $n$-dimensional vector space, and let $M\subset V\oplus V$ be an $n$-dimensional subspace. It is known that I can identify $M$ (non-uniquely) with a frame, i.e., with an injective linear operator
$$\left( \begin{array}{c} X \\ Y \end{array} \right):V\rightarrow V\oplus V$$ whose range is $M$ (here, $X,Y$ are $n\times n$ matrices).
Now, assume that $M=M_1\oplus M_2$ where $M_1,M_2$ are some subspaces of $V\oplus V$ (no specific restrictions about dimension). Is there some natural way to express the frame associated with $M$ using the "smaller" subspaces $M_1, M_2$?
Naively, $M_1, M_2$ don't have an associated frame (since their dimension is not $n$), but perhaps there is still some simple way to express the frame of $M$ using information about $M_1, M_2$?
If it helps somehow, I am especially interested in the case where the subspace $M$ is Lagrangian with respect to the standard symplectic form. But feel free to ignore that if you wish and just consider it as a linear algebra problem.
|
Does the frame of a direct sum of subspaces split somehow?
|
When the boat's speed was $45$ km/h, its engine suddenly stopped.
$7$ minutes after the engine stopped, the boat's speed was $12$ km/h.
It is known that on the boat, in the opposite direction to its movement, the frictional force of the water acts, which is proportional to the speed of the boat.
Find the law of the boat's motion in a river with a constant current of $3$ km/h in the direction of the boat's motion.
After how long will the speed of the boat be $4$ km/h?
My thoughts:
$$
\left\{
\begin{array}{c}
f'(x)=45-3f(x) \\
f'(7)=45-3f(7)=12 \rightarrow f(7)=11\\
\end{array}
\right.
$$
$$f(x)=15-4e^{21-3x}$$
so $$ f'(x)=12e^{21-3x}$$
and we need to solve $$f'(x)=4$$
so
$$12e^{21-3x}=4 \rightarrow x=7+\frac{\ln(3)}{3} = 7.366$$
|
My book states that it's not always true that:
$$E(\sum_{i=1}^{\infty} X_i) = \sum_{i=1}^{\infty} E(X_i)$$ and what makes it not true in general is this equality:
$$E(\sum_{i=1}^{\infty} X_i) = E(\lim_{n\to\infty}(\sum_{i=1}^{n}X_i)) \stackrel{?}{=} \lim_{n\to\infty}E(\sum_{i=1}^{n}X_i)$$
Two special cases that this can be true are:
1. $X_i$ are all nonnegative
2. $\sum_{i=1}^{\infty}E(|X_i|)$ is unbounded
I have no idea how to prove the relation in these two conditions. What am I missing here ?
|
>In right triangle $ABC$, we have $\angle ACB=90^{\circ}$, $AC=2$, and $BC=3$. Medians $AD$ and $BE$ are drawn to sides $BC$ and $AC$, respectively. $AD$ and $BE$ intersect at point $F$. Find the area of $\triangle ABF$.
<hr>
I aim to use the Shoelace Theorem to calculate the area of $\triangle ABF$, but I'm stuck on determining the coordinates of point $F$.
Could someone assist me in finding the coordinates of point $F$ so that I can proceed with calculating the area of $\triangle ABF$ using the Shoelace Theorem? Or is there another method altogether?
[1]: https://i.stack.imgur.com/xQrsV.png
|
Let $V$ be an $n$-dimensional vector space, and let $M\subset V\oplus V$ be an $n$-dimensional subspace. It is known that I can identify $M$ (non-uniquely) with a frame, i.e., with an injective linear operator
$$\left( \begin{array}{c} X \\ Y \end{array} \right):V\rightarrow V\oplus V$$ whose range is $M$ (here, $X,Y$ are $n\times n$ matrices).
Now, assume that $M=M_1\oplus M_2$ where $M_1,M_2$ are some subspaces of $V\oplus V$ (no specific restrictions about dimension). Is there some natural way to express the frame associated with $M$ using the "smaller" subspaces $M_1, M_2$?
Naively, $M_1, M_2$ don't have an associated frame (since their dimension is not $n$), but perhaps there is still some simple way to express the frame of $M$ using information about $M_1, M_2$?
Using some elementary arguments I can show the following:
[![enter image description here][1]][1]
(so $M$ splits into the direct sum of restricted images to $M_1, M_2$). But I am looking for something which is possibly a bit stronger.
If it helps somehow, I am especially interested in the case where the subspace $M$ is Lagrangian with respect to the standard symplectic form. But feel free to ignore that if you wish and just consider it as a linear algebra problem.
[1]: https://i.stack.imgur.com/OYSOv.png
|
For this question I am asked to find the *radius and centre-point* of the
circle of curvature for the following function:
> $β7.87e^{2.65x}$
I calculate the radius correctly with the formula: $R =\frac{1}{Ο}$
which is:
R = 0.9838228573
[working out for radius][1]
However, when it comes to finding *the centre-point* I have been getting it wrong every time.
I use this formula I found online to do so:
> $x+R\times\frac{dy}{dx}, y+R$
does anyone know another method I could use to find the centrepoint ?
I have seen vector variations which come to the right answer but I dont understand them very much :/
Thank you all ! for any help it is appreciated
[1]: https://i.stack.imgur.com/647Mo.png
|
Is there a way to define in a canonical way a $p$-adic valuation on the complex numbers or at least on the algebraic numbers, that extends the $p$-adic valuation on $\mathbb{Q}$? That is, is there a canonical embedding of $\overline{\mathbb{Q}}$ to $\overline{\mathbb{Q}_p}$?
To give a concrete example, take the polynomial $P(Y)=Y^2+2Y+3$ which has two complex roots $z_1$ and $z_2$ that are respectivelly $-1\pm\sqrt{-2}$. Over the $3$-adics the same polynomial has two roots in $\mathbb{Q}_3$ because $\sqrt{-2}$ is in $\mathbb{Q}_3$ (Proposition 3.4.3 in Gouvea's Book $p$-adic Numbers: an invertible element $b\in\mathbb{Z}_p^*$ is a square iff its reduction in $\mathbb{F}_p$ is a square). These roots are in fact in $\mathbb{Z}_3$, and with $x=\sqrt{-2}$ one has $x=\pm1$ mod $3\mathbb{Z}_3$ (they are two possibilities for quadratic residue of $1$ in $\mathbb{F}_3$) so that there is a root $r_1$ with $v(r_1)=1$ and a root $r_2$ with $v(r_2)=0$. So it is tempting to say that one of the complex roots $z_1$ or $z_2$ has valuation 1 and the other has valuation 1.
I guess all this stuff is not reasonable because the two complex roots are interchangeable from the point of view of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, whereas it is not the case for the two roots in $\mathbb{Q}_p$: $P(Y)$ is irreducible over $\mathbb{Q}$ and not over $\mathbb{Q}_3$. But maybe one can pass through $\mathbb{Q}_3[i][\sqrt{2}]$ where the $p$-adic valuation extends, and calculate $v(i\sqrt{2})$ (I'm not comfortable enough on $p$-adic extensions to do that quickly). But certainly the problem comes again in another form... Any ideas?
|
How to solve this limit with e power in both numerator and denominator?
|
I have $F(G(x))=G(x)$, where $x$ is any real number, and function $F$ can be computed if value $G(x)$ is given. I have also proved that function $G$ is monotone with respect to $x$ and $G(x)\in[0,1]$.
I want to obtain a mapping $G$ (or an algorithm G).
Is this a fixed-point problem? How to solve it, either mathematically or by algorithm?
|
$F(G(x))=G(x)$. what is this problem?
|
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following three equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Surprisingly, computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the solution of these equations is
$$
(u,v,w)= (0,0,0).
$$
[![enter image description here][1]][1]
How can I prove this claim?
I notice that these equations are equivalent to
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+w^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
[1]: https://i.stack.imgur.com/y5Tqw.jpg
|
A and B play sudoku until A has 3 consecutive win. Find expected value of the number of matches that they played. Known that there are three cases: win, tie, lose and the probability is uniform.
I've tried using recursion approach such that:
Assume f(x) is the probability that A has 3 consecutive win in x matches. We have:
$f(3) = (\frac{1}{3})^3$
$f(4) = (\frac{2}{3})(\frac{1}{3})^3$
$f(5) = (\frac{2}{3})(\frac{1}{3})^3$
$f(6) = (\frac{2}{3})(\frac{1}{3})^3$
For x > 6, assume $x = 3*n + m$ $(m = 1,2,3)$, we have:
$f(x) = (1-f(3))^{n-1}*f(m+3)$
Then, expected value is:
$E = 3*f(3)+4f(4)+5f(5)+6f(6)+\sum_{n=2}^{\infty}(3k+3)(1-f(3))^{n-1}f(6)+\sum_{n=2}^{\infty}(3k+1)(1-f(3))^{n-1}f(4)+\sum_{n=2}^{\infty}(3k+2)(1-f(3))^{n-1}f(5)=166$
The expected value is too large. Is my solution wrong?
|
Let $V$ be an $n$-dimensional vector space, and let $M\subset V\oplus V$ be an $n$-dimensional subspace. It is known that I can identify $M$ (non-uniquely) with a frame, i.e., with an injective linear operator
$$\left( \begin{array}{c} X \\ Y \end{array} \right):V\rightarrow V\oplus V$$ whose range is $M$ (here, $X,Y$ are $n\times n$ matrices).
Now, assume that $M=M_1\oplus M_2$ where $M_1,M_2$ are some subspaces of $V\oplus V$ (no specific restrictions about dimension). Is there some natural way to express the frame associated with $M$ using the "smaller" subspaces $M_1, M_2$?
Using some elementary arguments I can show the following:
[![enter image description here][1]][1]
(so $M$ splits into the direct sum of restricted images to $M_1, M_2$). But I am looking for something which is possibly a bit stronger - I'd rather not compute the entire frame on $M$ (and then take the restriction), but to compute things only on the smaller subspaces if possible. Naively, $M_1, M_2$ don't have an associated frame (since their dimension is not $n$), but perhaps there is still some simple way to express the frame of $M$ using only information about $M_1, M_2$?
If it helps somehow, I am especially interested in the case where the subspace $M$ is Lagrangian with respect to the standard symplectic form. But feel free to ignore that if you wish and just consider it as a linear algebra problem.
[1]: https://i.stack.imgur.com/OYSOv.png
|
I'm trying to evaluate the following integral, but I think I don't do something correctly.
$ \int\frac{\cos x}{1+x^2}\;dx\ $
I tried in 2 ways:
I defined $\gamma = \gamma_1 + \gamma_2 $, where $ \gamma_1 = [-R, R], \gamma_2 = R e^{it}, t \in [0, \pi] $.
so the integral on $\gamma_2$ approaches zero, and by residue theorem, we see that
$ \int_{\gamma} f(z)dz = 2\pi i Res(f, i) = 2\pi i \frac{cosz}{z+i} $.
Letting $ z=i$ does not give me the correct result, since $ \cos i = \frac{e + \frac{1}{e}}{2} $
The second way: by partial decomposition, I see that $$ \int\frac{\cos z}{1+z^2}\;dx\ = \int\frac{i\cos z}{2(z+i)}\;dx\ + \int\frac{-i\cos z}{2(z-i)}\;dx\ $$
and by Cauchy's theorem for integrals, the first integral is zero, and the latter is $ \pi i (-i) \cos i $, which is still not the correct result..
What do I miss here?
|
I have this integral: $\int_{-\infty}^{\infty}\frac{xe^{a - x}}{(1+e^{a-x})^2}dx$
I've tried to find its: $\int_{-\infty}^{\infty}\frac{xe^{a - x}}{(1+e^{a-x})^2}dx = | a-x = t| = -\int_{-\infty}^{\infty}\frac{(a-t)e^{t}}{(1+e^{t})^2}dx$, then I used integration by parts:
Let $u = a-t$ so $du = -dt$ and let $dv = \frac{e^{t}}{(1+e^{t})^2}dx$ so $v = \int_{}^{}\frac{e^{t}}{(1+e^{t})^2}dx = |1+e^t = s| = \int \frac{1}{s^2}dx =-\frac{1}{s} = - \frac{1}{1+e^t}$, then
-$\int_{-\infty}^{\infty}\frac{(a-t)e^{t}}{(1+e^{t})^2}dx = -(-\frac{a-t}{1+e^t}|^\infty_{-\infty} - \int_{-\infty}^{\infty}\frac{dx}{1+e^t})$ and here I have a problem: $\frac{a-t}{1+e^t}|^\infty_{-\infty} = \frac{a-\infty}{1+exp(\infty)} - \frac{a+\infty}{1+exp(-\infty)}$.
**As for me I was mistaken somewhere, but I don't understand where. I've tried to go another way - use $exp(a-x) = exp(a)*exp(-x)$ and there is the same problem.**
**QUESTION: where I was wrong?**
*Thanks for any help.*
|
Well I give an answer [here][1] and I want to know if it's true for a special example
*Answer :*
Not an answer just some speculation about gamma function and Bernstein's polynomial which is of independent interest here but could help .
Using fallaciously the inverse function of $f(x)$:
$$f\left(x\right)=e^{x^{4}-\ln\left(\frac{1}{y}\right)}!,x!=\Gamma(x+1),0<y<1$$
Then using Bernstein form it seems we have $\forall x>0$:
$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{f\left(\frac{k}{n}\right)n!}{k!\left(n-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(1+x^{\frac{1}{6}}\right)\right)^{\left(n-k\right)}=y!$$
*End of the answer :*
Now the special example :
Let $f(x)$ as above and $y=1/2$ have we for $0<x\leq 1$ :
>>$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{f\left(\frac{k}{n}\right)n!}{k!\left(n-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(1+x^{\frac{1}{6}}\right)\right)^{\left(n-k\right)}=\sqrt{\frac{\pi}{4}}$$
Is this special example true ?
Remark : I have no idea to tackle it rigourosly .
Addendum :
It's seems not always true using Askey-Kasper inequality for $x=0$ as we have :
Let :
$$f\left(x,y\right)=\frac{x!}{y!\left(x-y\right)!},h\left(x\right)=\sum_{n=0}^{p}\frac{\sum_{s=0}^{n}f\left(n+a,n-s\right)f\left(n+b,s\right)\left(\frac{x-1}{2}\right)^{s}\left(\frac{x+1}{2}\right)^{n-s}}{\frac{\left(n+b\right)!}{n!\left(b\right)!}},a>0,b>0,0<x\leq 1$$
$$m\left(x\right)=\sum_{k=0}^{P}\frac{r\left(\frac{k}{P}\right)P!}{k!\left(P-k\right)!}\left(\ln\left(x^{\frac{1}{6}}+1\right)\right)^{k}\left(1-\ln\left(x^{\frac{1}{6}}+1\right)\right)^{\left(P-k\right)},r\left(x\right)=h\left(e^{x^{4}-\ln\left(10\right)}\right)$$
We have some problem around zero $m'(0)=\infty$
[1]: https://math.stackexchange.com/questions/4884922/one-of-the-numbers-zeta5-zeta7-zeta9-zeta11-is-irrational/4890066#4890066
|
> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example $5$ and $13$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, we could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products of these three cases are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d = 1$ or has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we may get different sum representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is, the product
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
|
I don't really get what's the difference between them. What does each thing physically and mathematically signify?
Aren't both things just a dot product with the del operator?
|
What is gradient? What's the difference between gradient and divergence?
|
Mathematically, the **gradient** is a property of a scalar function $f:\mathbb R^n\to\mathbb R$, found as
$$\mathrm{grad} (f)=\nabla f=\begin{bmatrix}\frac{\partial f}{\partial x_1}\\\frac{\partial f}{\partial x_2}\\ \frac{\partial f}{\partial x_3}\\\vdots\end{bmatrix}.$$
In physical terms you can think of it as the equivalent of the derivative of a function of one variable. It is "the derivative" or "the slope" in higher dimensions, so to speak. For instance, for a function of two variables $f:\mathbb R^2\to\mathbb R$, which represents a surface when plotted, the gradient is a vector arrow that always points in the steepest direction from any point.
The **divergence** on the other hand is a property of a vector function $V:\mathbb R^n\to\mathbb R^n$, which more specifically is called a vector field, and is found as
$$\mathrm{Div}(\mathbf V)=\nabla \cdot \mathbf V=\frac{\partial f}{\partial x_1}+\frac{\partial f}{\partial x_2}+\frac{\partial f}{\partial x_3}+\cdots$$
Physically, if you think of a vector field as representing e.g. the wind velocity at every point, then the divergence can be thought of as indicating a local expansion rate (do the points tend to move away from or closer to each other at a point - i.e., does the density decrease or increase at a point?).
Both the gradient concept and the divergence concept can be defined using the nabla operator $\nabla$, which might be what you are referring to. But note that this operator is used in simple *scalar multiplication* in the case of the gradient because we there are dealing with a scalar function, whereas it is used in a *dot product* in the case of the divergence where we are dealing with a vector product. Also note have the outputs are entirely different types of objects that have different dimensions - the gradient is a vector whereas the divergence is a scalar. So I would not try to think of them as related.
Physically, scalar functions and vector fields should better be thought of as entirely different mathematical concepts that in turn can represent entirely different and essentially unrelated physical concepts, such as mountain surfaces vs. wind directions.
|
A group of *2n* individuals consisting of *n* couples, are randomly arranged at a round table. You are required to find an upper bound for the probability that none of the couples are seated next to each other.
**Solution:**
This is a combinatorial problem. Let's denote the total number of ways to arrange *2n* individuals around a round table as T, and the number of ways to arrange them such that no couples are seated next to each other as S. The probability that none of the couples are seated next to each other is then given by $P = \displaystyle\frac{S}{T}.$
1. Total arrangements (T): Since the table is round, we can fix one person and arrange the remaining 2n-1 people. This can be done in (2n-1)! ways.
2. Arrangements with no couples together (S): This is a bit trickier. We can think of each couple as a single entity first. So we have n entities to arrange, which can be done in (n-1)! ways (again, because the table is round). Now, within each couple, we have 2 people that can be arranged in 2! ways. Since we have *n* couples, the total number of arrangements is $(n-1)! \times (2!)^n$.
So, the probability $P =\displaystyle\frac{S}{T} = \frac{[(n-1)! * (2!)^n]}{(2n-1)!.}$
This is the exact probability, but author asked for an upper bound. An upper bound for this probability can be obtained by using the fact that $(n-1)! \leq n!$ and $(2n-1)! \geq (n!)^2$ for $n \geq 1.$ So, we have:
$P \leq \displaystyle\frac{[(n!)\times(2!)^n]}{(n!)^2} =\displaystyle\frac{(2^n)}{n!}.$
This is an upper bound for the probability that none of the couples are seated next to each other. Please note that this is a very loose upper bound, and the actual probability will be much less than this.
Is this answer correct?
Assuming in the above exercise, that *n* is large, how can we approximate the probability that exactly *k* of the couples are seated next to each other?
|
At first, it should be noted that Linear Programming (LP) **has benefited from** many mathematical tools, for a review, see e.g., [Algebraic and Topological Tools in Linear Optimization][1] and [Integer Programming and Number Theory][2].
Moreover, due to its widespread usage, LP has **become a strong driver** for the study of challenging math problems such as $d$-step conjecture for polyhedrons (see this Acta Mathematica paper: [The d-step conjecture for polyhedra of dimension $d<6$][3]) or the development of branches of mathematics such as [convex analysis][4] and tropical geometry (see Chapter 1.2 in the book [Introduction to Tropical Geometry][5] and pages 1031-1032 in [this paper][1]).
The relationship between LP and mathematics is not one-sided and **LP and its well-known extensions**, such as convex optimization and integer programming, has been applied in different fields of mathematics.
- Convex Optimization, a nice non-linear extension of LP that can be solved efficiently, has found many applications in **geometry** such as projection on a set, distance between sets, Euclidean distance and angle problems, extremal volume ellipsoids, etc. (see chapter 8 of the book [Convex Optimization][6])
- Integer programming, a discrete counterpart of LP that can be solved efficiently using LP-based branch-and-bound, has been also applied for analyzing and solving difficult **graph-theoretic** problems (see chapters on applications of the book [Combinatorial Optimization][7]). The [paper][8] pointed out by @MishaLavrov in a comment is an interesting usage of integer programming to disprove conjectures and improve several lower bounds in **extremal combinatorics**.
- Semidefinite Programming (SDP) is a very recent extension of LP, for which [an efficient algorithm designed by Farid Alizadeh][9]. In SDP, the order $\ge$ is replaced by $\succeq$. A number of recent papers provided direct or computer-assisted proofs based on **Flag Algebras**, allowing one to establish bounds through a double-counting argument by solving SDP models to prove various results in extremal combinatorics (see this [paper][10] and references therein).
- The **major theoretical breakthrough** that **LP models** with rational input data can **be solved in polynomial time** is a key result in **theoretical computer science** (see The New York Times paper [Leonid Khachiyan, 52; Helped to Advance Computer Math][11] and chapter 3 of the book [7]).
- Another interesting application of LP is in **probability theory** where the [Wasserstein distance][12] (used to compare different probability measures) can be obtained from a transportation problem when the sample space is finite (a well-known LP problem taught in undergrad LP courses, can be solved in polynomial time).
- The second application of LP in **probability theory** is to **derive sharp closed-form bounds** for the probability of the union of events or other quantity with partial information**, see these interesting papers [13], [14], [15], and references therein.
- LP enables us to obtain a [Nash equilibrium][16] for a two-player [zero-sum game][17] with finite strategy sets **in polynomial time**, which is of high theoretical importance as still no polynomial algorithm has discovered to find a Nash equilibrium of two-player general-sum game with finite strategy sets. In fact, this brilliant application of LP in **game theory** warmly opened the door to develop the amazing concept of [duality in optimization][18] (inspired by John von Neumann's min-max concept) with many theoretical applications in other fields.
- LP bounding has been used in the **sphere packing problem** (what fraction of $\mathbb R^n$ can be covered by congruent balls not intersecting except along their boundaries). Here, I list a number of papers published in $\color{red}{\text{Annals of Mathematics}}$, the most prestigious journal in mathematics, on sphere packing in which **linear programming** has been successfully used:
- [Universal optimality of the $E_8$ and Leech lattices and interpolation formulas][19],
- [The sphere packing problem in dimension 24][20]
- [New upper bounds on sphere packings I][21]
I quote the following statement from the third [paper][21]:
> **Linear programming bounds** (Delsarte, 1972) **are the most powerful
known technique** for producing upper bounds in **such problems**. In particular, Kabatiansky and Levenshtein (1978) use this technique to prove the best bounds known for sphere packing density in high dimensions.
[1]: https://www.ams.org/journals/notices/201907/rnoti-p1023.pdf
[2]: https://nyaspubs.onlinelibrary.wiley.com/doi/abs/10.1111/j.1749-6632.1981.tb51145.x
[3]: https://projecteuclid.org/journals/acta-mathematica/volume-117/issue-none/The-d-step-conjecture-for-polyhedra-of-dimension-d6/10.1007/BF02395040.full
[4]: https://en.wikipedia.org/wiki/Convex_analysis#:~:text=Convex%20analysis%20is%20the%20branch,A%203%2Ddimensional%20convex%20polytope.
[5]: https://web.archive.org/web/20140301003344id_/http://homepages.warwick.ac.uk/staff/D.Maclagan/papers/TropicalBook28.2.14.pdf
[6]: https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf
[7]: https://link.springer.com/book/10.1007/3-540-29297-7?page=2#toc
[8]: https://www.sciencedirect.com/science/article/pii/S0097316519301116
[9]: https://www.informs.org/Recognizing-Excellence/Award-Recipients/Farid-Alizadeh
[10]: https://ojs.aaai.org/index.php/AAAI/article/view/26470
[11]: https://www.nytimes.com/2005/05/23/obituaries/world/leonid-khachiyan-52-helped-to-advance-computer-math.html
[12]: https://en.wikipedia.org/wiki/Wasserstein_metric
[13]: https://www.sciencedirect.com/science/article/pii/S0166218X04001386
[14]: https://pubsonline.informs.org/doi/abs/10.1287/moor.2014.0657
[15]: https://epubs.siam.org/doi/abs/10.1137/21M1408294
[16]: https://en.wikipedia.org/wiki/Nash_equilibrium#:~:text=and%20wireless%20communications.-,History,produce%20to%20maximize%20its%20profit.
[17]: https://en.wikipedia.org/wiki/Zero-sum_game
[18]: https://en.wikipedia.org/wiki/Duality_(optimization)
[19]: https://annals.math.princeton.edu/2022/196-3/p03
[20]: https://annals.math.princeton.edu/2017/185-3/p08
[21]: https://annals.math.princeton.edu/2003/157-2/p09
|
I encountered very similar and related question. That is whether the following conditions are equivalent to each other,
(1) $f$ is $L$-smooth ($\|f(x) - f(y)\| \leq L \| x-y \|$)
(2) $|D_f(x,y)| \leq \frac{L}{2}\|x-y\|^2$
(3) $\|\nabla^2 f\| \leq L$.
Without otherwise stated, we consider Euclidean norm for both vectors and matrices. Here we define the Bregman distance $D_f(x,y) = f(x) - f(y) - \langle \nabla f(y), x-y\rangle$.
(1) $\iff$ (3) can be found in Lemma 1.2.2 in (Nesterov et.al 2018)
(1) $\implies$ (2) can be found in Lemma 1.2.3 in (Nesterov et.al 2018)
Then I find [smooth convex generalizations](https://web.stanford.edu/~sidford/courses/19fa_opt_theory/sidford_mse213_2019fa_chap_5_smooth_convex_gen.pdf)'s Lemma 8 shows that $|D_f (x,y)| \leq \frac{L}{2}\|x-y\|^2 \iff |z^\mathsf{T} \nabla^2 f(x) z| \leq L \|z\|^2$ ($z$ can be any vector), which means (2) $\iff$ (3).
So now, I wonder whether it holds that (1) $\iff$ (2) $\iff$ (3)?
At last, I think this question is asking whether (1) $\iff$ (2)?
Nesterov, Y. (2018). Lectures on convex optimization (Vol. 137, pp. 5-9). Berlin: Springer.
|
Is the $u$ substitution needed? Consider the following:
$\sqrt{\frac{a}{x}}-\sqrt{\frac{x}{a}}=\frac{a^{2}-1}{a}$
Now, re-write the RHS as $a-\frac{1}{a}$ and square both sides (this is the part that I'm not totally sure is valid) to give:
$\frac{a}{x}+\frac{x}{a}-1=a^{2}+\frac{1}{a^{2}}-1$
$\frac{a^{2}+x^{2}}{x}=\frac{a^{4}+1}{a}$
$a x^{2}-(a^{4}+1)x+a^{3}=0$
$(x-a^{3})(ax-1)=0$
$x=a^{3}$ or $x=\frac{1}{a}$
|
At first, it should be noted that Linear Programming (LP) **has benefited from** many mathematical tools, for a review, see e.g., [Algebraic and Topological Tools in Linear Optimization][1] and [Integer Programming and Number Theory][2].
Moreover, due to its widespread usage, LP has **become a strong driver** for the study of challenging math problems such as $d$-step conjecture for polyhedrons (see this Acta Mathematica paper: [The d-step conjecture for polyhedra of dimension $d<6$][3]) or the development of branches of mathematics such as [convex analysis][4] and tropical geometry (see Chapter 1.2 in the book [Introduction to Tropical Geometry][5] and pages 1031-1032 in [this paper][1]).
The relationship between LP and mathematics is not one-sided and **LP and its well-known extensions**, such as convex optimization and integer programming, has been applied in different fields of mathematics.
- Convex Optimization, a nice non-linear extension of LP that can be solved efficiently, has found many applications in **geometry** such as projection on a set, distance between sets, Euclidean distance and angle problems, extremal volume ellipsoids, etc. (see chapter 8 of the book [Convex Optimization][6])
- Integer programming, a discrete counterpart of LP that can be solved efficiently using LP-based branch-and-bound, has been also applied for analyzing and solving difficult **graph-theoretic** problems (see chapters on applications of the book [Combinatorial Optimization][7]). The [paper][8] pointed out by @MishaLavrov in a comment is an interesting usage of integer programming to disprove conjectures and improve several lower bounds in **extremal combinatorics**.
- Semidefinite Programming (SDP) is a very recent extension of LP, for which [an efficient algorithm designed by Farid Alizadeh][9]. In SDP, the order $\ge$ is replaced by $\succeq$. A number of recent papers provided direct or computer-assisted proofs based on **Flag Algebras**, allowing one to establish bounds through a double-counting argument by solving SDP models to prove various results in extremal combinatorics (see this [paper][10] and references therein).
- The **major theoretical breakthrough** that **LP models** with rational input data can **be solved in polynomial time** is a key result in **theoretical computer science** (see The New York Times paper [Leonid Khachiyan, 52; Helped to Advance Computer Math][11] and chapter 3 of the book [7]).
- Another interesting application of LP is in **probability theory** where the [Wasserstein distance][12] (used to compare different probability measures) can be obtained from a transportation problem when the sample space is finite (a well-known LP problem taught in undergrad LP courses, can be solved in polynomial time).
- The second application of LP in **probability theory** is to **derive sharp analytical bounds** for the probability of the union of events or other quantity with partial information, see these interesting papers [13], [14], [15], and references therein.
- LP enables us to obtain a [Nash equilibrium][16] for a two-player [zero-sum game][17] with finite strategy sets **in polynomial time**, which is of high theoretical importance as still no polynomial algorithm has discovered to find a Nash equilibrium of two-player general-sum game with finite strategy sets. In fact, this brilliant application of LP in **game theory** warmly opened the door to develop the amazing concept of [duality in optimization][18] (inspired by John von Neumann's min-max concept) with many theoretical applications in other fields.
- LP bounding has been used in the **sphere packing problem** (what fraction of $\mathbb R^n$ can be covered by congruent balls not intersecting except along their boundaries). Here, I list a number of papers published in $\color{red}{\text{Annals of Mathematics}}$, the most prestigious journal in mathematics, on sphere packing in which **linear programming** has been successfully used:
- [Universal optimality of the $E_8$ and Leech lattices and interpolation formulas][19],
- [The sphere packing problem in dimension 24][20]
- [New upper bounds on sphere packings I][21]
I quote the following statement from the third [paper][21]:
> **Linear programming bounds** (Delsarte, 1972) **are the most powerful
known technique** for producing upper bounds in **such problems**. In particular, Kabatiansky and Levenshtein (1978) use this technique to prove the best bounds known for sphere packing density in high dimensions.
[1]: https://www.ams.org/journals/notices/201907/rnoti-p1023.pdf
[2]: https://nyaspubs.onlinelibrary.wiley.com/doi/abs/10.1111/j.1749-6632.1981.tb51145.x
[3]: https://projecteuclid.org/journals/acta-mathematica/volume-117/issue-none/The-d-step-conjecture-for-polyhedra-of-dimension-d6/10.1007/BF02395040.full
[4]: https://en.wikipedia.org/wiki/Convex_analysis#:~:text=Convex%20analysis%20is%20the%20branch,A%203%2Ddimensional%20convex%20polytope.
[5]: https://web.archive.org/web/20140301003344id_/http://homepages.warwick.ac.uk/staff/D.Maclagan/papers/TropicalBook28.2.14.pdf
[6]: https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf
[7]: https://link.springer.com/book/10.1007/3-540-29297-7?page=2#toc
[8]: https://www.sciencedirect.com/science/article/pii/S0097316519301116
[9]: https://www.informs.org/Recognizing-Excellence/Award-Recipients/Farid-Alizadeh
[10]: https://ojs.aaai.org/index.php/AAAI/article/view/26470
[11]: https://www.nytimes.com/2005/05/23/obituaries/world/leonid-khachiyan-52-helped-to-advance-computer-math.html
[12]: https://en.wikipedia.org/wiki/Wasserstein_metric
[13]: https://www.sciencedirect.com/science/article/pii/S0166218X04001386
[14]: https://pubsonline.informs.org/doi/abs/10.1287/moor.2014.0657
[15]: https://epubs.siam.org/doi/abs/10.1137/21M1408294
[16]: https://en.wikipedia.org/wiki/Nash_equilibrium#:~:text=and%20wireless%20communications.-,History,produce%20to%20maximize%20its%20profit.
[17]: https://en.wikipedia.org/wiki/Zero-sum_game
[18]: https://en.wikipedia.org/wiki/Duality_(optimization)
[19]: https://annals.math.princeton.edu/2022/196-3/p03
[20]: https://annals.math.princeton.edu/2017/185-3/p08
[21]: https://annals.math.princeton.edu/2003/157-2/p09
|
Let $X\to \operatorname{Spec } K$ be a projective integral variety. Let $L|K$ be a field extension and consider an $L$ point $x\colon \operatorname{Spec} L\to X$ in $X$ (morphism over $K$).
Under which conditions on the extension $L|K$, the morphism $x \colon \operatorname{Spec} L\to X$ is projective or quasi-projective?
If $x$ is a closed point (i.e. $L|K$ is finite) then clearly the morphism is projective. But what about the other cases? For instance assume that the closure $\overline{\{x\}}$ is a proper subvariety of $X$, then I suspect that the morphism $x\colon \operatorname{Spec} L\to X$ must have some special properties.
|
Mathematically, the **gradient** is a property of a scalar function $f:\mathbb R^n\to\mathbb R$, found as
$$\mathrm{grad} (f)=\nabla f=\begin{bmatrix}\frac{\partial f}{\partial x^1}\\\frac{\partial f}{\partial x^2}\\ \frac{\partial f}{\partial x^3}\\\vdots\end{bmatrix}.$$
In physical terms you can think of it as the equivalent of the derivative of a function of one variable. It is "the derivative" or "the slope" in higher dimensions, so to speak. For instance, for a function of two variables $f:\mathbb R^2\to\mathbb R$, which represents a surface when plotted, the gradient is a vector arrow that always points in the steepest direction from any point.
The **divergence** on the other hand is a property of a vector function $V:\mathbb R^n\to\mathbb R^n$, which more specifically is called a vector field, and is found as
$$\mathrm{Div}(\mathbf V)=\nabla \cdot \mathbf V=\frac{\partial V^1}{\partial x^1}+\frac{\partial V^2}{\partial x^2}+\frac{\partial V^3}{\partial x^3}+\cdots$$
Physically, if you think of a vector field as representing e.g. the wind velocity at every point, then the divergence can be thought of as indicating a local expansion rate (do the points tend to move away from or closer to each other at a point - i.e., does the density decrease or increase at a point?).
Both the gradient concept and the divergence concept can be defined using the nabla operator $\nabla$, which might be what you are referring to. But note that this operator is used in simple *scalar multiplication* in the case of the gradient because we there are dealing with a scalar function, whereas it is used in a *dot product* in the case of the divergence where we are dealing with a vector product. Also note have the outputs are entirely different types of objects that have different dimensions - the gradient is a vector whereas the divergence is a scalar. So I would not try to think of them as related.
Physically, scalar functions and vector fields should better be thought of as entirely different mathematical concepts that in turn can represent entirely different and essentially unrelated physical concepts, such as mountain surfaces vs. wind directions.
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> Show that for any real random variable $X$ and any proper open subset $U$ of $\mathbb{R}$, we have $$\mathbb P(X \in U)=\sup\limits \left \{\mathbb P(X \in K): K \subseteq U \text { is compact} \right \}.$$
**My Attempt** $:$ Let $U \subseteq \mathbb R$ be open. Let $$\alpha : = \sup\limits \left \{\mathbb P (X \in K)\ :\ K \subseteq U\ \text {compact} \right \}.$$
For each $n \geq 1,$ consider the set $$K_n : = \left \{x \in U\ :\ d \left (x, U^c \right ) \geq n\ \text {and}\ \left \lvert x \right \rvert \leq n \right \}.$$
Then each $K_n$ is compact since the function $f : x \mapsto d \left (x, U^c \right )$ is
continuous and $$K_n = f^{-1} \left ( [n, \infty) \right ) \cap [-n,n].$$
Thus each $K_n,$ being the intersection of a closed set and a compact set
(in the Hausdorff space $\mathbb R),$ is compact.
Now for each $x \in U,$ there exists
$m_1 \in \mathbb N$ such that $\left \lvert x \right \rvert \leq m_1.$ Since $U^c$ is closed it follows that $d \left (x, U^c \right ) > 0$
and hence by Archimedean property there exists $m_2 \in \mathbb N$ such that
$d \left (x, U^c \right ) \geq \frac {1} {m_2}.$ Let $m_3 : = \max\limits \left \{m_1, m_2 \right \}.$ Then $x \in K_{m_3} \subseteq \bigcup\limits_{n=1}^{\infty} K_n.$ Since
$K_n \subseteq U$ for each $n \geq 1$ it follows that $$U = \bigcup\limits_{n=1}^{\infty} K_n.$$
Since finite union of compact sets is compact, replacing $K_n$ by
$K_n^{\prime} : = \bigcup\limits_{j=1}^{n} K_n \subseteq U,$ it follows that $U$ can be written as a countable
increasing union of compact sets $K_n^{\prime}.$ Then by the continuity of the
probability measure from below it follows that $$\mathbb P (X \in U) = \lim\limits_{n \to \infty} \mathbb P \left (X \in K_n^{\prime} \right).$$
Hence $\alpha \geq \mathbb P (X \in U).$ But since $\mathbb P (X \in K) \leq \mathbb P (X \in U),$ for each
compact set $K \subseteq U,$ we also have $\alpha \leq \mathbb P (X \in U).$ Thus $\alpha = \mathbb P (X \in U),$
as required. $\square$
Is it fine what I did? Thanks for reading.
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If I have an arbitrary vector $A = (a,b,c,0)$ how can I find a transformation matrix $M$ such that $M \times A = (0,1,0,0)$?
We can assume $A$ has a magnitude of $1$ if it helps simplify the derivation process.
The trivial case $A = (0,1,0,0)$ would cause $M$ to be the identity matrix.
If $A = (0,-1,0,0)$ then $M$ would be a 180 degree rotation matrix about the $x$ axis.
I heard of Rodrigues' rotation formula from [this question](https://math.stackexchange.com/questions/209768/transformation-matrix-to-go-from-one-vector-to-another) but I'm not sure how it would work in a 4 by 4 matrix.
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Below I will bring a passage from Heat Kernels by Wolfgang Arendt (Theorem 4.3.3, page 52). I need to understand it and write a more verbose report based on the chapter, however I am stuck at this part. I managed to repeat most of the steps, but the "this shows" (highlighted with **!!!**) sentence confuses me. How was the author able to do this argument? What does the observation do in this proof? I see that it has something to do with dual spaces and Banach adjoints ($\langle Ax,x^*\rangle = \langle x,A^*x^*\rangle$) and that the integral should be understood as a $L^1(\Omega)^* \cap L^\infty(\Omega)^*$ function (aka $\langle x, x^*\rangle$), but I can't understand where all this is going.
> Let $S \in \mathcal{L}(L^1(\Omega))$ be positive such that $\lVert S\rVert_{\mathcal{L}(L^1(\Omega))}, \lVert S\rVert_{\mathcal{L}(L^\infty(\Omega))} \leq M$.
>
> We have to show that there exists a unique operator $T \in \mathcal{L}(L^\infty(\Omega))$ such that $T^*f = Sf$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$. Then we let $S_\infty = T^*$. Uniqueness of T is easy to see. In order to prove its existence we note that the hypothesis implies that $S(L^1(\Omega) \cap L^\infty(\Omega)) \in S(L^1(\Omega) \cap L^\infty(\Omega))$ and $\lVert Sf\rVert_{L^p(\Omega)} \leq M\lVert f\rVert_{L^p(\Omega)}$ for $p = 1, \infty$ and for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$.
>
> Observe that a function $g \in L^\infty(\Omega)$ is in $L^1(\Omega) \cap L^\infty(\Omega)$ and $\lVert g \rVert_{L^1(\Omega)} \leq M$ if an only if
>
> $\lvert \int_\Omega g(x)\phi(x)dx \rvert \leq M\lVert \phi \rVert_{L^\infty(\Omega)}$
>
> for all $\phi \in L^1(\Omega) \cap L^\infty(\Omega)$. **(!!!)** This shows that the adjoint $S^* \in \mathcal{L}(L^\infty(\Omega))$ of $S$ leaves $L^1(\Omega) \cap L^\infty(\Omega)$ invariant and that $\lVert S^*f\rVert_{L^1(\Omega)} \leq M\lVert f\rVert_{L^1(\Omega)}$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$ **(!!!)**. By density of $L^1(\Omega) \cap L^\infty(\Omega)$ in $L^1(\Omega)$, there exists a unique operator $T \in \mathcal{L}(L^1(\Omega))$ such that $Tf = S^*f$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$. This implies that $T^*g = Sg$ for all $g \in L^1(\Omega) \cap L^\infty(\Omega)$.
|
May this [paper][1] answer your question?
*NB: there are typos in the last Corollary.*
[1]: https://ijpam.uniud.it/online_issue/202044/71%20RamiAlahmad.pdf
|
I'm trying to evaluate the following integral, but I think I don't do something correctly.
$ \int_{-\infty}^\infty\frac{\cos x}{1+x^2}\;dx\ $
I tried in 2 ways:
I defined $\gamma = \gamma_1 + \gamma_2 $, where $ \gamma_1 = [-R, R], \gamma_2 = R e^{it}, t \in [0, \pi] $, and letting R aproach infinity.
So the integral on $\gamma_2$ approaches zero, and by residue theorem, we see that
$ \int_{-\infty}^\infty\int_{\gamma} f(z)dz = 2\pi i Res(f, i) = 2\pi i \frac{cosz}{z+i} $.
Letting $ z=i$ does not give me the correct result, since $ \cos i = \frac{e + \frac{1}{e}}{2} $
The second way: by partial decomposition, I see that $$ \int_{-\infty}^\infty\frac{\cos z}{1+z^2}\;dx\ = \int_{-\infty}^\infty\frac{i\cos z}{2(z+i)}\;dx\ + \int_{-\infty}^\infty\frac{-i\cos z}{2(z-i)}\;dx\ $$
and by Cauchy's theorem for integrals, the first integral is zero, and the latter is $ \pi i (-i) \cos i $, which is still not the correct result..
What do I miss here?
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There is the following analague of natural density on the positive real numbers:
Suppose that $A\subseteq (0,\infty)$ is Lebesgue-measurable, then one can look at the limit (if it exists) $\frac{\lambda(A\cap(0,n))}{n}$ for $n\rightarrow\infty$, where $\lambda$ is the usual Lebesgue measure on $(0,\infty)$. My question is, if there exists a name for this concept, something like "real density" maybe instead of natural density?
|
That means if a one parameter differentiable group $\Phi(x,t):\mathbb{R}^d\times \mathbb{R}\to\mathbb{R}^d$ satisfies $$\Phi(\Phi(x_0,t_1),t_2)=\Phi(x_0,t_1+t_2),\Phi(x_0,0)=x_0$$
holds for $\forall x_0\in\mathbb{R}^d,t_1,t_2\in\mathbb{R}.$ Then is there exist a suitable function $F$ on $\mathbb{R}^d$ s.t. $\dot{\Phi(x(0),t)}=F(\Phi(x(0),t)),x(0)=x$?
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> What numbers can be written uniquely as a sum of two squares?
I was looking at sequence [A125022](https://oeis.org/A125022), which shows the numbers that can be uniquely written as a sum of two squares. Here are a few things that I noticed from the first numbers. We have $1$, $2$, $4$, $8$, $16$, $32$, $64$, $128$. It is then safe to assume that all numbers of the form $2^{s}$ can be written uniquely, where $s \in \mathbb{Z}_{+} \cup \{0\}$. Moreover, primes of the form $4k+1$, for example $5$ and $13$, also appear and, interestingly enough, $5^2$ and $13^{2}$ do not. So, we could also say that $p^{s}$ has a unique representation only when $s = 0$ or $s = 1$. If we analyze $A125022$ a bit more, we notice that $3^{2}$, $7^{2}$, $11^{2}$ are there, so we can conjecture that numbers of the form $q^{2}$ have a unique representation, where $q$ is a prime of the form $4k+3$. Furthermore, for reasons I will say later, I believe $d^{2}$, where $d$ has all of its prime factors of the form $4k+3$, can be uniquely represented as a sum of two squares. It is also possible to see that products of these three cases are in the sequence, for example $2^{2}\cdot 5$, $2 \cdot 5 \cdot 3^{2}$ and $2 \cdot 7^{2}$.
**Conjecture.** A number $n \in \mathbb{Z}_{+}$ can be written uniquely as a sum of two squares if, and only if, $n = 2^{s} d^{2} p^{e_1}$, where $s \in \mathbb{Z}_{+} \cup \{0\}$, $d = 1$ or $d$ has all of its prime divisors of the form $4k+3$, $p$ is a prime of the form $4k+1$ and $e_{1} \in \{0,1\}$.
It is known that a number can be written as a sum of two squares if, and only if, it can be written as $2^{s} t^{2} l$, where $s \in \mathbb{Z}_{+} \cup \{0\}$ and $l$ is a square-free positive integer with all of its prime factors of the form $4k+1$. Thus, we know the number $n$ we conjectured above can in fact be written as a sum of two squares. We only need to understand uniqueness. It is more natural to study these questions with the Gaussian integers, $\mathbb{Z}[i]$. If, for example, we have
$$n = a^{2} + b^{2} = (a+ib)(a-ib) = (\pi_1 \cdots \pi_k) (\overline{\pi_1} \cdots \overline{\pi_k}),$$
where the last expression is the factorization of $n$ in primes of $\mathbb{Z}[i]$, then we may get different sum representations of $n$ by exchanging, say, $\pi_j$ for $\overline{\pi_j}$. That is, the product
$$(\pi_1 \cdots \overline{\pi_j} \cdots \pi_n)(\overline{\pi_1} \cdots \pi_j \cdots \overline{\pi_n})$$
should yield a different sum when $\pi_j \neq \overline{\pi_j}$ and at least one of the other primes, say $\pi_i$, also satisfies $\pi_i \neq \overline{\pi_i}$. This does not seem to occur precisely for the numbers conjectured above, which makes me think those are the only numbers that can be uniquely represented.
*Question.* Is my guess correct or am I missing other numbers?
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Is $\textit{affine space}$ the same as $\textit{quotient space}$?
|
During my research I stumbled upon this system over $\mathbb{Z}_{2^n}$. Let $k=2^{n-1}-1$ then the system looks like
\begin{equation*}
\begin{cases}
x_1+2^{n-1}=d_k +x_k \\
x_2+2^{n-1}=d_1+x_1 \\
x_3+2^{n-1}=d_2+x_2 \\ \tag{1}
\dots \\
x_k+2^{n-1}=d_{k-1}+x_{k-1} \\
x_i \notin \{x_j,x_j + 2^{n-1}\}, i \neq j \\
d_i \notin \{d_j,-d_j\}, i \neq j,
\end{cases}
\end{equation*}
where '+' means addition in $\mathbb{Z}_{2^n}$ and $x_i,d_i$ are variables for i $\in \overline{1,k}$.
I've noticed, that system (1) without any conditions has solutions iff $d_1+d_2+\dots+d_k=2^{n-1}$, so my idea was to fix $d_1,d_2,\dots d_k$ and then the solution to the system without coniditons looks like
\begin{equation}
\begin{cases}
x_1=d_k +2^{n-1} \\
x_2=d_k+d_1 \\
x_3=d_k+d_1+d_2 +2^{n-1} \\ \tag{2}
\dots \\
x_{k-1}=d_k+d_1+\dots +d_{k-2} \\
x_k=0.
\end{cases}
\end{equation}
Now, let $c_i=d_i+2^{n-1}$, then if $d_i \notin \{d_j,-d_j\}, i \neq j$ it follows that $c_i \notin \{c_j,-c_j\}, i \neq j$. Then in terms of $c_i$ solution (2) look like
\begin{equation}
\begin{cases}
x_1=c_k \\
x_2=c_k+c_1 \\
x_3=c_k+c_1+c_2 \\ \tag{3}
\dots \\
x_{k-1}=c_k+c_1+\dots +c_{k-2} \\
x_k=0.
\end{cases}.
\end{equation}
So, solving of system (1) is equivalent to finding $c_1,\dots ,c_k$, such that $c_i \notin \{c_j,-c_j\}, i \neq j$ and all sums like $c_i+\dots +c_j, 1 \leq i \leq j \leq k-2$ are nonzero modulo $2^{n-1}$.
I've programmed it and was able to find some interesting candidates for $C=\{c_i\}$. It is $C=\{1,2,\dots,2^{n-2}-1,2^{n-1},2^{n-1}+1,\dots,2^{n-1}+2^{n-2}\}$ as for $n=4$ we have solutions like
$(c_1,c_2,\dots,c_7)=(3, 2, 1, 9, 10, 11, 12)$
$(c_1,c_2,\dots,c_7)=(1, 2, 3, 12, 11, 10, 9)$ and a couple of others.
For $n=5$ we have something like
$(c_1,c_2,\dots,c_{15})=(24, 19, 20, 18, 21, 22, 17, 23, 1, 2, 3, 4, 5, 6, 7)$
$(c_1,c_2,\dots,c_{15})=(1, 6, 23, 4, 19, 7, 24, 21, 22, 20, 3, 18, 2, 17, 5)$.
For $n=6$ my program unfortunately wasn't able to find any. So the question is, how to find solutions to (1) for any $n \in \mathbb{N}$ or at least prove its existence? Maybe there are other ways much simplier ways to solve (1) rather than mine? I'll appreciate any help.
|
>How to rigorously prove that $\sum_{n=1}^ \infty \frac{1}{4n-1} - \frac{1}{4n} =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}$$
$$f_N'(x) = \sum_{n=1}^ N x^{4n-2}- x^{4n-1}= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converge uniformly to be able to interchange the derivative and the summation but I don't think $f_N'$ converge uniformly because
$$f_N'(x) = \frac{-x^4}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ don't converge uniformly on $[0,1 )$.
Here I got stuck.
|
How to rigorously prove that $\sum\limits_{n=1}^ \infty \frac{1}{4n-1} - \frac{1}{4n} =\frac{\ln(64)- \pi}{8}$?
|
I'm performing a close critical study of David Nelson's *Penguin Dictionary of Mathematics* (4th ed., 2008).
Under the entry **hypergeometric differential equation**, it suggests the form:
$$x (1 - x) \dfrac {\mathrm d^2 \phi} {\mathrm d x^2} + [c - (a + b - 1) x] \dfrac {\mathrm d \phi} {\mathrm d x} - a b \phi = 0$$
However, everywhere else I look, I see it defined as:
$$x (1 - x) \dfrac {\mathrm d^2 \phi} {\mathrm d x^2} + [c - (a + b + 1) x] \dfrac {\mathrm d \phi} {\mathrm d x} - a b \phi = 0$$
I suspect, but would like to be certain, that Nelson's presentation is in fact incorrect, or whether it's a variant format which is equally acceptable as a representational format.
Before I report on this as an acual error, can it be confirmed that it is in fact wrong?
|
Form of Hypergeometric Differential Equation - possible mistake?
|
Suppose I have some tensor, for concreteness I will consider a rank 4 tensor with components $R_{abcd}$ where the indices run over 0,1,2,3 (my question arises in a physics context). Under what conditions is the existence of another tensor, call it $G_{abcd}$, which satisfies $G_{ab}^{~~~mn}G_{cdmn}=R_{abcd}$ guaranteed?
**Edit**: Indices are raised and lowered using the metric tensor : $G_{ab}^{~~~mn}=g^{cm}g^{dn}G_{abcd}$. Also, for my problem $R_{abcd}$ is the Riemann tensor, and I suppose this would place extra conditions on the existence of a suitable $G_{abcd}$ since it is defined in terms of the derivatives of the metric (in the torsion free case). This probably complicates things quite some, so for now I will leave $R_{abcd}$ to be any rank 4 tensor.
|
>How to rigorously prove that $\sum_{n=1}^ \infty \frac{1}{4n-1} - \frac{1}{4n} =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}$$
$$f_N'(x) = \sum_{n=1}^ N x^{4n-2}- x^{4n-1}= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converge uniformly to be able to interchange the derivative and the summation but I don't think $f_N'$ converge uniformly because
$$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ don't converge uniformly on $[0,1 )$.
Here I got stuck.
|
<h4>Question Summary (to Make Easier to Reference in the Answer)</h4>
$$
\text{If }x^2-16\sqrt{x}=12 \text{ what is the value of }f=x-2\sqrt{x}
\tag{Eq. 1}$$
<h4>Checking for Consistency by Plugging in the $f=x-2\sqrt{x}=2$</h4>
Check $ x - 2\sqrt{x}-2=0$ implies $(\sqrt{x}-1)^2=2+1=3$ so $\sqrt{x}=1+\sqrt{3}$ (only the positive solution holds since $\sqrt{x} \ge 0$. So then $x=(1+\sqrt{3})^2=1+2\sqrt{3}+3=4+2\sqrt{3}$. Then $x^2=16+12+16\sqrt{3}=28+16\sqrt{3}$ and $x^2 - 16\sqrt{x}-12=(28+16\sqrt{3})-16*(1+\sqrt{3})-12=(28-16-12)+(16-16)\sqrt{3}=0$. Thus indeed it must be so that $x-2*\sqrt{x}=2$.
<h4>Solving for $f=x-2\sqrt{x}$ starting with $-2\sqrt{x}$ and then simplifying further</h4>
To start, one can solve for $-2\sqrt{x}$ as follows (subtracting $x^2$ from the left and right of Equation 1, and then dividing both sides by $8$):
$$
\frac{(x^2-16\sqrt{x})-x^2}{8}
=\frac{(12)-x^2}{8} \underset{implies}\implies
\tag{Eq. 2}
$$
$$
\underset{implies}\implies -2\sqrt{x}=\frac{12-x^2}{8}
\underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}\\
\underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}
\tag{Eqs.3}$$
Now further simplify Equations 3, by multiply each side by $8$ and then by subtracting from each side $8*x$ so then:
$$
x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}
\underset{implies}\implies 8x - 16\sqrt{x}=8x+12+x^2 \\
\underset{implies}\implies - 16\sqrt{x}=12-x^2 \\
\underset{implies}\implies x^2 - 16\sqrt{x}-12=0 \tag{Eq. 4}$$
$$\underset{implies}\implies (\sqrt{x})^4 - 16\sqrt{x}-12=0 \tag{Eq. 5}
$$
The above Equation 5 is a [solvable fourth-degree equation according to the link][1], which makes its solution techniques a little bit more unfamiliar than a second order polynomial.
I am hoping to update the answer with the referenced technique, but also this link can be used directly to solve the problem even before I provide step-by-step instructions on the solution. The link has the author's request for algebraic steps to solve the problem.
Alternatively, there are on-line [solvers also for this equation as follows for instance][2]:
> [![solution for sqrt x][3]][3]
Since $\sqrt{x}>0$ and it is a [irrational solution greater than zero][4], the only applicable solution from that is $\sqrt{x}=1+\sqrt{3}$ as was to be proven.
<h4>Checking Line-by-Line Using $\sqrt{x}=1+\sqrt{3}$, $x=2*(2+\sqrt{3})$, and $x^2=28+16*\sqrt{3}$</h4>
From Equation 3, with $x=(\sqrt{x})^2
=\left(1+\sqrt{3}\right)^2=1+3+2\sqrt{3}=2*(2+\sqrt{3})$ and
$x^2=28+16*\sqrt{3}$ and $\sqrt{x}=1+\sqrt{3}$:
$$
-2\sqrt{x}=\frac{12-x^2}{8}
\underset{implies}\implies -2\left(1+\sqrt{3}\right)
=\frac{12-\left(1+\sqrt{3}\right)^4}{8} \\
=\frac{12-\left(4+2*\sqrt{3}\right)^2}{8} \\
=\frac{12-\left(28+16*\sqrt{3}\right)}{8}\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3a}$$
$$
x-2\sqrt{x}=\left(4+2\sqrt{3}\right)-2*\left(1+\sqrt{3}\right)
=2\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3b}
$$
$$
x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8} \\
\underset{implies}\implies
2= \frac{\displaystyle (8*2*(2+\sqrt{3}))+12-(28+16\sqrt{3})}
{\displaystyle 8} \\
\underset{implies}\implies
2= \frac{\displaystyle ((32+16\sqrt{3}))+12-(28+16\sqrt{3})}
{\displaystyle 8} \\
2= \frac{\displaystyle (32+12-28)}
{\displaystyle 8}
= \frac{\displaystyle (16)}
{\displaystyle 8} = 2\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3c}$$
<h4>Solving the Quartic 4th Degreee Polynomial $(\sqrt{x})^4 - 16\sqrt{x}-12=0$ from Equation 5</h4>
From Equation 5, the solutions for $\sqrt{x}$ can be written as solutions to the [Quartic 4th Degree Polynomial][5] $(\sqrt{x})^4 - 16\sqrt{x}-12=0$.
Since the coefficient for the leading term $(\sqrt{x})^4$ is 1, then generally it is possible to seek two second-degree polynomials such that:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2+c*(\sqrt{x})+d \right)=0
\tag{Eq. 6}$$
Expanding Equaiton 6, for each coefficient of $(\sqrt{x})^n$, the expanded term can be compared to the coefficients from Equation 6 and Equation 5. To start with, compare the expanded coefficient element for the expanded $(\sqrt{x})^3$ as
$(\sqrt{x})^3*(a+c)=0$, so immediately Equation 6 can be simplified (with $c=-a$) in Equation 7 as:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\tag{Eq. 7}$$
1. Observe that in Equation 7, that the roles of $b$ and $d$ are
entirely similar. Either $d>b$ so that $d=b_+$,
and $b=b_-$. Or $d<b$ so that $d=b_-$ and $b=b_+$. However,
because of the [Commutative Law of Algebra for Multiplication][6]
arbitrarily the symmetry can be broken, with the larger value
$b_+$ being placed on the factor on the left, namely,
$\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)$. And $b_-$ can be
placed on the factor on the right, namely
$\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)$. Hence:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b_+ \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+b_- \right)=0
\tag{Eq. 8}$$
2. Now consider the coefficient equations for $(\sqrt{x})^2$ as
$(\sqrt{x})^2*(b_+ + b_- - a^2)=0$ so $(b_+ + b_- - a^2)=0$.
3. Now consider the coefficient equations for $(\sqrt{x})^1$ as
$(\sqrt{x})^1*(a(b_- - b_+))=-16*(\sqrt{x})^1$ so
$(a*(b_- - b_+))=-16$ so $(a*(b_+ - b_-))=16$ implying that
$a>0$ since $b_+>b_-$ implies that $b_+-b_->0$,
also a positive value.
4. Now consider the coefficient equations for $(\sqrt{x})^0$ as
$(\sqrt{x})^0*(b_+*b_-)=-12*(\sqrt{x})^0$ so $b_+*b_-=-12$.
5. There are now three Equations, namely $(b_+ + b_- - a^2)=0$,
$(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$. And there are
three unknowns to be solved for, namely $a$, $b_+$, and $b_-$;
So the rearrangement of the three equations is accomplished in
Equation 9 as follows:
$$
a=\frac{\displaystyle 16}{\displaystyle b_+ - b_-}
=\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\underset{implies}\implies
a^2=
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2=b_+ + b_- = b_+ - \frac{12}{b_+}
\tag{Eq. 9a}$$
$$
\underset{implies}\implies
16^2=256=\left(b_+ - \frac{12}{b_+} \right)*
\left( b_+ + \frac{12}{b_+} \right)^2
=(4)*(8*8)=(4)*(8)^2
\tag{Eq. 9b}$$
$$\tag{Eqs. 9}$$
Since the number $256$ has a factor of $(4)^1$ and also has factors of $(8)^2$, substitute $(4)^1=\left(b_+ - \frac{12}{b_+} \right)^1$ and also $8^2=\left( b_+ + \frac{12}{b_+} \right)^2$ and then check for consistency. It might seem like a slight of hand to make these assignments. However, consider that $128=(2*2)*(2*2*2)*(2*2*2)$, and also the left term, namely $\left(b_+ - \frac{12}{b_+} \right)$ is less than the right term, namely $\left( b_+ + \frac{12}{b_+} \right)$ because $b_+>0$. Thus the number of $2$'s in the square root could be $2$. But that less than $256/4$. Similarly $2*2=4<256/4$. And finally, $2*2*2=8$ meets the requirement for
$2*2*2=\left( b_+ + \frac{12}{b_+} \right)$, that it is greater
than $2*2=\left(b_+ - \frac{12}{b_+} \right)$ and that because of the prime number decomposition that $256=2^8$ that this works, as
$256=2^8=2^4*\left(2*2*2\right)^2$.
$$
\left(b_+ - \frac{12}{b_+} \right)=4
\underset{implies}\implies
(b_+)^2-4*(b_+)=12
\\
\underset{implies}\implies
\left((b_+)-2 \right)^2=12+4=16
\underset{implies}\implies
b_+=4+2=6 \\
\tag{Eqs. 10a}$$
$$
\left(b_- - \frac{12}{b_-} \right)=4
\underset{implies}\implies
(b_-)^2-4*(b_-)=12
\\
\underset{implies}\implies
\left((b_-)-2 \right)^2=12+4=16
\underset{implies}\implies
b_-=-4+2=-2 \\
\tag{Eqs. 10b}$$
$$
\left(b_+ + \frac{12}{b_+}\right)=8=6+\frac{12}{6}=6+2\text{ is correct!}
\tag{Eqs. 10c}$$
$$
a=\frac{\displaystyle 16}{\displaystyle b_+ - b_- }
=\frac{\displaystyle 16}{\displaystyle (6) - (-2) }
=\frac{\displaystyle 16}{\displaystyle 8 } = 2
\tag{Eqs. 10d}$$
$$
\tag{Eqs. 10}
$$
Note that $b_+*b_-=(6)*(-2)=-12$ as was to be expected, as the greater value for $b_+$ is selected because $b_+>b_-$, and also the lesser value of the polynomial equation for $b_-$.
<h4>Solving the Second Degree Polynomial Derived from the 4th Degree Quartic</h4>
Finally, the polynomial can be chosen that has a [positive real root][7]. Really all polynomial combinations need to be considered. For a 4th degree polynomial like this one, there are two real roots where one is greater than zero and one is less. And there are another two that have the factor of $\sqrt{-1}$ which is not a solution for $\sqrt{x}$ which needs a positive real solution.
If the terms for $-1*\sqrt{x}a$ and $b=b_-$ then the root of the second degree polynomial below must be real and positive as to be desired. So Equation 7 is repeated with those conditions now here:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\\ \underset{implies}\implies
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\tag{Eq. 7}$$
$$
\left((\sqrt{x})^2-2*(\sqrt{x})-2 \right)=0
\\ \underset{implies}\implies
\left(\sqrt{x}-1\right)^2=2+1=3
\\ \underset{implies}\implies
\sqrt{x}=1+\sqrt{3} \text{ as was to be proven!}
\tag{Eqs. 11}$$
To highlight the solution now for $\sqrt{x}$ from the 4th degree polynomial quartic $(\sqrt{x})^4 - 16\sqrt{x}-12=0$, Equation 11 has the solution as:
$$
\boxed{\sqrt{x}=1+\sqrt{3} \text{ as was to be proven!}
}\tag{Eq. 12}
$$
<h4>Verifying the Consistency of the 4th Degree Polynomial Quartic</h4>
Now there is also consistency. From the online [Equation solver][2], there is also consistency that verifies this result:
[![online equation solver tractability][8]][8]
Since $a$ is positive, and $b_+>b_-$, the online equation solver has results for $a$, $b_+$, and $b_-$, namely $a$ =2, $b_+=6$, and
$b_-=-2$.
The three Equations, namely $(b_+ + b_- - a^2)=0$,
$(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$, can be tested against the online results to find any algebraic errors, applying
$a$ =2, $b_+=6$, and $b_-=-2$. There is no mistake is in Equations 9, and the values given by the online Equation Solver are correct, also yielding the desired consistent result:
1. $(b_+ + b_- - a^2)=0$ implies $6-2-4=0$, correct!
2. $(a*(b_+ - b_-))=16$ implies $2*(6-(-2))=2*8=16$, correct!
3. $(b_+*b_-)=-12$ implies $6*(-2)=-12$, correct!
So, now Equations 9 can be carefully checked:
$$\text{Checking Eqs. 9a:}\\
a=\frac{a}{b_+ + b_-}=\frac{16}{6+(-2)}=4 \text{ is correct!} $$
$$
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2
=b_+ + b_- = b_+ - \frac{12}{b_+}$$
$$
\underset{implies}\implies
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2=
\left(
\frac{\displaystyle 16}{6 + \frac{12}{6}}
\right)^2
=4=6-\frac{12}{6}=6-2=4 \text{ is correct!}
$$
[1]: https://math.stackexchange.com/questions/3658305/is-there-formula-that-solves-quartic-equation-ax4bxc-0
[2]: https://www.google.com/search?client=opera&q=solve%20%24x%5E4-16*x-12%3D0%24&sourceid=opera&ie=UTF-8&oe=UTF-8
[3]: https://i.stack.imgur.com/HsKeW.png
[4]: https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm
[5]: https://mathworld.wolfram.com/QuarticEquation.html
[6]: https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/economics/algebra/fundamental-laws-of-algebra.html
[7]: https://mathworld.wolfram.com/RealNumber.html#:~:text=The%20field%20of%20all%20rational,Wolfram%20Language%2C%20and%20a%20number
[8]: https://i.stack.imgur.com/MCqyu.png
|
I have the equation $x^2+xy+y^2=z^2$ to solve it in natural numbers.
The discriminant of it $D=4z^2-3b^2$ and must be perfect square.
I wrote Python program to get solutions for $1<x<100$ by enumeration.
def Solution():
A=[]
nMaximum=100
for a in range(1,nMaximum):
dTemp1a=3*a**2
for c in range(a+1, nMaximum):
dDiscriminant=4*c**2-dTemp1a
dTemp5=int(dDiscriminant**0.5)
if dTemp5**2!=dDiscriminant:
continue
dTemp6=(-1*a+dTemp5)/2
b=int(dTemp6)
if not CheckIfExists(A, c):
A.append([a,b,c])
return A
def CheckIfExists(arr, c):
bResult=False
for s in arr:
if s[2]==c:
bResult=True
break
return bResult
a = Solution()
print(len(a))
print(a)
[https://math.stackexchange.com/questions/1351994/three-variable-second-degree-diophantine-equation][1] doesn't explain how to get other solutions when we know the first solution $(3,5,7)$
Could you give me a hint ?
[1]: https://math.stackexchange.com/questions/1351994/three-variable-second-degree-diophantine-equation
|
How to find multiple solutions for 3 variable, 2 degree Diophantine equation?
|
>How to rigorously prove that $\sum_{n=1}^ \infty \frac{1}{4n-1} - \frac{1}{4n} =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}$$
$$f_N'(x) = \sum_{n=1}^ N x^{4n-2}- x^{4n-1}= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converge uniformly to be able to interchange the derivative and the summation but I don't think $f_N'$ converge uniformly because
$$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ don't converge uniformly on $[0,1 )$.
Here I got stuck but for some reason it works i.e $\int_0 ^1 \frac{x^2}{(1+x)(1+x^2)}= \frac{\ln(64)- \pi}{8} $ so the derivative could be interchanged with the summation here , but how ?
|
## Setup ##
So... I kinda handled most of my proof but I need help with some of the stuff I just kinda went with until it worked out. The problem relates to medicine and its decay in the body. We are given that the medicine will release over a period of $b$ hours and another dose is given at time $T$.
### Known Values ###
Decay Constant is 1. Each dose contains 1 gram of medicine. We know that $b=\frac{5}{4}$ & $T=\frac{5}{2}$. We are also given that $y(0)=0$ y being the amount of medicine in the body at time $t$. Finally we're given the simple equation $rate=rate_{in}-rate_{out}$.
Looking at our values we find that $\frac{4}{5}$ grams are released per hour over the course of $\frac{5}{4}$ hours. Using this and our value of T it's possible to make a piecewise function for the release $(rate_{in})$ of the medicine, $g(t)$. This function is "on" over the intervals $0<t<\frac{5}{4}$ & $\frac{5}{2}<t<\frac{15}{4}$ (note that $\frac{15}{4}$ just comes from $\frac{5}{2}+\frac{5}{4}$) and is "off" (to explain only having decay) between $\frac{5}{4}<t<\frac{5}{2}$ & $\frac{15}{4}<t<5$ (Idk if those should be $\le$ signs or not but that's less important right now).
I only know how to turn this into a rectangular window function like: $u(\frac{5}{4})-u(\frac{5}{2})$ and so on and so on but regardless... we can gather that because our $rate_{out}$ is reliant of the amount of grams present, that it's just *something* times our $y(t)$ function. Okay here's the two parts I need help on (three technically because I'm still not happy with my explanation for why the out function is $y(t)$:
## What I Need Help With ##
I set up an initial value problem that looks like this: $y'(t)=\frac{4}{5}g(t)-\frac{4}{5}y(t)$ and I'm pretty confident it's right.... but I don't know how to explain why I multiplied by $\frac{4}{5}$ (I just guessed based off of searches I had been doing but I've lost what I actually searched up) because the only thing I can think of is that because $\frac{4}{5}$ is $\frac{g}{h}$ that means that there's SOME kinda rate of change equal to $\frac{4}{5}$ but surely it can't be $\frac{dy}{dt}$ since then our IVP wouldn't work right? I guess it could also be a way of making our $rate=rate_{in}-rate_{out}$ equation EQUAL $\frac{4}{5}$ or something silly? Idk, this one probably needs more help than the next.
Also once I found that the Laplace transform of this solves to: $\frac{\frac{4}{5}G(s)}{s+\frac{4}{5}}$ I then basically tried to get $G(s)$ to equal the laplace of what I thought was the rectangular window function: $g(t)=(u(t)-u(t-\frac{5}{4}))+(u(t-\frac{5}{4})-u(t-\frac{5}{2}))$ but you can already see the problem right there. Stuff would cancel out. I remedied this by realizing that the equation through the next bounds $(\frac{5}{4}<t<\frac{5}{2})$ would have to pass through the last point given by the equation in the previous bounds... or in other words, to find the correct $y(t)$ we'd have to multiply say $y(t-b)\cdot(u(t-b)-u(t-c))$ by the result of $y(b)\cdot(1-u(b-b))$ (1 comes from $u(t)=1$) and so on and so forth and so I ended up getting:$y(t)=(1-e^{-\frac{4}{5}t})(u(t)-u(t-\frac{5}{4}))+(1-e^{-1})(e^{-\frac{4}{5}(t-\frac{5}{4})})((u(t-\frac{5}{4})-u(t-\frac{5}{2}))+(e^{-\frac{4}{5}(t-\frac{5}{2})})(1-(1-(e-1)e^{-2}))((u(t-\frac{5}{2})-u(t-\frac{15}{4}))+etc$
but see the problem is that we get that weird $1-(e-1)e^{-2}$ for the 3rd window which I can't come up with a good explanation for other than some weird shenanigans turning making one of our $u$ somehow equal $u(t)$? Like best I got is that because technically we're looking at the point $\frac{5}{2}$ for both sides you actually get 1-whatever because the step functions are interacting with each other???
Anyways if someone's willing to help just prove or AT LEAST EXPLAIN WITH WORDS why these things work that'd be much appreciated. Just something so I can justify what I did because while it did in fact work... it's been like a day since I wrote half of this stuff down and *for some reason I didn't write why it works so...*
### EDIT: MAJOR BREAKTHROUGH ON THE SECOND HALF ###
(Potentially)
(I think I've got the first half but it doesn't hurt to check so I'm leaving it), I still need some help explaining why it works BUT, I've found that for whatever reason, you can find the bit out in front of the second window (the $(1-e^{-1})$ part of $(1-e^{-1})(e^{-\frac{4}{5}(t-\frac{5}{4})})$ just by subtracting $e^{-\frac{4}{5}t}$ from $(e^{-\frac{4}{5}(t-\frac{5}{4})}$. It's just that the $(1-e^{-1})(e^{-\frac{4}{5}(t-\frac{5}{4})})$ is oversimplified, that entire equation is equal to $-e^{-\frac{4}{5}t}+e^{-\frac{4}{5}(t-\frac{5}{4})}$ The only problem is... where does that negative $e^{-\frac{4}{5}t}$ come from? My current way of thinking about this is that it's almost as if the first bound $u(t), u(t-5/4), etc$ is what's looked at during the inverse Laplace transform? Because all the unit step functions should be (in theory according to my IVP) multiplied by $1-e^{-\frac{4}{5}(t-a)}$ (a is $u(t-a)$) once the inverse Laplace transform is all said and done. However, if this were truly the case many things would cancel out before actually separating in order to do this leading me to believe that either the first $u(t-a)$ speaks for the entire rectangular window function OR it's something even deeper in the algebraic solving, something like:
$(-e^{-\frac{4}{5}t}+e^{-\frac{4}{5}(t-\frac{5}{4})})+(1-e^{-\frac{4}{5}\left(t-\frac{5}{2}\right)})$ (this one is specifically for the third window function and I found it to be correct). Thing is... WHY? Why does that work?
# New Update #
I think I've figured out a way to make this work actually. What if we just set $g(t)=0$ when $\frac{5}{4}<t<\frac{5}{2}$ and when $\frac{15}{4}<t<5$ since we said that $g(t)$ was our rate in right? Well I think we have that by just setting $g(t)=1$ for when the medicine is active we end up with the correct function THAT I'VE KNOWN ALL THIS TIME BTW BUT DOUBTED IT WOULD BE ACCEPTED: $\left(1-e^{-\frac{4}{5}t}\right)u\left(t\right)-\left(1-e^{-\frac{4}{5}\left(t-\frac{5}{4}\right)}\right)u\left(t-\frac{5}{4}\right)+\left(1-e^{-\frac{4}{5}\left(t-\frac{5}{2}\right)}\right)u\left(t-\frac{5}{2}\right)-\left(1-e^{-\frac{4}{5}\left(t-\frac{15}{4}\right)}\right)u\left(t-\frac{15}{4}\right)-\left(1-e^{-\frac{4}{5}\left(t-5\right)}\right)u\left(t-5\right)$.
I think that this just straight up works so... idk I'll leave this up but this is probably our answer.
|
I'm trying to evaluate the following integral, but I think I don't do something correctly.
I'm interested where my try failed, compared to other ways.
$ \int_{-\infty}^\infty\frac{\cos x}{1+x^2}\;dx\ $
I tried in 2 ways:
I defined $\gamma = \gamma_1 + \gamma_2 $, where $ \gamma_1 = [-R, R], \gamma_2 = R e^{it}, t \in [0, \pi] $, and letting R aproach infinity.
So the integral on $\gamma_2$ approaches zero, and by residue theorem, we see that
$ \int_{-\infty}^\infty\int_{\gamma} f(z)dz = 2\pi i Res(f, i) = 2\pi i \frac{cosz}{z+i} $.
Letting $ z=i$ does not give me the correct result, since $ \cos i = \frac{e + \frac{1}{e}}{2} $
The second way: by partial decomposition, I see that $$ \int_{-\infty}^\infty\frac{\cos z}{1+z^2}\;dx\ = \int_{-\infty}^\infty\frac{i\cos z}{2(z+i)}\;dx\ + \int_{-\infty}^\infty\frac{-i\cos z}{2(z-i)}\;dx\ $$
and by Cauchy's theorem for integrals, the first integral is zero, and the latter is $ \pi i (-i) \cos i $, which is still not the correct result..
What do I miss here?
|
we have $(X,F,\mu)$ be the measure space and let $M_{\mu}=\{p:X\to R \text{ measurable }: p>0 \mu -a.e and \int p d\mu =1\}$.
For each positive density $p\in M_{\mu}$ we define the Orlicz space $L^{\Phi}(p.\mu)$.
Then we have $B_{p}=\{u\in L^{\Phi_1}(p.\mu): E_{p}[u]=0\}$ where $E_{p}$ is the expectation.
Now we have for each $p\in M_{\mu}$ let $V_{p}$ be the open unit ball in $B_{p}$ that is $V_{p}=\{u \in B_{p}: ||u||_{\Phi_1,p}<1\}$. we define a map
$$e_{p}:V_{p}\to M_{\mu}$$ such that $$e_{p}(u)= e^{u-log E_{p}[e^u]}p.$$ This map is one-one.
The range of $e_p$ is $U_{p}$. The inverse image of $e_{p}$ on $U_{p}$ is the function $$s_{p}:q\in U_{p}\mapsto log\left(\frac{q}{p}\right)-E_{p}\left[log\left(\frac{q}{p}\right)\right]$$
Now it is given that $(U_p,s_p)$ for $p\in M_{\mu}$ is a chart. So basically here we are giving the manifold structure to $M_{\mu}$ now my question is that when we say chart that means $U_{p}$ will be the open set of $M_{\mu}$ but how do we show that? I mean $M_{\mu}$ is just a set there is no topology that we have defined on it then how we will show it. Please someone help me in this. Thanks.
|
Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, Prove that $\langle x, y \rangle = \overline{\langle y, x \rangle}.$
**My attempt:-**
$\overline{\langle y, x \rangle}=\overline{\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert y +i^k x\Vert^2}=\overline{\frac{1}{4}(\Vert y +x\Vert^2)+i\Vert y +i x\Vert^2-\Vert y - x\Vert^2-i\Vert y -ix\Vert^2)}=\frac{1}{4}(\overline{\Vert y +x\Vert^2)}-i\overline{\Vert y +i x\Vert^2}-\overline{\Vert y - x\Vert^2}+i\overline{\Vert y -ix\Vert^2}).$
Can I take the complex conjugate inside the norm?
|
During my research I stumbled upon this system over $\mathbb{Z}_{2^n}$. Let $k=2^{n-1}-1$ then the system looks like
\begin{equation*}
\begin{cases}
x_1+2^{n-1}=d_k +x_k \\
x_2+2^{n-1}=d_1+x_1 \\
x_3+2^{n-1}=d_2+x_2 \\ \tag{1}
\dots \\
x_k+2^{n-1}=d_{k-1}+x_{k-1} \\
x_i \notin \{x_j,x_j + 2^{n-1}\}, i \neq j \\
d_i \notin \{d_j,-d_j\}, i \neq j,
\end{cases}
\end{equation*}
where '+' means addition in $\mathbb{Z}_{2^n}$ and $x_i,d_i$ are variables for i $\in \overline{1,k}$.
I've noticed, that system (1) without any conditions has solutions iff $d_1+d_2+\dots+d_k=2^{n-1}$, so my idea was to fix $d_1,d_2,\dots d_k$ and then the solution to the system without coniditons looks like
\begin{equation}
\begin{cases}
x_1=d_k +2^{n-1} \\
x_2=d_k+d_1 \\
x_3=d_k+d_1+d_2 +2^{n-1} \\ \tag{2}
\dots \\
x_{k-1}=d_k+d_1+\dots +d_{k-2} \\
x_k=2^{n-1}.
\end{cases}
\end{equation}
Now, let $c_i=d_i+2^{n-1}$, then if $d_i \notin \{d_j,-d_j\}, i \neq j$ it follows that $c_i \notin \{c_j,-c_j\}, i \neq j$. Then in terms of $c_i$ solution (2) look like
\begin{equation}
\begin{cases}
x_1=c_k \\
x_2=c_k+c_1 \\
x_3=c_k+c_1+c_2 \\ \tag{3}
\dots \\
x_{k-1}=c_k+c_1+\dots +c_{k-2} \\
x_k=0.
\end{cases}.
\end{equation}
So, solving of system (1) is equivalent to finding $c_1,\dots ,c_k$, such that $c_i \notin \{c_j,-c_j\}, i \neq j$ and all sums like $c_i+\dots +c_j, 1 \leq i \leq j \leq k-2$ are nonzero modulo $2^{n-1}$.
I've programmed it and was able to find some interesting candidates for $C=\{c_i\}$. It is $C=\{1,2,\dots,2^{n-2}-1,2^{n-1},2^{n-1}+1,\dots,2^{n-1}+2^{n-2}\}$ as for $n=4$ we have solutions like
$(c_1,c_2,\dots,c_7)=(3, 2, 1, 9, 10, 11, 12)$
$(c_1,c_2,\dots,c_7)=(1, 2, 3, 12, 11, 10, 9)$ and a couple of others.
For $n=5$ we have something like
$(c_1,c_2,\dots,c_{15})=(24, 19, 20, 18, 21, 22, 17, 23, 1, 2, 3, 4, 5, 6, 7)$
$(c_1,c_2,\dots,c_{15})=(1, 6, 23, 4, 19, 7, 24, 21, 22, 20, 3, 18, 2, 17, 5)$.
For $n=6$ my program unfortunately wasn't able to find any. So the question is, how to find solutions to (1) for any $n \in \mathbb{N}$ or at least prove its existence? Maybe there are other ways much simplier ways to solve (1) rather than mine? I'll appreciate any help.
|
>How to rigorously prove that $\sum_{n=1}^ \infty( \frac{1}{4n-1} - \frac{1}{4n}) =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N (\frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n})$$
$$f_N'(x) = \sum_{n=1}^ N( x^{4n-2}- x^{4n-1})= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converges uniformly to be able to interchange the derivative and the summation, but I don't think $f_N'$ converges uniformly because
$$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ doesn't converge uniformly on $[0,1 )$.
Here I got stuck but for some reason it works, i.e., $\int_0 ^1 \frac{x^2}{(1+x)(1+x^2)}= \frac{\ln(64)- \pi}{8} $ so the derivative could be interchanged with the summation here, but how ?
|
How to rigorously prove that $\sum\limits_{n=1}^ \infty( \frac{1}{4n-1} - \frac{1}{4n} )=\frac{\ln(64)- \pi}{8}$?
|
I have the equation $x^2+xy+y^2=z^2$ to solve it in natural numbers.
The discriminant of it $D=4z^2-3y^2$ and must be perfect square.
I wrote Python program to get solutions for $1<x<100$ by enumeration.
def Solution():
A=[]
nMaximum=10**2
for x in range(1,nMaximum):
dTemp1a=3*x**2
for z in range(x+1, nMaximum):
dDiscriminant=4*z**2-dTemp1a
dTemp5=int(dDiscriminant**0.5)
if dTemp5**2!=dDiscriminant:
continue
dTemp6=(-1*x+dTemp5)/2
y=int(dTemp6)
if not CheckIfExists(A, z):
A.append([x,y,z])
return A
def CheckIfExists(arr, z):
bResult=False
for s in arr:
if s[2]==z:
bResult=True
break
return bResult
a = Solution()
print(len(a))
print(a)
# [3, 5, 7], [5, 16, 19], [6, 10, 14], [7, 8, 13] ...
[https://math.stackexchange.com/questions/1351994/three-variable-second-degree-diophantine-equation][1] doesn't explain how to get other solutions when we know the first solution $(3,5,7)$
Could you give me a hint ?
[1]: https://math.stackexchange.com/questions/1351994/three-variable-second-degree-diophantine-equation
|
How to check Pentagon axiom with induced associator of skeletal category?
|
>How to rigorously prove that $\sum_{n=1}^ \infty\left( \frac{1}{4n-1} - \frac{1}{4n}\right) =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \left(\frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}\right)$$
$$f_N'(x) = \sum_{n=1}^ N( x^{4n-2}- x^{4n-1})= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converges uniformly to be able to interchange the derivative and the summation, but I don't think $f_N'$ converges uniformly because
$$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ doesn't converge uniformly on $[0,1 )$.
Here I got stuck but for some reason it works, i.e., $\int_0 ^1 \frac{x^2}{(1+x)(1+x^2)}= \frac{\ln(64)- \pi}{8} $ so the derivative could be interchanged with the summation here, but how ?
|
How exactly does one detect if it is in a curved manifold?
|
Relationship for the real-value A and C given the medium B?
|
As I understand, in a function, the subset of the codomsin actually mapped to is called the range.
What about the domain? Is there a name for the subset of the domain actually used in a mapping?
|
Disclaimer : Not a direct answer but a methodology.
As your question can be understood as "how can I attack such an issue ?", I would like here to propose two natural tools for such questions involving angle bissectors in a triangle : [**trilinear coordinates**](https://en.wikipedia.org/wiki/Trilinear_coordinates) $(u:v:w)$ (abbreviation here : t.c.) and their use with [**isogonal conjugation**](https://en.wikipedia.org/wiki/Isogonal_conjugate#:~:text=In%20geometry%2C%20the%20isogonal%20conjugate,the%20isogonal%20conjugate%20of%20P.) $(u:v:w) \leftrightarrow (\tfrac{1}{u}:\tfrac{1}{v}:\tfrac{1}{w})$.
I will show it through a configuration of 8 points (see figure) *sharing some points with your own configuration*.
[![enter image description here][1]][1]
*Fig. 1.: Presentation in the case where $ABC$ is an equilateral triangle.*
This configuration is determined by a single point with t.c. $(u:v:w)$ in the following way (see the correspondence with Fig. 1 (and your own figure) :
$$\begin{cases}
(u:v:w)&\text{red disk}\\
(\tfrac{1}{u}:v:w)&\text{blue star ; your point O}\\
(u:\tfrac{1}{v}:w)&\text{green disk ; your point S}\\
(u:v:\tfrac{1}{w})&\text{yellow disk}\\
(u:\tfrac{1}{v}:\tfrac{1}{w})&\text{blue disk ; your point U}\\
(\tfrac{1}{u}:v:\tfrac{1}{w})&\text{green star ; your point Q}\\
(\tfrac{1}{u}:\tfrac{1}{v}:w)&\text{yellow star ; your point S}\\
(\tfrac{1}{u}:\tfrac{1}{v}:\tfrac{1}{w})&\text{red star}
\end{cases}$$
where two points with the same color are isogonal conjugates ; for example the yellow disk is conjugated with the yellow star (their t.c. are inverted componentwise).
**Remark :** one can consider this (2D!) points configuration as the perspective view of a cube represented with its 3 families of parallel edges prolongated until they meet resp. in $A,B,C$, playing the rΓ΄le of points at infinity.
Remark : the (extended) Morley configuration and its description in terms of t. c. can be found [here](https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem).
(to be continued)
[1]: https://i.stack.imgur.com/jkyRr.jpg
|
Suppose that $A\subseteq\mathbb{N}$. Then one can look at the function $f(n)=|\{0,...,n\}\setminus A|$ ($|S|$ denoting the cardinality of $S$ ). I am interested in the case when $f(n)\leq Cn^\alpha$ with $C>0$ and $\alpha<1$. In this case $A$ obviously has natural densitiy $1$. My question is: Is there already a name for this concept and if not how one "should" call sets with this property?
|
Say $r$ is an Asperger radius of $n$ if $(n-r,n+r)=(p^{a},q^{b})$ with $p$ and $q$ primes and integral $a$ and $b$ such that $ab=r$. If $r$ is a divisor of a prime number then of course $\min(a,b)=1$.
The famous twin prime conjecture can be reformulated as :
$1$ is an Asperger radius of infinitely many integers.
Are there infinitely many integers admitting at least one Asperger radius?
|
Let $X$ be a normed linear space over the field $\mathbb C$ with the norm $||.||$. Let $x,y \in X$,
Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, Prove that $\langle x, y \rangle = \overline{\langle y, x \rangle}.$
**My attempt:-**
$\overline{\langle y, x \rangle}=\overline{\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert y +i^k x\Vert^2}=\overline{\frac{1}{4}(\Vert y +x\Vert^2)+i\Vert y +i x\Vert^2-\Vert y - x\Vert^2-i\Vert y -ix\Vert^2)}=\frac{1}{4}(\overline{\Vert y +x\Vert^2)}-i\overline{\Vert y +i x\Vert^2}-\overline{\Vert y - x\Vert^2}+i\overline{\Vert y -ix\Vert^2}).$
Can I take the complex conjugate inside the norm?
|
As I understand, in a function, the subset of the codomain actually mapped to is called the range.
What about the domain? Is there a name for the subset of the domain actually used in a mapping?
|
<h4>Question Summary (to Make Easier to Reference in the Answer)</h4>
$$
\text{If }x^2-16\sqrt{x}=12 \text{ what is the value of }f=x-2\sqrt{x}
\tag{Eq. 1}$$
<h4>Checking for Consistency by Plugging in the $f=x-2\sqrt{x}=2$</h4>
Check $ x - 2\sqrt{x}-2=0$ implies $(\sqrt{x}-1)^2=2+1=3$ so $\sqrt{x}=1+\sqrt{3}$ (only the positive solution holds since $\sqrt{x} \ge 0$. So then $x=(1+\sqrt{3})^2=1+2\sqrt{3}+3=4+2\sqrt{3}$. Then $x^2=16+12+16\sqrt{3}=28+16\sqrt{3}$ and $x^2 - 16\sqrt{x}-12=(28+16\sqrt{3})-16*(1+\sqrt{3})-12=(28-16-12)+(16-16)\sqrt{3}=0$. Thus indeed it must be so that $x-2*\sqrt{x}=2$.
<h4>Solving for $f=x-2\sqrt{x}$ starting with $-2\sqrt{x}$ and then simplifying further</h4>
To start, one can solve for $-2\sqrt{x}$ as follows (subtracting $x^2$ from the left and right of Equation 1, and then dividing both sides by $8$):
$$
\frac{(x^2-16\sqrt{x})-x^2}{8}
=\frac{(12)-x^2}{8} \underset{implies}\implies
\tag{Eq. 2}
$$
$$
\underset{implies}\implies -2\sqrt{x}=\frac{12-x^2}{8}
\underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}\\
\underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}
\tag{Eqs.3}$$
Now further simplify Equations 3, by multiply each side by $8$ and then by subtracting from each side $8*x$ so then:
$$
x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}
\underset{implies}\implies 8x - 16\sqrt{x}=8x+12+x^2 \\
\underset{implies}\implies - 16\sqrt{x}=12-x^2 \\
\underset{implies}\implies x^2 - 16\sqrt{x}-12=0 \tag{Eq. 4}$$
$$\underset{implies}\implies (\sqrt{x})^4 - 16\sqrt{x}-12=0 \tag{Eq. 5}
$$
The above Equation 5 is a [solvable fourth-degree equation according to the link][1], which makes its solution techniques a little bit more unfamiliar than a second order polynomial.
I am hoping to update the answer with the referenced technique, but also this link can be used directly to solve the problem even before I provide step-by-step instructions on the solution. The link has the author's request for algebraic steps to solve the problem.
Alternatively, there are on-line [solvers also for this equation as follows for instance][2]:
> [![solution for sqrt x][3]][3]
Since $\sqrt{x}>0$ and it is a [irrational solution greater than zero][4], the only applicable solution from that is $\sqrt{x}=1+\sqrt{3}$ as was to be proven.
<h4>Checking Line-by-Line Using $\sqrt{x}=1+\sqrt{3}$, $x=2*(2+\sqrt{3})$, and $x^2=28+16*\sqrt{3}$</h4>
From Equation 3, with $x=(\sqrt{x})^2
=\left(1+\sqrt{3}\right)^2=1+3+2\sqrt{3}=2*(2+\sqrt{3})$ and
$x^2=28+16*\sqrt{3}$ and $\sqrt{x}=1+\sqrt{3}$:
$$
-2\sqrt{x}=\frac{12-x^2}{8}
\underset{implies}\implies -2\left(1+\sqrt{3}\right)
=\frac{12-\left(1+\sqrt{3}\right)^4}{8} \\
=\frac{12-\left(4+2*\sqrt{3}\right)^2}{8} \\
=\frac{12-\left(28+16*\sqrt{3}\right)}{8}\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3a}$$
$$
x-2\sqrt{x}=\left(4+2\sqrt{3}\right)-2*\left(1+\sqrt{3}\right)
=2\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3b}
$$
$$
x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8} \\
\underset{implies}\implies
2= \frac{\displaystyle (8*2*(2+\sqrt{3}))+12-(28+16\sqrt{3})}
{\displaystyle 8} \\
\underset{implies}\implies
2= \frac{\displaystyle ((32+16\sqrt{3}))+12-(28+16\sqrt{3})}
{\displaystyle 8} \\
2= \frac{\displaystyle (32+12-28)}
{\displaystyle 8}
= \frac{\displaystyle (16)}
{\displaystyle 8} = 2\text{ is correct! }
{\unicode{x2714}}
\tag{Eqs. 3c}$$
<h4>Solving the Quartic 4th Degreee Polynomial $(\sqrt{x})^4 - 16\sqrt{x}-12=0$ from Equation 5</h4>
From Equation 5, the solutions for $\sqrt{x}$ can be written as solutions to the [Quartic 4th Degree Polynomial][5] $(\sqrt{x})^4 - 16\sqrt{x}-12=0$.
Since the coefficient for the leading term $(\sqrt{x})^4$ is 1, then generally it is possible to seek two second-degree polynomials such that:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2+c*(\sqrt{x})+d \right)=0
\tag{Eq. 6}$$
Expanding Equaiton 6, for each coefficient of $(\sqrt{x})^n$, the expanded term can be compared to the coefficients from Equation 6 and Equation 5. To start with, compare the expanded coefficient element for the expanded $(\sqrt{x})^3$ as
$(\sqrt{x})^3*(a+c)=0$, so immediately Equation 6 can be simplified (with $c=-a$) in Equation 7 as:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\tag{Eq. 7}$$
1. Observe that in Equation 7, that the roles of $b$ and $d$ are
entirely similar. Either $d>b$ so that $d=b_+$,
and $b=b_-$. Or $d<b$ so that $d=b_-$ and $b=b_+$. However,
because of the [Commutative Law of Algebra for Multiplication][6]
arbitrarily the symmetry can be broken, with the larger value
$b_+$ being placed on the factor on the left, namely,
$\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)$. And $b_-$ can be
placed on the factor on the right, namely
$\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)$. Hence:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b_+ \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+b_- \right)=0
\tag{Eq. 8}$$
2. Now consider the coefficient equations for $(\sqrt{x})^2$ as
$(\sqrt{x})^2*(b_+ + b_- - a^2)=0$ so $(b_+ + b_- - a^2)=0$.
3. Now consider the coefficient equations for $(\sqrt{x})^1$ as
$(\sqrt{x})^1*(a(b_- - b_+))=-16*(\sqrt{x})^1$ so
$(a*(b_- - b_+))=-16$ so $(a*(b_+ - b_-))=16$ implying that
$a>0$ since $b_+>b_-$ implies that $b_+-b_->0$,
also a positive value.
4. Now consider the coefficient equations for $(\sqrt{x})^0$ as
$(\sqrt{x})^0*(b_+*b_-)=-12*(\sqrt{x})^0$ so $b_+*b_-=-12$.
5. There are now three Equations, namely $(b_+ + b_- - a^2)=0$,
$(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$. And there are
three unknowns to be solved for, namely $a$, $b_+$, and $b_-$;
So the rearrangement of the three equations is accomplished in
Equation 9 as follows:
$$
a=\frac{\displaystyle 16}{\displaystyle b_+ - b_-}
=\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\underset{implies}\implies
a^2=
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2=b_+ + b_- = b_+ - \frac{12}{b_+}
\tag{Eq. 9a}$$
$$
\underset{implies}\implies
16^2=256=\left(b_+ - \frac{12}{b_+} \right)*
\left( b_+ + \frac{12}{b_+} \right)^2
=(4)*(8*8)=(4)*(8)^2
\tag{Eq. 9b}$$
$$\tag{Eqs. 9}$$
Since the number $256$ has a factor of $(4)^1$ and also has factors of $(8)^2$, substitute $(4)^1=\left(b_+ - \frac{12}{b_+} \right)^1$ and also $8^2=\left( b_+ + \frac{12}{b_+} \right)^2$ and then check for consistency. It might seem like a slight of hand to make these assignments. However, consider that $128=(2*2)*(2*2*2)*(2*2*2)$, and also the left term, namely $\left(b_+ - \frac{12}{b_+} \right)$ is less than the right term, namely $\left( b_+ + \frac{12}{b_+} \right)$ because $b_+>0$. Thus the number of $2$'s in the square root could be $2$. But that less than $256/4$. Similarly $2*2=4<256/4$. And finally, $2*2*2=8$ meets the requirement for
$2*2*2=\left( b_+ + \frac{12}{b_+} \right)$, that it is greater
than $2*2=\left(b_+ - \frac{12}{b_+} \right)$ and that because of the prime number decomposition that $256=2^8$ that this works, as
$256=2^8=2^2*\left(2*2*2\right)^2$.
$$
\left(b_+ - \frac{12}{b_+} \right)=4
\underset{implies}\implies
(b_+)^2-4*(b_+)=12
\\
\underset{implies}\implies
\left((b_+)-2 \right)^2=12+4=16
\underset{implies}\implies
b_+=4+2=6 \\
\tag{Eqs. 10a}$$
$$
\left(b_- - \frac{12}{b_-} \right)=4
\underset{implies}\implies
(b_-)^2-4*(b_-)=12
\\
\underset{implies}\implies
\left((b_-)-2 \right)^2=12+4=16
\underset{implies}\implies
b_-=-4+2=-2 \\
\tag{Eqs. 10b}$$
$$
\left(b_+ + \frac{12}{b_+}\right)=8=6+\frac{12}{6}=6+2\text{ is correct!}
\tag{Eqs. 10c}$$
$$
a=\frac{\displaystyle 16}{\displaystyle b_+ - b_- }
=\frac{\displaystyle 16}{\displaystyle (6) - (-2) }
=\frac{\displaystyle 16}{\displaystyle 8 } = 2
\tag{Eqs. 10d}$$
$$
\tag{Eqs. 10}
$$
Note that $b_+*b_-=(6)*(-2)=-12$ as was to be expected, as the greater value for $b_+$ is selected because $b_+>b_-$, and also the lesser value of the polynomial equation for $b_-$.
<h4>Solving the Second Degree Polynomial Derived from the 4th Degree Quartic</h4>
Finally, the polynomial can be chosen that has a [positive real root][7]. Really all polynomial combinations need to be considered. For a 4th degree polynomial like this one, there are two real roots where one is greater than zero and one is less. And there are another two that have the factor of $\sqrt{-1}$ which is not a solution for $\sqrt{x}$ which needs a positive real solution.
If the terms for $-1*\sqrt{x}a$ and $b=b_-$ then the root of the second degree polynomial below must be real and positive as to be desired. So Equation 7 is repeated with those conditions now here:
$$
\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)*
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\\ \underset{implies}\implies
\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0
\tag{Eq. 7}$$
$$
\left((\sqrt{x})^2-2*(\sqrt{x})-2 \right)=0
\\ \underset{implies}\implies
\left(\sqrt{x}-1\right)^2=2+1=3
\\ \underset{implies}\implies
\sqrt{x}=1+\sqrt{3} \text{ as was to be proven!}
\tag{Eqs. 11}$$
To highlight the solution now for $\sqrt{x}$ from the 4th degree polynomial quartic $(\sqrt{x})^4 - 16\sqrt{x}-12=0$, Equation 11 has the solution as:
$$
\boxed{\sqrt{x}=1+\sqrt{3} \text{ as was to be proven!}
}\tag{Eq. 12}
$$
<h4>Verifying the Consistency of the 4th Degree Polynomial Quartic</h4>
Now there is also consistency. From the online [Equation solver][2], there is also consistency that verifies this result:
[![online equation solver tractability][8]][8]
Since $a$ is positive, and $b_+>b_-$, the online equation solver has results for $a$, $b_+$, and $b_-$, namely $a$ =2, $b_+=6$, and
$b_-=-2$.
The three Equations, namely $(b_+ + b_- - a^2)=0$,
$(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$, can be tested against the online results to find any algebraic errors, applying
$a$ =2, $b_+=6$, and $b_-=-2$. There is no mistake is in Equations 9, and the values given by the online Equation Solver are correct, also yielding the desired consistent result:
1. $(b_+ + b_- - a^2)=0$ implies $6-2-4=0$, correct!
2. $(a*(b_+ - b_-))=16$ implies $2*(6-(-2))=2*8=16$, correct!
3. $(b_+*b_-)=-12$ implies $6*(-2)=-12$, correct!
So, now Equations 9 can be carefully checked:
$$\text{Checking Eqs. 9a:}\\
a=\frac{a}{b_+ + b_-}=\frac{16}{6+(-2)}=4 \text{ is correct!} $$
$$
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2
=b_+ + b_- = b_+ - \frac{12}{b_+}$$
$$
\underset{implies}\implies
\left(
\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}}
\right)^2=
\left(
\frac{\displaystyle 16}{6 + \frac{12}{6}}
\right)^2
=4=6-\frac{12}{6}=6-2=4 \text{ is correct!}
$$
[1]: https://math.stackexchange.com/questions/3658305/is-there-formula-that-solves-quartic-equation-ax4bxc-0
[2]: https://www.google.com/search?client=opera&q=solve%20%24x%5E4-16*x-12%3D0%24&sourceid=opera&ie=UTF-8&oe=UTF-8
[3]: https://i.stack.imgur.com/HsKeW.png
[4]: https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm
[5]: https://mathworld.wolfram.com/QuarticEquation.html
[6]: https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/economics/algebra/fundamental-laws-of-algebra.html
[7]: https://mathworld.wolfram.com/RealNumber.html#:~:text=The%20field%20of%20all%20rational,Wolfram%20Language%2C%20and%20a%20number
[8]: https://i.stack.imgur.com/MCqyu.png
|
In [this][1] question it is proved in the answers that If $f :[0 , \infty ) \to \mathbb{R}$ and $f $ is bounded on every $(a,b)$ such that $a<b <\infty$, prove that $\lim\limits_{x \to \infty }(f(x+1) - f(x) )=l$ implies $\lim\limits_{x \to \infty }\frac{f(x)}{x}=l$.
This condition looks suspiciously similar to Stolz-Cesaro theorem see [this][2] for a proof of using Stolz-Cesaro theorem .
----
My question is : If If $f,g :[0 , \infty ) \to \mathbb{R}$ $\lim\limits_{x\to\infty}\frac{f(x+1)-f(x)}{g(x+1)-g(x)}=l$ what are sufficient conditions for $f,g$ to make $\lim\limits_{x\to\infty}\frac{f(x)}{g(x)}=l$?
Of course when $f$ and $g $ are both differentiable functions on $\mathbb{R}$ and if $\lim\limits_{x \to \infty } g(x)= \infty$ this condition sufficient to make $\lim\limits_{x\to\infty}\frac{f(x)}{g(x)}=l$, but this hypothesis is too strong is there a weaker hypothesis. and I want to generalise this result.
I conjectured this more general hypothesis : If $f$ is bounded on every $(a,b)$ such that $a<b <\infty$, $g$ is a continuous increasing function on $\mathbb{R}$ then $\lim\limits_{x\to\infty}\frac{f(x)}{g(x)}=l$.
I couldn't prove this result but I think it is true.
My question is this conjecture true ? If it was true is there a weaker hypothesis ? If it wasn't true, Is there a weaker hypothesis than $f$ and $g $ are both differentiable functions on $\mathbb{R}$ and ]$\lim\limits_{x \to \infty } g(x)= \infty$?
[1]: https://math.stackexchange.com/questions/1642662/prove-that-lim-x-to-infty-fracfxx-l-if-lim-x-to-infty-f
[2]: https://math.stackexchange.com/questions/4858423/can-stolz-cesaro-theorem-be-applied-to-this-problem-if-lim-limits-x-to-infty?noredirect=1&lq=1
|
>How to rigorously prove that $\sum\limits_{n=1}^ \infty\left( \frac{1}{4n-1} - \frac{1}{4n}\right) =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \left(\frac{x^{4n-1}}{4n-1} - \frac{x^{4n}}{4n}\right)$$
$$f_N'(x) = \sum_{n=1}^ N( x^{4n-2}- x^{4n-1})= x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$$
I need to show that $x^{4}\left(\frac{1-x}{x^2} \right)\frac{x^{4N+4}-1 }{x^4-1}$ converges uniformly to be able to interchange the derivative and the summation, but I don't think $f_N'$ converges uniformly because
$$f_N'(x) = \frac{-x^2}{(1+x)(1+x^2)}\cdot (x^{4N+4}-1) $$ and $(x^{4N+4}-1)$ doesn't converge uniformly on $[0,1 )$.
Here I got stuck but for some reason it works, i.e., $\int_0 ^1 \frac{x^2}{(1+x)(1+x^2)}= \frac{\ln(64)- \pi}{8} $ so the derivative could be interchanged with the summation here, but how ?
|
I am trying to show that the polynomial $y-x^2-x^3$ is reducible in the formal power series ring $k[[x,y]]$. I am attempting the question by finding a polynomial in $k[[x,y]]$ which is the square root of $x^2+x^3$. In order to find the square root I wrote the general polynomial in $k[[x,y]]$, $$a_{00}+a_{10}x+a_{01}y+a_{20}x^2 ...........$$ Took it's square and equated the coefficients to the coefficients of $x^2+x^3$. I got the following system of equations
$$a_{00}^2=0$$
$$2a_{10} a_{00}=0$$
$$2a_{20} a_{00}+{a_{10}}^2=1 $$
and so on. But this system doesnot have a solution. I am sure that the root does exist. What am I doing wrong?
|
# Problem
Compute
$$
\oint_L\frac{xdy-ydx}{4x^2+y^2}
$$
where $L$ is a circle centered at $(1, 0)$ with a radius of $R > 1$, and the direction of $L$ is counterclock-wise.
# Solution
To by pass the singular point at $(0, 0)$ which is inside the circle $L$ surrounds, adding an auxiliary curve:
$$
C:4x^2+y^2 = \delta^2
$$
The integral can then be computed by applying Green's theorem. However, different choices of the direction of $C$ produce different answers. If we choose counterclock-wise:
$$
\oint_L\frac{xdy-ydx}{4x^2+y^2}=\oint_{L+C}-\oint_{C}=-\oint_{C}=-\pi
$$
οΌ if we choose clockwise:
$$
\oint_L\frac{xdy-ydx}{4x^2+y^2}=\oint_{L+C^{-1}}-\oint_{C^{-1}}=\oint_{C}=\pi
$$
What I got wrong here?
|
Consider a subgroup of the real number (additive subgroup) and a positive homomorphism $\tau$ from the subgroup to $R$. Positive homomorphism means it maps positive elements to positive elements. I believe that $\tau$ has to be a positive scalar multiple ie $\tau(x)=cx$ but I am not sure how to show this. This is easy if we are looking at a positive homomorphism from $R$ to $R$ since we can show that $\tau$ maps every element to the scalar multiple of $\tau(1)$ and then we inductively show this for the integers, rational numbers, and real numbers. However an additive subgroup of $R$ may not contain the integers and it may not even contain $1$. So how should I show this?
|
Why is it not possible to define the necessity operator internally $\Box: \Omega \to \Omega$ in a topos?
|
A straight line has, in the traditional sense of the term, two directions; we can mark a starting point and from it we observe two directions, or in other words, to reach any point on a straight line we can do it only in two directions.
I heard that there is a topological demonstration that it is not possible to algebraically construct a straight line with more than two directions.
Do any of you know that demonstration and where can I find it?"
|
There are two ways to interpret the question. First, is an allocation in the core for the initial assignment of houses? Second, is the allocation in the core if every agent owns their assigned house. For the second question, just run the top trading cycle algorithm starting from the allocation (that seems to be your answer, too). If nothing changes, you have a core allocation. For the first question, just run the top trading cycle algorithm from the initial assignment of houses and compare. With strict preferences, there is always a unique weak core allocation; this is part of Theorem 2 of
> Roth, Alvin E., and Andrew Postlewaite. "[Weak versus strong
> domination in a market with indivisible goods.][1]" *Journal of
> Mathematical Economics* 4.2 (1977): 131-137.
[1]: https://doi.org/10.1016/0304-4068(77)90004-0
|
Consider a subgroup of the real number (additive subgroup) and a positive homomorphism $\tau$ from the subgroup to $\mathbb R$. Positive homomorphism means it maps positive elements to positive elements. I believe that $\tau$ has to be a positive scalar multiple ie $\tau(x)=cx$ but I am not sure how to show this. This is easy if we are looking at a positive homomorphism from $R$ to $R$ since we can show that $\tau$ maps every element to the scalar multiple of $\tau(1)$ and then we inductively show this for the integers, rational numbers, and real numbers. However an additive subgroup of $R$ may not contain the integers and it may not even contain $1$. So how should I show this?
|
Let $X$ be a normed linear space over the field $\mathbb C$ with the norm $\|\cdot \|$. Let $x,y \in X$. Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$. Prove that $\langle x, y \rangle = \overline{\langle y, x \rangle}.$
**My attempt:**
\begin{align*}
\overline{\langle y, x \rangle} &= \overline{\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert y +i^k x\Vert^2} \\
&= \overline{\frac{1}{4}(\Vert y +x\Vert^2)+i\Vert y +i x\Vert^2-\Vert y - x\Vert^2-i\Vert y -ix\Vert^2)} \\
&= \frac{1}{4}(\overline{\Vert y +x\Vert^2)}-i\overline{\Vert y +i x\Vert^2}-\overline{\Vert y - x\Vert^2}+i\overline{\Vert y -ix\Vert^2}).
\end{align*}
Can I take the complex conjugate inside the norm?
|
The following is a particular case of rank theorem.
Theorem 1( Rank theorem for surjective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is surjective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$
and for all $x=(x^1,...,x^m)( \in\varphi(U)$, and
$\psi\circ F \circ \varphi^{β1}(x) = (x^1,...,x^n)$
which is used to prove:
**The regular value theorem** If $F : M \to N$ is an embedding, then $F (M)$ is an embedded
submanifold of $N$
Proof. Let $p \in M$. Since $F$ is an immersion, by Theorem 1 there are charts $(U,\varphi)$ around $p$ and $(V,\psi )$ around $F(p)$ such that $F(U) \subseteq V$ and for all
$x \in \phi(U)$,
$\psi\circ F \circ \varphi^{β1}(x) = (x; 0_{nβm})\tag{4}$
By applying translations we can assume that $\varphi(p) = 0_m$ and $\psi(F(p)) = 0_n$. Let $\varepsilon > 0$ so that
$B_{\varepsilon}(0_m) \subseteq \varphi(U)$ and $B_{\varepsilon}(0_n) \subseteq \psi(V )$
Then $B_{\varepsilon}(0_m) \times \{0_{n-m}\} \subseteq B_{\varepsilon}(0_n) \subseteq \psi(V)\tag{5}$
Because $F:M\to F(M) $is a homeomorphism, then $F(\varphi^{β1}(B_{\varepsilon}(0_m))) \subseteq F(M)$ open in the subspace topology. So there is an open subset $W \subseteq N$ such that
$F(\varphi^{β1}(B_{\varepsilon}(0_m))) =W\cap F(M) \tag{ 6}$
And because of (4),
$\psi(W\cap F(M)) = (\psi\circ F \circ \varphi^{β1}(x))(B_{\varepsilon}(0_m))) \subseteq B_{\varepsilon}(0_m) \times \{0_{n-m}\}\tag{ 7}$
Let $\tilde {W}=W \cap \psi^{-1}(B_{\varepsilon}(0_n))$
So $(\tilde {W}, \psi|_{\tilde W}) $ is a chart around $F(p)$ with
$\psi|_{\tilde {W}}(\tilde{W} \cap F(M)) $
$= \psi|_{\tilde{W}}( W \cap \psi^{-1}(B_{\varepsilon}(0_n)) \cap F(M))$
$=\psi (W \cap F(M)) \tag{8}$
$=B_{\varepsilon}(0_m)\times \{0_{n-m}\}\tag{9}$
$=\psi|_{\tilde {W}}(\tilde {W})\cap (\Bbb R^{m}\times\{0_{n-m}\})\tag{10}$
1) Is there is a typo in (5)? I think it should be "=" instead of "\subseteq"
because then I think that's what it's being used in (9).
2)**I can't figure out what is going on in the last chain of equalities**.The preceding expressions should be used somehow , but it's not clear how. How does $\psi^{-1}(B_{\varepsilon}(0_n))$ dissapear in (8)? How do I get (9)?(That should be clear if I'm right in question (1)) and how do I go from there to (10)?
The definition of embedded submanifolds that is being used in this last part is
|
We are given that a, b, and c are positive real numbers such that:
a + b + c = abc - 2
We want to prove the following inequality:
$sqrt(1/(1+a)) + sqrt(1/(1+b)) + sqrt(1/(1+c)) <= sqrt(3)$
**Can anyone provide a proof for this inequality under the condition that a, b, and c are positive?**
|
Suppose that $M$ is a smooth manifold. Let $\omega$ be an $n-$ form on $M$ with compact support. Then we define $\int_M\omega$ using partitions of unity. If $M$ is covered by a single chart $h:M\to \mathbb R^n$, then we define $\int_M\omega:= \int_{\mathbb R^n} (h^{-1})^\ast \omega$, where $\ast$ denotes pullback. $\tag 1$
But often the following definition is stated:
$\int_{\gamma} \omega := \int_{[0,1]} \gamma^\ast \omega$, where $\gamma:[0,1]\to M$ is a smooth curve and $\omega$ is a $1$- form on $M$. $\tag 2$
My questions are:
$(a)$ what is the source of the definition in $(2)$?
$(b)$ Does this somehow follow from the definition in $(1)$?
I think the answer to $(b)$ is no because taking the definition in $(1)$ to be a general definition, the term $\int_{\gamma} \omega$ makes sense iff $\gamma[0,1]$ is a $1$- manifold but that's not the case in general: Smooth image of a $1$- manifold is not necessarily a manifold.
That brings me back to $(a)$. I didn't find the definition $(2)$ neither in Tu's book nor in Lee's book nor in Spivak's.
Can anyone please direct me to where the definition $(2)$ has been stated?
|
Given constant matrices $A_1\in\mathbb{R}^{1\times l}$ and $A_2\in\mathbb{R}^{1\times l}$, and constants $b_i$, $i=1,\cdots,n$. Cosider the following mixed inter programm(MIP) with decision variable $c_i\in\{0,1\}$, and $X=[x_1,\cdots,x_n]\in\mathbb{R}^{l\times n}$ with $x_i\in\mathbb{R}^{l}$ for $i=1,\cdots,n$.
Objective: min $\sum_{1}^{n} |c_iA_1x_i|$
Constraints: $A_2x_i<c_ib_i$; $\;\;$ $\sum_{1}^nc_i>=1$; $\;\;$ $c_i\in\{0,1\}$.
This problem is untractable because the objective function contains the product of the decision variables $c_i$ and $x_i$. Is it possible to derive a tractable formulation for this problem?
|
The following is a particular case of rank theorem.
Theorem 1( Rank theorem for surjective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is surjective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$
and for all $x=(x^1,...,x^m)( \in\varphi(U)$, and
$\psi\circ F \circ \varphi^{β1}(x) = (x^1,...,x^n)$
which is used to prove:
**The regular value theorem** If $F : M \to N$ is an embedding, then $F (M)$ is an embedded
submanifold of $N$
Proof. Let $p \in M$. Since $F$ is an immersion, by Theorem 1 there are charts $(U,\varphi)$ around $p$ and $(V,\psi )$ around $F(p)$ such that $F(U) \subseteq V$ and for all
$x \in \phi(U)$,
$\psi\circ F \circ \varphi^{β1}(x) = (x; 0_{nβm})\tag{4}$
By applying translations we can assume that $\varphi(p) = 0_m$ and $\psi(F(p)) = 0_n$. Let $\varepsilon > 0$ so that
$B_{\varepsilon}(0_m) \subseteq \varphi(U)$ and $B_{\varepsilon}(0_n) \subseteq \psi(V )$
Then $B_{\varepsilon}(0_m) \times \{0_{n-m}\} \subseteq B_{\varepsilon}(0_n) \subseteq \psi(V)\tag{5}$
Because $F:M\to F(M) $is a homeomorphism, then $F(\varphi^{β1}(B_{\varepsilon}(0_m))) \subseteq F(M)$ open in the subspace topology. So there is an open subset $W \subseteq N$ such that
$F(\varphi^{β1}(B_{\varepsilon}(0_m))) =W\cap F(M) \tag{ 6}$
And because of (4),
$\psi(W\cap F(M)) = (\psi\circ F \circ \varphi^{β1}(x))(B_{\varepsilon}(0_m))) \subseteq B_{\varepsilon}(0_m) \times \{0_{n-m}\}\tag{ 7}$
Let $\tilde {W}=W \cap \psi^{-1}(B_{\varepsilon}(0_n))$
So $(\tilde {W}, \psi|_{\tilde W}) $ is a chart around $F(p)$ with
$\psi|_{\tilde {W}}(\tilde{W} \cap F(M)) $
$= \psi|_{\tilde{W}}( W \cap \psi^{-1}(B_{\varepsilon}(0_n)) \cap F(M))$
$=\psi (W \cap F(M)) \tag{8}$
$=B_{\varepsilon}(0_m)\times \{0_{n-m}\}\tag{9}$
$=\psi|_{\tilde {W}}(\tilde {W})\cap (\Bbb R^{m}\times\{0_{n-m}\})\tag{10}$
**1)Is there is a typo in (5)?** I think it should be "$=$" instead of "$\subseteq$"
because then I think that's what it's being used in (9).
2)**I can't figure out what is going on in the last chain of equalities**.The preceding expressions should be used somehow , but it's not clear how. How does $\psi^{-1}(B_{\varepsilon}(0_n))$ dissapear in (8)? How do I get (9)? and how do I go from there to (10)?
The definition of embedded submanifolds that is being used in this last part is:
DEF Let $M$ be a smooth manifold of dimension $m$, and $S β M$. Then
$S$ is an embedded submanifold of dimension $k$ if for every $p \in S $ there is a chart $(U, \varphi)$ of $M$ around $p$ is such that
$\varphi(U \cap S) = (\Bbb R^k \times \{0_{m-k}\}) \cap \varphi(U)$
|
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying
$$abc\ne 0.$$
Consider the following equations of $(u,v,w)\in\mathbb{R}^3$:
$$
(-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw,
$$
$$
(-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT,
$$
$$
(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT,
$$
where $$T= -au-bv-cw.$$
Surprisingly, computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the solution of these equations is
$$
(u,v,w)= (0,0,0).
$$
[![enter image description here][1]][1]
How can I prove this claim? I tried many times but fail...
I notice that these equations are equivalent to
$$
3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+w^2+2vT}{-b} = \frac{2uT}{-a},
$$
which gives that (if $u\ne 0$ )
$$
T = \frac{a}{c}\frac{vw}{u}.
$$
Can I use this to find something different?
[1]: https://i.stack.imgur.com/y5Tqw.jpg
|
We are given that a, b, and c are positive real numbers such that:
$a + b + c = abc - 2$
We want to prove the following inequality:
$\sqrt{\frac{1}{1+a}}+\sqrt{\frac{1}{1+b}} + \sqrt{\frac{1}{1+c}} \leq \sqrt{3}$
**Can anyone provide a proof for this inequality under the condition that a, b, and c are positive?**
|
The following is a particular case of rank theorem.
Theorem 1( Rank theorem for surjective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is surjective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$
and for all $x=(x^1,...,x^m)( \in\varphi(U)$, and
$\psi\circ F \circ \varphi^{β1}(x) = (x^1,...,x^n)$
which is used to prove:
**The regular value theorem** If $F : M \to N$ is an embedding, then $F (M)$ is an embedded
submanifold of $N$
Proof. Let $p \in M$. Since $F$ is an immersion, by Theorem 1 there are charts $(U,\varphi)$ around $p$ and $(V,\psi )$ around $F(p)$ such that $F(U) \subseteq V$ and for all
$x \in \phi(U)$,
$\psi\circ F \circ \varphi^{β1}(x) = (x; 0_{nβm})\tag{4}$
By applying translations we can assume that $\varphi(p) = 0_m$ and $\psi(F(p)) = 0_n$. Let $\varepsilon > 0$ so that
$B_{\varepsilon}(0_m) \subseteq \varphi(U)$ and $B_{\varepsilon}(0_n) \subseteq \psi(V )$
Then $B_{\varepsilon}(0_m) \times \{0_{n-m}\} \subseteq B_{\varepsilon}(0_n) \subseteq \psi(V)\tag{5}$
Because $F:M\to F(M) $is a homeomorphism, then $F(\varphi^{β1}(B_{\varepsilon}(0_m))) \subseteq F(M)$ open in the subspace topology. So there is an open subset $W \subseteq N$ such that
$F(\varphi^{β1}(B_{\varepsilon}(0_m))) =W\cap F(M) \tag{ 6}$
And because of (4),
$\psi(W\cap F(M)) = (\psi\circ F \circ \varphi^{β1}(x))(B_{\varepsilon}(0_m))) \subseteq B_{\varepsilon}(0_m) \times \{0_{n-m}\}\tag{ 7}$
Let $\tilde {W}=W \cap \psi^{-1}(B_{\varepsilon}(0_n))$
So $(\tilde {W}, \psi|_{\tilde W}) $ is a chart around $F(p)$ with
$\psi|_{\tilde {W}}(\tilde{W} \cap F(M)) $
$= \psi|_{\tilde{W}}( W \cap \psi^{-1}(B_{\varepsilon}(0_n)) \cap F(M))$
$=\psi (W \cap F(M)) \tag{8}$
$=B_{\varepsilon}(0_m)\times \{0_{n-m}\}\tag{9}$
$=\psi|_{\tilde {W}}(\tilde {W})\cap (\Bbb R^{m}\times\{0_{n-m}\})\tag{10}$
**1)Is there is a typo in (7)?** I think it should be "$=$" instead of "$\subseteq$"
because then I think that's what it's being used to go from (8) to (9).
2)**I can't figure out what is going on in the last chain of equalities**.The preceding expressions should be used somehow , but it's not clear how. How does $\psi^{-1}(B_{\varepsilon}(0_n))$ dissapear in (8)? How do I get (9)? and how do I go from there to (10)?
The definition of embedded submanifolds that is being used in this last part is:
DEF Let $M$ be a smooth manifold of dimension $m$, and $S β M$. Then
$S$ is an embedded submanifold of dimension $k$ if for every $p \in S $ there is a chart $(U, \varphi)$ of $M$ around $p$ is such that
$\varphi(U \cap S) = (\Bbb R^k \times \{0_{m-k}\}) \cap \varphi(U)$
|
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