Saturday, September 11, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


Can we construct reduced group scheme which is same group structure given an abstract group?

Posted: 11 Sep 2021 08:10 PM PDT

Sorry for my bad English.

Let $G$ be an abstract group (if necessary finite), and $k$ be an algebraically closed field.

Now is there a group scheme $X$ over $k$ such that group of $k$-valued point $X(k)=\operatorname {Hom}_k(\operatorname {Spec} k, X)$ is isomorphic to $G$?

Especially when $G$ is finite, is there such X as reduced?

(I mean "scheme over $k$" as separated finite type over $k$.)

Simple proof $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$ using Tannery's Theorem

Posted: 11 Sep 2021 08:09 PM PDT

The question of proving $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$ has been posed several times.

The answers seem to be overly complex - eg (link).

My previous question (link) was not answered, but was challenged regarding the interchange of limit operators.

Here I attempt to use Tannery's Theorem (link) to justify the interchange of limit operators.

Here $\zeta(s)=\sum 1/n^s$ for complex $s=\sigma+it$, and $\sigma>1$.


Question: Is the following simple argument correct?

Tannery's Theorem says that

$$ \lim_{\sigma \rightarrow +\infty} \sum 1/n^s = \sum \lim_{\sigma \rightarrow +\infty} 1/n^s$$

if $|1/n^s| \leq M_n$ and $\sum M_n < \infty$.

In our case $|1/n^s| \leq 1/n^{\sigma}$ and $\sum 1/n^{\sigma} < \infty$ for $\sigma>1$. Thus, the conditions for Tannery's theorem are met.

Therefore $\zeta(s)\rightarrow1$ as $\sigma\rightarrow +\infty$.


I am having a tough time understanding how the two integrals are equal to one another

Posted: 11 Sep 2021 08:09 PM PDT

enter image description here

I figured it was a typo and it meant to say uv^3 but I didn't want to assume my professor was wrong.

Proving the triangle inequality for Euclidean metric

Posted: 11 Sep 2021 08:05 PM PDT

I am trying to solve this below problem.

Consider $\mathbb{R}^n$ with the metric $d(x,y) = \left(\sum\limits_{i=1}^n \left(x_i - y_i\right)^2 \right)^{1/2}$. Prove the triangle inequality.

I first need to prove the Cauchy-Schwarz inequality, which states that for $x,y \in \mathbb{R}^n$, we have \begin{align*} \left(\sum\limits_{i=1}^n x_i y_i \right)^2 \leq \left(\sum\limits_{i=1}^n x_i^2 \right) \cdot \left(\sum\limits_{i=1}^n y_i^2 \right). \end{align*}

The only way I know how to prove this is by "proceeding backwards," and arguing that every step is reversible. My attempt: \begin{align*} \left(\sum\limits_{i=1}^n x_i y_i \right)^2 \leq \left(\sum\limits_{i=1}^n x_i^2 \right) \cdot \left(\sum\limits_{i=1}^n y_i^2 \right) & \iff 0 \leq \left(\sum\limits_{i=1}^n x_i^2 \right) \cdot \left(\sum\limits_{i=1}^n y_i^2 \right) - \left(\sum\limits_{i=1}^n x_i y_i \right)^2 \\ & \iff 0 \leq 2 \left(\sum\limits_{i=1}^n x_i^2 \right) \cdot \left(\sum\limits_{i=1}^n y_i^2 \right) - 2\left(\sum\limits_{i=1}^n x_i y_i \right)^2 \\ & \iff 0 \leq 2 \sum\limits_{i=1}^n x_i^2 \sum\limits_{i=1}^n y_i^2 - 2 \sum\limits_{i=1}^n x_i y_i \sum\limits_{i=1}^n x_i y_i \\ & \iff 0 \leq \sum\limits_{i=1}^n x_i^2 \sum\limits_{j=1}^n y_i^2 + \sum\limits_{j=1}^n x_j^2 \sum\limits_{i=1}^n y_i^2 - 2 \sum\limits_{i=1}^n x_i y_i \sum\limits_{i=1}^n x_i y_i \\ & \iff 0 \geq \sum\limits_{i=1}^n \sum\limits_{j=1}^n \left(x_i y_j + x_j y_i \right)^2 \end{align*} As the final line is true since a sum of non-negative terms is negative, our starting statement is also true.

The steps of the proof are a bit obscure to me. I know we're working toward the final statement, ultimately, and the way to. build that is to break apart the sum in the third line and notice that the actual index we use doesn't matter, so I need only index $x$ with $i$ and $j$ and $y$ with $i$ and $j$. If anyone knows how to make this proof more intuitive or more elegant, I'd appreciate.

Onto the triangle inequality:

I need to show that for $x,y,z \in \mathbb{R}^n$, we have: $$d(x,y) \leq d(x,z) + d(z,y) \text{ or } \left(\sum\limits_{i=1}^n (x_i - y_i)^2 \right)^{1/2} \leq \left(\sum\limits_{i=1}^n (x_i - z_i)^2 \right)^{1/2} + \left(\sum\limits_{i=1}^n (z_i - y_i)^2 \right)^{1/2}.$$

We start with the square of both sides above and again derive a true statement: \begin{align*} \sum\limits_{i=1}^n (x_i - y_i)^2 & \leq \sum\limits_{i=1}^n \left(x_i^2 - 2x_i y_i + y_i^2\right) \\ & = \sum\limits_{i=1}^n x_i^2 - 2 \sum\limits_{i=1}^n x_i y_i + \sum\limits_{i=1}^n y_i^2 \end{align*} I need to work $z_i$ in here somewhere, and I assume what I need to do is add and subtract $z_i$ for each $i$ somewhere. An alternative is to start with the full statement, square both sides, and derive a true statement.

Any hints on how to proceed would be appreciated.

Differential function on $(a,b)$ satisfying $f’(x)+f(x)^2 +1 \geq 0$.

Posted: 11 Sep 2021 08:01 PM PDT

Prove that if $f$ is differential function on $(a,b)$ and if $\lim_{x \to a^+} f(x) = + \infty$, $\lim_{x \to b^-} f(x) = - \infty$, and $f'(x) + f(x)^2 + 1 \geq 0$ for all $x \in (a,b)$, then $b-a \geq \pi$.

My intuition is to compare the function $f$ with $-\tan x$ because $(-\tan (x))' + \tan^2 (x) + 1 = -\sec^2 x +\sec^2 x=0$. But, I'm not sure how to start the proof.

Expressing absolute value intervals in terms of inequalities

Posted: 11 Sep 2021 07:51 PM PDT

The question below is in my textbook about ınequalities of absolute values. The question Express the interval in terms of an inequality involving absolute value. is asked. I do not understand how the answers correlate with the questions above. How would I be able to get the answers for them?

Examples:

1.[-2, 2]  2.(0,4)  3.[-1, 8]  

Answers:

1. |x| ≤ 2    2. |x - 2| < 2  3. |x - 3.5| ≤ 4.5  

Complex function theory

Posted: 11 Sep 2021 07:38 PM PDT

Let $p(z)= a_7 z^7+a_6 z^6+...+a_1 z+a_0$ be a polynomial such that $a_7,a_6, a_5, a_3, a_1$ are even, $a_4$ and $a_0$ odd).

how can I prove that $|p(z)|\ge 1$ for all $|z|=1$?

If $k[x] \to k[y]$ maps $x \mapsto f(y)$ where $f(y)$ is a polynomial of degree $n$, then $k[y]$ is a finitely generated $k[x]$-module?

Posted: 11 Sep 2021 07:33 PM PDT

Suppose $k[x] \to k[y]$ maps $x \mapsto f(y)$ where $f(y)$ is a polynomial of degree $n$.

We want to show that $k[y]$ is a finitely generated $k[x]$-module generated by $1,y, \dots, y^{n-1}$.

I believe it suffices to show that $y^n$ can be written as a linear combination of $1, y, \dots, y^{n-1}$ with coefficients in $k[x]$. If we have this, then we have enough to rewrite any polynomial $g(y)$ in terms of the $1, y, \dots, y^{n-1}$.

However, I am unable to figure this out.

How do we get started?

Probability and Statistics Question: Four Sided Die [closed]

Posted: 11 Sep 2021 07:40 PM PDT

This is the question I need help on: A four sided die is rolled 15 times. Find the probability that each side of the die appears at least once during the 15 rolls.

This was the hint I was given: Consider the events Ai = The number i does not occur during the 15 rules and use the complement formula

I have tried to do 1- ((15(4 3)(4 3)(4 3)(4 3))/15^4) but the hint leads me to believe that I need to use the law of total probability.

I need help understanding the fundamental theorem of algebra and more importantly how it generalizes to the quaternions [closed]

Posted: 11 Sep 2021 08:10 PM PDT

The quaternions are described by i^2 = j^2 = k^2 = ijk = -1

ij = -ji , ik = -ki , jk = -kj this is anticommutativity

define x=1 + i + j + k

x^2 = x + ix + jx + kx

x^2 = x + i(1 + i + j + k) + j(1 + i + j + k) + k(1 + i + j + k)

x^2 = x + i + i^2 + ij + ik + j + ji + j^2 + jk + k + ki +kj +k^2

x^2 = x + i + j + k - 1 - 1 - 1

x^2 = x + 1 + i + j + k - 1 - 3

x^2 = 2x - 4

x^2 -2x + 4 = 0

(x - Re[z] - iIm[z])(x - Re[z] + iIm[z]) = 0

(x - Re[z])^2 - (iIm[z])^2 = 0

x^2 - 2Re[z]x + Re[z]^2 + Im[z]^2 = 0

x^2 - 2Re[z]x + |z|^2 = 0 , set x=z

x^2 - 2Re[x]x + |x|^2 = 0 this function is true for all values of x

when i plug this function into wolfram alpha it just pops out true. If this makes sense than any number with the same real component and the same magnitude would be solutions to the same equation. How does this relate back to the fundamental theorem of algebra. When ever its presented it always says that a polynomial with real co-efficient will always have n roots were n is its degree. I kind of don't understand and would like some help with this stuff please. So simply why is this going on?

Information-based function approximation. Series of bits instead of series of real/complex coefficients

Posted: 11 Sep 2021 07:13 PM PDT

There are multiple series and functional bases (e.g. Taylor, Fourier) that allow approximating a function.

In some sense, such approximations associate a function with an infinite sequence of real or complex numbers (e.g. coefficients). Also, there are often theorems that bound the approximation errors for a truncated series.

One problem I have with these approaches is that they use real numbers which contain infinite amount of information.

I'm interested in an approximation technique where the function is associated with an infinite series of bits. Ideally there should be a theorem that bounds the approximation errors for a truncated series of bits.

Are there any information-based approximation methods?

Compact continuous functions and traces in sobolev spaces?

Posted: 11 Sep 2021 07:05 PM PDT

In Evans book the definition of a Sobolev space, in terms of compactly continuous functions is given as above. Can anyone clarify being "0" (zero) on the boundary is only valid for derivatives up to k-1? The definition given in the book by Evans book is as follows. Thanks!

DEFINITION. We denote by $W_{0}^{k, p}(U)$ the closure of $C_{c}^{\infty}(U)$ in $W^{k, p}(U)$ Thus $u \in W_{0}^{k, p}(U)$ if and only if there exist functions $u_{m} \in C_{c}^{\infty}(U)$ such that $u_{m} \rightarrow u$ in $W^{k, p}(U)$. We interpret $W_{0}^{k, p}(U)$ as comprising those functions $u \in W^{k, p}(U)$ such that $" D^{\alpha} u=0$ on $\partial U$ " for all $|\alpha| \leq k-1$

Can someone please provide a proof for the derivatives up to K-1 being zero?

How to evaluate the limit of $f$ and find the domain of $f$?

Posted: 11 Sep 2021 08:03 PM PDT

Let this function : $$ f\left( x\right) =\dfrac{\sqrt{2x}-\sqrt{2}}{x^{3}-5x^{2}+8x-4} $$ How can I determine the domain of $f$ ?

How can I evaluate algebrically the following in its simplest and exact form?

$$ \lim _{x\rightarrow 1}f\left( x\right) $$

I tried using the quadratic formula but the gradient was negative. What should do to solve this?

Jordan decomposition of Idempotent matrix.

Posted: 11 Sep 2021 07:08 PM PDT

Matrix A $\in$ $\mathbb{R}^{n\times n}$ is idempotent if $A^{2} = A$. Describe the Jordan form of A.

How do I do this?

I am able to decompose a matrix to its Jordan form given that the matrix contains numeric entries. But I am having hard time figuring out the correct way to approach the solution in a general case such as this question. Thanks!

$5y''+6y'=x^2+5x +3$ What is the $ y_p$ here? Can someone link to proof?

Posted: 11 Sep 2021 07:38 PM PDT

$5y''+6y'=x^2+5x +3$ What is the $ y_p$ here?

So I tried the usual. when the right hand side is polynomial degree of 2, my $ y_p = Ax^2 + Bx + C$. I then differentiate once and twice and sub them back in. But this time it doesn't work. I think it is because the left hand side is missing a constant term.

So for example in general,

$ay'' + by' + cy =$ some polynomial, where $c = 0$

Can someone help out? I don't want the answer. I just want to know how do you get $y_p$? What is the function form? How do you derive it?

Can someone prove it (ie what y_p) is or link a proof to me? Also, how do you mathematically express "some polynomial"?

Find the inverse of AB without using dot product

Posted: 11 Sep 2021 07:36 PM PDT

Given $A=\left[\begin{array}{cc}5 & -5 \\ -5 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}-4 & 7 \\ -5 & 3\end{array}\right]$ Find out the value of $(A B)^{-1}$ if possible without using the product $A B$. If not possible, explain the reason.

Calculate the diagonal $d$ of three equal squares inscribed into a triangle whose sides $a$, $b$ and $c$ (and interior angles) are given

Posted: 11 Sep 2021 07:16 PM PDT

We have a triangle, whose sides $a$, $b$ and $c$ (and interior angles $\alpha$, $\beta$ and $\gamma$) are given. Into this triangle three equally sized squares with a side length $r$ and a diagonal $d$ are inscribed as shown by the figure below.

I am searching for a formula depending on the triangle's properties ($a$, $b$, $c$, $\alpha$, $\beta$, $\gamma$) that calculates $d$.

A few basic thoughts about angular relationships have come to my mind:

  • $360=2\cdot90+\beta+\frac{\gamma'}{2}+90+\frac{\alpha'}{2}=2\cdot90+\gamma+\frac{\alpha'}{2}+90+\frac{\beta'}{2}=2\cdot90+\alpha+\frac{\beta'}{2}+90+\frac{\gamma'}{2}$
  • $90=\alpha+\frac{\beta'}{2}+\frac{\gamma'}{2}=\beta+\frac{\gamma'}{2}+\frac{\alpha'}{2}=\gamma+\frac{\alpha'}{2}+\frac{\beta'}{2}$
  • $180=2\alpha+\beta'+\gamma'=2\beta+\gamma'+\alpha'=2\gamma+\alpha'+\beta'$

and

  • $\alpha'+\beta'+\gamma'=90$
  • $\alpha+\beta+\gamma=180$

I decompose the triangle sides as follows (see the illustration below):

  • $c=c_a+c'+c_b$
  • $a=a_b+a'+a_c$
  • $b=b_c+b'+b_a$

Diagonal and side length of the three (equally sized) squares satisfy $d=r\sqrt{2}$.

Maybe using trigonometric relationships we can deduce an elegant formula that yields $d$ directly when inputting the triangle's parameters.

enter image description here

How to prove that in PA there exists a formula exp for exponentiation such that exp(x,y+1)=exp(x,y)•x for all x,y (also nonstandard)?

Posted: 11 Sep 2021 07:07 PM PDT

I tried with Gödel's Beta function or with some definitions but i couldn't achieve a result like that. I also should prove that a such formula is functional i.e. 1)for all x,y exists z=exp(x,y); 2)for all x exists only one y s.t. exp(x,y); Can someone help me?

Intuition for why a rectangle with a fixed perimeter has a maximal area when L=W

Posted: 11 Sep 2021 07:25 PM PDT

If you have a rectangle with perimeter $P$, then the length and width you should choose to maximize the area are such that $L=W$. Can somebody give me an intuitive reason on why this is? Thanks

Orientation of sphere bundle

Posted: 11 Sep 2021 07:28 PM PDT

In Bott&Tu's book Differential forms in algebraic topology. They give two definitions of orientation of sphere bundle.(p114)

A sphere bundle with fiber $S^n$, is said to be orientable if for each fiber $F_x$, it is possible to choose a generator $[\sigma_x]$ of $H^n(F_x)$ satisfying the local compatibility condition: around any point there is a neighborhood $U$ and a generator $[\sigma_U]$ of $H^n(E|_U)$ such that for any $x$ in $U$, $[\sigma_U]$ restricted to the fiber $F_x$ is the chosen generator $[\sigma_x]$; equivalently, there is an open cover ${U\alpha}$ of $M$ and generators {$[\sigma_{U_\alpha}]$} of $H^n(E|_U)$ so that $[\sigma_{U_\alpha}]$ = $[\sigma_{U_\beta}]$ in $H^n(E|_{U_{\alpha\beta}})$.

However in p117, they claimed that for a given good cover of $M$, {$\mathfrak U$}, a sphere bundle is orientable if and only if there exists generators on this given good cover,{$[\sigma_{\mathfrak U}]$}, satisfying $[\sigma_{U_\alpha}]$ = $[\sigma_{U_\beta}]$ in $H^n(E|_{U_{\alpha\beta}})$. It is a stronger condition, but they gave no explanation. I want to know how to prove it.

What is the pushforward measure of the Wiener measure on $C^0([0,1])$ when we apply Legendre Fenchel transform?

Posted: 11 Sep 2021 07:28 PM PDT

Let us consider a Brownian motion on $[0,1]$ denoted by $(W_t)_{t \in [0,1]}$.

Let us consider the function $f_W(p) = \sup_{0 \le t \le 1 } (pt - W_t)$

What is the distribution of $f_W(p)$ for a fixed $p$?

More generally, what is the measure we obtain on the set of convex functions when we pushforward the Wiener measure on $C^0([0,1])$ by using the Legendre-Fenchel transform?

A question about Euler class in Bott&Tu's book

Posted: 11 Sep 2021 07:50 PM PDT

In the beginning of the chapter 11 of Bott&Tu's book Differential Forms in Algebraic Topology, we want to find a global closed form of a sphere bundle $E\rightarrow M$ which restricts to a generator of the cohomology on each fiber. In their book, suppose {$U_\alpha$} is a good cover of $M$, they found a global form of $E$, which restricts to the $d$-cohomology class of $E|_{U_\alpha}$ on all $U_\alpha$, and they claimed that this global closed form restricrs to a generator of the cohomology on each fiber. I wonder why, wish someone could help.

PS: This problem is equivalent to a generator of the cohomology of a sphere bundle over a manifold which is diffeomorphic to $\mathbb R ^n$ restricts to a generator of the cohomology on each fiber.

How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$

Posted: 11 Sep 2021 07:20 PM PDT

I tried like this:

Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln(y+2)\Rightarrow x=\dfrac{\ln(y+2}{2\ln a}$

Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0}-2=-1.$

But we I put each and every this assumption in the given expression, then I get hanged due to $x^x.$ How to use algebra or any other easy procedure to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$.

On showing that the indicator function $f(x) = \begin{cases}1:& x \in (s, t)\\0:& \text{ else }\end{cases}$ is Riemann integrable over $[a, b]$

Posted: 11 Sep 2021 08:10 PM PDT

Let $f(x) = \begin{cases}1:& x \in (s, t)\\0:& \text{ else }\end{cases}$ for $x \in [a, b]$ with $a, b, s, t \in \mathbb{R}, a \leq s < t \leq b$. I have to show that 1.) $f$ is Riemann integrable on $[a, b]$ and 2.) $\int_a^b f = t - s$. Currently I think that I've proven the first step (although it could be more rigorous), but my attempt for the second one is at best handwavy. Therefore I'm asking for any comments/possible improvements for the first and hints/guide for the second.

1.) To show that $f$ is Riemann integrable, we have to show that the upper $U(f, [a, b]) = \inf_{P} U(f, P, [a, b])$ and lower $L(f, [a,b]) = \sup_{P}L(f, P, [a, b])$ Riemann integrals of $f$ over $[a, b]$ coincide, where $P$ is a partition over $[a, b]$ and $U(f, P, [a, b]$ is the upper Riemann summ (resp. lower Riemann sum). To this end, let $\epsilon > 0$ be arbitrary and define a partition $P$ of $[a, b]$ as $x_0 = a, x_n = b$ and $x_i - x_{i-1} = \frac{b - a}{n}$. Choose large enough $n$ such that i.) $\frac{2(b- a)}{n} < \epsilon$, ii.) for some $1 \leq j \leq n$ and $0 \leq k < n$ we have that $x_{j - 1} < s, t < x_{j + k}$ and $s < x_p < t$ for $p = j, j+1,\dots,j-1 + k$.

Then as $U(f, [a, b]) - L(f, [a, b]) \leq U(f, P, [a, b]) - L(f, P, [a, b])$ for any partition $P$ by definition of the upper and lower Riemann integrals, we have that $U(f, [a, b]) - L(f, [a, b]) \leq \sum_{i = 1}^n(x_i - x_{i-1}\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) \Longleftrightarrow$

$U(f, [a, b]) - L(f, [a, b]) \leq \sum_{i = 1}^{j - 1}(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) + \sum_{i = j}^{j + k}(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) + \sum_{i = j + k + 1}^n(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) \Longleftrightarrow $

$U(f, [a, b]) - L(f, [a, b]) \leq \sum_{i = 1}^{j - 1}(x_i - x_{i-1})(0 - 0) + \sum_{i = j}^{j + k}(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) + \sum_{i = j + k + 1}^n(x_i - x_{i-1})(0 - 0) \Longleftrightarrow $

$U(f, [a, b]) - L(f, [a, b]) \leq \sum_{i = j}^{j + k}(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) \Longleftrightarrow $

$U(f, [a, b]) - L(f, [a, b]) \leq (x_j - x_{j-1})(\sup_{[x_{j-1}, x_j]}f - \inf_{[x_{j-1}, x_j]}f) + \sum_{i = j + 1}^{j + k - 1}(x_i - x_{i-1})(\sup_{[x_{i-1}, x_i]}f - \inf_{[x_{i-1}, x_i]}f) + (x_{j + k} - x_{j-1 + k})(\sup_{[x_{j-1 + k}, x_{j + k}]}f - \inf_{[x_{j-1 + k}, x_{j + k}]}f)\Longleftrightarrow $

$U(f, [a, b]) - L(f, [a, b]) \leq (x_j - x_{j-1})(1 - 0) + \sum_{i = j + 1}^{j + k - 1}(x_i - x_{i-1})(1 - 1) + (x_{j + k} - x_{j-1 + k})(1 - 0) = \frac{2(b - a)}{n} < \epsilon$.

Hence we've shown that for any $\epsilon > 0$ there exists a partition such that $U(f, P, [a,b]) - L(f, P, [a, b]) < \epsilon$. Thus $U(f, [a, b]) = L(f, [a, b]$, so that $f$ is Riemann integrable.

Then for the step 2.): As $U(f, [a, b]) = L(f, [a, b]) = \int_a^b f$, we may only consider the upper Riemann integral $U(f, [a, b]) = \inf_P U(f, P, [a, b])$. As $f$ is zero everywhere else except on the open interval $(s, t)$, the upper Riemann sum $(f, P, [a, b])$ is the lower the less the partition intervals $[x_{i-1}, x_i]$ contain anything else from the set $[a, b]\setminus (s, t)$. Therefore the infimum partition will partition the interval $[a, b]$ approximately into three parts: $[a, s], [s, t], [t, b]$, so that the upper Riemann sum will be $0\cdot (s - a) + 1\cdot (t - s) + 0 \cdot (b - t) = t - s$. Hence $\int_a^b f = t - s$.

Universal constructions and characterisations of the tropical semiring among all (idempotent, topological) semirings

Posted: 11 Sep 2021 07:28 PM PDT

Each of the rings $\mathbf{Q}$, $\mathbf{Z}_p$, $\mathbf{Q}_p$, $\mathbf{R}$ admits a universal construction: the rationals are the field of fractions of the integers: $\mathbf{Q}:=\mathrm{Frac}(\mathbf{Z})$; the $p$-adic integers $\mathbf{Z}_p$ and $p$-adic numbers $\mathbf{Q}_p$ are the completion of the rings $\mathbf{Z}$ and $\mathbf{Q}$ with respect to the $p$-adic valuation; the ring $\mathbf{R}$ is the completion of $\mathbf{Q}$ with respect to the infinite valuation.

Some of these rings also have natural characterisations among some specific class of rings. For example, $\mathbf{R}$ is the terminal archimedean field, while $\mathbf{Q}$ is the initial characteristic zero field.

Are there universal constructions or characterisations for the tropical semiring $\mathbf{T}$?

To be more precise, there are four different semirings that are usually called "the" tropical semiring:

  1. the set $\mathbf{N}\cup\{\infty\}$ equipped with $\max$ as addition and $+$ as multiplication;
  2. the set $\mathbf{R}\cup\{\infty\}$ with the same operations;
  3. each of the above examples, but with $-\infty$ instead of $\infty$ and with $\min$ instead of $\max$ as the addition.

However, this last point does not matter: the semirings $\mathbf{N}\cup\{\infty\}$ and $\mathbf{N}\cup\{-\infty\}$ are canonically isomorphic, and similarly for the version involving $\mathbf{R}$. So it suffices to focus on $\mathbf{N}\cup\{\infty\}$ and $\mathbf{R}\cup\{\infty\}$ only.

Now, the semirings $\mathbf{R}\cup\{\infty\}$ and $\mathbf{N}\cup\{\infty\}$ are both idempotent semirings, meaning that $a+a=a$ for each element $a$ in these rings. Also, $\mathbf{R}\cup\{\infty\}$ carries a topology making addition and multiplication into continuous morphisms, turning it into a topological semiring; it is defined as the unique topology making the map $-\log:\mathbf{R}_{\geq0}\to\mathbf{T}$ into a homeomorphism. Similarly, the ring $\mathbf{N}\cup\{\infty\}$ carries the order topology.

So, my questions are:

  1. Are there universal constructions of $\mathbf{N}\cup\{\infty\}$ and $\mathbf{R}\cup\{\infty\}$ as
    • semirings?
    • topological semirings?
  2. Are there characterisations of $\mathbf{N}\cup\{\infty\}$ and $\mathbf{R}\cup\{\infty\}$ as universal among:
    • all semirings?
    • idempotent semirings?
    • topological semirings?

(Bonus question: what interesting topologies are there on $\mathbf{N}\cup\{\infty\}$ making it into a topologial semiring, besides the order topology? In particular, does the one-point compactification topology work?)


So far, we have the following universal characterisations of $\mathbf{N}\cup\{\infty\}$:

  • Schweber. The initial idempotent semiring with a nonzero additively indecomposable annihilator.
  • Dave. The semiring of ideals of any discrete valuation ring.

Solving definite Integrals given the value of a part of the Integral

Posted: 11 Sep 2021 07:34 PM PDT

I have a problem that how I can apply integration by parts on in the form $$\int _{b}^{a} udv$$ I have been given

$$\int _{b}^{a} dv=\frac{\sqrt{\pi }}{2}$$

If I apply the integration by parts formula I get

$$I\ =\ [ uv]_{b}^{a} -\int _{b}^{a} vdu$$

Is the following expression true?

$$[ uv]_{b}^{a} =[ u]_{b}^{a} \times [ v]_{b}^{a}$$

Does connection of vector bundle always take values in Lie Algebra

Posted: 11 Sep 2021 07:41 PM PDT

It is true that if connection $\omega$ in a vector bundle is $\mathfrak{g}$-valued ($\mathfrak{g}$ being Lie Algebra of the structure Lie Group $G$) in a patch $U$, then it will be $\mathfrak{g}$-valued in all other patches due to the transformation law:

\begin{align} \omega_V = c^{-1}_{UV} \omega_U c_{UV} + c^{-1}_{UV} d c_{UV}. \end{align}

However, I'm trying to understand if it must take values in Lie algebra in order to be a connection?

Non-differentiable points in a graph: Vertical Tangent vs Vertical Asymptote

Posted: 11 Sep 2021 08:02 PM PDT

Can someone help me understand the difference between the two?

Vertical tangent at $x = a$: $f$ is continuous at a but $f'(a)$ blows up. · How is this different from a vertical asymptote?

Visually stunning math concepts which are easy to explain

Posted: 11 Sep 2021 08:03 PM PDT

Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges

Posted: 11 Sep 2021 07:38 PM PDT

Find examples of two series $\sum a_n$ and $\sum b_n$ both of which diverge but for which $\sum \min(a_n, b_n)$ converges. To make it more challenging, produce examples where $a_n$ and $b_n$ are positive and decreasing.

Edit: This problem is taken verbatim from Exercise 2.7.11 on page 68 of Abbott's Understanding analysis.

No comments:

Post a Comment