When is $||Ax|{|}_{2}\le ||x|{|}_{2}$ true for all $x$?

Araceli Clay
2022-07-02
Answered

When is $||Ax|{|}_{2}\le ||x|{|}_{2}$ true for all $x$?

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asked 2022-06-22

How to determine bounds on one variable in a system of inequalities?

I am interested in the point of 'cross-over' between a generalised harmonic number where the denominator of the summand is raised to a power, and a non-exponential harmonic sum operating on some subset of the natural numbers.

For example, take the generalised harmonic number ${H}_{x}^{(k)}=\sum _{n=1}^{x}\frac{1}{{n}^{k}}$, and a harmonic number operating only on odd denominators ${G}_{x}=\sum _{n=1}^{x}\frac{1}{2n-1}$.

Clearly, there exist values of x,k such that ${G}_{x}<{H}_{x}^{(k)}$ and values such that ${H}_{x}^{(k)}<{G}_{x}$. Thus there exists a value $c={G}_{{x}_{0}}$ such that

${G}_{{x}_{0}}=c<{H}_{{x}_{0}}^{(k)}=\sum _{n=1}^{x}\frac{1}{{n}^{k}}$

and

${H}_{{x}_{0}+2}^{(k)}<{G}_{{x}_{0}+2}=c+\frac{1}{2{x}_{0}+1}+\frac{1}{2{x}_{0}+3}$

or

${H}_{{x}_{0}+2}^{(k)}-c<\frac{1}{2{x}_{0}+1}+\frac{1}{2{x}_{0}+3}$

The values of $c,{x}_{0},k$ are obviously co-dependent. I am searching for a way to solve for ${x}_{0}$ or at least put bounds on it.

I am interested in how to approach this algebraically rather than numerically. This is a single simple example of $G$ and I want to be able to explore how to solve such problems generally, for whatever pattern of $G$ I choose (provided it's formulable!).

Algebraically, how do I put bounds on ${x}_{0}$ in terms of $c,k$?

I am interested in the point of 'cross-over' between a generalised harmonic number where the denominator of the summand is raised to a power, and a non-exponential harmonic sum operating on some subset of the natural numbers.

For example, take the generalised harmonic number ${H}_{x}^{(k)}=\sum _{n=1}^{x}\frac{1}{{n}^{k}}$, and a harmonic number operating only on odd denominators ${G}_{x}=\sum _{n=1}^{x}\frac{1}{2n-1}$.

Clearly, there exist values of x,k such that ${G}_{x}<{H}_{x}^{(k)}$ and values such that ${H}_{x}^{(k)}<{G}_{x}$. Thus there exists a value $c={G}_{{x}_{0}}$ such that

${G}_{{x}_{0}}=c<{H}_{{x}_{0}}^{(k)}=\sum _{n=1}^{x}\frac{1}{{n}^{k}}$

and

${H}_{{x}_{0}+2}^{(k)}<{G}_{{x}_{0}+2}=c+\frac{1}{2{x}_{0}+1}+\frac{1}{2{x}_{0}+3}$

or

${H}_{{x}_{0}+2}^{(k)}-c<\frac{1}{2{x}_{0}+1}+\frac{1}{2{x}_{0}+3}$

The values of $c,{x}_{0},k$ are obviously co-dependent. I am searching for a way to solve for ${x}_{0}$ or at least put bounds on it.

I am interested in how to approach this algebraically rather than numerically. This is a single simple example of $G$ and I want to be able to explore how to solve such problems generally, for whatever pattern of $G$ I choose (provided it's formulable!).

Algebraically, how do I put bounds on ${x}_{0}$ in terms of $c,k$?

asked 2022-07-01

How to solve this system of linear inequalities?

$\begin{array}{rl}{\theta}_{1}+{\theta}_{2}& \ge 0\\ 2{\theta}_{1}+k{\theta}_{2}& \ge 0\\ {\theta}_{1}+3{\theta}_{2}& \le 0\end{array}$

$\begin{array}{rl}{\theta}_{1}+{\theta}_{2}& \ge 0\\ 2{\theta}_{1}+k{\theta}_{2}& \ge 0\\ {\theta}_{1}+3{\theta}_{2}& \le 0\end{array}$

asked 2022-07-03

Let $\mathrm{\Delta}$ be an indecomposable root system in a real inner product space $E$, and suppose that $\mathrm{\Phi}$ is a simple system of roots in $\mathrm{\Delta}$, with respect to an ordering of $E$. If $\mathrm{\Phi}=\{{\alpha}_{1},\dots ,{\alpha}_{l}\}$, prove that

${\alpha}_{1}+\cdots +{\alpha}_{l}\in \mathrm{\Delta}$

I know that any positive root $\gamma $ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

${\alpha}_{1}+\cdots +{\alpha}_{l}\in \mathrm{\Delta}$

I know that any positive root $\gamma $ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?

asked 2022-06-27

If

$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$

where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$

Further if at least one of $a,b,c$ is a proper fraction, prove that:

(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$

(ii) $abc>-1$

$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$

where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$

Further if at least one of $a,b,c$ is a proper fraction, prove that:

(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$

(ii) $abc>-1$

asked 2022-05-21

Find the number of solutions to

$a-{b}^{2}\ge \frac{1}{4},b-{c}^{2}\ge \frac{1}{4},c-{d}^{2}\ge \frac{1}{4},d-{a}^{2}\ge \frac{1}{4}$

$a-{b}^{2}\ge \frac{1}{4},b-{c}^{2}\ge \frac{1}{4},c-{d}^{2}\ge \frac{1}{4},d-{a}^{2}\ge \frac{1}{4}$

asked 2022-06-21

Let $(A,\mathfrak{m}\mathfrak{)}$ be a local Noetherian ring and let ${x}_{1},\dots ,{x}_{d}$ be a system of parameters, i.e. $\mathfrak{m}=({x}_{1},\dots ,{x}_{d})$. Then

$\mathrm{dim}A/({x}_{1},\dots ,{x}_{i})=d-i$

$i=1,\dots ,d$

I know just a few basic facts about dimension theory. I think I can prove the inequality $\le $ via Krull's Hauptidealsatz in this way: the maximal ideal of $A/({x}_{1},\dots ,{x}_{i})$ is

${\mathfrak{m}}_{i}:=(\overline{{x}_{i+1}},\dots ,\overline{{x}_{d}}).$

So it must be $\mathrm{h}\mathrm{t}({\mathfrak{m}}_{i})\le d-i$.

But how to prove the other inequality? I think I should do it by induction, but I cannot understand how to begin. So, if what I said so far is right, my question is: how can I prove that

$\mathrm{dim}A/({x}_{1})\ge d-1?$

$\mathrm{dim}A/({x}_{1},\dots ,{x}_{i})=d-i$

$i=1,\dots ,d$

I know just a few basic facts about dimension theory. I think I can prove the inequality $\le $ via Krull's Hauptidealsatz in this way: the maximal ideal of $A/({x}_{1},\dots ,{x}_{i})$ is

${\mathfrak{m}}_{i}:=(\overline{{x}_{i+1}},\dots ,\overline{{x}_{d}}).$

So it must be $\mathrm{h}\mathrm{t}({\mathfrak{m}}_{i})\le d-i$.

But how to prove the other inequality? I think I should do it by induction, but I cannot understand how to begin. So, if what I said so far is right, my question is: how can I prove that

$\mathrm{dim}A/({x}_{1})\ge d-1?$

asked 2022-07-10

$\{\begin{array}{l}{y}^{2}-3\ge 0\\ 16{y}^{4}-96{y}^{2}\ge 0\end{array}$

the solution for the first inequality is $y\le -\sqrt{3}$ or $y\ge \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6}\le y\le \sqrt{6}$. Then for my result the solution for the system is

Then for my result the solution for the system is or $\sqrt{3}\le y\le \sqrt{6}$

the solution for the first inequality is $y\le -\sqrt{3}$ or $y\ge \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6}\le y\le \sqrt{6}$. Then for my result the solution for the system is

Then for my result the solution for the system is or $\sqrt{3}\le y\le \sqrt{6}$