For questions about or involving the absolute value function.

learn more… | top users | synonyms

0
votes
0answers
30 views

Sketch subset of $\mathbb{C}$ which satisfies $|z-3-4i|=5$

I proceeded by plotting $z$ on the complex plane, and the modulus of $z-3-4i$: From this I deduced that: ...
0
votes
1answer
20 views

Integral of reciprocal of absolute value

I am having trouble with the following question. For which values of $n, p$ is the integral $\int_{\mathbb{R}^n}\frac{1}{|x|^p}$ convergent? I need to derive a relationship between $n$ and $p$, i.e. ...
2
votes
3answers
51 views

Inequality involving the min function

I'm trying to prove the following inequality: $$ \left|y_{1}\land x_{1}-y_{2}\land x_{2}\right|\leq\left|y_{1}-y_{2}\right|+\left|x_{1}-x_{2}\right|, $$ where $x\land y=\mbox{min}(x,y)$. By ...
0
votes
1answer
54 views

Function inequalities

I want to resolve this inequality, any help? $$\left\lvert\frac {2+\sin (x)}{x+4} \right\rvert<k$$ for $k>0$. (I am sorry for my English. It's not my first language)
1
vote
1answer
42 views

What is the relation between the square root of the sum of squares and the sum of the absolute values?

I want to prove that $\sqrt{\sum a_{i}^{2}} \geq \sum \left | a_{i} \right |$, is it possible ?
1
vote
6answers
56 views

Solve for the values of $x$ in $|x+k|=|x|+k$ where $k$ is a positive real number

The question asks me for which values of the real number $x$ is $|x+k|=|x|+k$ where $k$ is a positive real number. How do I go about this? Can I square both sides to get rid of the absolute value ...
0
votes
1answer
25 views

Does there exist a perfect square in the form: $|x^2+52x|$, where $x\in \Bbb Z$? [closed]

Does there exist a perfect square in the form: $|x^2+52x|$, where $x\in \Bbb Z$? $x<0,x\neq-52$
1
vote
3answers
80 views

Does |x| = |y| requires checking conditions while solving?

I am trying to solve this equation $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ (specifically, find set of all points $(x, y)$ satisfying equation). $\lvert 2x \rvert = \lvert x - 2 + y \rvert$ is ...
1
vote
2answers
34 views

Velleman exercise 6 in section 4.3

I am stuck on exercise 6 section 4.3, in Daniel J. Velleman's book "How To Prove It". I just need to prove the following, but cannot do it. The free variables $r$ and $s$ are arbitrary positive real ...
2
votes
1answer
42 views

The time derivative of the absolute value of a gradient.

I am interested in finding out the time rate of change of the absolute value of the density gradient, such that the directional change of the density gradient does not affect the final sign of the ...
3
votes
1answer
28 views

Evaluate x in this absolute value form equation [closed]

|x-1|+|x-2|=|x-3| Can you show me the solution to this equation?
1
vote
1answer
37 views

Graph the solutions of $ | z-2| + |z+2| < 5 \quad z\in \mathbb{C} $ [closed]

I really don't get how to solve this kind of equations and inequalities on complex numbers. Can someone solve this as an example, or others similars to teach me how to do it please? Thanks a lot.
1
vote
0answers
33 views

Usefulness of absolute value in optimization algorithms

In a course of Optimization Algorithms at university, professor said that in every algorithm the objective/object function/function cost is defined as: $$f(\bar x)=\lvert x_0 - g(\bar x)\rvert^{2}$$ ...
3
votes
1answer
254 views

The case $x < - 3$ in the absolute value equation $|x + 3| + |x - 2| = 5$

In the absolute value equation $|x + 3| + |x - 2| = 5$, why do we replace $|x + 3|$ by $-x - 3$ rather than $3 - x$ when $-\infty < x < -3$? $$|x+3|+|x-2|=5$$ What is the result set? ...
1
vote
1answer
35 views

What are the solutions to the inequality $|x|(ax+1)<2$?

Solve absolute value inequalities depending on the parameter $a$. $$|x|(ax+1)<2$$ In the first case where we have $x>0,\;a>0$ we get: $$ax^2+x-2<0$$ I get that $x$ is in the interval ...
7
votes
3answers
84 views

Solving modulus inequality $|x - 1| + |x - 6|\le11$ geometrically

Find all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\le11$$ is true. I know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection ...
1
vote
4answers
60 views

Absolute value, and multiplying by x

How would you solve $$\left|\frac{-2}{x}\right|<1?$$ I did $-1<-2/x<1$ but my teacher always tells me not to multiply by variables, so what do I do?
-4
votes
2answers
38 views

Is there any complex value for $x$ where $|x| < 0$?

What I'm really asking is if I get to a point in a calculation where I have $|x| = -4$, do I say There is no solution for $x$ or do I say There is no solution for $x ∈ ℝ$
0
votes
1answer
37 views

Magnitude of complex function

I was going through an example in a book and and it says to take the magnitude of the function. What it shows is $$X(\omega)=\frac 1{\alpha+j\omega} \implies |X(\omega)|=\frac 1{\alpha^2+\omega^2}$$ ...
1
vote
1answer
290 views

A proof that $|x+yi|=\sqrt{x^2+y^2}$, based on the given the conditions

If we attempt to define $|x+yi|$ by following conditions: $|x|=|xi|=x\operatorname{sgn}(x)$ (implicitly meaning the result will always be $\ge 0$) $|xz|=|x||z|$ $|z^x|=|z|^x$ for $x \in ...
-1
votes
2answers
43 views

Determine if the function is injective. [closed]

prove that $f(x)=\frac{1+|4x+1|}{2}$ is injective or not, thanks. I can think of counterexamples of it being not injective, but only with non-integers, but $x,y$ must be integers. Suppose that ...
1
vote
1answer
37 views

$\lim_{x \to -1} \frac{|x + 1|} { x^2 + 2x + 1}$?

Is the solution: "no limit" Because: $$\frac{|x + 1| }{ (x^2 + 2x + 1)} = \frac{1}{(x + 1)}$$ Or is factoring different for $|x + 1|$?
2
votes
1answer
35 views

Range of values which satisfy this inequality

Consider the following inequality: $$|f(a)| = \left|\frac{1}{2}(a \pm \sqrt{a^{2}-2})\right| \leq 1$$ I got this inequality while doing stability analysis of a fixed point of a certain discrete ...
-2
votes
1answer
53 views

Absolute value proof for $|x−y| = ||x|−|y||$ [closed]

How would you prove $|x−y| = ||x|−|y||$ please?
0
votes
0answers
12 views

determine the absolute difference in bits between using an indexed colour

Given the following 4 x 4 image, determine the absolute difference in bits between using an indexed colour (where the index uses the lowest number of bits possible, but the colour is represented as ...
0
votes
1answer
22 views

Proving the existence of a inequality (concerning with distance of numbers) from other three inequalities

This is an exercise from Terence Tao's analysis 1 book, chapter of integers and rational numbers. Let $x$, $y$ and $w$ be rational numbers, and let $d(a,b):=|a-b|$ for any rationals $a$ and $b$. If ...
0
votes
2answers
41 views

finding median with cumulative distribution function (absolute value)

I am currently working on distribution with density function $$f(x)=\begin{cases} \frac{2}{5}|x-2|,& \text{0 ≤ x ≤ 3} \\\\0 & \text{otherwise}\end{cases}$$ I have found that cumulative ...
0
votes
1answer
22 views

Proving an inequality given some conditions.

I would like to prove the statement: If $|a| > |b|$, with $a > 0$, where $a$ and $b$ are real numbers, then $|a + a^{2}| > |b + b^{2}|$. I am fairly certain that this claim is true. ...
2
votes
0answers
38 views

Cases in which a certain inequality holds true.

I previously asked this question on this forum, and have been demonstrated counterexamples to the claim that $|a| > |b|$ implies $\big|\frac{b+b^{2}}{a+a^{2}}\big| < 1$, which I had previously ...
1
vote
3answers
38 views

$f(x)=x+\frac{1}{e^x+1}$. Prove that for any $x,y$ : $|f(x)-f(y)|\leq|x-y|$

I feel like this question is related to the Mean value theorem, but the absolute value interferes with it. I get to: $$\frac{|f(x)-f(y)|}{|x-y|}\leq 1$$ And from there I want to prove that the ...
4
votes
3answers
69 views

Proving or disproving that an inequality implies another inequality.

I am wondering if $|a| > |b|$ implies $|\frac{b+b^{2}}{a+a^{2}}| < 1$, where $a$ and $b$ are real numbers. I have tested numerically with many cases and I have found this to be true in all of my ...
2
votes
1answer
54 views

Definition of absolute value of complex number

The definition I see everywhere for the absolute value of a complex number is: Let $z=x+iy$ then $|z| := \sqrt{x^2+y^2}$. But the square root operation is multivalued in complex analysis. So while ...
1
vote
3answers
64 views

Prove that a specific inequality holds

Let $n \in \mathbb{N}$. Let $z_1, \ldots, z_n$ and $w_1, \ldots, w_n$ be complex numbers such that $$ \sum_{j = 1}^n |w_j|^2 \leq 1 $$ and $$ \left| \sum_{j = 1}^n z_j w_j \right| \leq 1 $$ Show that ...
1
vote
2answers
63 views

Can you explain why $x>\frac 23$ is not a solution to this inequality

$|2x + 2| + |x - 1| > 3$ why can't $x>\frac 23$ be a part of the solution? Thanks for your help!
1
vote
0answers
20 views

Univalent triangle inequality [duplicate]

$|Z_1| = | \frac{v(1+\alpha) + \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$ Triangle inequality |x+y|=|x|+|y| Where x= $\frac{v(1+\alpha)}{2}$ and $y= \frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}$ I've been ...
0
votes
1answer
40 views

Triangle inequality univalent

$|Z_1| = | \frac{v(1+\alpha)+ \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$ I know that using triangle inequality method $|Z_1|$ is: $|Z_1|= |\frac{v(1+\alpha)}{2}| + ...
2
votes
1answer
48 views

There are no archimedean function fields

Definition: a field $L\supseteq K$ is called a function field over $K$ if the extension $L|K$ is finitely generated, regular and of transcendence degree $1$. In the book "Topics in the theory of ...
0
votes
3answers
34 views

Show $\lim_\limits{n\rightarrow \infty} \sqrt{n}(|c+\frac{d}{\sqrt{n}}|-|c|) = d* sign(c)$

Consider the sequence of real numbers $\sqrt{n}\left(\left|c+\frac{d}{\sqrt{n}}\right|-|c|\right)$ with $c,d \in \mathbb{R}$ and $c\neq 0$. Could you help me to show that $$\lim_\limits{n\rightarrow ...
3
votes
1answer
44 views

Integration with double absolute value

How to prove that? $$ \int\left|\sqrt{|x|}\right|\ \mathrm{d}x = \frac{1}{3} x \left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x) + \sqrt{x} + \sqrt{-x}\right) $$ I cannot ...
4
votes
1answer
72 views

Problem with inequality: $ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}$

Prove that for for all $p,q\in \mathbb{Z}$, $q>0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| > \frac{1}{3q^2}. $$ To be honest, I do not know where to start - any help would be ...
3
votes
4answers
54 views

What can be said about the relationship between the complex numbers $\lvert z\rvert^n$ and $\lvert z^n\rvert$?

I've been playing around with this for a while without much progress. More precisely, I suppose, I'd like to know if one always less than or equal to the other? The fact that one never sees this in ...
6
votes
1answer
70 views

Basic question $|x^2| < 9$

I have a rather basic question. Let's assume that $|x^2| < 9$, where $x\in \mathbb{R}$. Then everyone knows that $x \in$ (-3,3). However, I have trouble arriving at the answer based on basic ...
0
votes
0answers
28 views

Is it always possible to define an absolute value in an ordered field?

I am trying to show (not sure if possible) that i can generalize all basic arithmetic operations between limits of sequences of real numbers to any ordered field, so i need to build a generalized ...
1
vote
1answer
15 views

Trouble with an inequality between magnitudes of complex numbers

We are supposed to show that $$|ab^* + a^*b| \leq 2|ab|$$ where a and ba re complex numbers and a* and b* are their respective conjugates (so $a = x_1+iy_1$, $a^* = x_1-iy_1$, $b = x_2+iy_2$, $b^* = ...
1
vote
0answers
28 views

Derivative of the maximum of a function on a interval

My question is as follows: Given a function $f: [-h,\infty) \rightarrow \mathcal{R}$ and the maximum function given by \begin{equation} \max_{s\in [-h,0]} |f(t+s)| \end{equation} for $t\geq0$. Then ...
3
votes
3answers
51 views

Find the area of the region described by $|5x|+|6y| \le 30 $

Find the area of the region described by $|5x|+|6y| \le 30 $ (where $|z|$ denotes the absolute value of $z$). My effort Imagining a number line and interpreting the problem as the request to ...
0
votes
2answers
36 views

Inequality with absolute value.

Show that $\forall a,b\in \mathbb{R}$: $$ \left| \frac{a}{1+a^2} - \frac{b}{1+b^2} \right| \leq |a-b| $$ Being honest, I do not know where to start (apart from common denominator form) and would ...
0
votes
2answers
38 views

If $\left| x \right| \ge \left| y \right|$ then Can we say $\left| {x + y} \right| \ge \left| x \right| - \left| y \right|$?

Let $x,y\in \mathbb{R}$ and $\left| x \right| \ge \left| y \right|$. Can we say $\left| {x + y} \right| \ge \left| x \right| - \left| y \right|$?
0
votes
1answer
24 views

solve and skecth $\log{|z|}=-2\arg(z)$

Ive asked this question a week ago, but nobody managed to answer but it is doing my heading from then. I know usually You demand some initial work done on the question but I just dont know how to ...
0
votes
1answer
28 views

Double integral over the set with an absolute value of $y$

I need to calculate an integral over the set: $$D \colon 0\leq x\leq \pi\text{ and }|y|\leq x$$ from the set (definite integral) $D \int \cos(y)dA$ I don't understand what $|y| \leq x$ means. Can ...