For questions about or involving the absolute value function.

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0
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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
40 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
73 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
56 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
65 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!
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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}| + |\frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}...
2
votes
1answer
49 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
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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
46 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
73 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 appreciated....
3
votes
4answers
57 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
29 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
31 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
37 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
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2answers
40 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
35 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 ...
1
vote
1answer
60 views

Double integration over function with absolute values

I have having difficulty in how to solve the following double integral problem involving absolute values and the assumption that $\alpha > 1$: $\iint_{-\infty}^{+\infty} \frac{1}{1+|x|^\alpha} \...
0
votes
1answer
39 views

Does every non-archimedean absolute value on field take value in $\mathbb{Q}$

Let $K$ be a field, a non-archimedean absolute value is defined to be a map $K\to \mathbb{R}$ satisfying $|x|=0\Rightarrow x=0$, $|x|\cdot|y|=|xy|$ and $|x+y|\leq\max(|x|,|y|)$. Is there an example ...
0
votes
3answers
92 views

Why is $(\sqrt{x^2})$ equal to $|x|$ [duplicate]

I don't understand why you have to write the absolute value sign when solving for the square root of $x$ squared. Shouldn't the answer automatically be positive? Why is the absolute value sign ...
6
votes
5answers
382 views

Absolute value and max/min function: why $a + b + |a - b|=2\max(a,b)$? [duplicate]

I am being told that $a + b + |a - b|$ is equal to $2\max(a,b)$. What is the reasoning behind this?
3
votes
2answers
65 views

Prove $|x|^2$ = $x^2$

My first attempt at this proof divided into 2 cases, one where $x^2$ is greater than or equal to 0, and another where $x^2$ is less than 0. For the first case, I said that the definition of absolute ...
1
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5answers
65 views

$\left|x\right| < \left|\tan(x)\right|$ close to $0$

I was trying to prove this inequality $\left|x\right| < \left|\tan(x)\right|$ in a neighborhood of $0$. I tried splitting into the four cases opening the modulus but still wasnt able to solve it. I ...
0
votes
1answer
25 views

Why/how do I ignore these absolute values while using Variation of Parameters?

I'm given the initial value problem: $x' = \frac{3}{t}x + e^{3t}$, $x(1) = 2$ Using the variation of parameters formula, I end up with: $2e^{3 \ln|t|} + e^{3 \ln|t|} \int^{t}_{1}e^{-3\ln|s|} e^{3s} ...
5
votes
3answers
115 views

What is the fastest way to find the range of functions having modulus: $f(x) = |x+3| - |x+1| - |x-1| + |x-3|$

While solving problems I saw a question in which I was supposed to find the range of a function $$f(x) = |x+3| - |x+1| - |x-1| + |x-3|$$ I know the way in which I can take different cases of $x$ ...
0
votes
1answer
43 views

Complex numbers converge if their absolute values and arguments converge

Let the sequence $\{z_n\}_{n>0}$ and $w \not=0$ be such that $|z_n| \to |w|$ and $\operatorname{Arg}(z_n) \to \operatorname{Arg}(w)$. Show that $z_n \to w$. My proof: $z_n= |z_n|e^{i \arg(z_n)} \...
3
votes
2answers
115 views

Prove $||a| - |b||$ is less than or equal to $|a-b|$

I was given the hint to split it into two cases ($|a| - |b|$ being positive and negative) and then use the triangle inequality. However, since the triangle inequality says that $|a+b|$ is less than or ...
5
votes
1answer
59 views

Prove that $|x|^2$ = $x^2$.

This is what I did, but I'm not sure if it's a good enough proof: Since $|x|$ is equal to $x$ when $x$ is greater than or equal to 0, and is equal to $-x$ when $x$ is less than 0, I said that $|x|^2$ ...
0
votes
2answers
46 views

Absolute Value Rational Inequalities

Ok so I have the following two inequalities: \begin{equation} \left| \frac{x+6}{x-2}\right| \leq 4 \end{equation} and \begin{equation} \frac{x^2-1}{\left| x+2\right|} \leq 3(1-x) \end{...
-2
votes
1answer
103 views

How to prove that $\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$? [closed]

Prove that $$\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$$ I couldn't prove this. Please suggest a solution.
1
vote
4answers
65 views

Prove $\lvert x\rvert$ = $\lvert-x\rvert$ for all real numbers $x$ [closed]

Been at this one for a long time. I'm trying to use the fact that $|x|$ = $x$ if $x$ is greater than or equal to 0, and $|x|$ = $-x$ if $x$ is less than 0. Then I want to split the proof into these 2 ...
0
votes
2answers
32 views

Piecewise from Rational Absolute Value Function

How would one separate a function like the following into piecewise? $$f(x)={\left|4-x\right|\over{\left|x-4\right|}}$$ I've been taught that with a rational function with an absolute value in the ...
4
votes
3answers
73 views

Proving $\max$ of $a, b$.

How do I prove that $$\max{\{a, b\}} = \frac{a + b + \left | a - b \right |}{2}$$ I have no idea how to even start the proof, any idea / intuition that can get me started is greatly appreciated. ...
0
votes
3answers
42 views

Square divided by absolute value

First time posting on Math SE, with kind of a basic algebra question. Question Does the relation: $$\dfrac{(ab)^2}{|ab|} = \left|ab\right|$$ with $a,b \in \mathbb{R_{\ne 0}}$ always hold? It seems ...
0
votes
0answers
35 views

Function with absolute value in denominator - limits

f(x)=(x-1)/(|2-x|-1) |2-x|= { |2-x|; x < +2} {-|2-x|; x >= +2} State domain, range and the equations of the asymptotes. D(f)= {x | x > 3 or x < 3} R(f)= {y | y > 1 or y <= -1} ...
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0answers
36 views

Show that absolute value satisfies triangle inequality, how? [duplicate]

I wish to show that given $a,b,c \in \mathbb{R}$, the following holds: $|a - c| \leq |a-b| + |b-c|$ Using the definition $| x | = \max\{x, -x\}$ I can't seem to be able to show that $|a - c| \leq ...
2
votes
6answers
98 views

Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$

I was given this question and asked to find $x$: $$\left| \frac{x}{x+4} \right|<4$$ I broke this into three pieces: $$ \left| \frac{x}{x+4} \right| = \left\{ \begin{array}{ll} \...
2
votes
2answers
63 views

Evaluating the Fourier coefficients of $abs(x)$

Let's get started: $$\hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} |x|e^{-inx} dx$$ since $|x|$ is an even function: $$= \frac{1}{\pi}\int_0^{\pi} xe^{-inx} dx$$ Integration by parts yields: $$e^{-inx}\...
0
votes
0answers
26 views

Continuity properties of an example function $f:\mathbb{R}^n\to\mathbb{R}$

Consider the function $f:\mathbb{R}^n\to\mathbb{R}$ defined as follows: $$ f(x)=\begin{cases} ||x||^2 & \text{if $||x||\le 1$,}\\ 1/||x||^2 & \text{if $||x||> 1$,} \end{cases} $$ where $||...
0
votes
3answers
46 views

Nested absolute-value inequality

I try to solve a problem in two ways, but the results are not the same. Method 1. $$\lvert \lvert x \rvert + x \rvert \le 2$$ For $x < 0$, we have $\lvert x \rvert = -x$. Therefore: $$\lvert -x+...
0
votes
1answer
29 views

Modulus of Two Complex Numbers, Squared

I have a very silly question to ask! I have $|z_{1} + z_{2}|^2 = |z_{1}|^2+|z_{2}|^2+2|z_{1}||z_{2}|\cos{\theta}$, where $z_{1}$ and $z_{2}$ are complex numbers. For the life of me I cannot ...
1
vote
1answer
56 views

Prove those inequalities are true [duplicate]

I want to prove that those inequalities are true for $a, b ∈ R$: $$ |a + b| ≤ |a| + |b| $$ $$ ||a| − |b|| ≤ |a − b| $$ $$ |a − b| ≤ |a − c| + |c − b| $$ Now I can see that they are true, and I could ...
0
votes
1answer
36 views

Proving inequality with absolute value [duplicate]

How can I show the following inequality for any real numbers $x,y,z$? $$\frac{|x-z|}{1+|x-z|}\le \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}.$$ The triangle inequality could be useful, but I am ...
0
votes
1answer
22 views

Which of the two following solutions is correct for absolute value of this expression?

I'm currently going through Spivak and ran across this problem, but i see a difference in my answer and the answer that i'm checking it again. The problem is to eliminate the absolute value signs in $...