Absolute value Definition and 368 Threads

I am reading Manfred Stoll's Book: "Introduction to Real Analysis" ... and am currently focused on Chapteer 4: Limits and Continuity ... I need some help with an inequality involving absolute values in Example 4.1.2 (a) ... Example 4.1.2 (a) ... reads as follows: In the above text we read ...

I am reading Manfred Stoll's Book: "Introduction to Real Analysis" ... and am currently focused on Chapteer 4: Limits and Continuity ... I need some help with an inequality involving absolute values in Example 4.1.2 (a) ... Example 4.1.2 (a) ... reads as follows:In the above text we read ...

Rewrite each statement using absolute values. 1. The number y is less than one unit from the number t. y < | t - 1 | 2. The sum of the distances of a and b from the origin is greater than or equal to the distance of a + b from the origin. | a + b - 0 | > or = | a + b - 0 | Is this correct...

Rewrite each statement using absolute values. 1. The distance between x and 4 is at least 8. | x - 4 | > or = 8 Can this also be expressed as | 4 - x | > or = 8? If so, why? 2. The distance between x^3 and -1 is at most 0.001. | x^3 -(-1) | < or = 0.001 Can this also be expressed as | - 1...

The | x | = x when x > or = 0. The | x | = - x when x < 0. Rewrite the following expression in a form that does not contain absolute value. | x + 3 | + 4 | x + 3 |, where x < -3 -(x + 3) + 4 -(x + 3) -x - 3 + 4 - x - 3 -2x - 6 + 4 -2x - 2 Correct?

Rewrite each statement using absolute values. 1. The distance between x and 4 is at least 8. Work: | x - 4 | > or = 8 Correct? Why must we write greater than or equal to for AT LEAST statements? 2. The distance between x^3 and -1 is at most 0.001. Work: | x^3 - (-1) | < or = 0.001...

Homework Statement How close is x to x_0 (x_0 \neq 0) so that 2. Homework Equations The Attempt at a Solution I tried to use absolute value properties:- \epsilon \lt \frac{\sqrt{x_0^2+1}}{x_0^3} - \frac{\sqrt{x^2+1}}{x^3} \lt \epsilonBy adding in the three sides, we...

Hello, I'm a student of electrical engineering. This task appeared in one of the past exams. I've been using the procedure I believe should yield the correct result, however, it turns out I was wrong. Could somebody please check out where the mistake lays in my calculations? Homework Statement...

Precalculus by David Cohen 3rd Edition Chapter 1, Section 1.2. Question 68, page 11. Before typing the textbook question, I must say that I have not been able to find a satisfactory answer to absolute value equations that equal a negative number. Question: Explain why there are no real...

Homework Statement Hi guys, I am going to calculate the following integral: $$\int_0^{f_c+f_m} |Y(f)|^2\, df$$ where:$$Y(f)=\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0) + e^{-j(\omega \tau_l + \theta_m)} \delta(\omega +...

An absolute value property is $$\lvert a \rvert \geq b \iff a\leq-b \quad \text{ or } \quad a\geq b,$$ for ##b>0##. Is this true for the case ##a=0##? I mean if ##a=0, \lvert a \rvert =0## so ##0 \geq b##. But ##b## is supposed to be ##b>0##, so we have a contradiction. How can this property...

If we define ##|a| = \sqrt{a^2}##, then why can't we do something like ##\sqrt{a^2} = (\sqrt{a})^2 = a##? Or equivalently ##\sqrt{a^2} = (a^2)^{1/2} = a^{2/2} = a##? Isn't this a contradiction? Also, how would this relate to showing that ##\sqrt{|a|} = |\sqrt{a}|## is true or false?

I was reading Bransden's Quantum Mechanics 2nd edition, chapter 2 page 61. There,it says "It should be noted that since E =hv (v for nu), the absolute value of the frequency has no physical significance in Quantum mechanics..." Why is that? Isn't this a contradiction?

I would like to ask you why the author does not use absolute value of y instead of y? Source: Mathematical Methods in the Physical Sciences by Mary L. Boas Thank you.

Hello good evening to all, I was studying here and got stuck with this. I solved the integral and got [x+sin(x) -1] and that´s the farthest that I got. I would appreciate the help.

Homework Statement X and Y 2 real numbers / |x| <1 and |y|<1 Prove that |x+y|<|xy+1|Homework EquationsThe Attempt at a Solution |x+y|<2 I couldn't prove that |xy+1| >2 And couldn't find a way to solve the problem Please help

Homework Statement I am trying to evaluate double integral ∫∫D (|y - x2|)½ D: -1<x<1, 0<y<2 Homework Equations None The Attempt at a Solution I know that in order to integrate with the absolute value I have to split the integral into two parts: y>x^2−−−>√y−x2 y>x^2−−−>√y−x2 I just can't...

I have the following equation: ##\left | r-5 \right | = \left | r+2 \right |##. What is a general, analytical way that I can solve equations like these? I always get stumped when trying to solve them...

I understand that this determines a probability, but of what exactly for a single photon? The probability that the photon will be detected on a surface where the photon is pumped, e.g. where on the surface the laser is aimed?

If $\left| a \right| \le b$, then $-b\le a\le b$. Let $a,b \in\Bbb{R}$ The definition of the absolute value is $ \left| x \right|= x, x\ge 0$ and $\left| x \right|=-x, x< 0$, where x is some real number. Case I:$a\ge 0$, $\left| a \right|=a>b$ Case II: a<0, $\left| a \right|=-a<b$the solution...

Homework Statement why we need to make x as absolute value ? as we can see, the original is x , why we we need to make x as absolute value ? is the working wrong ? Homework EquationsThe Attempt at a Solution

Homework Statement ##\displaystyle (x+3)\frac{dy}{dx} = y - 2##, where x is not 3 and y is not 2. Homework EquationsThe Attempt at a Solution ##\displaystyle (x+3)\frac{dy}{dx} = y - 2## ##\displaystyle \frac{dy}{y-2} = \frac{dx}{x+3}## ##\displaystyle \int \frac{dy}{y-2} = \int...

We begin with this definition of a hyperbola. \left(\overline{F_1 P}-\overline{F_2 P}=2 a\right)\land a>0 Perform a few basic algebraic manipulations. \sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a \sqrt{(c+x)^2+y^2}=2 a+\sqrt{(x-c)^2+y^2} (c+x)^2+y^2=4 a^2+4 a \sqrt{(x-c)^2+y^2}+(x-c)^2+y^2...

Let $\lim_{{k}\to{\infty}}d\left({x}_{m\left(k\right)},{x}_{m\left(k\right)-1}\right)=\varepsilon$ and $\lim_{{k}\to{\infty}}d\left({x}_{n\left(k\right)},{x}_{m\left(k\right)}\right)=\varepsilon$...Can we say that...

Solve the following inequality. Represent your answer graphically: ## |z-1| + |z-5| < 4 ## Homework Equations ## z = a + bi \\ |x+y| \leq |x| + |y| ## Triangle inequality The Attempt at a Solution ## |z-1| + |z-5| < 4 \\ \\ x = z-1 \ \ , \ \ y = z-5 \\ \\ |z-1+z-5| \leq |z-1| + |z-5| \\...

Homework Statement Solve: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4}##. Homework EquationsThe Attempt at a Solution I am trying to solve the following limit: ##\displaystyle \lim_{x\rightarrow 2} \frac{\left | x^2 + 3x + 2 \right |}{x^2 - 4} =...

1. Homework Equations Solving Polynomial Inequalities The Attempt at a Solution Then I used the property of absolute value inequality to get rid of it. But I really don't know if I'm doing the right step. Is this correct? So that I could separate them in two cases and find the...

Homework Statement http://imgur.com/RlIdmFh http://imgur.com/3dnLK3m Homework Equations |x| = -x? The Attempt at a Solution I'm trying to make sense of this definition in my book because they are trying to prove the triangle inequality(second link), yet it keeps saying that the absolute value...

Specifically when doing integration problems. I know the indef integral of cosx/sinx+1 is ln(sinx+1) + C, but absvalue is not required here. I think it's because the sine fn must be >= 0 or it's undefined? What about in other cases, is there a general rule to know when to use absvalue? Thanks

If ##\sqrt{x^2} = |x|##, why ##\sqrt{16} ≠ |4|## instead of 4 (please see below image)?

How does one solve an equation with two absolute value functions as below My algebra book does not show how to solve with two abs functions. 2|4x-1| = 3|4x+2| I thought this might work.. |4x-1|/|4x+2| = 3/2 then |(4x-1)/(4x+2)| = 3/2 and solve the normal way..

Sketch the region in the plane consisting of all points (x,y) such that |x-y|+|x|-|y|<=2

Homework Statement [/B]Homework Equations The Attempt at a Solution I've highlighted two equations on the screenshot. How did it proceed from the first to the second? I'm actually confused with the absolute values. What is the idea behind getting rid of the first absolute value(1-5v^2) while...

Can we integrate: $$\int_a^b |x| dx$$ using an antiderivative of ##|x|##, namely ##\frac{1}{2} x |x|##, instead of splitting up the integration interval? I know this is not particularly useful for integrals such as: $$\int_{-5}^5 |t^3 - 8| dt$$ However, for absolute value functions with linear...

how do i evaluate say a definite integral from [-3,3] of |x+1|? Any advice? I'm so confused.

Let $[a,b]$ be a closed real interval. Let $f:[a,b] \to \mathbb{C}$ be a continuous complex-valued function. Then $$\bigg|\int_{b}^{a} f(t)dt \ \bigg| \leq \int_{b}^{a} \bigg|f(t)\bigg| dt,$$ where the first integral is a complex integral, and the second integral is a definite real integral...

Determine the minimum of $y$ where $y=|x-1|+|2x-1|+|3x-1|+\cdots+|nx-1|$.

So I was helping my sister on homework and there was this problem: 2 abs(2x + 4) +1 > or equal to -3 teacher told her to ignore the -3 and just set it equal to zero. Soo should you? This question got me confused. can't you just go about solving, bringing the 1 to the left and then dividing by 2...

The derivative of ##|f(x)|## with respect to ##x## is ##f'(x)## for ##f(x) > 0## and ##-f'(x)## for ##f(x) < 0##. However, it is undefined wherever the value of the function is zero. I was wondering, though, if the product of this "undefined derivative" and zero is zero.

For the problem of differentiating ##y = x^5(3x-1)^3## using logarithmic differentiation, the solution provides the first step as rewriting the functions as ##\left |y \right | = \left | x \right |^5 \cdot \left | 3x-1 \right |^3##. This confuses me. First, how are we, mathematically, able to...

Show that the equation $|4x-5|-|3x+1|+|5-x|+|1+x|=0.99$ has no solutions.

I know this is an elementary question, but it has been some time since I multiplied exponentials, and with imaginary terms combined with absolute values, things get muddled up so easy that I want to clear this up So if I have $$ \Psi (x,t) = c_{1} \psi_{1} e^{- \frac {i E_{1}}{\hbar} t} +...

Homework Statement Make the design of a circuit returning the absolute value of a number of 3 bits . The input and output must be signed , use the complement 2. Show your approach and draw your track using logic gates. I seriously do not understand this at all. What am I supposed to do...

is e^{2 \ln{|x|}} = |x^2| or x^2?

Homework Statement For f(x) = abs(x^3 - 9x), does f'(0) exist? The Attempt at a Solution [/B] The way I tried to solve this question was to find the right hand and left hand derivative at x = 0. Right hand derivative = (lim h--> 0+) f(h) - f(0) / h = (lim h--> 0+) abs(h^3 - 9h) / h...

How do I do this? I have tried a few methods and end up getting x values that don't work when placed back into the equation.

Evaluate \lim_{n \to \infty} \int_0^1 | \sin(nx)| \ dx.

Question: True or False If x^2<4 then |x|<=2 My solution: I get -2<x<2 when I solve the problem so it should be false. Yet the text says its true? Is this a mistake? If |x| is equal to 2 then it should be a closed interval, not an open interval which seems to be correct to me.

I'm working on a ODE with initial conditions y(2)=4 and y'(2)=1/3. I solved it to be y=\frac{c_1}{|x-6|^8} + c_2|x-6|^{\frac{2}{3}}. How do I apply the second initial condition? I'm stuck at taking the derivative.