In mathematics, the absolute value or modulus of a real number x, denoted |x|, is the non-negative value of x without regard to its sign. Namely, |x| = x if x is positive, and |x| = −x if x is negative (in which case −x is positive), and |0| = 0. For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.
Generalisations of the absolute value for real numbers occur in a wide variety of mathematical settings. For example, an absolute value is also defined for the complex numbers, the quaternions, ordered rings, fields and vector spaces. The absolute value is closely related to the notions of magnitude, distance, and norm in various mathematical and physical contexts.
How does a norm differ from an absolute value? For example, is
\|\mathbf{x}\| = \sqrt{x_1^2 + \cdots + x_n^2}
any different than
|\mathbf{x}| = \sqrt{x_1^2 + \cdots + x_n^2}
??
I am to find the imaginary part, real part, square, reciprocal, and absolut value of the complex function:
y(x,t)=ie^{i(kx-\omega t)}
y(x,t)=i\left( cos(kx- \omega t)+ i sin(kx- \omega t) \right)
y(x,t)=icos(kx- \omega t)-sin(kx- \omega t)
the imaginary part is cos(kx- \omega t)
the...
Hi all, I have the following question.
Are the following spaces homeomorphic in the real number space with absolute value topology?
1) [a,b) and (a,b]
2) (a,b) and (r,s)U(u,v) where r < s < u < v.
For 1), I got that they are not homeomorphic because it fails the topological property that...
Hi
So I am a little confused about the absolute value sign here. Is the derivative right:
[x-1/|x-1|]e^x + |x-1|e^x
But what do I do when I set this to zero? Do I get two expressions, one for x>1 and one for x<1 ? (thus removing the absolute value signs)
Is it correct that |\sin x|=-\sin x as x\rightarrow 0-?
It's clear for a function like y=x, but I ask the question since the sine oscillates from positive to negative values, so for different x's the abs. value is either pos. or neg. Or do I only need to consider values of x that are between...
"Simple" absolute value problem with inequalities
OK...Im totally stuck and could use some help :)
given...for all e>0, d>0...the following holds
|x-a|<d => |f(x) - f(a)| < e
where f(x) = sqrt(x)
how do I find d in terms of e?
Thanks in advance
hello all
Iv been working on a lot of integrability questions and I am having trouble with this problem
let f be integrable on [a,b] then show that |f| is integrable and that
|\int_{a}^{b}f|\le \int_{a}^{b}|f|
now this is what i know
\int_{a}^{b^U}f =\int_{a_{L}}^{b}f= \int_{a}^{b}f...
(|2x-3| + x) / (x^2 - 3x + 2) < 1
|(x^2 - 5x + 4) / ( x^2 - 4)| =< 1
Can somebody help me with this quadratic inequalities, please... If you have time, please also give me an idea of how you solved them... Thank you!
int(int(abs(x-y)*6*x^2*y)) the range of x and y are 0,1. Normally i'd check to split it up and change the limits, but i think my brain is broken because I'm not seeing it at the moment.
simple question that i need to know how to do for stats without using maple :P