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