Recent content by radiator

Thanks Mark, Really appreciate it :)

so that would justify them to be equivalent ? since the absolute value would mean that -1/2 < x/\sqrt{2} < 1/2

why don't we put the argument x/√2 in an absolute value and have it |\frac{x}{\sqrt{2}}|< \frac{1}{2} I am kind of confused now! how do we prove them to be the same?

Note: I think I solved this while writing this topic, did not want to scrap it! if you think its wrong let me know! I am trying to manipulate the rectangular function with different arguments and came across a confusing one Trying to show: \prod (x^2) = \prod (\frac{x}{\sqrt{2}}) Recall that...

Thanks for the correction. I guess I understand the justification.

something I often see without justification in my physics books. What is the justification for the following convolution property (pulling the derivative inside the integral) (f*g)^\prime = \frac{d}{dx} \int f(x) g(x-u) dx = \int f(x) \frac{d}{dx} g(x-u) dx = f(x) * g(x)^\prime

Thanks so much Mark, this clarified it

Thanks Mark, I think I am missing a basic principle here about the ≥ relations if I have a x^2 - 1 \geq 0 then solving for x is x^2 - 1 = 0 \rightarrow x = \pm 1 so in the case of positive one x\geq 1 and for negative one it changes to x \leq -1 and the positive it remains...

so for x^2-1>0 I have H(x^2-1) = 1 for x>1 and x>-1 ( so its equal to 1 from -1 to infinity) and H(x^2-1) = 0 for x<1 and x<-1 ( so its equal to 0 from 1 to negative infinity) so why would I choose only : equals 1 for x>1 and x<-1, what about the other conditions?

Hello, I am not where this question goes, its not part of a homework either! I am trying to figure out how to plot the heaviside (unit step) with such an expression H(x^2-1) so I do this:H(x^2-1) = 1 for x^2-1>0 -> x>+- 1 and H(x^2-1) = 0 for x^2-1<-0 -> x<-+1 But this tells me only that it...

You were right only missed a factor of 3 by 1.153x10^11, the rest check your calculator. I know its been a while :cry:

Thanks very much arildno :)

I get it, but just to further understand more suppose again I have f(x) = x^2 , which means the f_e(x) = x^2 and f_o(x) = 0, similarly f(x) = x^3 gives f_e(x) = 0 and f_o(x) = x^3 which is graphically a reflection around the y-axis and the origin, respectively. but for absolute x it would...

Thanks arildno, I suppose if I have a step function, the even part will be equal to odd part? i.e. f(x) = (H(x) + 0)/2 + (H(x) - 0 )/2 since H(x) = 0 for x<0

How would you decompose a given function to its even and odd parts? let's say you have f(x)=e^ix, and would like to know the even and odd parts of it? how do you proceed? Thank you