Recent content by Seda

Ugh, still have 1 sign error tripping me up... going by my definitions.. replacing only g*(x) ∫f(x)g*(x)dx = ∫f(x)[∫G(p)exp[2∏ipx]dp]dx = ∫f(x)[∫G*(p)exp[-2∏ipx]dp]dx = ∫G*(p)[∫f(x)exp[-2∏ipx]dx]dp In that last line if the negative wasn't there I could replace it with F(P) from y...

Aha, I believe I have to use FT of g*(x) = G*(-x) Fixes my sign issue I believe.

My instructor has the 2∏ in the exponential. Yes I've seen cases where it is infront of the integral as a 1/(2∏). I don't fully understand the (non) difference, just sticking with what I've seen.

Do I have my signs backwards? When taking the Fourier transform on a function do I have a negative exponential? My class notes don't match will other things I've found...

Homework Statement If F(p) and G(p) are the Fourier transforms of f(x) and g(x) respectively, show that ∫f(x)g*(x)dx = ∫ F(p)G*(p)dp where * indicates a complex conjugate. (The integrals are from -∞ to ∞) Homework Equations F(p) = ∫f(x)exp[2∏ipx]dx G(p) = ∫g(x)exp[2∏ipx]dx G*(p) =...

But my equation doesn't even have an x. So that's where I get tangled up. I'm missing something obvious. What I mean is, no matter what I substitute, I'm not doing to be able to determine what to replace for dx.

Let u= mx ; du = mdx (1/m)∫f(u/m)δ(u-c)du Zero everywhere except where u=c therefore .. sorry, I'm stuck

A "simple" application of dirac delta "shift theorem"...help Homework Statement show that for a, b, c, d positive: δ(a/b-c/d) = bdδ(ad-bc) Homework Equations ∫f(x)δ(x-a)dx = f(a) The Attempt at a Solution Ok so I start with ∫δ(a/b-c/d)f(x)dx But I am not sure how to apply the shift...

Um...yeah. I wasn't thinking that way. But that's true. Thanks

Doh...well I'd rather be thinking too hard than thinking too little I guess. Thanks for your help! I'd hate to ask for more of your time, but do you have a hint for the second problem?

yeah, I just wrote it in a way to overly-clearly include the -√3 and -i also, even though it could be considered implicit. As for the "hard direction" am I wording this right?: Obviously (√3 + i) ∈ Q(√3+ i) Therefore (√3 + i)^3 = 8i ∈ Q(√3+ i). Similarily, (-8)(-i) ∈ Q(√3+ i). So i, -i ∈...

Aw cool it's 8i! I'll try to figure that in... Yeah sorry, I caught the sign error. Is the way i figured Q(√3+i) ⊆ Q(√3, -√3, i, -i) ok?

2+2i√3 ...? It shows that the degree is the same for both?

PROBLEM 1: SOLVED THANKS TO POSTER DICK [SIZE="3"][SIZE="2"]How can I prove these two field extensions are equal? Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers. I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy: Let m ∈ Q(√3+i). Therefore, m =...

Well, it's the answer given in the book, it may be a typo. I wanted to make sure I wasn't doing anything stupid. Thanks.