Recent content by mathmajor314

Ugh! Why didn't I think of that!? Now I've got it. Thank you for all of your help; I appreciate it!

Oops, it should be \frac{F(k)(1+k^2)}{9+k^2}. I know that the inverse transform of \frac{1}{9+k^2} is \frac{exp(-3|k|}{6} so I guess that's where that 3 comes from. I'm still stuck on getting F(k) by itself though.

I rearranged it and found that too but I can't figure out what to do with \int \frac{F(k) (k^2+9)}{k^2+1} dk .

Clearly, U(k) = \frac{F(k)}{1+4(\frac{2}{1+k^2})}. Next, I think I am supposed to apply the inverse Fourier transform version of the Convolution Theorem, which states "If the functions f(x), g(x) have Fourier transforms F(k), G(k), respectively, then the inverse Fourier transform of...

I'm sorry, the original equation should have a 4 in front of the integral sign. Also, in my book, the problem only has u instead of u(x) on the left hand side. And yes, that is the correct Fourier transform of g(x). Taking the Fourier transform of the left hand side, I know that the Fourier...

The problem: Solve the integral equation \int\stackrel{\infty}{-\infty}exp(-abs(x-y))u(y)dy+u=f(x) for -\infty<x<\infty. The solutions say "Use the convolution theorem to find u(x)=f(x)-\frac{4}{3}\intf(t)exp(-3abs(x-t))dt." The Convolution Theorem in my book states "If the functions f(x)...

Yes, I emailed my professor and it was supposed to be ">=" instead of "=". After changing his typo, I figured it out. Thanks!

Let L be a self-adjoint operator satisfying <Lf,f>=0. Show that \sigma(L)\subseteq[0,\infty). I know that L being self-adjoint implies that <Lf,f>=<\lambdaf,f>=\lambda<f,f>=\lambdanorm(f). And <Lf,f>=<f,L*f>=<f,Lf>. I'm not sure where to go from here though. Thank you in advance for any help!