Fourier tells us that we can write a "pulse" by summing up sinusoidal functions: $f(x) = \int_{-\infty}^{\infty} a(k)e^{ikx}dk$ If we were to compute $f(x) = \int_{-\infty}^{\infty} a(k)e^{ik(x-vt)}dk$ where $v$ is a known constant, what would we get? 1. $f(x)$ 2. $f(vt)$ 3. $f(x-vt)$ 4. Something complicated! 5. ??? Note: * Correct Answer: C

## Announcements * HW 11 is posted * Looks long, but 2 questions are roughly the same... * Graded HW 9, Quiz 5, and HW 10 will be returned Wednesday; sorry!

Fourier tells us that we can write a "pulse" by summing up sinusoidal functions: $f(x) = \int_{-\infty}^{\infty} a(k)e^{ikx}dk$ If we were to compute $f(x) = \int_{-\infty}^{\infty} a(k)e^{ik(x-vt)}dk$ where $v$ is a known constant, what would we get? 1. $f(x)$ 2. $f(vt)$ 3. $f(x-vt)$ 4. Something complicated! 5. ??? Note: * Correct Answer: C

Fourier tells us that we can write a "pulse" by summing up sinusoidal functions: $f(x) = \int_{-\infty}^{\infty} a(k)e^{ikx}dk$ If we were to compute $f(x) = \int_{-\infty}^{\infty} a(k)e^{ik(x-v(k)t)}dk$ where $v(k)$ is function, what would we get? 1. $f(x)$ 2. $f(vt)$ 3. $f(x-vt)$ 4. Something more complicated! 5. ??? Note: * Correct Answer: D