Math 414-501 — Spring 2020

Assignment 6 - Due Friday, 3/6/2020.

• Read sections 2.1 and 2.2.
• Problems.
  1. Chapter 2 exercises: 1, 2, 4.
  2. Find the Fourier transform of $f(t) = e^{-|t|}$. In addition, use this transform and the properties listed in Theorem 2.6 to find the Fourier transforms of the following functions:
     1. $te^{-|t|}$ (Use #2.)
     2. $e^{-2|t-3|}$ (#6 and #7)
     3. ${\rm sign}(t)e^{-|t|}$ (Hint: differentiate $e^{-|t|}$; use #4.)
     4. $(1+(t-2)^2)^{-1}$  (Hint: How is this function related to $\hat f(\lambda)$, where $f(t)=e^{-|t|}$? Once you've gotten this, use #6.)

  3. Find the Fouirer transforms of these functions.
     1. $g(t) = \left\{\begin{array}{cl} 1 & \text{if }-1 \le t \le 2 \\ 0 & \text{otherwise}. \end{array} \right.$

     2. $h(t) = \left\{\begin{array}{cl} -1 & \text{if }-3 \le t \le 0 \\ 1 & \text{if }\ 0 < t \le 3 \\ 0 & \text{otherwise}. \end{array} \right.$

Points and solutions.

1. (25 pts.) Chapter 2, #2. \[ \hat f(\lambda)= (2\pi)^{-1/2} \int_{-\infty}^\infty f(t)e^{-i\lambda t}dt = (2\pi)^{-1/2}\int_{-\pi}^\pi \underbrace{\sin(3t)}_{odd}e^{-i\lambda t}dt= (2\pi)^{-1/2}i\int_{-\pi}^\pi \sin(3t)\sin(\lambda t)dt\] Usiing the trig identity $\sin(a)\sin(b) = \frac{1}{2}\big(\cos(a-b)-\cos(a+b)\big)$, we have, if $|\lambda| \not= 3$, then \[ \begin{aligned} \int_{-\pi}^\pi \sin(3t)\sin(\lambda t)dt &= \frac{1}{2}\int_{-\pi}^\pi \bigg(\cos((3-\lambda)t) - \cos( (3+\lambda)t)\bigg)dt\\ &= \frac{2\sin((3-\lambda)\pi)}{2(3-\lambda)}- \frac{2\sin((3+\lambda)\pi)}{2(3+\lambda)} \\ &=-\sin(\lambda \pi)\big(\frac{1}{3+\lambda}+ \frac{1}{3-\lambda}\big) =-\frac{6\sin(\lambda \pi)}{9-\lambda^2}. \end{aligned} \] Putting everything together and doing some algebra, we see that \[ \hat f(\lambda)= -3i\sqrt{\frac{2}{\pi}}\frac{\sin(\lambda \pi)}{9-\lambda^2} \]
2. (25 pts. for $\hat f$.)     Since $f(t)$ is even, \[ \hat f(\lambda) = 2(2\pi)^{-1/2} \int_0^\infty e^{-t}\cos(\lambda t)dt. \] This is a standard integral and is fairly easy to do. Actualy, there are other ways to do it: \[\begin{aligned} f(\lambda) &= (2\pi)^{-1/2} \int_{-\infty}^\infty e^{-|t| -i\lambda t}dt\\ &= (2\pi)^{-1/2} \int_{-\infty}^0 e^{t-i\lambda t}dt+ (2\pi)^{-1/2} \int_0^\infty e^{-t +i\lambda t}dt \\ &= (2\pi)^{-1/2}\big(\frac{1}{1-i\lambda}+ \frac{1}{1+i\lambda}\big) = \sqrt{\frac{2}{\pi}}\frac{1}{1+\lambda^2} \end{aligned}\] (If they get the factors with $\pi$ wrong, just take a point off.)
   1. (15 pts.) By #2, $\mathcal F(te^{-|t|}) = i \frac{d}{d\lambda}\big(\sqrt{\frac{2}{\pi}}\frac{1}{1+\lambda^2}\big)= -i\sqrt{\frac{2}{\pi}}\frac{2\lambda}{(1+\lambda^2)^2}$
   2. Skip.
   3. Skip.
   4. (15 pts.) This one is slighty tricky. By the inversion theorem (or what I called the quartet in class), $ e^{-|\lambda|}=\mathcal F(\sqrt{\frac{2}{\pi}}(1+t^2)^{-1})$, so $ \sqrt{\frac{\pi}{2}}e^{-|\lambda|}=\mathcal F((1+t^2)^{-1})$. By #6, $\mathcal F((1+(t-2)^2)^{-1}) = e^{-2i\lambda}\sqrt{\frac{\pi}{2}}e^{-|\lambda|}$

3. (20 pts.) Grade part (b). \[\begin{aligned} \mathcal F(h)=&\frac{1}{\sqrt{2\pi}}\int_{-3}^0 (-1)e^{-i\lambda t}dt + \int_{0}^3 e^{-i\lambda t}dt\\ &=\frac{1}{-i\lambda\sqrt{2\pi}}\big(-(1-e^{3i\lambda})+(e^{-3i\lambda}-1) \big)\\ &=\frac{2}{-i\lambda\sqrt{2\pi}}\big(\cos(3\lambda)-1\big) =\frac{-2i}{\lambda\sqrt{2\pi}}\big(1-\cos(3\lambda)\big) \end{aligned}\]