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--- math121a-f23:october_4_wednesday [2023/10/04 17:03]
pzhou
+++ math121a-f23:october_4_wednesday [2026/02/21 14:41] (current)
@@ Line 5: / Line 5: @@
   * Laplace transform. Given $f(t)$, on $t \in [0, \infty)$, we want to write $f(t) = \int_{c + i \R} e^{pt} F(p) dp$.
   * Gamma function, Stirling Formula
+===== What is Fourier Transformation =====
+When we solve equation of the form
+$$ \frac{df(x)}{dx} = \lambda f(x), $$
+the general soluition is
+$$ f(x) = c e^{\lambda x}. $$
+We say $e^{\lambda x}$ is an eigenfunction for the operator $d/dx$ with eigenvalue $\lambda$, here $\lambda$ can be any complex number.
+  * For $\lambda$ purely imaginary, $\lambda = i k$, the function $e^{i k x}$, which has constant size $|e^{ikx}| = 1$.
+  * For $\lambda$ purely real, $\lambda \in \R$, the function $e^{\lambda x}$, which is real, and has exponential growth (if $\lambda > 0$) or exponential decay.
+  * For $\lambda$ general complex number, say $\lambda = a + i b$ for $a,b\in\R$, then $e^{\lambda x} = e^{ax} e^{ibx} $ has both oscillation factor $e^{ibx}$ and exponential growth/decay $e^{ax}$.
+If you give me a function $f(x)$, we can try to write it as a linear combination of these easy to understand pieces $e^{\lambda x}$. Say
+$$ f(x) = \int_{C} c(\lambda) e^{\lambda x} d\lambda $$
+The benefit for doing this? If a differential operator $d/dx$ walk to it, we would have
+$$ (d/dx) f(x) = \int_{\lambda} c(\lambda) (d/dx)  e^{\lambda x} d\lambda = \int_{\lambda} c(\lambda) \lambda e^{\lambda x} d\lambda. $$
+Of course, you would immediately complain:
+  * what is this integration contour $C$? Is it along the real axis of $\lambda$? Is it along the imaginary axis of $\lambda$?
+  * Why we can switch the order of integration and differentiation?
+Let's keep these questions in our head, we will return to those.
+So, what is Fourier transformation? If you go on wiki, or open some textbook, you see the following definition: given a function $f(x)$ (with some condition on it), we can define
+$$ F(p) := \int_{x \in \R} f(x) e^{-ipx} dx. $$
+This is called the Fourier transformation.
+Let's throw in some input and see what we get.
+===== Example 1 =====
+$$ f(x) = \cosh(x) = (e^{x} + e^{-x} )/2. $$
+This function grows to infinity when $x \to \pm \infty$, indeed, when $x \to +\infty$, $e^x$ dominate, and $x \to -\infty$ we have $e^{-x}$ dominate. Either way, you blow up. Good luck when integrating $\int_\R f(x) dx$.
+But, let's try to do the Fourier transformation anyway. We get
+$$ F(p) = \int_x f(x) e^{-ipx} dx = (1/2) \int_\R e^{(1-ip)x} dx + (1/2) \int_\R e^{(-1-ip)x} dx. $$
+OK, we cannot continue.
+==== Lesson 1 ====
+$f(x)$ need to have sufficient 'decay' near infinity for the Fourier transformation to make sense. More precisely, a sufficient condition is that $f(x)$ is 'absolutely integrable', which means
+$$ \int_\R |f(x)|dx < \infty. $$
+Given this condition, we know
+$$ |F(p)| = | \int_\R e^{-ipx} f(x) dx | \leq \int_\R |e^{-ipx} f(x)| dx  = \int_\R |f(x)| dx < \infty. $$
+so we are safe.
+===== Example 2 =====
+The Gaussian. $f(x) = e^{-x^2}. $ It decays pretty nicely in both directions.
+Since we learned our lesson, we need to do a check first
+$$ \int_\R e^{-x^2} dx = \sqrt{\pi}. $$
+Great, it is finite.
+Now, let's Fourier transform.
+$$ F(p) = \int e^{-ipx - x^2} dx = \int_\R e^{- (x-ip/2)^2 - p^2/4} dx = e^{-p^2/4} \int_{u \in -ip/2+\R} e^{-u^2} du $$
+where we used the new variable $u = -ip/2+x$.
+We can move the contour for $u$ as long as the integral is convergent. Recall from last Wednesday's lecture, $e^{-u^2}$ decays fast to $0$ when $|u| \to \infty$ in the sectors $\arg(u) \in (-\pi/4, \pi/4)$ and $\arg(u) \in \pi + (-\pi/4, \pi/4)$. So, we are going to move the integration contour of $u$ from the shifted real line $-ip/2+\R$ back to $\R$. We get
+$$ \int_{u \in -ip/2+\R} e^{-u^2} du  = \int_{u \in \R} e^{-u^2} du = \sqrt{\pi}.$$
+Thus, we get
+$$ F(p) = \sqrt{\pi} e^{-p^2/4}.$$
+Success!
+===== Example 3 =====
+Let $f(x) = 1/(1+x^2)$. Let's check that $f(x)$ is absolutely integrable first
+$$ \int_\R f(x) dx = \int_{C_R} f(z) dz = 2\pi i Res_{z=i} f(z) = 2\pi i \frac{1}{2i } = \pi. $$
+where $C_{+,R}$ is the contour consist of two part
+  * $[-R, R]$
+  * a semi-circle in the upper half plane of radius $R$.
+Equivalently, we can choose a different contour $C_{-,R}$, which also consist of two part
+  * $[-R, R]$
+  * a semi-circle in the **lower** half plane of radius $R$.
+Note that when we go around $C_{-,R}$ it goes around the singularity of $f(z)$ at $z=-i$ in the 'negative' (clockwise) direction, hence we get
+$$ \int_{C_{-,R}} f(z) dz = (-2\pi i) Res_{z=-i} f(z) = -2\pi i \frac{1}{-2i} = \pi. $$
+Whew! the two minus sign cancels, we get the same answer.
+Now, let's compute the little variation
+$$ F(p) = \int_\R e^{-ipx} f(x) dx. $$
+We first take the 'truncation' approximation, which recovers the original limit under the limit $R \to \infty$.
+$$ F(p) = \lim_{R \to \infty} \int_{-R}^R  e^{-ipx} f(x) dx. $$
+Then, we try to 'close the contour', by connecting the point $R$ to $-R$ along some way. Due to this exponential factor $e^{-ipz}$, it matters whether we close the contour in the upper half-plane, or the lower one. Indeed, we have
+$$ |e^{-ipz}| = e^{Re(-ip(x+iy))} = e^{Re(-ipx + py)} = e^{py} = e^{p Im(z)}. $$
+So,
+  * if $p \geq 0$, to avoid $|e^{-ipz}|$ blow up to infinity, we need to keep $Im(z)$ bounded from above on the contour.
+  * if $p \leq 0$, to avoid $|e^{-ipz}|$ blow up to infinity, we need to keep $Im(z)$ bounded from below on the contour.
+Thus, for $p \geq 0$, we can complete the by the lower-half-semicircle $C_{-,R}$. We get
+$$ (p\geq 0) F(p) = \int_{C_{-,R}} \frac{e^{-ipz}}{1+z^2}  dz = (-2\pi i) Res_{z=-i} \frac{e^{-ipz}}{1+z^2} = \pi  e^{-p}. $$
+Similarly, for $p \geq 0$, we get
+$$ (p\leq 0) F(p) = \int_{C_{+,R}} \frac{e^{-ipz}}{1+z^2}  dz = (2\pi i) Res_{z=i} \frac{e^{-ipz}}{1+z^2} = \pi  e^{p}. $$
+We can write the result in a cool way as
+$$ F(p) = \pi e^{-|p|}. $$
+So when $|p| \to \infty$, we see $F(p) \to 0$ very quickly.
+====== Exercise ======
+warm up:Does the function $f(x)=1$ admits Fourier transformation? Why? How about $f(x) = 1 / (1+|x|)$?
+Find the Fourier transformation of the following functions.
+$$ f(x) = \begin{cases}
+& 0<x<1 \cr
+& \text{else}
+\end{cases}
+$$
+$$ f(x) = \begin{cases}
++x & -1<x<0 \cr
+-x & 0<x<1 \cr
+& \text{else}
+\end{cases}
+$$

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