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--- math121b:03-20 [2020/03/19 23:27]
pzhou
+++ math121b:03-20 [2020/03/19 23:42] (current)
pzhou
@@ Line 18: / Line 18: @@
 And to leading order, we can replace $f(z)$ by its value at the critical point.
+** Example **
+Physicists ususally use $1/\hbar$ as $\lambda$. Here are some sample computation.
+$$ \int_\R e^{-\frac{1}{\hbar} x^2/2} dx = \sqrt{2\pi \hbar} $$
+$$ \int_\R x e^{-\frac{1}{\hbar} x^2/2} dx = 0 $$
+$$ \int_\R x^2 e^{-\frac{1}{\hbar} x^2/2} dx = \hbar^{3/2} \int_\R u^2 e^{- u^2/2} du = \hbar^{3/2} \sqrt{2\pi} $$
+The point is that, having $x^2$ in the pre-factor will make the integral much smaller (indeed, the $x^2$ factor is killing the peak of $e^{-x^2/(2\hbar)}$ at $x=0$.)
+Another trick to compute the last equality is that
+$$ \int_\R x^2 e^{- \lambda x^2/2} dx = -2 \frac{d}{d\lambda} \int_\R  e^{- \lambda x^2/2} dx = -2 \frac{d}{d\lambda} \sqrt{2\pi / \lambda} = \sqrt{2\pi} \lambda^{-3/2} $$
 ----
@@ Line 50: / Line 61: @@
 where the last step uses trig identity $\sin(A) = \cos(\pi/2 - A)$, so that we agree with Boas.
+Here is a picture of the 'ideal contour', where the phase $S(u)$ has 'constant phase'.
+{{ :math121b:steepest_descent_method.png?nolink&400 |}}
+It is made with [[https://www.wolframalpha.com/input/?i=contourplot%5B+Im+%5B%28x%2BI+y%29+-+1%2F%28x%2BI+y%29%5D+%2C+%7Bx%2C+-2%2C2%7D%2C+%7By%2C+-2%2C2%7D%5D | Wolfram Alpha.]]
+------
-Here is a [[https://www.wolframalpha.com/input/?i=contourplot%5B+Im+%5B%28x%2BI+y%29+-+1%2F%28x%2BI+y%29%5D+%2C+%7Bx%2C+-2%2C2%7D%2C+%7By%2C+-2%2C2%7D%5D | picture]] of the 'ideal contour', where the phase $S(u)$ has 'constant phase'.
+This finishes Chapter 12. I did not talk about the Airy function, Hermite function, Laguerre functions. I will leave those
-{{ :math121b:steepest_descent_method.png?nolink&600 |}}
+as reading materials, and post homework questions about them.

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