Commit 54bcdc4c authored by Tommy Rushton's avatar Tommy Rushton

Bring things closer into line with target notebook.

parent c005bd82
......@@ -8,7 +8,7 @@
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/lib/js/jquery.stickytableheaders.js"></script>
#+HTML_HEAD: <script type="text/javascript" src="http://www.pirilampo.org/styles/readtheorg/js/readtheorg.js"></script>
# #+PROPERTY: header-args :session :exports both
#+PROPERTY: header-args :session :exports both
* Asking the math library
My computer tells me that $\pi$ is /approximatively/
......@@ -38,8 +38,7 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
* Using a surface fraction argument
A method that is easier to understand and does not make use of the
$\sin$ function is based on the fact that if $X \sim U(0,1)$ and $Y \sim U(0,1)$, then $P[X^2 + Y^2 \leq 1] = \pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo Method" on Wikipedia]]). The following code uses this approach:
A method that is easier to understand and does not make use of the $\sin$ function is based on the fact that if $X\sim U(0,1)$ and $Y\sim U(0,1)$, then $P[X^2+Y^2 \leq 1] = \pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach:
#+begin_src python :results output file :session :var matplot_lib_filename="figure_pi_mc2.png" :exports both
import matplotlib.pyplot as plt
......@@ -65,8 +64,7 @@ print(matplot_lib_filename)
[[file:figure_pi_mc2.png]]
It is then straightforward to obtain a (not really good) approximation
to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller
than 1:
to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:
#+begin_src python :results value :session :exports both
4*np.mean(accept)
......
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