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43800f9eebbeb989c35b6b37774edfea
mooc-rr
Commits
1fac9617
Commit
1fac9617
authored
Jul 06, 2025
by
Anton Y.
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task try 6
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e06f6a7c
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toy_document_orgmode_python_en.org
module2/exo1/toy_document_orgmode_python_en.org
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module2/exo1/toy_document_orgmode_python_en.org
View file @
1fac9617
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@@ -31,7 +31,8 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
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@@ -31,7 +31,8 @@ theta = np.random.uniform(size=N, low=0, high=pi/2)
#+end_src
#+end_src
* Using a surface fraction argument
* 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
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
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:
Wikipedia]]). The following code uses this approach:
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@@ -55,9 +56,8 @@ plt.savefig(matplot_lib_filename)
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@@ -55,9 +56,8 @@ plt.savefig(matplot_lib_filename)
print(matplot_lib_filename)
print(matplot_lib_filename)
#+end_src
#+end_src
It is then straightforward to obtain a (not really good) approximatin
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
to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1:
than 1:
#+begin_src python :results output :session *python* :exports both
#+begin_src python :results output :session *python* :exports both
4*np.mean(accept)
4*np.mean(accept)
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