Skip to content
Projects
Groups
Snippets
Help
Loading...
Help
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
M
mooc-rr
Project
Project
Details
Activity
Releases
Cycle Analytics
Repository
Repository
Files
Commits
Branches
Tags
Contributors
Graph
Compare
Charts
Issues
0
Issues
0
List
Board
Labels
Milestones
Merge Requests
0
Merge Requests
0
CI / CD
CI / CD
Pipelines
Jobs
Schedules
Charts
Wiki
Wiki
Snippets
Snippets
Members
Members
Collapse sidebar
Close sidebar
Activity
Graph
Charts
Create a new issue
Jobs
Commits
Issue Boards
Open sidebar
f38e78a5c454cc0b5545f6b1ac3e32b6
mooc-rr
Commits
3c1a9589
Commit
3c1a9589
authored
Sep 30, 2025
by
f38e78a5c454cc0b5545f6b1ac3e32b6
Browse files
Options
Browse Files
Download
Email Patches
Plain Diff
second commit
parent
dc6d5e0d
Changes
1
Hide whitespace changes
Inline
Side-by-side
Showing
1 changed file
with
8 additions
and
17 deletions
+8
-17
toy_notebook_en.ipynb
module2/exo1/toy_notebook_en.ipynb
+8
-17
No files found.
module2/exo1/toy_notebook_en.ipynb
View file @
3c1a9589
...
...
@@ -4,18 +4,9 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"# <center>toy_notebook_en</center>\n",
"\n",
"<center>March 28, 2019</center>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 1 On the computation of $π$\n",
"## 1.1 Asking the maths library\n",
"My computer tells me that $π$ is *approximatively*"
"# On the computation of $π$\n",
"## Asking the maths library\n",
"My computer tells me that $\\pi$ is *approximatively*"
]
},
{
...
...
@@ -40,7 +31,7 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"##
1.2
Buffon’s needle\n",
"## Buffon’s needle\n",
"Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**"
]
},
...
...
@@ -73,9 +64,9 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"##
1.3
Using a surface fraction argument\n",
"A method that is easier to understand and does not make use of the
sin
function is based on the\n",
"fact that if $X
∼ U(0, 1)$ and $Y ∼ U(0, 1)$, then $P[X^2 + Y^2 ≤ 1] = π
/4$ (see [\"Monte Carlo method\"\n",
"## Using a surface fraction argument\n",
"A method that is easier to understand and does not make use of the
$\\sin$
function is based on the\n",
"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 [\"Monte Carlo method\"\n",
"on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
]
},
...
...
@@ -117,7 +108,7 @@
"metadata": {},
"source": [
"\n",
"It is then straightforward to obtain a (not really good) approximation to $
π
$ by counting how\n",
"It is then straightforward to obtain a (not really good) approximation to $
\\pi
$ by counting how\n",
"many times, on average, $X^2 + Y^2$\n",
"is smaller than 1:"
]
...
...
Write
Preview
Markdown
is supported
0%
Try again
or
attach a new file
Attach a file
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment