Solution 4

parent 0d9ec61d
...@@ -12,7 +12,7 @@ ...@@ -12,7 +12,7 @@
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Asking the maths library\n", "## Asking the maths library\n",
"My computer tells me that $\\pi$ is approximatively" "My computer tells me that $\\pi$ is *approximatively*"
] ]
}, },
{ {
...@@ -37,13 +37,13 @@ ...@@ -37,13 +37,13 @@
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## Buffalo'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__" "Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the __approximation__"
] ]
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 9, "execution_count": 10,
"metadata": {}, "metadata": {},
"outputs": [ "outputs": [
{ {
...@@ -52,7 +52,7 @@ ...@@ -52,7 +52,7 @@
"3.128911138923655" "3.128911138923655"
] ]
}, },
"execution_count": 9, "execution_count": 10,
"metadata": {}, "metadata": {},
"output_type": "execute_result" "output_type": "execute_result"
} }
...@@ -63,21 +63,21 @@ ...@@ -63,21 +63,21 @@
"N = 10000\n", "N = 10000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n", "x = np.random.uniform(size=N, low=0, high=1)\n",
"theta = np.random.uniform(size=N, low=0, high=pi/2)\n", "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum((x+np.sin(theta))>1)/N)\n" "2/(sum((x+np.sin(theta))>1)/N)"
] ]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"## 1.3 Using a surface fraction argument" "## Using a surface fraction argument"
] ]
}, },
{ {
"cell_type": "markdown", "cell_type": "markdown",
"metadata": {}, "metadata": {},
"source": [ "source": [
"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 [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). 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 [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
] ]
}, },
{ {
...@@ -99,7 +99,7 @@ ...@@ -99,7 +99,7 @@
} }
], ],
"source": [ "source": [
"%matplotlib inline\n", "%matplotlib inline \n",
"import matplotlib.pyplot as plt\n", "import matplotlib.pyplot as plt\n",
"\n", "\n",
"np.random.seed(seed=42)\n", "np.random.seed(seed=42)\n",
......
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