diff --git a/cours.md b/cours.md
index 3a77f121ec07cca4181e239360e1573d1855c476..46801d0865802183c61b6bc7fa017759a98e7732 100644
--- a/cours.md
+++ b/cours.md
@@ -131,13 +131,13 @@ import numpy as np
import matplotlib.pyplot as plt
def f(x):
- return x**3
+ return x**3
def df(x):
- return 3*x**2
+ return 3*x**2
def approx_df(x0,x1):
- return (f(x1)-f(x0))/(x1-x0)
+ return (f(x1)-f(x0))/(x1-x0)
dx = 0.1
xmin = -2
@@ -218,22 +218,22 @@ import numpy as np
import matplotlib.pyplot as plt
import math
-def f(x):
- return np.exp(x)
+def f(x):
+ return np.exp(x)
def p(x):
- return 1+x
+ return 1+x
-def df(x):
- return np.exp(x)
+def df(x):
+ return np.exp(x)
x0 = 0
dx = np.arange(0,0.3,0.05)
x = np.array([])
e = np.array([])
for delta in dx:
- x = np.append(x, x0 + delta)
- e = np.append(e, f(x0 + delta)-p(x0 + delta))
+ x = np.append(x, x0 + delta)
+ e = np.append(e, f(x0 + delta)-p(x0 + delta))
plt.figure(1)
@@ -262,12 +262,12 @@ sera proportionnelle à $(x-x_0)^{N+1}$, qu'on écrit souvent comme $\mathcal{O}
$$
f(x)=\sum_{n=0}^N\frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n+\mathcal{O}((x-x_0)^{N+1}).
$${#eq:taylor_err}
-On dit que **l'ordre** de l'approximation est de $N+1$ (l'erreur va décroître comme $(x-x_0)^{N+1}$).
+On dit que **l'ordre** de l'approximation est de $N+1$ (l'erreur va décroître comme $(x-x_0)^{N+1}$).
# L'approximation des dérivées
-De ce que nous venons de voir dans la section précédente, il est possible d'approximer les dérivées de fonctions à l'aide de calculs assez simples.
-Par ailleurs, ces estimations numériques permettent également de travailler sur des échantillons de fonctions inconnues (dans le cas du traitement d'un
+De ce que nous venons de voir dans la section précédente, il est possible d'approximer les dérivées de fonctions à l'aide de calculs assez simples.
+Par ailleurs, ces estimations numériques permettent également de travailler sur des échantillons de fonctions inconnues (dans le cas du traitement d'un
signal par exemple, ou la pixellisation d'une image).
{#fig:sin_noise width=70%}
@@ -277,17 +277,18 @@ Sur la @fig:sin_noise, on voit un signal échantillonné (en noir) et sa dériv
numériquement. Nous étudieront également l'erreur effectuée lors de ces approximations.
Afin de créer la @fig:sin_noise nous avons utilisé le listing suivant
+
```{.python}
import numpy as np
import matplotlib.pyplot as plt
import math
def f(x): # sinus + bruit
- return np.sin(x) + np.random.rand(len(x))*0.1
+ return np.sin(x) + np.random.rand(len(x))*0.1
def approx_df(x,dx): # forward finite diff
- tmp = f(x)
- return (np.roll(tmp,-1)-tmp) / dx
+ tmp = f(x)
+ return (np.roll(tmp,-1)-tmp) / dx
dx = 0.2
xmin = -math.pi
@@ -307,8 +308,8 @@ plt.show() # plot on the screen
## Les dérivées numériques (différences divisées)
Comme nous l'avons vu dans le chapitre précédent, nous pouvons approximer les dérivées
-en mesurant le taux de variation d'une fonction sur une distance $h>0$ "petite". Nous avons
-écrit l'@eq:diff_div
+en mesurant le taux de variation d'une fonction sur une distance $h>0$ "petite". Nous avons
+écrit l'@eq:diff_div
$$
f'(x_0)\cong \frac{f(x_0+h)-f(x_0)}{h},
$$
@@ -333,25 +334,26 @@ et la **différence finie centrale d'ordre 2**
$$
f'(x_0)\cong \frac{f(x_0+h/2)-f(x_0-h/2)}{h}.
$$
-Pour chacune de ces méthodes vous pouvez observer l'approximation de la dérivée de
-la fonction $f(x)=x^3$ ($f'(x)=3x^2$) sur la @fig:finite_diff. On peut constater que la méthode
-centrale d'ordre 2 est bien plus précises que les deux autres. Cela est dû à l'ordre de chacune de ces méthodes: alors que la méthode centrale est d'ordre 2, celles en avant et en arrière sont d'ordre 1 uniquement.
-Comme pour le cas du développement de Taylor l'ordre est en fait représentatif
+Pour chacune de ces méthodes vous pouvez observer l'approximation de la dérivée de
+la fonction $f(x)=x^3$ ($f'(x)=3x^2$) sur la @fig:finite_diff. On peut constater que la méthode
+centrale d'ordre 2 est bien plus précises que les deux autres. Cela est dû à l'ordre de chacune de ces méthodes: alors que la méthode centrale est d'ordre 2, celles en avant et en arrière sont d'ordre 1 uniquement.
+Comme pour le cas du développement de Taylor l'ordre est en fait représentatif
de la précision de l'approximation. Dans la section suivante nous allons justement discuter des erreurs.
{#fig:finite_diff width=70%}
-La @fig:finite_diff est obtenue avec le listing suivant
+La @fig:finite_diff est obtenue avec le listing suivant
+
```python
import numpy as np
import matplotlib.pyplot as plt
import math
def f(x):
- return x**3
+ return x**3
def df(x):
- return 3*x**2
+ return 3*x**2
dx = 0.2
xmin = 0
@@ -375,20 +377,19 @@ plt.show()
## Évaluation de l'erreur
-Lorsqu'on effectue des approximations de fonctions,
-il est nécessaire d'avoir une idée de la valeur de l'erreur
+Lorsqu'on effectue des approximations de fonctions,
+il est nécessaire d'avoir une idée de la valeur de l'erreur
qu'on fait. Le but de cette sous section est d'introduire une méthode
d'étude de l'erreur d'une approximation.
-Il est très difficile en général de connaître exactement l'erreur de l'approximation
-pour un $h$ donné. En revanche, on peut connaître l'ordre de l'erreur, c'est-à-dire
+Il est très difficile en général de connaître exactement l'erreur de l'approximation
+pour un $h$ donné. En revanche, on peut connaître l'ordre de l'erreur, c'est-à-dire
comment l'erreur va dépendre de $h$. Pour ce faire nous allons utiliser le développement
de Taylor et ce que nous connaissons de l'erreur de l'approximation en série de Taylor.
-
### Méthode de différences finies vers l'avant d'ordre 1
-Dans cette sous section nous essayons de déterminer l'erreur que nous commettons lorsque
+Dans cette sous section nous essayons de déterminer l'erreur que nous commettons lorsque
nous calculons la dérivée d'une fonction à l'aide des différences finies vers l'avant d'ordre un.
On veut en fait calculer
$$
@@ -411,27 +412,28 @@ On constate donc que l'erreur est d'ordre un, soit $\mathcal{O}(h)$ (proportionn
Cela veut dire que lorsque nous divisons $h$ par deux, l'erreur va environ décroître d'un facteur 2.
A l'inverse si $h$ est multiplié par deux l'erreur sera également multipliée par un facteur 2.
Néanmoins, il faut également ce rendre compte que le facteur $h$ est multiplié par un nombre, $f''(\xi)$, $\xi\in[x,x+h]$.
-Lorsque $h$ va varier, ce nombre variera aussi potentiellement. On aura donc que cette évaluation de l'erreur
+Lorsque $h$ va varier, ce nombre variera aussi potentiellement. On aura donc que cette évaluation de l'erreur
sera **bornée** par $\mathcal{K}\cdot h$, où $\mathcal{K}=\max_{h} f''(\xi),\forall h$.
Une illustration de ce comportement peut se voir sur la @fig:deriv_avant_1. On peut y voir l'évolution du calcul de l'approximation
de la dérivée de la fonction $f(x)=x^3$ ($f'(x)=3x^2$). On constate que l'erreur n'est pas exactement une droite, mais qu'elle est bornée
par une droite dont la pente est la valeur maximale de $f''(\xi)$.
-![Calcul de la dérivée de $f(x)=x^3$ (à gauche) ainsi que de l'erreur (à droite). On peut constater
-que l'erreur est bornée par une droite donc la pente est donnée par le maximum des deuxièmes dérivées
+![Calcul de la dérivée de $f(x)=x^3$ (à gauche) ainsi que de l'erreur (à droite). On peut constater
+que l'erreur est bornée par une droite donc la pente est donnée par le maximum des deuxièmes dérivées
de $f(x)$ sur l'intervalle $[0,0.5]$.](figs/deriv_avant_1.svg){#fig:deriv_avant_1 width=70%}
La @fig:deriv_avant_1 est obtenue à l'aide du listing suivant
+
```python
import numpy as np
import matplotlib.pyplot as plt
-def f(x):
- return x**3
+def f(x):
+ return x**3
-def df(x):
- return 3 * x**2
+def df(x):
+ return 3 * x**2
x0 = 0.0
dx = np.arange(0.05,0.5,0.05)
@@ -439,10 +441,10 @@ x = np.array([])
deriv = np.array([])
e = np.array([])
for delta in dx:
- x = np.append(x, x0 + delta)
- dd = (f(x0 + delta)-f(x0))/delta
- deriv = np.append(deriv, dd)
- e = np.append(e, abs(df(x0) - dd))
+ x = np.append(x, x0 + delta)
+ dd = (f(x0 + delta)-f(x0))/delta
+ deriv = np.append(deriv, dd)
+ e = np.append(e, abs(df(x0) - dd))
max_d2 = np.max(e/dx) # calcul du max des deuxièmes dérivées
@@ -470,20 +472,20 @@ plt.show()
---
-Exercice (Méthode de différences finies vers l'arrière d'ordre 1) +.#
+Exercice (Méthode de différences finies vers l'arrière d'ordre 1) +.#
-En vous inspirant de ce que nous venons de voir, montrez que
+En vous inspirant de ce que nous venons de voir, montrez que
la méthode de différences finies vers l'arrière est bien d'ordre 1.
---
-A présent, on peut également évaluer l'erreur totale que nous commettons lorsque nous évaluons la dérivée
+A présent, on peut également évaluer l'erreur totale que nous commettons lorsque nous évaluons la dérivée
avec une approximation. Lorsque nous évaluons la dérivée sur un ensemble de $q$ points $\{x_i\}_{i=0}^{q-1}$, avec
$x_i=x_0+i\cdot h$, on a une fonction d'erreur sur chaque point qui est donnée par
-$$
+$$
E(x_i)=\left(f'(x_i)-\frac{f(x_i+h)-f(x_i)}{h}\right)^2.
$$
-Cette définition est un peu différente de celle vue précédemment où nous utilisons une valeur absolue.
+Cette définition est un peu différente de celle vue précédemment où nous utilisons une valeur absolue.
Cela est dû au fait qu'il est beaucoup plus aisé de manipuler des carrés que la valeur absolue (en particulier
sa dérivée est continue contrairement à la valeur aboslue) mais les deux méthodes sont une bonne indication de l'erreur d'une méthode numérique.
On peut donc calculer l'erreur totale, $E_\mathrm{tot}$ en sommant l'erreur sur chaque point,
@@ -512,7 +514,7 @@ On voit donc que l'erreur totale sera constante avec l'augmentation de $q$ (ou l
{#fig:err_deriv_avant width=70%}
-Sur la @fig:err_deriv_avant, on peut voir l'erreur totale, ainsi que l'erreur moyenne. Comme attendu l'erreur totale vaut environ un et est constante
+Sur la @fig:err_deriv_avant, on peut voir l'erreur totale, ainsi que l'erreur moyenne. Comme attendu l'erreur totale vaut environ un et est constante
en fonction de $h$. On voit également que l'erreur moyenne augmente linéairement avec $h$.
La @fig:err_deriv_avant est obtenue à l'aide du listing suivant
@@ -522,29 +524,29 @@ import numpy as np
import matplotlib.pyplot as plt
import math
-def f(x):
- return np.sin(x)
+def f(x):
+ return np.sin(x)
-def df(x):
- return np.cos(x)
+def df(x):
+ return np.cos(x)
-def app(x, h):
- return (f(x+h)-f(x)) / h
+def app(x, h):
+ return (f(x+h)-f(x)) / h
def comp_error_sqr(xmin, xmax, h):
- x = np.arange(xmin, xmax, h)
- approx = app(x,h)
- exact = df(x)
+ x = np.arange(xmin, xmax, h)
+ approx = app(x,h)
+ exact = df(x)
- error = (approx - exact)**2
- return x, error
+ error = (approx - exact)**2
+ return x, error
def comp_error_moy(error):
- q = error.size
- return np.sum(np.sqrt(error) / q)
+ q = error.size
+ return np.sum(np.sqrt(error) / q)
def comp_error_tot(error):
- return np.sum(np.sqrt(error))
+ return np.sum(np.sqrt(error))
@@ -579,9 +581,9 @@ e_tot = np.array([])
e_moy = np.array([])
for dx in h:
- x, error = comp_error_sqr(xmin, xmax, dx)
- e_tot = np.append(e_tot, comp_error_tot(error))
- e_moy = np.append(e_moy, comp_error_moy(error))
+ x, error = comp_error_sqr(xmin, xmax, dx)
+ e_tot = np.append(e_tot, comp_error_tot(error))
+ e_moy = np.append(e_moy, comp_error_moy(error))
plt.figure(2)
@@ -613,21 +615,21 @@ En substituant ces deux relations dans la formule ci-dessus, il vient
E_\mathrm{tot}&=\sqrt{\sum_{i=0}^{q-1}\left(f'(x_i)-\frac{hf'(x_0)+\frac{h^3}{48}\left(f'''(\xi)+f'''(\eta)\right)}{h}\right)^2},\nonumber\\
&=\frac{h^2}{48}\sqrt{\sum_{i=0}^{q-1}\left(\left(f'''(\xi)+f'''(\eta)\right)\right)^2}\leq \mathcal{H}\cdot h,
\end{align}
-oû $\mathcal{H}$ est la valeur maximale de $f'''(\xi)+f'''(\eta)$. De même il vient que
+oû $\mathcal{H}$ est la valeur maximale de $f'''(\xi)+f'''(\eta)$. De même il vient que
$$
E_\mathrm{moy}=\leq \mathcal{H}\cdot h^2.
$$
On a donc que la formule des différences finies centrées est d'ordre 2, notée $\mathcal{O}(h^2)$. L'erreur moyenne décroissant proportionnellement à $h^2$: lorsqu'on divise $h$ par 2, on divise l'erreur par 4.
-{#fig:err_deriv_avant_centree width=70%}
-On peut voir ce comportement sur la @fig:err_deriv_avant_centree que l'erreur est beaucoup plus faible à résolution équivalente pour
+On peut voir ce comportement sur la @fig:err_deriv_avant_centree que l'erreur est beaucoup plus faible à résolution équivalente pour
les différences finies centrées par rapport aux différences finies en avant. De plus la pente de l'erreur est plus faible
pour les différences finies centrées. Elle est en fait de pente $2$ en échelle log-log, contrairement aux différences finies centrées en avant
qui est de pente un.
-{#fig:err_cos_ac width=70%}
On constate également sur la @fig:err_cos_ac que l'erreur sur chaque point est beaucoup plus faible poru les différences finies centrées, par rapport à celle en avant.
@@ -641,40 +643,40 @@ import numpy as np
import matplotlib.pyplot as plt
import math
-def f(x):
- return np.sin(x)
+def f(x):
+ return np.sin(x)
-def df(x):
- return np.cos(x)
+def df(x):
+ return np.cos(x)
-def app(x, h):
- return (f(x+h)-f(x)) / h
+def app(x, h):
+ return (f(x+h)-f(x)) / h
-def app_c(x, h):
- return (f(x+h/2)-f(x-h/2)) / h
+def app_c(x, h):
+ return (f(x+h/2)-f(x-h/2)) / h
def comp_error_sqr(xmin, xmax, h):
- x = np.arange(xmin, xmax, h)
- approx = app(x,h)
- exact = df(x)
+ x = np.arange(xmin, xmax, h)
+ approx = app(x,h)
+ exact = df(x)
- error = (approx - exact)**2
- return x, error
+ error = (approx - exact)**2
+ return x, error
def comp_error_sqr_c(xmin, xmax, h):
- x = np.arange(xmin, xmax, h)
- approx = app_c(x,h)
- exact = df(x)
+ x = np.arange(xmin, xmax, h)
+ approx = app_c(x,h)
+ exact = df(x)
- error = (approx - exact)**2
- return x, error
+ error = (approx - exact)**2
+ return x, error
def comp_error_moy(error):
- q = error.size
- return np.sum(np.sqrt(error) / q)
+ q = error.size
+ return np.sum(np.sqrt(error) / q)
def comp_error_tot(error):
- return np.sum(np.sqrt(error))
+ return np.sum(np.sqrt(error))
@@ -714,13 +716,13 @@ e_moy = np.array([])
e_moy_c = np.array([])
for dx in h:
- x, error = comp_error_sqr(xmin, xmax, dx)
- e_tot = np.append(e_tot, comp_error_tot(error))
- e_moy = np.append(e_moy, comp_error_moy(error))
+ x, error = comp_error_sqr(xmin, xmax, dx)
+ e_tot = np.append(e_tot, comp_error_tot(error))
+ e_moy = np.append(e_moy, comp_error_moy(error))
- x, error = comp_error_sqr_c(xmin, xmax, dx)
- e_tot_c = np.append(e_tot_c, comp_error_tot(error))
- e_moy_c = np.append(e_moy_c, comp_error_moy(error))
+ x, error = comp_error_sqr_c(xmin, xmax, dx)
+ e_tot_c = np.append(e_tot_c, comp_error_tot(error))
+ e_moy_c = np.append(e_moy_c, comp_error_moy(error))
plt.figure(2)
@@ -749,7 +751,7 @@ Saurez-vous écrire la formule d'ordre $\mathcal{O}(h^3)$ pour calculer la déri
Lorsque nous utilisons les formules précédentes sur un ordinateur, nous voudrions utiliser un $h$ aussi petit que possible afin d'être aussi précis que possible.
Néanmoins, plus $h$ devient petit, plus nous risquons des erreurs numériques. En effet, sur un ordinateur la précision numérique
à disposition est finie. En effet, les nombres à virgules flottantes se représentent en 32 ou 64 bits typiquement, soit une précision d'environ 7 chiffres significatifs et
-16 respectivement. Ici, nous allons essayer de déterminer quelle est la valeur minimale de $h$ que nous pouvons utiliser en fonction du nombre de chiffres significatifs
+16 respectivement. Ici, nous allons essayer de déterminer quelle est la valeur minimale de $h$ que nous pouvons utiliser en fonction du nombre de chiffres significatifs
que nous avons à disposition sur un ordinateur.
En d'autres termes soit un nombre réel $a$, et sa représentation sur un ordinateur, $\bar{a}$, la différence entre les deux est la précision numérique
@@ -757,7 +759,7 @@ de la machine et peut s'écrire comme
$$
|h-\bar{h}|\sim |h|\cdot 10^{-N},
$$
-où $N$ est le nombre de chiffres significatifs de la machine. On a donc que $f(x_0+h)$ et $f(x_0)$ sont connus à
+où $N$ est le nombre de chiffres significatifs de la machine. On a donc que $f(x_0+h)$ et $f(x_0)$ sont connus à
une précision
\begin{align}
f(x_0+h)-\overline{f(x_0+h)}\sim |f(x_0+h)|\cdot 10^{-N}\sim |f(x_0)|\cdot 10^{-N},\\
@@ -767,7 +769,7 @@ En utilisant cette information sur la formule de la dérivée en avant par exemp
$$
f'(x_0)\cong \frac{f(x_0+h)-f(x_0)}{h}\sim \frac{2\cdot |f(x_0)|\cdot 10^{-N}}{h}.
$$
-Si $h\gg 2\cdot |f(x_0)|\cdot 10^{-N}$ la précision numérique n'aura pas d'impact sur l'évaluation de la dérivée. En revanche, si
+Si $h\gg 2\cdot |f(x_0)|\cdot 10^{-N}$ la précision numérique n'aura pas d'impact sur l'évaluation de la dérivée. En revanche, si
$h\sim 10^{-N}$, l'erreur d'arrondi aura un effet dramatique sur le résultat.
## Dérivées d'ordre deux
@@ -783,7 +785,7 @@ On peut à présent utiliser les formules de différences finies en avant sur $f
f'(x_0+h/2)&=\frac{f(x_0+h/2+h/2)-f(x_0+h/2-h/2)}{h}=\frac{f(x_0+h)-f(x_0)}{h},\\
f'(x_0-h/2)&=\frac{f(x_0-h/2+h/2)-f(x_0-h/2-h/2)}{h}=\frac{f(x_0)-f(x_0-h)}{h}.
\end{align}
-On peut constater que la première formule revient à calculer $f'(x_0)$ avec la formule de différences finies en avant,
+On peut constater que la première formule revient à calculer $f'(x_0)$ avec la formule de différences finies en avant,
alors que la seconde est celle en arrière. On peut donc réécrire l'@eq:deriv_deux comme
$$
f''(x_0)=\frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2}.
@@ -807,8 +809,6 @@ Déterminer la formule pour $f''(x_0)$ en utilisant la formule de la dérivée e
---
-
-
## Résumé
Dans ce chapitre, nous avons vu comment approximer le calcul de dérivées numériquement et prédire l'ordre de l'erreur de ces approximations.
@@ -822,8 +822,8 @@ dans un signal ou dans une image. Cela est très utile si nous souhaitons rééc
ou dépixeliser une image (un peu comme dans la série "les experts"). On appelle **techniques d'interpolation**
les méthodes utilisées pour résoudre ce genre de problème.
-Dans la @fig:interp_experts nous pouvons voir 3 versions d'une image. La première pixelisée (tout à gauche),
-la seconde (au centre) rééchantillonnée de façon naïve, et la troisième (à droite)
+Dans la @fig:interp_experts nous pouvons voir 3 versions d'une image. La première pixelisée (tout à gauche),
+la seconde (au centre) rééchantillonnée de façon naïve, et la troisième (à droite)
avec une méthode plus avancée.
{#fig:points width=100%}
-Le but à présent est de déterminer une fonction $f$ définie
+Le but à présent est de déterminer une fonction $f$ définie
dans tout l'intervalle $[x_0,x_{n-1}]$ et passant par tous les points $y_i$ (voir @fig:points_int).
{#fig:points_int width=100%}
@@ -857,7 +857,7 @@ où $p_m(x_i)=y_i$ ($i=0,...,n-1$), avec $m\in\natural$ et où tout le jeu consi
## Idée naive
-L'idée la plus simple que nous pouvons avoir pour déterminer un polynôme interpolant est
+L'idée la plus simple que nous pouvons avoir pour déterminer un polynôme interpolant est
d'écrire le système d'équations suivant
\begin{align}
p_m(x_0)=y_0&=a_0+a_1x_0+\dots+a_mx_0^m,\\
@@ -869,7 +869,7 @@ On se retrouve donc avec un système de $n$-équations à $m+1$ inconnues.
---
-Question +.#
+Question +.#
Quel est l'ordre du polynôme qu'on peut faire passer par tous les $n$ points $x_i,y_i$ (si tous les $x_i$ et $y_i$ sont différents les uns des autres)?
@@ -881,7 +881,7 @@ peut être plus petit que le nombre de points.
---
-Question +.#
+Question +.#
Pouvez-vous trouver de tels exemples?
@@ -891,37 +891,37 @@ Dans le cas général, nous essayons de résoudre le système de $n$ équations
\begin{equation}
\mat{X}\cdot \vec a=\vec y,
\end{equation}
-où $\vec a$ est l'inconnue (nous recherchons les coefficients du polynôme) et
+où $\vec a$ est l'inconnue (nous recherchons les coefficients du polynôme) et
où $\mat{X}$, $\vec y$ et $\vec x$ sont donnés par
\begin{equation}
-\mat{X} = \begin{pmatrix}
- 1 & x_{0} & x_{0}^2 & \cdots & x_0^{n-1} \\
- 1 & x_{1} & x_{1}^2 & \cdots & x_1^{n-1} \\
+\mat{X} = \begin{pmatrix}
+ 1 & x_{0} & x_{0}^2 & \cdots & x_0^{n-1} \\
+ 1 & x_{1} & x_{1}^2 & \cdots & x_1^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-1}
\end{pmatrix},
\vec y=
\begin{pmatrix}
- y_0\\
- y_1\\
- \vdots\\
- y_{n-1}
+ y_0\\
+ y_1\\
+ \vdots\\
+ y_{n-1}
\end{pmatrix},
\vec a=
\begin{pmatrix}
- a_0\\
- a_1\\
- \vdots\\
- a_{n-1}
+ a_0\\
+ a_1\\
+ \vdots\\
+ a_{n-1}
\end{pmatrix}.
\end{equation}
-Afin de déterminer les coefficients $a_i$, nous devons résoudre ce système d'équations linéaires, qui possède une solution unique
+Afin de déterminer les coefficients $a_i$, nous devons résoudre ce système d'équations linéaires, qui possède une solution unique
seulement si $\det(\mat{X})\neq 0$. Nous n'allons pas résoudre ce système de façon générale, car cela serait une solution beaucoup trop
-couteuse.
+couteuse.
---
-Question +.#
+Question +.#
Quel est le degré du polynôme d'interpolation qui passe par $2$ points?
@@ -948,15 +948,15 @@ p_1(x)=x.
Exercice (La parabole) +.#
-Déterminer le polyôme d'interpolation passant par les points
+Déterminer le polyôme d'interpolation passant par les points
$(x_0,y_0)=(0,1)$, $(x_1,y_1)=(1,0)$, et $(x_2,y_2)=(-1,0)$.
---
-La méthode naïve bien que très simple à écrire est beaucoup trop compliquée à résoudre. En effet, cela implique en général de résoudre
-un système d'équations linéaire où $\mat{X}$ est une matrice dont tous les coefficients sont non-nuls. On doit donc employer des méthodes
-comme la décomposition $LU$ pour y parvenir. Grâce à différentes autres techniques que nous pouvons voir dans ce qui suit,
-nous pouvons grandement nous faciliter la tâche en écrivant le système d'équations sous une forme beaucoup plus
+La méthode naïve bien que très simple à écrire est beaucoup trop compliquée à résoudre. En effet, cela implique en général de résoudre
+un système d'équations linéaire où $\mat{X}$ est une matrice dont tous les coefficients sont non-nuls. On doit donc employer des méthodes
+comme la décomposition $LU$ pour y parvenir. Grâce à différentes autres techniques que nous pouvons voir dans ce qui suit,
+nous pouvons grandement nous faciliter la tâche en écrivant le système d'équations sous une forme beaucoup plus
agréable à résoudre.
En fait si on écrit le polynôme $p_m(x)$ sous la forme[^2]
@@ -971,18 +971,18 @@ p_{n-1}(x_1)&=b_0+b_1(x_1-x_0)=y_1,\\
p_{n-1}(x_{n-1})&=b_0+b_1(x_{n-1}-x_0)+b_2(x_{n-1}-x_0)(x_{n-1}-x_1)(x_{n-1}-x_2)\nonumber\\
&+\cdots b_{n-1}(x_{n-1}-x_0)(x_{n-1}-x_1)\cdots(x_{n-1}-x_{n-2}) =y_{n-1}.
\end{align}
-On a donc un système d'équations à résoudre de la forme
+On a donc un système d'équations à résoudre de la forme
$$
\mat{X}'\cdot \vec b=\vec y,
$$
où $\mat{X}'$ est une matrice triangulaire inférieure (seule la partie sous la diagonale est non-nulle)
\begin{equation}
-\mat{X}'' =
-\begin{pmatrix}
- 1 & 0 & 0 & \cdots & 0 \\
- 1 & x_{1}-x_0 & 0 & \cdots & 0 \\
- \vdots & \vdots & \ddots & \vdots & 0\\
- 1 & x_{n-1}-x_0 & (x_{n-1}-x_0)(x_{n-1}-x_1) & \cdots & (x_{n-1}-x_0)(x_{n-1}-x_1)\cdots(x_{n-1}-x_{n-2})
+\mat{X}'' =
+\begin{pmatrix}
+ 1 & 0 & 0 & \cdots & 0 \\
+ 1 & x_{1}-x_0 & 0 & \cdots & 0 \\
+ \vdots & \vdots & \ddots & \vdots & 0\\
+ 1 & x_{n-1}-x_0 & (x_{n-1}-x_0)(x_{n-1}-x_1) & \cdots & (x_{n-1}-x_0)(x_{n-1}-x_1)\cdots(x_{n-1}-x_{n-2})
\end{pmatrix},
\end{equation}
Le système d'équation est donc beaucoup plus simple à résoudre.
@@ -1001,10 +1001,10 @@ On a donc que $p_1(x)$ peut se mettre sous la forme
$$
p_1(x)=y_0+(x-x_0)f'(x_0).
$$
-Ce polynôme ressemble terriblement au développement de Taylor de $f(x)$ proche de $x_0$
+Ce polynôme ressemble terriblement au développement de Taylor de $f(x)$ proche de $x_0$.
Si à présent, nous avons $3$ points à disposition, soit
-$(x_0,y_0)$, $(x_1,y_1)$, et $(x_2,y_2)$, nous pouvons faire passer un polynôme de degré deux par ces points. Comme nous connaissons déjà le polynôme de degré un passant par $(x_0,y_0)$ et $(x_1,y_1)$, nous
+$(x_0,y_0)$, $(x_1,y_1)$, et $(x_2,y_2)$, nous pouvons faire passer un polynôme de degré deux par ces points. Comme nous connaissons déjà le polynôme de degré un passant par $(x_0,y_0)$ et $(x_1,y_1)$, nous
allons écrire $p_2(x)$ comme $p_1(x)$ auquel nous allons ajouter une correction
$$
p_2(x)=p_1(x)+b_2(x-x_0)(x-x_1),
@@ -1038,7 +1038,7 @@ On constate que $b_2$ est en fait une approximation de la deuxième dérivée d'
f'(x_2)&\cong\frac{y_2-y_1}{x_2-x_1},\\
f'(x_0)&\cong\frac{y_1-y_0}{x_1-x_0},
\end{align}
-où à la première ligne on évalue la dérivée "en arrière" et vers l'avant sur la deuxième. Il nous reste à voir qu'avec ces définitions
+où à la première ligne on évalue la dérivée "en arrière" et vers l'avant sur la deuxième. Il nous reste à voir qu'avec ces définitions
$b_2$ est donnée par
$$
b_2=\cong\frac{f'(x_2)-f'(x_0)}{x_2-x_0}\cong f''(x_0),
@@ -1056,7 +1056,7 @@ y[x_i]&=y_i,\\
&\vdots\nonumber\\
\delta^n y[x_{i_0},x_{i_1},\dots,x_{i_n}]&=\frac{\delta^{n-1} y[x_{i_1},\dots,x_{i_{n}}]-\delta^{n-1} y[x_{i_0},\dots,x_{i_{n-1}}]}{x_{i_n}-x_{i_0}}.
\end{align}
-Avec cette définition, on peut définir le polynôme interpolant
+Avec cette définition, on peut définir le polynôme interpolant
passant par les points $(x_i,y_i)$, pour $i=0,..,n$ avec la **formule de Newton**, comme
\begin{align}
p_n(x)&=y[x_0]+(x-x_0)\delta y[x_0,x_1]+\dots\nonumber\\
@@ -1073,16 +1073,16 @@ Soient les points $(x_i,y_i)$ donnés par $x_0=0$, $x_1=1$, $x_2=-1$, $y_0=0$, $
### Erreur de l'interpolation
-Nous avons vu dans la section précédente comment écrire un polynôme d'interpolation passant par $n+1$ points, $(x_i,y_i)$, avec $x_i\in[a,b]$, pour $a,b\in\real$. Si nous supposons que
+Nous avons vu dans la section précédente comment écrire un polynôme d'interpolation passant par $n+1$ points, $(x_i,y_i)$, avec $x_i\in[a,b]$, pour $a,b\in\real$. Si nous supposons que
$$
y_i=f(x_i),
$$
-où $f$ est une fonction
+où $f$ est une fonction
$$
f:[a,b]\rightarrow\real,
$$
dérivable au moins $n$ fois et que
-les points $(x_i,y_i)$ ne sont en fait qu'un échantillonnage de $f(x)$.
+les points $(x_i,y_i)$ ne sont en fait qu'un échantillonnage de $f(x)$.
Nous pouvons nous poser la question de ce que vaudra la différence entre $p_n(x)$ et $f(x)$.
En particulier, est-ce que si $n\rightarrow\infty$ on a bien que $p_n(x)\rightarrow f(x)$?
@@ -1091,22 +1091,21 @@ La réponse à cette question n'est pas aisée, pour s'en rendre compte prenons
f(x)&=\sin(x),\ x\in[-5,5],\\
g(x)&=\frac{1}{1+x^2},\ x\in[-5,5].
\end{align}
-Pour la fonction $f(x)$, on voit sur la @fig:sin_interp_n que lorsqu'on fait passer le nombre de point d'interpolation de $n=5$ à $n=20$,
+Pour la fonction $f(x)$, on voit sur la @fig:sin_interp_n que lorsqu'on fait passer le nombre de point d'interpolation de $n=5$ à $n=20$,
le polynôme d'interpolation s'approche de plus en plus de $\sin(x)$ jusqu'à ce qu'on ne puisse plus distinguer de différences.
Bonne nouvelle donc!
-{#fig:sin_interp_n width=100%}
-
En revanche pour la fonction $g(x)$, on voit sur la @fig:one_over_one_x2 qu'en augmentant $n=5$ à $n=20$, le polynôme d'interpolation
-s'approche de plus en plus de $g(x)$ proche du centre du domaine ($x\cong0$), mais que sur les bords,
+s'approche de plus en plus de $g(x)$ proche du centre du domaine ($x\cong0$), mais que sur les bords,
l'interpolation devient de plus en plus mauvaise. Ceci est donc une mauvaise nouvelle!
-{#fig:one_over_one_x2 width=100%}
-En fait comme vous pouvez le constater ici, nous ne sommes pas assurés d'avoir une convergence du polynôme d'interpolation vers la fonction
+En fait comme vous pouvez le constater ici, nous ne sommes pas assurés d'avoir une convergence du polynôme d'interpolation vers la fonction
$f(x)$ en général. Il se trouve que la différence entre le polynôme d'interpolation, $p_n(x)$ et $f(x)$ est donnée par[^3]
$$
f(x)-p_n(x)=(x-x_0)\cdot (x-x_1)\cdot \dots \cdot (x-x_n)\cdot \frac{f^{(n+1)}(\xi)}{(n+1)!},
@@ -1116,12 +1115,12 @@ $$
f^{(2n)}=\pm\sin(x)\in[-1,1],\quad f^{(2n+1)}=\pm\cos(x)\in[-1,1].
$$
On a donc que la différence va tendre vers zéro pour $n\rightarrow\infty$.
-En revanche pour $g(x)=1/(1+x^2)$, les dérivées se comportent beaucoup moins bien. Par exemple,
+En revanche pour $g(x)=1/(1+x^2)$, les dérivées se comportent beaucoup moins bien. Par exemple,
$$
g^{(10)}=\frac{3628800\cdot(11 x^{10}-165 x^8+462 x^6-330 x^4+55 x^2-1))}{(x^2+1)^{11}}.
$$
-On voit donc que $g^{(10)}= 11!\cdot \frac{(x^{10}-15 x^8+42 x^6-30 x^4+5 x^2-\frac{1}{11})}{(x^2+1)^{11}}$. Cette relation se généralise et
-on a que
+On voit donc que $g^{(10)}= 11!\cdot \frac{(x^{10}-15 x^8+42 x^6-30 x^4+5 x^2-\frac{1}{11})}{(x^2+1)^{11}}$. Cette relation se généralise et
+on a que
$$
g^{(n)}\sim (n+1)!.
$$
@@ -1138,8 +1137,8 @@ Une solution est d'utiliser l'interpolation de Chebyshev que nous ne verrons pas
### Erreur d'arrondi
-Une autre source d'erreur qui peut apparaître lorsque nous faisons des interpolations, est l'erreur d'arrondi (comme nous l'avions discuté
-dans le chapitre précédent pour les dérivées, voir @sec:arroni_dervi).
+Une autre source d'erreur qui peut apparaître lorsque nous faisons des interpolations, est l'erreur d'arrondi (comme nous l'avions discuté
+dans le chapitre précédent pour les dérivées, voir @sec:arroni_dervi).
Sans entrer dans les détails de l'évaluation de l'effet de l'erreur d'arrondi, on peut simplement le constater sur l'exemple de $f(x)=\sin(x)$. Nous savons de ce que nous avons discuté dans la section précédente, que l'interpolation de $f(x)=\sin(x)$ va bien se passer, car ses dérivées sont bornées. Sur la @fig:sin_arrondi, on constate que lorsqu'on augmente le nombre de points d'interpolation, on a soudain une augmentation dramatique de l'erreur.
@@ -1152,7 +1151,7 @@ pour s'affranchir du phénomène d'oscillations de Runge. Pour ce faire, il suff
### Interpolation par intervalles de degré un
-On considère toujours un ensemble $N$ de points équidistants
+On considère toujours un ensemble $N$ de points équidistants
entre deux bornes $a,b\in\real$ et $x_i=a+h\cdot i$, avec $h=\frac{b-a}{N}$. Sur chacun des points $x_i$, nous avons une fonction $f(x_i)$ passant par les points $y_i$
$$
f(x_i)=y_i.
@@ -1161,11 +1160,11 @@ On veut construire une fonction $f_h(x)$, continue, qui passe par les points $(x
$$
f_h(x_i)=y_i,\mbox{ et } f_h(x)\in p_1(x),\ x\in[x_i,x_{i+1}],\ i=1,..,N-1.
$$
-Un exemple d'une telle fonction peut se voir sur la @fig:sin_deg1.
+Un exemple d'une telle fonction peut se voir sur la @fig:sin_deg1.
![L'interpolation par intervalle sur la fonction $f(x)=\sin(x)$ (en noir) sur $n=10$ points et $f_h(x)$ (en rouge) qui est un polynôme de degré un par morceaux sur $[x_i,x_{i+1}]$.](figs/sin_deg1.svg){#fig:sin_deg1 width=50%}
-Ici, comme on peut le voir sur le listing ci-dessous qui a été utilisé pour générer la @fig:sin_deg1, l'interpolation par intervalle de degré un est l'interpolation par défaut utilisée par la librairie python, `matplotlib` lors de l'affichage de graphiques.
+Ici, comme on peut le voir sur le listing ci-dessous qui a été utilisé pour générer la @fig:sin_deg1, l'interpolation par intervalle de degré un est l'interpolation par défaut utilisée par la librairie python, `matplotlib` lors de l'affichage de graphiques.
```{.python}
import numpy as np
@@ -1173,10 +1172,10 @@ import matplotlib.pyplot as plt
import math
def echant(x0, x1, n, f):
- dx = (x1-x0)/n
- x = np.arange(x0,x1+dx/10,dx)
- # return (x,np.sin(x))
- return (x,f(x))
+ dx = (x1-x0)/n
+ x = np.arange(x0,x1+dx/10,dx)
+ # return (x,np.sin(x))
+ return (x,f(x))
n = 5
x0 = -5
@@ -1200,14 +1199,13 @@ plt.show()
---
-Remarque +.#
+Remarque +.#
De façon générale cette fonction sera continue, mais pas dérivable (comme on peut le constater sur la @fig:sin_deg1) .
---
-
-Nous voulons à présent savoir si notre fonction $f_h(x)\rightarrow f(x)$, pour $h\rightarrow 0$ (ce qui est équivalent à $N\rightarrow \infty$).
+Nous voulons à présent savoir si notre fonction $f_h(x)\rightarrow f(x)$, pour $h\rightarrow 0$ (ce qui est équivalent à $N\rightarrow \infty$).
En fait un théorème que nous ne démontrerons pas, nous dit que
@@ -1246,7 +1244,7 @@ $$
f_h(x)=a_ix+b_i,\quad x\in[x_i,x_{i+1}].
$$
-On peut même faire un exemple pour voir si cela est observable dans un cas particulier. Soit $f(x)=\sin(x)$, le listing suivant calcule l'erreur pour $N=10, ..., 100$ par pas de $10$.
+On peut même faire un exemple pour voir si cela est observable dans un cas particulier. Soit $f(x)=\sin(x)$, le listing suivant calcule l'erreur pour $N=10, ..., 100$ par pas de $10$.
```{.python}
import numpy as np
@@ -1254,53 +1252,53 @@ import matplotlib.pyplot as plt
import math
def echant(x0, x1, n, f):
- dx = (x1-x0)/n
- x = np.arange(x0,x1+dx/10,dx)
- return (x,f(x))
+ dx = (x1-x0)/n
+ x = np.arange(x0,x1+dx/10,dx)
+ return (x,f(x))
def interp_piece(x, x_p1, y, y_p1, dx):
- a = (y_p1 - y) / (x_p1 - x)
- b = - (x*y_p1 - x_p1 * y) / (x_p1-x)
+ a = (y_p1 - y) / (x_p1 - x)
+ b = - (x*y_p1 - x_p1 * y) / (x_p1-x)
- loc_x = np.arange(x, x_p1, dx)
+ loc_x = np.arange(x, x_p1, dx)
- return (loc_x, loc_x * a + b)
+ return (loc_x, loc_x * a + b)
def interp(x0, x1, n1, n2, f):
- dx = (x1 - x0) / n1
- all_x = np.array([])
- all_y = np.array([])
- for i in range(n1):
- x = x0 + i*dx
- x_p1 = x0 + (i+1)*dx
+ dx = (x1 - x0) / n1
+ all_x = np.array([])
+ all_y = np.array([])
+ for i in range(n1):
+ x = x0 + i*dx
+ x_p1 = x0 + (i+1)*dx
+
+ loc_dx = (x_p1 - x) / n2
+ (loc_x, loc_y) = interp_piece(x, x_p1, f(x), f(x_p1), loc_dx)
- loc_dx = (x_p1 - x) / n2
- (loc_x, loc_y) = interp_piece(x, x_p1, f(x), f(x_p1), loc_dx)
+ if np.size(loc_x) > n1: # sometimes ranges goes too far
+ loc_x = loc_x[0:n1]
+ loc_y = loc_y[0:n1]
- if np.size(loc_x) > n1: # sometimes ranges goes too far
- loc_x = loc_x[0:n1]
- loc_y = loc_y[0:n1]
+ all_x = np.append(all_x, loc_x)
+ all_y = np.append(all_y, loc_y)
- all_x = np.append(all_x, loc_x)
- all_y = np.append(all_y, loc_y)
-
- return (all_x, all_y)
+ return (all_x, all_y)
res = np.array([])
error = np.array([])
for n in range(10,100,10):
- x0 = -5
- x1 = 5
- (x,y) = echant(x0, x1, n, np.sin)
+ x0 = -5
+ x1 = 5
+ (x,y) = echant(x0, x1, n, np.sin)
- (x,y) = interp(x0, x1, n, n, np.sin)
+ (x,y) = interp(x0, x1, n, n, np.sin)
- n *= n
- (x_a,y_a) = echant(x0, x1, n, np.sin)
+ n *= n
+ (x_a,y_a) = echant(x0, x1, n, np.sin)
- error = np.append(error, np.sqrt(np.sum((y_a[0:n]-y)**2) / n))
- res = np.append(res, math.sqrt(n))
+ error = np.append(error, np.sqrt(np.sum((y_a[0:n]-y)**2) / n))
+ res = np.append(res, math.sqrt(n))
plt.loglog(res, error, 'ko-', label="Error")
plt.loglog(res, res**(-2), 'g-', label="$N^{-2}$")
@@ -1324,13 +1322,13 @@ Déterminer la forme de la fonction d'interpolation de degré deux par morceaux.
### Spline
-Les fonctions interpolantes par morceaux comme définies ci-dessus ont un désavantage: elles possèdent des
-dérivées qui sont discontinues. Pour palier à ce problème, on a créé une autre catégorie
-de polynômes d'interpolation: les *spline*. Le but est de trouver une fonction, $s:[a,b]\rightarrow \real$, qui passe par les points $(x_i,y_i)$
+Les fonctions interpolantes par morceaux comme définies ci-dessus ont un désavantage: elles possèdent des
+dérivées qui sont discontinues. Pour palier à ce problème, on a créé une autre catégorie
+de polynômes d'interpolation: les *spline*. Le but est de trouver une fonction, $s:[a,b]\rightarrow \real$, qui passe par les points $(x_i,y_i)$
et qui est deux fois continuement différentiable (elle est dérivable partout deux fois et les dérivées sont continues)[^4].
-Il se trouve qu'une telle fonction est une fonction cubique par morceaux sur chaque intervalle $[x_i,x_{i+1}]$.
-Notons le polynôme défini sur $[x_i,x_{i+1}]$ comme $s_i(x)$. Nous voulons qu'il passe par les points $(x_i,y_i)$ et
+Il se trouve qu'une telle fonction est une fonction cubique par morceaux sur chaque intervalle $[x_i,x_{i+1}]$.
+Notons le polynôme défini sur $[x_i,x_{i+1}]$ comme $s_i(x)$. Nous voulons qu'il passe par les points $(x_i,y_i)$ et
$(x_{i+1},y_{i+1})$ et que sa dérivée soit continue. Notons $p_i$ la dérivée sur les points $x_i$ (la pente de la tangente). On a donc quatre conditions
sur l'intervalle $[x_i,x_{i+1}]$ que $s_i(x)$ doit satisfaire
\begin{align}
@@ -1347,21 +1345,21 @@ s_i(x)&=y_i+\delta y[x_{i+1},x_i](x-x_i)+c(x-x_i)(x-x_{i+1})\nonumber\\
&\quad\quad+d\left((x-x_i)^2(x-x_{i+1})+(x-x_i)(x-x_{i+1})^2\right),
\end{align}
où $c$ et $d$ sont deux coefficients à déterminer. Le deuxième terme (proportionnel à $(x-x_i)(x-x_{i+1})$) est la forme standard des différences divisées de Newton, et le troisième est le terme s'annulant en $x_i$ et $x_{i+1}$ mais étant symétrique
-(on ne veut pas de terme du type $(x-x_i)^2(x-x_{i+1})$ car cela romprait la symétrie). Avec les conditions $s_i'(x_i)=p_i$ et $s'_i(x_{i+1})=p_{i+1}$, on peut trouver
-les coefficients $b$ et $c$ en résolvant
+(on ne veut pas de terme du type $(x-x_i)^2(x-x_{i+1})$ car cela romprait la symétrie). Avec les conditions $s_i'(x_i)=p_i$ et $s'_i(x_{i+1})=p_{i+1}$, on peut trouver
+les coefficients $c$ et $d$ en résolvant
\begin{align}
\delta y[x_{i+1},x_i]-c \left( x_{{i+1}}-x_{{i}} \right) +d \left( x_{{i+1}}-x_{{i}}
\right) ^{2}=p_{{i}},\\
\delta y[x_{i+1},x_i]+c \left( x_{{i+1}}-x_{{i}} \right) +d \left( x_{{i+1}}-x_{{i}}
\right) ^{2}=p_{{i+1}}.
\end{align}
-On trouve aisément que
+On trouve aisément que
\begin{align}
-b&=\frac{1}{2}\left(\frac{p_{i+1}-p_i}{x_{i+1}-x_i}\right),\\
-c&=\frac{1}{2}\frac{1}{(x_{i+1}-x_i)^2}\left(p_{i+1}+p_i-2\delta y[x_{i+1},x_i]\right).
+c&=\frac{1}{2}\left(\frac{p_{i+1}-p_i}{x_{i+1}-x_i}\right),\\
+d&=\frac{1}{2}\frac{1}{(x_{i+1}-x_i)^2}\left(p_{i+1}+p_i-2\delta y[x_{i+1},x_i]\right).
\end{align}
-On constate donc que $b$ est proportionnel à la deuxième dérivée d'une fonction ayant pour dérivée $p_{i}$ et $p_{i+1}$ en $x_i$ et $x_{i+1}$ et que
-$c$ est proportionnel à la deuxième dérivée d'une fonction ayant pour dérivée $p_{i}$, $p_{i+1}$, et $\delta y[x_{i+1},x_i]$ en $x_i$, $x_{i+1}$, et $x_{i+1/2}$.
+On constate donc que $c$ est proportionnel à la deuxième dérivée d'une fonction ayant pour dérivée $p_{i}$ et $p_{i+1}$ en $x_i$ et $x_{i+1}$ et que
+$d$ est proportionnel à la deuxième dérivée d'une fonction ayant pour dérivée $p_{i}$, $p_{i+1}$, et $\delta y[x_{i+1},x_i]$ en $x_i$, $x_{i+1}$, et $x_{i+1/2}$.
On a que $s_i(x)$ est donné par
\begin{align}
s_i(x)&=y_i+\delta y[x_{i+1},x_i](x-x_i)+\frac{1}{2(x_{i+1}-x_i)}\left(p_{i+1}-p_i\right)(x-x_i)(x-x_{i+1})\nonumber\\
@@ -1394,6 +1392,214 @@ $p_0$ et $p_N$ qui doivent être fixés de façon ad-hoc (et on restera donc ave
Pour chacune de ces conditions cela va imposer une valeur pour $p_0$ et $p_N$ (dans le cas 2 elles sont imposées explicitement, dans les deux autres cas elles sont déduites).
+# Optimisation
+
+## Le voyageur de commerce
+
+Le problème du voyageur du commerce est assez simple à formuler. Soit une liste de villes et les distances entre chaque paire connue. Quelle est
+le parcours passant par toutes les villes une seule fois et arrivant dans la ville de départ minimisant la distance totale parcourue?
+
+---
+
+Exemple +.#
+
+Soient les villes $A$, $B$, et $C$ comme sur la @fig:voyage_commerce. On peut comparer tous les chemins possibles
+
+{#fig:voyage_commerce width=70%}
+
++------+------+------+------+------+------+
+| ABC | ACB | BAC | BCA | CAB | CBA |
++------+------+------+------+------+------+
+| 6 km | 6 km | 6 km | 6 km | 6 km | 6 km |
++------+------+------+------+------+------+
+
+On constate d'abord que tous les trajets donnent le même résultat.
+Essayons de comprendre pourquoi. La ville de départ ne change pas la longueur du chemin. On a donc que les chemins
+"cycliques" ABC, BCA, CAB sont les mêmes chemins. De plus, comme les chemins peuvent être parcourus dans les deux sens,
+on a également CBA, ACB et BAC sont les mêmes chemins. Il nous reste qu'un seul chemin possible.
+
+---
+
+---
+
+Exercice +.#
+
+{#fig:voyage_commerce_4 width=70%}
+
+1. Combien de chemins possibles différents existe-t-il pour quatre villes $A$, $B$, $C$, et $D$?
+2. Quel est le chemin le plus cours, si les distances entre les villes sont comme celles représentées sur la @fig:voyage_commerce_4?
+3. Pouvez-vous généraliser le nomre de chemin à $N$ villes?
+
+---
+
+Le nombre total de chemin entre $n$ villes est de $n!$. Comme la ville de départ ne change pas le chemin à emprunter, on peut l'enlever du compte total,
+il nous en reste donc $(n-1)!$. Finalement comme la direction dans laquelle nous parcourons le chemin ne change pas la distance,
+nous pouvons encore diviser ce résultat par deux. Le nombre total de chemins existants est de $(n-1)!/2$. En supposant que le
+calcul pour 5 villes prend $10\mu$s à calculer sur un ordinateur, on peut voir sur le tableau ci-dessous le temps que cela prendrait pour
+10-25 villes. On constante le phénomène *d'explosion combinatoire* rendant le problème incalculable dans un temps raisonnable.
+
++--------------+-----------------+---------------------+
+| Nb de villes | Nb possibilités | Temps (est.) |
++:============:+:===============:+:===================:+
+| 5 | 12 | $10\mu$s |
++--------------+-----------------+---------------------+
+| 10 | 181440 | 0.15s |
++--------------+-----------------+---------------------+
+| 15 | $4.4\cdot10^{10}$ | 10h |
++--------------+-----------------+---------------------+
+| 20 | $6\cdot10^{16}$ | 1600 ans |
++--------------+-----------------+---------------------+
+| 25 | $3\cdot10^{23}$ | 8 milliard d'années |
++--------------+-----------------+---------------------+
+
+La force brute est donc totalement hors de question pour résoudre ce type de problème. En fait il n'existe aucun algorithme **exact**
+calculant la solution en temps polynomial. Les meilleurs algorithmes de ce types sont limités à 200'000 villes jusqu'à aujourd'hui. Néanmoins, il existe différentes heuristiques qui nous donnent une solution approchée (qui se trouve à 2-3% de la solution optimale avecune très grande probabilité)
+et qui fonctionnent avec des millions de villes.
+
+Le problème du voyageur du commerce est un problème **d'optimisation dicrète**, et en particulier **d'optimisation combinatoire**.
+Ce genre de problème est en général beaucoup plus difficile à résoudre que les
+problèmes d'optimisation continue bien que beaucoup plus simples à énoncer.
+Une des difficulté est le cadre théorique (la formulation mathématique générale d'un problème d'optimisation discret) qui est
+assez difficile à appréhender et qui peut être très différent entre les différents problèmes
+d'optimisations.
+
+Le voyageur du commerce peut s'écrire mathématiquement sous la formue suivante.
+Supposons qu'il y ait $N$ villes qui sont numérotées de $1$ à $N$.
+Soit $d_{ij}$ la distance entre la ville $i$ et la ville $j$ (on a que $d_{ij}d_{ji}$) pour $i,j\in 1,..,N$.
+Soit $x_{ij}$ définiti comme
+$$x_{ij}=
+ \left\{\begin{array}{ll}
+ $1,$&$\mbox{ si le chemin va de la ville } i { à la ville }j$\\
+ $0,$&$\mbox{ sinon.}$
+ \end{array}\right.$$
+
+On cherche donc
+\begin{align}
+&\min \sum_{i=1}^N\sum_{j=1, j\neq i}^N x_{ij}d_{ij}, \mbox{ tel que},\\
+&\sum_{i=1,i\neq j}^N x_{ij}=1,\ \forall i=1,..,N,\\
+&\sum_{j=1,neq i}^N x_{ij}=1,\ \forall j=1,..,N.
+\end{align}
+La première contrainte nous dit qu'on ne peut arriver sur une ville $j$ que depuis une seule ville $i$ (impossible d'y arriver de deux façons différentes). La deuxième contrainte
+nous dit que depuis chaque ville on ne peut partir qu'en direction d'une seule autre ville. Ces conditions (déjà assez complexes à formuler) ne suffisent pas pour définir le problème du voyageur de commerce. Il faut y ajouter deux contraintes supplémentaires
+afin de garantir qu'il n'y a qu'un parcours qui couvre toutes les villes en y passant qu'une fois et que le parcours n'est pas une combinaison d'au moins deux parcours disjoints parcourant ensemble toutes les villes. Cette contrainte supplémentaire peut s'écrire (cette contrainte n'est pas exprimable de façon unique)
+\begin{align}
+&u_1=1,\\
+2\leq u_i\leq n,\ \forall i\neq 1,\\
+&u_i-u_j+N\cdot x_{ij}\leq N-1,\quad i\neq 1,\ \j\neq 1,
+\end{align}
+où $u_i$ est une variable supplémentaire associée à chaque ville dont le seul but est
+d'empêcher des cycles à l'intérieur du parcours (imposer qu'on passe qu'une seule fois par chaque ville).
+Il est possible de s'en convaincre de la façon suivante. Cette contrainte supplémentaire nous impose que pour chaque
+$x_{ij}=1$, on a que $u_j\geq u_i+1$. Par conséquent si on rentre dans une partie du parcours où on
+rentre dans des sous-tours, les $u_i$ vont s'incrémenter à l'infini. Bien que les $u_i$ rajoutent des variables au problèmes (et pourraient le complexifier à première vue), on constate que les $u_i$ numérotent
+les villes dans le parcours et peuvent aider à formuler un ordre ou la proximité d'une ville dans la suite.
+
+## La régression linéaire
+
+Lors d'une régression linéaire, le but est de trouver la droite, $y(x)=a\cdot x + b$, qui passe au mieux au travers d'un nuage de $N$ points $(x_i, y_i)$,
+$i=1,...,N$ (voir @fig:reg).
+
+{#fig:reg width=70%}
+
+Pour déterminer l'équation de cette droite, nous devons donc trouver les coefficients $a$ et $b$ tels que la droite
+passe au plus proche des points. Nous devons d'abord définir ce que signifie mathématiquement "passe au mieux par au travaers du nuage de points".
+Une façon de mesurer la "qualité" d'une droite est de mesurer la somme des distances au carré entre les points $(x_i,y_i)$ et
+la droite $y(x)=a\cdot x + b$ pour des valeurs de $a$ et $b$ données, soit
+$$
+E(a,b)=\sum_{i=1}^N (y(x_i)-y_i)^2.
+$$
+Nous cherchons par conséquent à minimiser $E(a,b)$ sous la contrainte que $y(x)$ est une droite. Pour simplifier encore plus le problème mathématique,
+nous pouvons rajouter comme contrainte que la droite $y(x)$ passe par le point $(0,0)$, on a donc que $y(x)=a\cdot x$ (l'ordonnée à l'origine est nulle, $b=0$) et que
+$$
+E(a)=\sum_{i=1}^N (y(x_i)-y_i)^2,
+$$
+est indépendant de $b$. En résumé nous cherchons à résoudre le problème mathématique
+\begin{align}
+&\min_{a\in\real} E(a) = \min_{a \in\real} \sum_{i=1}^N (y(x_i)-y_i)^2,\\
+&\mbox{où }y(x)=a\cdot x, \quad \mbox{(contrainte)}.
+\end{align}
+On peut réécrire la fonction $E(a)$ comme
+\begin{align}
+E(a)&=\sum_{i=1}^N \left(y^2(x_i)-2\cdot y_i\cdot y(x_i)+y_i^2\right)=\sum_{i=1}^N \left(a^2\cdot x_i^2-2\cdot a\cdot x_i\cdot y_i+y_i^2\right),\nonumber\\
+ &=a^2\sum_{i=1}^Nx_i^2 + 2a\sum_{i=1}^Nx_iy_i+\sum_{i=1}^Ny_i^2.
+\end{align}
+Les $x_i$ et $y_i$ étant connus, nous cherchons $a$, tel que $E(a)$ soit minimal. $E(a)$ est en fait l'équation d'une parabole: elle a la forme
+$$
+E(a)=B\cdot a^2-2C\cdot a + D,
+$$
+avec $B=\sum_{i=1}^Nx_i^2$, $C=\sum_{i=1}^Nx_iy_i$, et $D=\sum_{i=1}^N y_i^2$. $B$ étant forcément positif cette parabole sera **convexe** et donc
+nous sommes assurés qu'il existe un minimum pour $E(a)$. Une façon de déterminer $a$, tel que $E(a)$ est minimal est d'utiliser la dérivée.
+On a l'équation $E'(a)=0$ à résoudre:
+\begin{align}
+E'(a)&=0,\nonumber\\
+2\cdot B\cdot a-2\cdot C&=0,\nonumber\\
+a &= \frac{C}{B}=\frac{\sum_{i=1}^Nx_iy_i}{\sum_{i=1}^Nx_i^2}.
+\end{align}
+
+---
+
+Exemple +.#
+
+Soient les 4 points $(0, 0.1)$, $(1, 0.3)$, $(2, 0.3)$ et $(3, 0.4)$. La fonction d'erreur $E(a)$ s'écrit
+$$
+E(a)=14\cdot a^2-4.2\cdot a + 0.35.
+$$
+On peut la représenter comme sur la @fig:e_a et on constate qu'elle possède un minimum proche de $a=0$.
+
+![La fonction $E(a)=14a^2-4.2a+0.35$ pour $a\in[-1,1]$. On voit bien qu'elle possède un minimum proche de $a=0$.](figs/e_a.svg){#fig:e_a width=70%}
+
+En résolvant $E'(a)=0$, on obtient $a=4.2/24=0.15$. On a que l'équation de la droite passant par $(0,0)$ et au plus proche de nos 4 points est
+$$
+y(x)=0.15\cdot x.
+$$
+On peut observer le résultat de la régression sur la @fig:regression_ex, où on voit les 4 points (en noir), ainsi que la droite obtenue (en trait bleu).
+
+{#fig:regression_ex width=70%}
+
+---
+
+La régression linéaire est un problème **d'optimisation continu**. On voit que contrairement au problème du voyageur du commerce,
+l'ensemble des solutions est $a\in\real$ et non une suite discrète de villes. Ce genre de problème, bien que possédant un espace de recherche infini,
+est bien souvent plus simple à résoudre, car il possède un cadre théorique mieux défini.
+
+Pour le résoudre, nous avons commencé, comme pour le problème du voyageur du commerce par faire un modèle mathématique.
+Nous avons construit une fonction à minimiser, $E(a)$, et ajouté une contraite, la forme de $y(x)$. Puis, il a suffit de trouver le minimum de $E(a)$
+sous la contrainte et le tour était joué.
+
+## L'optimisation mathématique
+
+Suite à ces deux exemples, nous allons essayer de définir de façon assez théorique comment formuler mathématiquement un probélème d'optimisation.
+Il existe comme on vient de le voir deux types disctincts de problèmes d'optimisation:
+
+1. L'optimisation continue.
+2. L'optimisation discrète (ou optimisation combinatoire).
+
+### L'optimisation continue
+
+Soit $f:\real^n\rightarrow\real$ une fonction objectif (ou fontion de coût), on cherche $\vec x_0\in\real^n$, tel que $f(\vec x_0)\leq f(\vec x)$ pour $\vec x$ certaines conditions: les contraintes qui sont en général des égalités strictes ou des inégalités qui peuvent s'exprimer de la façon suivante.
+Soient $m$ fonctions $g_i:\real^n\rightarrow\real$
+\begin{align}
+&g_i(\vec x)\leq 0,\quad i=1,...,m.
+\end{align}
+Si $m=0$ on a à faire à un problème d'optimisation sans contraintes. On peut résumer tout cela comme
+\begin{align*}
+&\min_{\vec x\in\real^n}f(\vec x),\\
+&g_i(\vec x)\leq 0,\quad i=1,...,m,\\
+&\mbox{pour }m\geq 0.
+\end{align*}
+Les contraintes limites l'espace des solutions et forment un sous-ensemble, noté $A$, de $\real^n$ ($A\subseteq\real^n$).
+
+Une des difficultés pour déterminer le minimum d'une fonction coût est l'existence de plusieurs minima locaux.
+Un **minimum local**, $\vec x^\ast\in A$, est tel que pour une région proche de $\vec x^\ast$, on a que $f(\vec x)\geq f(\vec x^\ast)$. Une fonction particulièrement problématique où il est
+très difficile de trouver le minimum global est une fonction de Ackley. En une dimension, elle est de la forme (voir la @fig:ackley)
+$$
+f(x)=-20e^{-0.2*\sqrt{0.5x^2}}-e^{0.5(\cos(2\pi x))}+e+20.
+$$
+
+{#fig:ackley width=70%}
+
[^1]: On pourrait, de façon similaire, utiliser la formule de différences finies en avant ou en arrière (ou un mélange des deux).
[^2]: Comme ce polynôme passe par les points $(x_0,y_0)$, $(x_1,y_1)$, ..., $(x_m,y_m)$, il est unique, c'est donc exactement le même que celui exprimé avec les $\{a_i\}_{i=0}^m$.
[^3]: En fait $\frac{f^{(n+1)}(\xi)}{(n+1)!}=\delta^{n+1}[x_0,\dots,n_n]$.
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new file mode 100644
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index 0000000000000000000000000000000000000000..f8ef05f762f896c944641d39bf44fb70167f730d
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diff --git a/python/ackley.py b/python/ackley.py
new file mode 100644
index 0000000000000000000000000000000000000000..6445ae4a9c09c860d8d7821d6a48d46082c42040
--- /dev/null
+++ b/python/ackley.py
@@ -0,0 +1,10 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+N=30
+x = np.arange(-30,30+1/N,1/N)
+
+y = -20*np.exp(-0.2*np.sqrt(0.5*x**2))-np.exp(0.5*(np.cos(2*np.pi*x)))+np.exp(1)+20
+
+plt.plot(x,y)
+plt.show()
\ No newline at end of file