Solution-TestTP3-2025

Contenu

• Exercice 1 (data)
• Exercice 2 (série convergeante)
• Exercice 3 (Table)
• Exercice 4 (for + break)

Exercice 1 (data)¶

Ecrire le résultat de la commande print :

#data =[2, 5, 4, 6]  #V1
data =[2, 8, 3, 2]   #V2
n = len(data)
a = sum(data[i] for i in range(n))/n
b = sum(val for val in range(n))/n
c = sum(data)/n
print(f'{a = }, {b = }, {c = }')

a = . . . . . . , b = . . . . . , c = . . . . .

In [1]:

data =[2, 5, 4, 6]   #Variante 1
data =[2, 8, 3, 2]    #Variante 2

n = len(data)

a = sum(data[i] for i in range(n))/n
b = sum(val for val in range(n))/n   # rien à voir avec data
c = sum(data)/n

print(f'{a = }, {b = }, {c = }')

a = 3.75, b = 1.5, c = 3.75

Exercice 2 (série convergeante)¶

1. Ecrire en une seule exression compacte, avec sum() une commande qui permet de calculer la somme suivante : $$ V1 : S = \sum_{k=0}^n \frac{1}{k^2 + \sqrt{k+1}} \quad,\quad V2 : S = \sum_{k=0}^n \frac{1}{\sqrt{2+k} + k^2 } $$
2. Réécrire l'expression en utilisant une boucle for

In [2]:

n = 1_000_000
print(sum(1/(k*k + (k+1)**0.5) for k in range(n))) # Variante 1
print(sum(1/((2+k)**0.5 + k*k) for k in range(n))) # Variante 2

s = 0   # Variante 1
for k in range(n):
    s += 1/(k*k + (k+1)**0.5)
print(s)

s = 0   # Variante 2
for k in range(n):
    s += 1/((2+k)**0.5 + k*k)
print(s)



# La série converge vers 1.946538....

1.9465375317080276
1.5945707892406802
1.946537531708018
1.5945707892406684

In [3]:

# Liste des 10 premières valeurs (non demandé)
n=1_000_000
l = [1/(k*k + (k+1)**0.5) for k in range(n)]

print([f'{v:0.2f}' for v in l[:10]])
sum(l)

['1.00', '0.41', '0.17', '0.09', '0.05', '0.04', '0.03', '0.02', '0.01', '0.01']

Out[3]:

1.9465375317080276

Exercice 3 (Table)¶

Le code suivant permet de construire un tableau table de n lignes et m colonnes.

table = []
for i in range(4):
    table.append([])
    for j in range(2):
        table[i].append((i+1)+0.1*(j+1))

Ecrire ce que vont afficher les commandes suivantes :

+ `print(len(table)) . .  . . . . . . . . . .`  
+ `print(len(table[0])) . . . . . . . . . . . .`  
+ `print(table[2]) . . . . . . . . . . . . . . .`
+ `print(table[2][1]) . . . . . . . . . . . . . . .`

In [4]:

# Variante 1
table = []
for i in range(4):
    table.append([])
    for j in range(2):
        table[i].append( (i+1) + 0.1 * (j+1) )
print(f'{len(table)    = }')
print(f'{len(table[1]) = }')
print(f'{table[2]      = }')
print(f'{table[3][1]   = }')

len(table)    = 4
len(table[1]) = 2
table[2]      = [3.1, 3.2]
table[3][1]   = 4.2

In [5]:

# Variante 2
table = []
for i in range(5):
    table.append([])
    for j in range(2):
        table[i].append( (i+1) * 10 + (j+1) )
#[print(l) for l in table]


print(f'{len(table)    = }')
print(f'{len(table[2]) = }')
print(f'{table[3]      = }')
print(f'{table[4][1]   = }')

len(table)    = 5
len(table[2]) = 2
table[3]      = [41, 42]
table[4][1]   = 52

Exercice 4 (for + break)¶

Le code ci-dessous permet de calculer la somme de quelques nombres impairs. Ecrire ce que va afficher la commande print

s = 0
for i in range(5):
    j = 2*i+1
    if j>5 : break
    s += j
print(f'{i = }, {j = }, {s = }')

i = . . . . ., j = . . . . ., s = . . . . .

In [6]:

# Variante 1
s = 0
for i in range(5):
    j = 2*i+1
    if j>5 : break
    s += j
print(f'{i = }, {j = }, {s = }')

i = 3, j = 7, s = 9

In [7]:

# Variante 2
s = 0
for i in range(6):
    j = 2*(i+1)
    if j>6 : break
    s += j
print(f'{i = }, {j = }, {s = }')

i = 3, j = 8, s = 12