114 lines
2.1 KiB
Python
114 lines
2.1 KiB
Python
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######Exercice 1 ######
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from math import*
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##Question 1:##
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def multiplesde7(n):
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l=[]
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for i in range(n):
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if i%7==0:
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l+=[i]
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return l
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print(multiplesde7(100))
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##Question 2##
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Un nombre (n) est premier si aucun nombre inferieur a sqrt(n) en divise (n).
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On vérifie ici ce théoreme en testant sucessivement la divisibilité des (i) par (n) avec (i) variant de 2 a sqrt(n)
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##Question 3##
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def isprime(n):
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s=int(sqrt(n))
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t=True
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i=2
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while i<=s and t==True:
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if n% i==0:
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t=False
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i+=1
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return(t)
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def listepremier(n):
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l=[]
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for i in range(2,n+1):
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if isprime(i):
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l+=[i]
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return l
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##Question 4##
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def premier():
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p=[]
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n=2
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while len(p)!=100:
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if isprime(n):
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p+=[n]
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n+=1
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return p
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##Question 5##
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def jumeaux():
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l=[]
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for i in range(2,3000):
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if isprime(i) and isprime(i+2):
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l+=[(i,i+2)]
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return l
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##Question 6#
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n=0
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while 1<2:
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a=2**(2**n) +1
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if isprime(a) == False:
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print(False, n)
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break
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n+=1
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## Question 7 BONUS (à traiter plus tard) ##
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######Exercice 2######
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##Question 1##
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a=1
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print(a)
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for i in range(100):
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b=(1/2)*(a+(2/a))
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print(b)
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a=b
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##Question 2##
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a=1
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print(a)
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for i in range(10):
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b=(1/2)*(a+(2/a))
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print((b)**2 - 2)
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a=b
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# les valeurs tendent très rapidement vers 0 car sqrt(2)**2 - 2 = 0
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##Question 3#
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a=1
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l=[1]
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for i in range(10):
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b=(1/2)*(a+(2/a))
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l+=[b**2-2]
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a=b
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i=0
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while 1<2:
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if l[i]<0.00002
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print(i)
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break
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i+=1
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# On cherche a encadrer |Un**2 -2|
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#Sachant que la condition d'arrêt de départ est :
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#|Un-sqrt(2)|<10**-5
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#Or x->sqrt(x) est 1/2 lispsiptienne => |f(y)-f(x)|<1/2 |x-y| On pose x=Un**2 et y=2=(sqrt(2))**2
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#Ainsi, on a |Un -sqrt(2)|<1/2|Un**2-2|
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#Or on souhaite avoir |Un -sqrt(2)|<10**-5. Une proposition plus forte consiste à dire:
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# |Un-sqrt(2)|<1/2|Un**2-2|<10**-5
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#<=> 2*|Un-sqrt(2)|<|Un**2-2|<2*10**-5
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#D'ou la condition d'arrêt : |Un**2-2|<2*10**-5
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