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### Quelques livres d'exercices corrigés pour la licence de maths

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### Licence STS Mention MATH L3 ATN

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x = cos(x) = 0 f (x) = 1 n x2] ; [n2 2 2 2
1cos(x)> 0 f (x) = ! 0 fn 2 nn cosxe n!+1
J =I = [ ; ]
2 2

f : [ ; ] ! R
2 2
0 x> 0;
x ! 1 x =
2
P
x = f (x)n2
] ; [ x cosx> 0 n02 2 P P21 1 n 1n n ! 0 f (x)0 2 2 2 n 2n cosx n cosxn ne e n!+1
p
2K = [ ; ] cosx n nn0 04 4 2 P
1 1
pf (x) f Kn 2 n2n 2=2 ne
fn
s
20 n sinxf (x) = n2 0n n cosxe P
0 1 0n n x2 K f (x) f (x)0 2n nn
K s K
P
0f (x)n
0s(x)
P 20 sinx 0 n sinxs(x) x2 [0; ] s(x) = 2n cosxe 4 e
sinx sinx 1 10s(x) cosx< 1) >
cosx cosxe e e e
f [ 4;4]
Z X 1
Sf(x) = b sinnx b = f(x)sinnxdx:n n
n1
R2f(x)sinnx b = x(x )sinnx dxn 0
Z 2
b = (2x )cosnxdx:n
n 0
Z Z h i2 2 cosnx
(2x )cosnxdx = sinnxdx =
n n n 00 0
4n n n2n
8
b = 0 b =2n 2n+1 3(2n+1)
f f(x) = Sf(x)
x2R
P n( 1) 8x = Sf(x) 32 n0 (2n+1) 2 f =
2 4
n 3X ( 1)
= :
3(2n+1) 32
n0
Z 2X Xb1 322n+12f (x)dx = = :
2 62 2 (2n+1) n0 n0
Z Z 5 4 2 3 5x 2x x 2 2 2f (x)dx = 2 x (x ) (x)dx = 2 + = :
5 4 3 15 0 0
6X 1
= :
6(2n+1) 960
n0
n+1 2(n+1)x n 1
j j ! jxj
2 n n!+1(n+1) 1 nx
R = 1
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2n2 n 1

n n nx 1 x x
= :
2n 1 2 n+1 n 1
P P n1 xnx = = ln(1
n0 1 x n1 n
x) x2] 1;1[

n n+1 n 2X X Xx 1 x 1 x 1 x
= = = ln(1 x) x
n+1 x n+1 x n x 2
n2 n2 n3
n n 1 nX X Xx x x
=x =x = xln(1 x):
n 1 n 1 n
n2 n2 n1
0 2f(x) 1 ln(1 x) x 1 1 1 1 x 2= 1 +xln(1 x) = + ln(1 x)( 1 1=x )
x 2 x 2 2 2 1 x x

21 1+x
f(x) = 1+x+ ln(1 x)
2 x
P
n nx = 1 > 12n2 n 1 n 1
n n 1
= >
2n 1 (n 1)(n+1) n+1
P
n
2n2 n 1
P n( 1) n
x = 1 2n2 n 1
oin3remarqueMathq,s?rieonaussiobtienttlelaainsis?ried?duitAud'o?onOnour.etLicenceg?n?raltoutergenours?riepAu.NotonsEnergenremarquanoint.queiOns?rietsurobtienenaestdonctermeond'unepdiveutte,?crirelaontg?om?triqueps?rieersla2.testt?grandivinte.qu'enpelonst2008uequi,conobtergenenparlacrit?rePconcalculerer-onpquelesestaltern?estermeRappestlavd?rivte?critleOnde?evdegence.our(sons?riesg?n?ralvtermeettendcedernier0).3.