CCP 07 Maths 1, filière MP (4h) EXERCICE

f2→f(x, y) = x+y
(1 + x2)(1 + y2)
F= [0,1] ×[0,1] f F
= sup
(x,y)∈F
f(x, y)
Ω =]0,1[×]0,1[
M= 3√3
8
f3√3
8
x∈]0,+∞[t7→ −ttx−1]0,+∞[
x∈]0,+∞[ Γ(x) = Z+∞
0
−ttx−1dt
x∈]0,+∞[ Γ(x+1) Γ(x) Γ(n)
n
]1,+∞[ζ(x) =
+∞
P
n=1
1
nx
ζ(2) = π2
6ζ(4) = π4 90 p ζ(p)qπpq
ζ(p)
ζ(p)
n x x > 1Rn(x) =
+∞
P
k=n+1
1
kx=ζ(x)−
n
P
k=1
1
kx
n x x > 1Rn(x)61
(x−1)nx−1
p>2ε > 0n
n
P
k=1
1
kp−ζ(p)
6ε
ζ(7) 10−6
(fn) [a, b]f
[a, b] Zb
a
fn(x) dx!Zb
a
f(x) dx

Zb
a
f(x) dx
(fn) [0,1]
f[0,1] Z1
0
fn(x) dx
Z1
0
f(x) dx
fn(x)
(fn) [0,1]
Z1
0
fn(x) dxZ1
0
f(x) dx(fn)
[0,1]
(fn)n>1I= [0,+∞[
fn(x) = xn−x
n!
[a, b]I
I
(fn)I
f I
p x ∈I|f(x)|61 + |fp(x)|
f I
ZI
fn(x) dxZI f(x) dx `(I)
I
(fn)I
f I ϕ
I n x ∈I|fn(x)|6ϕ(x)f
ZI
fn(x) dxZI f(x) dx
fn
I f I
I
(fn)f
lim
n→+∞Z+∞
0
sin( x
n)
1 + x2dx
Pfn[a, b] [a, b]
PZb
a
fn(x) dx
+∞
X
n=0 Zb
a
fn(x) dx=Zb
a +∞
X
n=0
fn(x)!dx
a0
2+P
n>1
[ancos(nx) + bnsin(nx)] (an)
(bn)
2π

2π
P1
√nsin(nx)
2π
n p Z2π
0
cos(px) cos(nx) dx=
Z2π
0
sin(px) sin(nx) dx=(0n6=p
π n =p6= 0 Z2π
0
sin(px) cos(nx) dx= 0
PfnI
f I PZI|fn(x)|dx f
IPZI fn(x) dx
+∞
X
n=0 ZI
fn(x) dx=ZI +∞
X
n=0
fn(x)!dx
Pan
Panxn
n!n>0
f
x7→ f(x)−x[0,+∞[Z+∞
0
f(x)−xdx
(P(−1)nxn)n>0
I= [0,1[
(P(−1)nxn)n>0I= [0,1[
(PZ1
0
(−1)nxndx)n>0
+∞
X
n=0 Z1
0
(−1)nxndx=Z1
0 +∞
X
n=0
(−1)nxn!dx
PfnI
f I
fnI f I
n x ∈I Sn(x) =
n
P
k=0
fk(x)
(Sn)
(PZI fn(x) dx)n>0
+∞
P
n=0 ZI
fn(x) dx=ZI +∞
X
n=0
fn(x)!dx
Z+∞
0
t3
t−1dt
t∈]0,+∞[1
t−1=
−t
1−−t
uλ
uλ=8πhc
λ5
1
hc
kBλT −1

h kBc λ
T
u
u=Z+∞
0
uλdλ
M M u M =c
4u
M=σT 4σ=2π5(kB)4
15h3c2
x x > 1Z+∞
0
tx−1
t−1dtΓ(x)ζ(x)
Z+∞
0
t
t−1dtZ+∞
0
t6
t−1dt