Calcul de développement par équation di érentielle

p∈N
f(x) =
+∞
X
n=0 n+p
pxn
f(x) (1 −x)f0(x)
x7→ Z+∞
0
e−t2sin(tx) dt
f:x7→ px+√1 + x2
f(x) = arcsin x
√1−x2
f
f(0) = 0
f
f:x7→ arccos x
√1−x2
x7→ sh (arcsin x)
f:x7→ ex2/2Zx
0
e−t2/2dt
f
f
f:x > −17→ Z+∞
1
e−t
x+tdt
f
f(x) = ch(x) cos(x)
f
y(4) + 4y= 0
k > 0
f(x) = Z1
0
tksin(xt) dt
fR
fR
∀x∈R, xf0(x)+(k+ 1)f(x) = sin x
xy0+ (k+ 1)y= sin x

(E): ty0+y= 3t2cos(t3/2)
v(E)
(E)R∗
+
v

n+p
p=n(n−1) . . . (n−p+ 1) p!∼1 p!np
f
]−1 ; 1[ fC∞
f0(x) =
+∞
X
n=1 n+p
pnxn−1
(1 −x)f0(x) =
+∞
X
n=0
(n+ 1)n+p+ 1
pxn−
+∞
X
n=0
nn+p
pxn=
+∞
X
n=0
αnxn
αn= (n+ 1)n+p+ 1
p−nn+p p
αn= (n+p+ 1)n+p
p−nn+p p= (p+ 1)n+p p
(1 −x)f0(x)=(p+ 1)f(x)
(1 −x)y0= (p+ 1)y
]−1 ; 1[
y(x) = C
(1 −x)p+1
C∈R
f(0) = 1
f(x) = 1
(1 −x)p+1
sin(tx) =
+∞
X
k=0
(−1)k
(2k+ 1)!t2k+1x2k+1
Z+∞
0
t2k+1e−t2=k!
2
Z+∞
0
(−1)k
(2k+ 1)!t2k+1x2k+1
dt≤k!
2(2k+ 1)! |x|2k+1
Z+∞
0
e−t2sin(tx) dt=
+∞
X
k=0
(−1)kk!
2(2k+ 1)!x2k+1
x∈R
t7→ e−t2sin(tx)R+
d
dxe−t2sin(tx)
≤te−t2
t7→ te−t2R+
f:x7→ Z+∞
0
e−t2sin(tx) dt
C1
f0(x) = Z+∞
0
te−t2cos(tx) dt
f0(x) = 1
2−1 2xf(x)
fR
2y0+xy = 1
f f(0) = 0

2y0+xy = 1 y(0) = 0
g:x7→
+∞
X
k=0
(−1)kk!
2(2k+ 1)!x2k+1
R= +∞
f g R
2y0+xy = 1 y(0) = 0
f
g
f(x) =
+∞
X
k=0
(−1)kk!
2(2k+ 1)!x2k+1
x∈R
fC∞
f0(x) = 1
2√1 + x2f(x)
f00(x) = −x
2(1 + x2)3/2f(x) + 1
2√1 + x2f0(x)
f
(1 + x2)y00(x) + xy0(x)−1
4y(x) = 0
y(0) = 1 y0(0) = 1/2
PanxnR > 0S
x∈]−R;R[
S(x) =
+∞
X
n=0
anxn, S0(x) =
+∞
X
n=1
nanxn−1
S00(x) =
+∞
X
n=2
n(n−1)anxn−2=
+∞
X
n=0
(n+ 2)(n+ 1)an+2xn
(1 + x2)S00(x) + xS0(x)−S(x)/4=0
+∞
X
n=0 (n+ 2)(n+ 1)an+2 + (n2−1/4)anxn= 0
∀n∈N, an+2 =−1
4
(2n+ 1)(2n−1)
(n+ 2)(n+ 1) an
S(0) = 1 S0(0) = 1/2
a2p=(−1)p
24p−1
(4p−2)!
((2p)!)((2p−1)!) a2p+1 =(−1)p 24p
(4p−1)!
(2p+ 1)!(2p−1)!
Pa2px2pPa2p+1x2p+1
an+2
an
=(2n+ 1)(2n−1)
4(n+ 2)(n+ 1) →1
R≥1
(1 + x2)y00(x) + xy0(x)−1
4y(x)=0
y(0) = 1 y0(0) = 1/2
f
f
f]−1 ; 1[
(1 −x2)f0(x)−xf(x)=1 f(0) = 0
x7→ 1
√1−x2]−1 ; 1[
x7→ arcsin x
]−1 ; 1[ f
f f
f(x) =
+∞
X
n=0
anx2n+1

f0(x) = P+∞
n=0 (2n+ 1)anx2n
(1−x2)f0(x)−xf(x) =
+∞
X
n=0
(2n+ 1)anx2n−
+∞
X
n=0
(2n+ 1)anx2n+2 −
+∞
X
n=0
anx2n+2
(1 −x2)f0(x)−xf(x) = a0+
+∞
X
n=0
((2n+ 3)an+1 −(2n+ 2)an)x2n+2 = 1
a0= 1 ∀n∈N, an+1 =2n+ 2
2n+ 3an
an=22n(n!)2
(2n+ 1)!
x6= 0
an+1x2n+3
anx2n+1 
=4(n+ 1)2x2
(2n+ 3)(2n+ 2) →x2
R= 1
f
R≥1
f(x) =
+∞
X
n=0
anxn]−1 ; 1[
f f
(x2−1)y0+xy −1=0
(x2−1)f0(x) + xf(x)−1 = −(a1+ 1) +
+∞
X
n=1
(nan−1−(n+ 1)an+1)xn
a1=−1∀n≥1, an+1 =n
n+ 1an−1
a0=f(0) = π/2
a2p=(2p−1)
2p× ··· × 1 2a0=(2p)! (2pp!)2
π
2a2p+1 =2p 2p+ 1 ··· 2 3a1=−(2pp!)2 (2p+ 1)!
f:x7→ sh (arcsin x)
f
(1 −x2)y00 −xy0−y= 0
y(0) = 0 y0(0) = 1
PanxnR > 0S
S]−R;R[
a0= 0, a1= 1 ∀n∈N, an+2 =n2+ 1
(n+ 2)(n+ 1)an
a2p= 0 a2p+1 =Qp
k=1 (2p−1)2+ 1
(2p+ 1)!
Panxn
x6= 0 up=a2p+1x2p+1
up+1
up
→ |x|2
S
(1 −x2)y00 −xy0−y= 0
y(0) = 0 y0(0) = 1
f
]−1 ; 1[