(un)
un+1 =f(un)f I
f(un)
f
①f(un)u1−u0
②f(u2n) (u2n+1)
u1> u0
P(n)un+1 >unP(k)uk+1 >ukf
uk+1 >uk=⇒f(uk+1)>f(uk) =⇒uk+2 >uk+1
pn=u2nin=u2n+1
pn+1 =f◦f(pn)in+1 =f◦f(in)
f fof p i p
u2>u0f u36u1i16i0i
R♥C
(un)
unI n
①∀x∈I, f(x)>x(un)
②∀x∈I, f(x)6x(un)
①I f u0∈I
P(n)n n ∈N
P(n) : un+1 >un
u1=f(u0)>u0P(0)
k∈NP(k)
uk+2 =f(uk+1)>uk+1 P(k+ 1)
nNun+1 >un
(un)②
R♥C
f(un)
un+1 =f(un)
f I (un)ℓ
f f(ℓ) = ℓ
(un)ℓlim
n→+∞un= lim
n→+∞un+1 =ℓ f
I ℓ lim
x→ℓf(x) = f(ℓ)
lim
n→+∞f(un) = f(ℓ) (un)f(ℓ) = ℓ
R♥C
a f f
a ǫ lim
x→aǫ(x) = 0
f(x) = f(a) + (x−a)f′(a) + ǫ(x)
x a ǫ(x) (x−a)f(x)f(a)
R♥C
fRf′=ff(0) = 1
fRf′=f(0) = 1
∀x∈Rf(−x)f(x) = 1 f
φ∀x∈Rφ(x) = f(−x)f(x)φ′= 0
φ(0) = 1
f g R
f′=f g′=g f(0) = g(0) = 1
ψRψ=f
gψ′= 0 ψ(0) = 1
ψ f =g
R♥C
a b n
exp(−a) = 1
exp(a)
exp(a+b) = exp(a)×exp(b)
exp(a−b) = exp(a)
exp(b)
exp(a×n) = (exp(a))n
n
b∈Rχ x b
χ(x) = exp(x+b)
exp(x).
χ′= 0 χ(0) = exp(b)xexp(x+b) = exp(x) exp(b)
b
aP(n) exp(a×n) = (exp(a))n
nP(n)n∈Nexp(a)6= 0 exp(a)0= 1 = exp(0)
R♥C
Cexp
∀x∈R,ex>x+ 1
φ φ(x) = ex−x−1
RφRφ′(x) = ex−1
ex−1>0⇐⇒ ex>e0⇐⇒ x>0
φ φ(0) = 0 φ
R
R♥C
z z′C
z=z
z+z′=z+z′
zz′=z×z′
1
z′=1
z′z′6= 0
z
z′=¯z
z′z′6= 0
zn= (¯z)nn∈Z, z ∈C∗
z z′z=a+ ib z′=a′+ ib′
zz′= (a+ ib)(a′+ ib′) = aa′−bb′+ i(ab′+a′b) = aa′−bb′−i(ab′+a′b)
z×z′= (a−ib)(a′−ib′) = ···=aa′−bb′−i(ab′+a′b) = zz′
1
z′=z′
z′z′=1
z′z′×z′
z′z′∈R+
1
z′=1
z′z′×z′=1
z′z′×z′=1
z′z′×z′=1
z′
z
z′=z×1
z′
n∈N∗
n∈N∗z−n=1
zn
z−n= 1/zn= 1/zn= 1/zn=1
zn
= (z−1)n=z−n
R♥C
z z′C
|¯z|=|z|
|zz′|=|z| × |z′|
|1/z′|= 1/|z′|pour z′6= 0
|z/z′|=|z|/|z′|pour z′6= 0
|zn|=|z|npour n ∈Z∗
|zz′|=pzz′×zz′=pzz′×zz′=pzz ×z′z′=√zz ×pz′z′=|z| × |z′|
R♥C
a b
cos(a+b) = cos acos b−sin asin b
sin(a+b) = sin acos b+ sin bcos a
z z′
①arg zz′= arg z+ arg z′[2π]
②arg ¯z=−arg z[2π]
③arg 1
z=−arg z[2π]
④arg z
z′= arg z−arg z′[2π]
⑤arg zn=narg z[2π]
z z′
z=|z|(cos θ+ i sin θ)z′=|z′|(cosθ′+ i sin θ′)
①
①zz′=|z|(cos θ+ i sin θ)× |z′|(cosθ′+ i sin θ′)
zz′=|z| × |z′|(cos θcos θ′−sin θsin θ′+ i(sin θcos θ′+ sin θ′cos θ))
=|z| × |z′|(cos(θ+θ′) + i sin(θ+θ′))
2π
|zz′|=|z| × |z′|arg zz′=θ+θ′= arg z+ arg z′[2π]
②(Ox)M z M′
¯z M M′
(~u ;−−−→
OM′) = −(~u ;−−→
OM)②
③1/z = (1/|z|2)ׯz. ① ② 1/|z|2∈R+arg(1/|z|2) = 0
④① ③ z/z′=z×(1/z′)
⑤n①
R♥C
A B zAzB
C D
①arg(z−−→
AB ) = arg (zB−zA) = (~u ;−−→
AB)
②arg zB−zA
zD−zC= (−−→
CD;−−→
AB)
①M(zB−zA)−−→
OM =−−→
AB
arg (zB−zA) = (~u ;−−→
OM) = (~u ;−−→
AB)
②
arg zB−zA
zD−zC= arg(zB−zA)−arg(zD−zC)
= arg(z−−→
AB )−arg(z−−→
CD)
= (~u ;−−→
AB)−(~u ;−−→
CD)
= (−−→
CD;~u ) + (~u ;−−→
AB)
= (−−→
CD;−−→
AB)
R♥C
A(xA;yA;zA)P
ax +by +cz +d= 0 d(A;P)
d(A;P) = |axA+byA+czA+d|
√a2+b2+c2
d(A;P) = AH H A PBP
−−→
AB ·~n =−−→
AH ·~n
AH =|−−→
AB ·~n |
k~n k
d=−(axB+byB+czB)
R♥C
R
y′=ay fk
fk(x) = keax
k
①fkk
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