Approfondissement no 3et5: Partie entière et fractionnaire

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x x bxck∈Z
bxc=k6x<k+ 1 = bxc
x{x}
{x}=x− bxc.
bAc=B
bAc=B
B
B6A A < B + 1.
bAc6BbAc>B
bxc6x < bxc+ 1 x−1<bxc6x.
x∈R
{x} ∈ [0; 1[.
x∈Rbxcx
{x} {x}=x−bxclog z z > 0
log z=ln z
ln 10
∀z > 0,10log z=z
∀z > 0,∀z0>0,log(z.z0) = log z+ log z0
∀a∈R,log(10a) = a
log(100) = 2 log(√10) = 1
2

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x
x= 10{log x}.10blog xc
x
x > 0 (10{log x},blog xc) (α, n)
[1; 10[×Zx=α.10n
x > 0γ=b10{log x}cγ∈ {1; 2; ...; 9}
γ x
k16k69pk= log 1 + 1
k9
P
k=1
pk= 1
(pk){1; 2; ...; 9}
X
Y={log X}[0; 1[
k∈ {1; 2; ...; 9} b10Yc=k⇐⇒ k610Y< k + 1
Γ = b10{log X}cX
Γ
Y g R
(h1)g M a0∈R
(h2)g]− ∞;a0] [a0; +∞[
y∈Rn∈Z{y}={y−n}
n∈Z{Y} {Y−n}
˜g Y − ba0c
˜g˜a0∈[0,1[
˜g(h1) (h2) ˜a0a0
X
f X
Y Z
Y=bXcZ={X}=X− bXc
Y
Y+ 1
Y
∀t∈]− ∞; 0[, FZ(t)=0 ∀t∈[1; +∞[, FZ(t)=1.
t∈[0; 1[ FZ(t)P(n6X6n+t)
t∈[0; 1[ n
P(n6X6n+t)
∀t∈[0; 1[, FZ(t) = 1−e−t
1−e−1.
Z Z