Thème 1 Feuilles de TD No 5 Corrigé

BEN AHMED MOHSEN
omega.center.cp@gmail.com

Méthodes
Quantitatives

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Variables Aléatoires Discrètes

Thème
❶

Vecteurs Aléatoires Discrets

Feuille de Travaux Dirigés No 3
Sujet No 5/ Corrigé
Exercice 1 :
Énoncé
𝑺𝒐𝒊𝒆𝒏𝒕 𝑿 𝒆𝒕 𝒀 𝒅𝒆𝒖𝒙 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒂𝒍é𝒂𝒕𝒐𝒊𝒓𝒆𝒔 𝒊𝒏𝒅é𝒑𝒆𝒏𝒅𝒂𝒏𝒕𝒆𝒔 𝒅é𝒇𝒊𝒏𝒊𝒆𝒔 𝒔𝒖𝒓 𝛀, 𝓕, 𝓟 ,
𝒊𝒏𝒅é𝒑𝒆𝒏𝒅𝒂𝒏𝒕𝒆𝒔, 𝒅𝒆 𝒎ê𝒎𝒆 𝒍𝒐𝒊 𝒈é𝒐𝒎é𝒕𝒓𝒊𝒒𝒖𝒆 𝒔𝒖𝒓 ℕ∗ 𝒅𝒆 𝒑𝒂𝒓𝒂𝒎è𝒕𝒓𝒆 𝒑 =
1)
2)
3)
4)
5)

𝟏
.
𝟐

𝑪𝒂𝒍𝒄𝒖𝒍𝒆𝒓 𝑷 𝒀 ≥ 𝑿
𝑪𝒂𝒍𝒄𝒖𝒍𝒆𝒓 𝑷 𝒀 = 𝑿
𝑫é𝒅𝒖𝒊𝒓𝒆 𝑷 𝒀 > 𝑋
𝑪𝒂𝒍𝒄𝒖𝒍𝒆𝒓 𝑷 𝒀 < 𝑋
𝑶𝒏 𝒏𝒐𝒕𝒆 𝑼 = 𝐦𝐚𝐱 𝑿, 𝒀 𝒆𝒕 𝑽 = 𝐦𝐢𝐧 𝑿, 𝒀 . 𝑪𝒂𝒍𝒄𝒖𝒍𝒆𝒓 𝒑𝒐𝒖𝒓 𝒕𝒐𝒖𝒕 𝒖, 𝒗 ∈ ℕ∗ 𝟐 𝒍𝒂 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕é
𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 . 𝑬𝒏 𝒅é𝒅𝒖𝒊𝒓𝒆 𝒍𝒆𝒔 𝒍𝒐𝒊𝒔 𝒅𝒆 𝑼 𝒆𝒕 𝑽 .

Corrigé
1)
𝑿 𝒆𝒕 𝒀𝒔𝒐𝒏𝒕 𝒅𝒆𝒖𝒙 𝒗. 𝒂. 𝒊𝒅𝒆𝒏𝒕𝒊𝒒𝒖𝒆𝒔 𝒆𝒕 𝒊𝒏𝒅é𝒑𝒆𝒏𝒅𝒂𝒏𝒕𝒆𝒔 𝒅𝒆 𝒍𝒂 𝒍𝒐𝒊 𝒈é𝒐𝒎é𝒕𝒓𝒊𝒒𝒖𝒆 𝒆𝒕 𝓖

𝒅𝒐𝒏𝒄 ∶ 𝑷 𝑿 = 𝒙 =

𝑷 𝒀 =𝒚 =

𝟏−

𝟏 𝒙−𝟏 𝟏
𝟏
𝟏−
× = 𝒙 , 𝒔𝒊 𝒙 ∈ 𝑿 𝛀 = ℕ∗
𝟐
𝟐 𝟐
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

𝟏 𝒚−𝟏 𝟏
𝟏
× = 𝒚 , 𝒔𝒊 𝒚 ∈ 𝒀 𝛀 = ℕ∗
𝟐
𝟐 𝟐
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

𝟏
, 𝒐𝒏 𝒂
𝟐

𝟏 ;

𝟐

𝒆𝒕 ∀ 𝒙, 𝒚 ∈ ℕ∗ 𝟐 , 𝑷 𝑿 = 𝒙, 𝒀 = 𝒚 = 𝑷 𝑿 = 𝒙 𝑷 𝒀 = 𝒚 =

𝟏
𝟐

𝒙+𝒚

𝟑

𝑳’é𝒗é𝒏𝒆𝒎𝒆𝒏𝒕 𝒀 ≥ 𝑿 𝒑𝒆𝒖𝒕 ê𝒕𝒓𝒆 𝒅é𝒄𝒐𝒎𝒑𝒐𝒔é 𝒆𝒏 𝒍𝒂 𝒓é𝒖𝒏𝒊𝒐𝒏 𝒅𝒆𝒔 é𝒗é𝒏𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏𝒄𝒐𝒎𝒑𝒂𝒕𝒊𝒃𝒍𝒆𝒔 ∶
𝑿 = 𝒌, 𝒀 ≥ 𝒌 , 𝒌 ∈ ℕ∗
1

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com

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+∞

𝑪𝒆 𝒒𝒖𝒆 𝒍’𝒐𝒏 𝒑𝒆𝒖𝒕 𝒕𝒓𝒂𝒅𝒖𝒊𝒓𝒆 𝒑𝒂𝒓 : 𝒀 ≥ 𝑿 =

𝑿 = 𝒌, 𝒀 ≥ 𝒌 ,
𝒌=𝟏

𝒂𝒗𝒆𝒄 𝑿 = 𝒌𝟏 , 𝒀 ≥ 𝒌𝟏 ∩ 𝑿 = 𝒌𝟐 , 𝒀 ≥ 𝒌𝟐 = ∅ , ∀𝒌𝟏 ≠ 𝒌𝟐
+∞

𝑷𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 , 𝑷 𝒀 ≥ 𝑿 = 𝑷

+∞

𝑿 = 𝒌, 𝒀 ≥ 𝒌

=

𝑷 𝑿 = 𝒌, 𝒀 ≥ 𝒌
𝒌=𝟏

𝒌=𝟏

𝑶𝒓 𝒑𝒂𝒓 𝒊𝒏𝒅é𝒑𝒆𝒏𝒅𝒂𝒏𝒄𝒆 𝒅𝒆𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝑿 𝒆𝒕 𝒀 𝟑 , 𝒐𝒏 𝒂 ∶
+∞

+∞

𝑷 𝒀≥𝑿 =

𝟏
𝑷 𝒀≥𝒌
𝟐𝒌

𝑷 𝑿=𝒌 𝑷 𝒀≥𝒌 =
𝒌=𝟏

𝒌=𝟏

𝑪𝒂𝒍𝒄𝒖𝒍𝒐𝒏𝒔 𝒅’𝒂𝒃𝒐𝒓𝒅 𝑷 𝒀 ≥ 𝒌 ∶
+∞

+∞

𝑷 𝒀≥𝒌 =

𝑷 𝒀=𝒚 =
𝒚=𝒌

𝒚=𝒌

𝟏
=
𝟐𝒚

+∞

𝒚=𝒌

𝟏
𝟐

+∞

𝒚

; 𝑶𝒓 𝒑𝒐𝒖𝒓 𝒒 < 1, 𝑜𝑛 𝑎 ∶
𝒏=𝒍

𝒒𝒍
𝑼𝒏 =
𝟏−𝒒

𝟏 𝒌
𝟏 𝒌
𝟏
𝑨𝒊𝒏𝒔𝒊 , 𝑷 𝒀 ≥ 𝒌 = 𝟐 = 𝟐 = 𝒌−𝟏
𝟏
𝟏
𝟐
𝟏−𝟐
𝟐
+∞

𝟏
𝑷 𝒀≥𝒌
𝟐𝒌

𝒆𝒕 𝑷 𝒀 ≥ 𝑿 =
𝒌=𝟏
+∞

𝟏
𝟐𝒌

=
𝒌=𝟏
+∞

=
𝒌=𝟏
+∞

=
𝒌=𝟏

𝟏
𝟐𝒌−𝟏

𝟏
𝟐𝟐𝒌−𝟏
𝟏
𝟐−𝟏

+∞

=𝟐
𝒌=𝟏
+∞

=𝟐
𝒌=𝟏

𝟏
𝟐𝟐𝒌
𝒌

𝟏
𝟐𝟐
𝟏
𝟒

𝒌

2

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com
𝟏 𝟏
=𝟐 𝟒
𝟏
𝟏−𝟒
=𝟐×

Téléphone: (+216) 97 619191 / 54 619191
https://web.facebook.com/OMEGACENTER2014

𝟒 𝟏
×
𝟑 𝟒
𝑫′ 𝒐ù 𝑷 𝒀 ≥ 𝑿 =

𝟐
𝟑

2)
𝑳’é𝒗é𝒏𝒆𝒎𝒆𝒏𝒕 𝒀 = 𝑿 𝒑𝒆𝒖𝒕 ê𝒕𝒓𝒆 𝒅é𝒄𝒐𝒎𝒑𝒐𝒔é 𝒆𝒏 𝒍𝒂 𝒓é𝒖𝒏𝒊𝒐𝒏 𝒅𝒆𝒔 é𝒗é𝒏𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏𝒄𝒐𝒎𝒑𝒂𝒕𝒊𝒃𝒍𝒆𝒔 ∶
𝑿 = 𝒌, 𝒀 = 𝒌 , 𝒌 ∈ ℕ∗
+∞

𝑪𝒆 𝒒𝒖𝒆 𝒍’𝒐𝒏 𝒑𝒆𝒖𝒕 𝒕𝒓𝒂𝒅𝒖𝒊𝒓𝒆 𝒑𝒂𝒓 : 𝒀 = 𝑿 =

𝑿 = 𝒌, 𝒀 = 𝒌 ,
𝒌=𝟏

𝒂𝒗𝒆𝒄 𝑿 = 𝒌𝟏 , 𝒀 = 𝒌𝟏 ∩ 𝑿 = 𝒌𝟐 , 𝒀 = 𝒌𝟐 = ∅ , ∀𝒌𝟏 ≠ 𝒌𝟐
+∞

𝑷𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 , 𝑷 𝒀 = 𝑿

=𝑷

+∞

𝑿 = 𝒌, 𝒀 = 𝒌

=

𝑷 𝒀=𝑿

=
𝒌=𝟏

𝟏
𝟒

𝒌

=

𝟏
𝟒

𝟏

𝑷 𝑿 = 𝒌, 𝒀 = 𝒌 =
𝒌=𝟏

𝒌=𝟏
+∞

+∞

+∞

=

𝟏
𝟐𝟐𝒌

𝟐

𝒌+𝒌

𝟏

, 𝒅é𝒇𝒊𝒏𝒊𝒆 𝒔𝒖𝒓 ℕ∗

𝒌=𝟏

𝒌=𝟏

𝟏

𝟏
𝟏−𝟒

=

𝟏 𝟒
×
𝟒 𝟑

𝑫′ 𝒐ù 𝑷 𝒀 = 𝑿

=

𝟏
𝟑

3) 𝒀 é𝒕𝒂𝒏𝒕 𝒖𝒏𝒆 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆 𝒂𝒍é𝒂𝒕𝒐𝒊𝒓𝒆 𝒅𝒊𝒔𝒄𝒓è𝒕𝒆 𝒅𝒆 𝒍𝒂 𝒍𝒐𝒊 𝒈é𝒐𝒎é𝒕𝒓𝒊𝒒𝒖𝒆 𝓖
𝑷𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 , 𝑷 𝒀 ≥ 𝑿 = 𝑷 𝒀 > 𝑋 + 𝑷 𝒀 = 𝑿

𝟐

⇒𝑷 𝒀>𝑋 =𝑷 𝒀≥𝑿 −𝑷 𝒀=𝑿

𝑫′ 𝒐ù 𝑷 𝒀 > 𝑋 =

=

𝟐 𝟏
−
𝟑 𝟑

𝟏
𝟑

4)
+∞

𝑷 𝒀<𝑋 =

+∞

𝟏
𝑷 𝒀<𝑘
𝟐𝒌

𝑷 𝑿=𝒌 𝑷 𝒀<𝑘 =
𝒌=𝟏

𝒌=𝟏

𝒂𝒗𝒆𝒄 , 𝑷 𝒀 < 𝑘 = 𝟏 − 𝑷 𝒀 ≥ 𝒌 = 𝟏 −

𝟏
𝟐𝒌−𝟏
3

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com

Téléphone: (+216) 97 619191 / 54 619191
https://web.facebook.com/OMEGACENTER2014

+∞

𝟏
𝟐𝒌

𝒑𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 , 𝑷 𝒀 < 𝑋 =
𝒌=𝟏
+∞

𝒌=𝟏

=
𝒌=𝟏
+∞

=
𝒌=𝟏
+∞

=
𝒌=𝟏

𝟏
−
𝟐𝒌
𝟏
−
𝟐𝒌

+∞

𝒌=𝟏

=
𝒌=𝟏

𝟏
𝟐

𝟏
𝟐

𝟏
𝟏
× 𝟐𝒌
−𝟏
𝟐
𝟐

𝒌=𝟏
+∞

𝟏
𝟐𝟐

𝒌=𝟏

+∞

𝒌

−𝟐
𝒌=𝟏

𝟏

𝟏
𝟏−𝟐

−𝟐

=𝟏− 𝟐×
=𝟏−

𝟏
𝟐𝟐𝒌−𝟏

+∞

𝟏
−𝟐
𝟐𝒌

+∞

=

𝟏
𝟐𝒌−𝟏

𝟏
𝟏
− 𝟐𝒌−𝟏
𝒌
𝟐
𝟐

=
+∞

𝟏−

𝟏
𝟒

𝒌

𝟏
𝟒

𝒌

𝟏

𝟏
𝟏−𝟒

𝟏 𝟒
×
𝟒 𝟑

𝟐
𝟑
𝑫′ 𝒐ù 𝑷 𝒀 < 𝑋 =

𝟏
𝟑

Remarque
𝑶𝒏 𝒗é𝒓𝒊𝒇𝒊𝒆 𝒃𝒊𝒆𝒏 𝒒𝒖𝒆 : 𝑷 𝒀 < 𝑋 + 𝑷 𝒀 = 𝑿 + 𝑷 𝒀 > 𝑋 = 𝟏
5) 𝑳𝒐𝒊 𝒅𝒖 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒆𝒕 𝒅𝒖 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 ∶ 𝑺𝒐𝒊𝒆𝒏𝒕 𝑿 𝒆𝒕 𝒀 𝒅𝒆𝒖𝒙 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒂𝒍é𝒂𝒕𝒐𝒊𝒓𝒆𝒔 𝒔𝒖𝒓
𝜴, 𝓕, 𝓟 . 𝑬𝒏 𝒑𝒐𝒔𝒂𝒏𝒕 𝒁 = 𝐦𝐚𝐱 𝑿, 𝒀 𝒆𝒕 𝑻 = 𝐦𝐢𝐧 𝑿, 𝒀 , 𝒐𝒏 𝒂 𝒁 𝝎 , 𝑻 𝝎 ⊂ 𝑿 𝝎 ∪ 𝒀 𝝎
∀𝒙 ∈ 𝒁 𝝎 , 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 = 𝒙 = 𝑷 𝑿 = 𝒙 ∩ 𝒀 < 𝑥

+𝑷 𝒀=𝒙 ∩ 𝑿≤𝒙

∀𝒙 ∈ 𝑻 𝝎 , 𝑷 𝐦𝐢𝐧 𝑿, 𝒀 = 𝒙 = 𝑷 𝑿 = 𝒙 ∩ 𝒀 > 𝑥

+𝑷 𝒀=𝒙 ∩ 𝑿≥𝒙

𝑰𝒍 𝒆𝒔𝒕 𝒆𝒏 𝒇𝒂𝒊𝒕 𝒔𝒐𝒖𝒗𝒆𝒏𝒕 𝒑𝒍𝒖𝒔 𝒋𝒖𝒅𝒊𝒄𝒊𝒆𝒖𝒙 𝒅𝒆 𝒑𝒂𝒔𝒔𝒆𝒓 𝒑𝒂𝒓 𝒍𝒂 𝒇𝒐𝒏𝒄𝒕𝒊𝒐𝒏 𝒅𝒆 𝒓é𝒑𝒂𝒓𝒕𝒊𝒕𝒊𝒐𝒏 ∶
𝑷 𝐦𝐚𝐱 𝑿, 𝒀 ≤ 𝒖 = 𝑷 𝑿 ≤ 𝒖 ∩ 𝒀 ≤ 𝒖
4

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
Téléphone: (+216) 97 619191 / 54 619191
omega.center.cp@gmail.com
https://web.facebook.com/OMEGACENTER2014
∗
𝑶𝒏 𝒂 ∶ 𝑿 𝛀 = 𝒀 𝛀 = ℕ 𝒆𝒕 𝑼 = 𝐦𝐚𝐱 𝑿, 𝒀 ⇒ 𝑼 𝛀 = ℕ∗
𝒅𝒆 𝒎ê𝒎𝒆 𝑿 𝛀 = 𝒀 𝛀 = ℕ∗ 𝒆𝒕 𝑽 = 𝐦𝐢𝐧 𝑿, 𝒀 ⇒ 𝑽 𝛀 = ℕ∗
𝑷𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 ∀ 𝒖, 𝒗 ∈ ℕ∗ 𝟐 , 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 = 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 ≤ 𝒖, 𝐦𝐢𝐧 𝑿, 𝒀 ≥ 𝒗
=𝑷 𝑿≤𝒖 ∩ 𝒀≤𝒖 ; 𝑿≥𝒗 ∩ 𝒀≥𝒗
𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 = 𝑷 𝑿 ≤ 𝒖 ∩ 𝒀 ≤ 𝒖 ∩ 𝑿 ≥ 𝒗 ∩ 𝒀 ≥ 𝒗
𝑶𝒓 𝑿 𝒆𝒕 𝒀 𝒅𝒆𝒖𝒙 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒂𝒍é𝒂𝒕𝒐𝒊𝒓𝒆𝒔 𝒊𝒏𝒅é𝒑𝒆𝒏𝒅𝒂𝒏𝒕𝒆𝒔 , 𝒅𝒐𝒏𝒄 ,
𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 = 𝑷 𝑿 ≤ 𝒖

𝑷 𝒀≤𝒖

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 = 𝟏 − 𝑷 𝑿 > 𝑢

𝑷 𝑿≥𝒗

𝟏−𝑷 𝒀> 𝑢

𝑷 𝒀≥𝒗

𝑷 𝑿≥𝒗

𝑷 𝒀≥𝒗

𝟏
∗
𝟏
𝑶𝒓 𝒐𝒏 𝒗𝒊𝒆𝒏𝒕 𝒅𝒆 𝒅é𝒎𝒐𝒏𝒕𝒓𝒆𝒓 𝒒𝒖𝒆 𝒔𝒊 𝒁 ↝ 𝓖
, 𝒂𝒍𝒐𝒓𝒔 , 𝑷 𝒁 = 𝒛 = 𝟐𝒛 , 𝒔𝒊 𝒛 ∈ ℕ 𝒆𝒕
𝟐
𝟎 , 𝒔𝒊 𝒏𝒐𝒏
𝑷 𝒁 ≥𝒛 =

𝑷 𝒁 >𝑧 =

𝟏
𝟐𝒛−𝟏

⇒𝑷 𝒁 >𝑧 =

𝟏

𝟐

−𝑷 𝒁 =𝒛 =
𝒛−𝟏

𝟏

𝟐

−
𝒛−𝟏

𝟏
𝟏
𝟏
𝟏
=
×
𝟏
−
=
𝟐𝒛 𝟐𝒛−𝟏
𝟐
𝟐𝒛

𝟏
𝟐𝒛

𝑨𝒊𝒏𝒔𝒊 , 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 = 𝟏 − 𝑷 𝑿 > 𝑢 ² 𝟏 − 𝑷 𝒀 > 𝑢
= 𝟏−

𝑷 𝑿≥𝒗

𝑷 𝒀≥𝒗

𝟏
𝟏
𝟏
𝟏
𝟏− 𝒖
𝒖
𝒗−𝟏
𝒗−𝟏
𝟐
𝟐 𝟐
𝟐

𝑫′ 𝒐ù ∀ 𝒖, 𝒗 ∈ ℕ∗ 𝟐 , 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 =

𝟏−

𝟏
𝟐𝒖

𝟐

𝟏
𝟐𝒗−𝟏

𝑷 𝒂 < 𝑋 ≤ 𝒃, 𝒄 < 𝑌 ≤ 𝒅 = 𝑭𝑿,𝒀 𝒃, 𝒅 + 𝑭𝑿,𝒀 𝒂, 𝒄 − 𝑭𝑿,𝒀 𝒂, 𝒅 − 𝑭𝑿,𝒀 𝒃, 𝒄

𝑳𝒂 𝒇𝒐𝒏𝒄𝒕𝒊𝒐𝒏 𝒅𝒆 𝒓é𝒑𝒂𝒓𝒕𝒊𝒕𝒊𝒐𝒏 𝒄𝒐𝒏𝒋𝒐𝒊𝒏𝒕𝒆 𝒅’𝒖𝒏 𝒗𝒆𝒄𝒕𝒆𝒖𝒓 𝒂𝒍é𝒂𝒕𝒐𝒊𝒓𝒆

𝑿
𝒑𝒆𝒓𝒎𝒆𝒕 𝒅𝒆 𝒅é𝒕𝒆𝒓𝒎𝒊𝒏𝒆𝒓
𝒀

𝒍𝒆𝒔 𝒇𝒐𝒏𝒄𝒕𝒊𝒐𝒏𝒔 𝒅𝒆 𝒓é𝒑𝒂𝒓𝒕𝒊𝒕𝒊𝒐𝒏 𝒅𝒆𝒕𝒐𝒖𝒕𝒆𝒔 𝒍𝒆𝒔 𝒎𝒂𝒓𝒈𝒆𝒔 𝒅𝒆 𝑿 𝒆𝒕 𝒀 ∶
𝑭𝑿 𝒙 = 𝒍𝒊𝒎 𝑭𝑿,𝒀 𝒙, 𝒚 = 𝑭𝑿,𝒀 𝒙, +∞ = 𝑷𝑿 𝑿 ≤ 𝒙
𝒚→+∞

𝑭𝒀 𝒚 = 𝒍𝒊𝒎 𝑭𝑿,𝒀 𝒙, 𝒚 = 𝑭𝑿,𝒀 +∞, 𝒚 = 𝑷𝑿 𝒀 ≤ 𝒚
𝒙→+∞

𝒍𝒊𝒎 𝑭𝑿,𝒀 𝒙, 𝒚 = 𝑭𝑿,𝒀 −∞, 𝒚 = 𝒍𝒊𝒎 𝑭𝑿,𝒀 𝒙, 𝒚 = 𝑭𝑿,𝒀 𝒙, −∞ = 𝟎

𝒙→−∞

𝒚→−∞

𝒍𝒊𝒎 𝒍𝒊𝒎 𝑭𝑿,𝒀 𝒙, 𝒚 = 𝑭𝑿,𝒀 +∞, +∞ = 𝟏

𝒙→+∞ 𝒚→+∞

5

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com

Téléphone: (+216) 97 619191 / 54 619191
https://web.facebook.com/OMEGACENTER2014

𝒖 +∞

𝒖

𝑶𝒏 𝒂 ∶ 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 − 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 =

𝑷 𝑼 = 𝒓, 𝑽 = 𝒔

−

𝑷 𝑼 = 𝒓, 𝑽 = 𝒔

𝒓=𝟏 𝒔=𝒗

𝒖

+∞

𝒓=𝟏 𝒔=𝒗+𝟏

+∞

=

+∞

𝑷 𝑼 = 𝒓, 𝑽 = 𝒔 −

𝑷 𝑼 = 𝒓, 𝑽 = 𝒔

𝒓=𝟏 𝒔=𝒗
𝒖

𝒔=𝒗+𝟏

𝒗

=

𝑷 𝑼 = 𝒓, 𝑽 = 𝒔
𝒓=𝟏 𝒔=𝒗
𝒖

=

𝑷 𝑼 = 𝒓, 𝑽 = 𝒗
𝒓=𝟏

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 − 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 = 𝑷 𝑼 ≤ 𝒖, 𝑽 = 𝒗

𝟏

𝒖

𝒖−𝟏

𝑷 𝑼 ≤ 𝒖, 𝑽 = 𝒗 − 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 = 𝒗 =

𝑷 𝑼 = 𝒓, 𝑽 = 𝒗

𝑷 𝑼 = 𝒓, 𝑽 = 𝒗

−

𝒓=𝟏

𝒓=𝟏

𝑷 𝑼 ≤ 𝒖, 𝑽 = 𝒗 − 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 = 𝒗 = 𝑷 𝑼 = 𝒖, 𝑽 = 𝒗

𝟐

𝑷 𝑼 = 𝒖, 𝑽 = 𝒗 = 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 − 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 − 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 − 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 + 𝟏
𝑷 𝑼 = 𝒖, 𝑽 = 𝒗 = 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 − 𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 − 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 + 𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 + 𝟏

 𝑳𝒐𝒊 𝒎𝒂𝒓𝒈𝒊𝒏𝒂𝒍𝒆 𝒅𝒆 𝑼 = 𝐦𝐚𝐱 𝑿, 𝒀 ∶

 𝟏è𝒓𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆
+∞

𝑷 𝑼=𝒖 =

𝑷 𝑼 = 𝒖, 𝑽 = 𝒗
𝒗=𝟏
+∞

=

+∞

+∞

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 −
𝒗=𝟏
+∞

=

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 −
𝒗=𝟏

𝟏−
𝒗=𝟏

𝟏
= 𝟏− 𝒖
𝟐

𝟏
𝟐𝒖

𝟐 +∞
𝒗=𝟏

𝟏
=𝟒 𝟏− 𝒖
𝟐

𝟏
=𝟑 𝟏− 𝒖
𝟐

−

𝟐𝒗−𝟏
𝟐
𝟐𝒗

𝟐

𝟐 +∞
𝒗=𝟏

𝟐 +∞
𝒗=𝟏

𝟏−
𝒗=𝟏

𝟏
− 𝟏− 𝒖
𝟐

𝟏
𝟐𝒗

𝟐

𝟏
𝟐𝒗

𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 +
𝒗=𝟏

+∞

𝟐

𝟏

+∞

𝟏
𝟐𝒖

𝟐 +∞
𝒗=𝟏

𝟏
− 𝟏− 𝒖
𝟐

𝟏
𝟐𝒗

−

−𝟑 𝟏−

𝟏−
𝒗=𝟏

𝟐

− 𝟏−

𝟐 +∞

𝟐

+∞

𝟐

𝟏
𝟐𝒗

𝒗=𝟏

𝟏
𝟐𝒖−𝟏

𝟏
𝟐𝒗

𝟏

𝟐𝒖−𝟏

𝟐𝒗−𝟏

𝟐𝒖−𝟏

𝒗=𝟏

𝟐

−𝟒 𝟏−

𝒗=𝟏

6

𝟏
𝟐𝒗

𝟐
𝟐𝒗

+∞

𝟐

𝟏

𝟐 +∞

𝟏

𝟐 +∞

𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 + 𝟏
𝒗=𝟏

+
𝒗=𝟏

𝟐

𝟏
𝟐𝒖−𝟏

𝒗=𝟏

𝟐 +∞

𝟏

+ 𝟏−
𝟐 +∞

𝟏−

𝟐𝒖−𝟏

𝟏
𝟐𝒗

𝒗=𝟏

𝟏

𝟏
𝟐𝒗

𝟐𝒖−𝟏
𝟏
𝟐𝒗

𝟐

+ 𝟏−

𝟐

𝟐

𝟏
𝟐𝒖−𝟏

𝟐 +∞
𝒗=𝟏

𝟏
𝟐𝒗

𝟐

𝟐

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com
𝟐 +∞

𝟏
=𝟑 𝟏− 𝒖
𝟐
+∞

= 𝟑
𝒗=𝟏

𝟏
𝟒

𝒗=𝟏
𝒗

𝟏 𝟏
= 𝟑× 𝟒
𝟏
𝟏−𝟒
= 𝟑×

𝟏 𝟒
×
𝟒 𝟑

𝟏
𝟒

𝒗

−𝟑 𝟏−

𝟏
𝟏− 𝒖
𝟐

𝟐

𝟏
𝟏− 𝒖
𝟐

𝟐

𝟏−

𝟐𝒖−𝟏

− 𝟏−

=

𝟏
𝟐𝒖+𝟏 − 𝟑
𝟐𝟐𝒖

𝒗

𝟐𝒖−𝟏
𝟐

𝟏
𝟐𝒖−𝟏

𝟏
𝟏
− 𝟏 + 𝒖−𝟏
𝒖
𝟐
𝟐

𝟏
𝟏
𝟏
×
−𝟏
+
×
×
𝟐𝒖
𝟐−𝟏
𝟐𝒖

𝒗=𝟏

𝟏
𝟒

𝟐

𝟏

− 𝟏−

=

𝟐 +∞

𝟏

Téléphone: (+216) 97 619191 / 54 619191
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𝟏−

𝟏
𝟏
+ 𝟏 − 𝒖−𝟏
𝒖
𝟐
𝟐

𝟐 × 𝟐𝒖 − 𝟏 −

𝟏
𝟐−𝟏

𝟐𝒖+𝟏 − 𝟑
, 𝒔𝒊 𝒖 ∈ 𝑼 𝛀 = ℕ∗
𝟐𝒖
𝒅′ 𝒐ù , 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 = 𝒖 = 𝑷 𝑼 = 𝒖 =
𝟐
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

 𝟐è𝒎𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆
∀𝒖 ∈ 𝑼 𝛀 = ℕ∗ , 𝑷 𝑼 = 𝒖 = 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 = 𝒖 = 𝑷 𝑿 = 𝒖 ∩ 𝒀 < 𝒖 + 𝑷 𝒀 = 𝒖 ∩ 𝑿 ≤ 𝒖
= 𝑷 𝑿=𝒖 𝑷 𝒀<𝒖 + 𝑷 𝒀=𝒖 𝑷 𝑿≤𝒖
=

𝟏
𝟏−𝑷 𝒀≥𝒖
𝟐𝒖

=

𝟏
𝟐𝒖

=

𝟏 𝟐𝒖−𝟏 − 𝟏 𝟐𝒖 − 𝟏
+
𝟐𝒖
𝟐𝒖−𝟏
𝟐𝒖

𝟏−

𝟏

+

𝟐𝒖−𝟏

+ 𝟏−

𝟏
𝟏−𝑷 𝑿>𝑢
𝟐𝒖
𝟏
𝟐𝒖

𝟏 𝟐𝒖 − 𝟐 𝟐𝒖 − 𝟏
= 𝒖
+
𝟐
𝟐𝒖
𝟐𝒖
=

𝟏
𝟐 × 𝟐𝒖 − 𝟑
𝟐𝟐𝒖

𝟐𝒖+𝟏 − 𝟑
, 𝒔𝒊 𝒖 ∈ 𝑼 𝛀 = ℕ∗
𝑶𝒏 𝒓𝒆𝒕𝒓𝒐𝒖𝒗𝒆 ∶ 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 = 𝒖 = 𝑷 𝑼 = 𝒖 =
𝟐𝟐𝒖
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

7

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
omega.center.cp@gmail.com
 𝟑è𝒎𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆

Téléphone: (+216) 97 619191 / 54 619191
https://web.facebook.com/OMEGACENTER2014

∀𝒖 ∈ 𝑼 𝛀 = ℕ∗ , 𝑷 𝑼 = 𝒖 = 𝑭𝑼 𝒖 − 𝑭𝑼 𝒖 − 𝟏 = 𝑷 𝑼 ≤ 𝒖 − 𝑷 𝑼 ≤ 𝒖 − 𝟏
𝑶𝒓 𝑷 𝑼 ≤ 𝒖 = 𝑷 𝐦𝐚𝐱 𝑿, 𝒀 ≤ 𝒖 = 𝑷 𝑿 ≤ 𝒖 ∩ 𝒀 ≤ 𝒖

= 𝑷 𝑿≤𝒖 𝑷 𝒀≤𝒖
𝑿 𝒆𝒕 𝒀 ,𝟐𝒗.𝒂. 𝒊𝒅𝒆𝒏𝒕𝒊𝒒𝒖𝒆𝒔 𝒅𝒆𝓖

𝟐

𝑷 𝑼≤𝒖 = 𝟏−𝑷 𝑿>𝑢
𝑨𝒊𝒏𝒔𝒊 , 𝑷 𝑼 = 𝒖 = 𝟏 −

𝟏
𝟐𝒖

𝟏
= 𝟏− 𝒖
𝟐
𝟐

− 𝟏−

𝟐

⇒𝑷 𝑼≤𝒖−𝟏 = 𝟏−
𝟐

𝟏

=

𝟐𝒖−𝟏

= 𝑷 𝑿≤𝒖

𝟐

𝟏
𝟐

𝟐

𝟏
𝟐𝒖−𝟏

𝟐𝒖+𝟏 − 𝟑
𝟐𝟐𝒖

𝟐𝒖+𝟏 − 𝟑
, 𝒔𝒊 𝒖 ∈ 𝑼 𝛀 = ℕ∗
𝑷 𝐦𝐚𝐱 𝑿, 𝒀 = 𝒖 = 𝑷 𝑼 = 𝒖 =
𝟐𝟐𝒖
𝟎 , 𝒔𝒊 𝒏𝒐𝒏
 𝑳𝒐𝒊 𝒎𝒂𝒓𝒈𝒊𝒏𝒂𝒍𝒆 𝒅𝒆 𝑽 = 𝐦𝐢𝐧 𝑿, 𝒀 ∶

 𝟏è𝒓𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆
+∞

𝑷 𝑽=𝒗 =

𝑷 𝑼 = 𝒖, 𝑽 = 𝒗
𝒖=𝟏
+∞

=

+∞

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 −
𝒖=𝟏

=

=

+∞

𝟏
𝟐𝟐𝒗−𝟐

𝒖=𝟏

𝟐𝟐𝒗−𝟐

𝒖=𝟏

𝟏
𝟏− 𝒖
𝟐

𝟏
= 𝟐𝒗−𝟐 − 𝟐𝒗
𝟐
𝟐

𝑷 𝑽=𝒗

𝟒
𝟏
= 𝟐𝒗 − 𝟐𝒗
𝟐
𝟐
𝟑
= 𝟐𝒗
𝟐

+∞

𝒖=𝟏

𝟐

𝟏
𝟏− 𝒖
𝟐

+∞

𝟏

+∞

𝒖=𝟏
+∞

𝒖=𝟏

+∞

𝑷 𝑼 ≤ 𝒖, 𝑽 ≥ 𝒗 + 𝟏 −
𝒖=𝟏

𝟏

𝑷 𝑽=𝒗

+∞

𝑷 𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 +
𝒖=𝟏

𝟏
− 𝟐𝒗
𝟐

+∞

𝒖=𝟏

𝟐

− 𝟏−

𝟏
𝟏− 𝒖
𝟐
𝟏

𝟐𝒖−𝟏

𝟐

𝟐

−

𝟏
− 𝟐𝒗
𝟐

𝑼 ≤ 𝒖 − 𝟏, 𝑽 ≥ 𝒗 + 𝟏
𝒖=𝟏

𝟏
𝟐𝟐𝒗−𝟐
+∞

𝒖=𝟏

+∞

𝟏−
𝒖=𝟏

𝟏
𝟏− 𝒖
𝟐

𝟏
𝟐𝒖−𝟏

𝟐

𝟏
+ 𝟐𝒗
𝟐

𝟐

− 𝟏−

𝟏

+∞

𝟏−
𝒖=𝟏

𝟏

𝟐

𝟐𝒖−𝟏

𝟐

𝟐𝒖−𝟏

𝟐𝒖+𝟏 − 𝟑
𝟐𝟐𝒖

𝟐𝒖+𝟏 − 𝟑
𝟐𝟐𝒖

𝟐𝒖+𝟏 − 𝟑
𝟐𝟐𝒖

8

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶

BEN AHMED MOHSEN
Téléphone: (+216) 97 619191 / 54 619191
omega.center.cp@gmail.com
https://web.facebook.com/OMEGACENTER2014
𝒖+𝟏
𝟐
−𝟑
, 𝒔𝒊 𝒖 ∈ 𝑼 𝛀 = ℕ∗ 𝒆𝒔𝒕 𝒖𝒏𝒆 𝒍𝒐𝒊 𝒅𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕é
𝟐𝒖
𝑶𝒓 𝑷 𝑼 = 𝒖 =
𝟐
𝟎 , 𝒔𝒊 𝒏𝒐𝒏
+∞

⇒

+∞

𝑷 𝑼 =𝒖 =
𝒖=𝟏

𝒖=𝟏

𝒑𝒂𝒓 𝒍𝒂 𝒔𝒖𝒊𝒕𝒆 , 𝑷 𝑽 = 𝒗

𝟐𝒖+𝟏 − 𝟑
=𝟏
𝟐𝟐𝒖
=

𝟒
𝟏
− 𝟐𝒗 =
𝟐𝒗
𝟐
𝟐

𝟑
∗
𝒅 𝒐ù 𝑷 𝐦𝐢𝐧 𝑿, 𝒀 = 𝒗 = 𝑷 𝑽 = 𝒗 = 𝟐𝟐𝒗 , 𝒔𝒊 𝒗 ∈ 𝑽 𝛀 = ℕ
𝟎 , 𝒔𝒊 𝒏𝒐𝒏
′

 𝟐è𝒎𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆
∀𝒗 ∈ 𝑽 𝛀 = ℕ∗ , 𝑷 𝑽 = 𝒗 = 𝑷 𝒎𝒊𝒏 𝑿, 𝒀 = 𝒗 = 𝑷 𝑿 = 𝒗 ∩ 𝒀 > 𝒗 + 𝑷 𝒀 = 𝒗 ∩ 𝑿 ≥ 𝒗
= 𝑷 𝑿=𝒗 𝑷 𝒀>𝒗 + 𝑷 𝒀=𝒗 𝑷 𝑿≥𝒗
=
=

𝟏
𝟏
𝟏
𝟏
× 𝒗 + 𝒗 × 𝒗−𝟏
𝒗
𝟐
𝟐
𝟐
𝟐
𝟏
𝟐
+ 𝟐𝒗
𝟐𝒗
𝟐
𝟐

𝟑
∗
𝑶𝒏 𝒓𝒆𝒕𝒓𝒐𝒖𝒗𝒆 𝒂𝒊𝒏𝒔𝒊 ∶ 𝑷 𝐦𝐢𝐧 𝑿, 𝒀 = 𝒗 = 𝑷 𝑽 = 𝒗 = 𝟐𝟐𝒗 , 𝒔𝒊 𝒗 ∈ 𝑽 𝛀 = ℕ
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

 𝟑è𝒎𝒆 𝒎é𝒕𝒉𝒐𝒅𝒆
∀𝒗 ∈ 𝑽 𝛀 = ℕ∗ , 𝑷 𝑽 ≥ 𝒗 = 𝑷 𝒎𝒊𝒏 𝑿, 𝒀 ≥ 𝒗 = 𝑷 𝑿 ≥ 𝒗 ∩ 𝒀 ≥ 𝒗

= 𝑷 𝑿≥𝒗 𝑷 𝒀≥𝒗 = 𝑷 𝑿≥𝒗

𝟐

𝟏
𝑿 𝒆𝒕 𝒀 ,𝟐𝒗.𝒂. 𝒊𝒅𝒆𝒏𝒕𝒊𝒒𝒖𝒆𝒔 𝒅𝒆𝓖 𝟐

∗

∀𝒗 ∈ 𝑽 𝛀 = ℕ , 𝑷 𝑽 ≥ 𝒗 =

𝟏
𝟐𝒗−𝟏

𝟐

𝟏
⇒𝑷 𝑽≥𝒗+𝟏 = 𝒗
𝟐

𝒐𝒓, 𝑷 𝑽 = 𝒗 = 𝑷 𝑽 ≥ 𝒗 − 𝑷 𝑽 ≥ 𝒗 + 𝟏 ⇒ 𝑷 𝑽 = 𝒗 =

𝟏

𝟐

𝟐

−
𝟐𝒗−𝟐

𝟏
𝟒
𝟏
=
−
𝟐𝟐𝒗 𝟐𝟐𝒗 𝟐𝟐𝒗

𝟑
∗
𝑷 𝐦𝐢𝐧 𝑿, 𝒀 = 𝒗 = 𝑷 𝑽 = 𝒗 = 𝟐𝟐𝒗 , 𝒔𝒊 𝒗 ∈ 𝑽 𝛀 = ℕ
𝟎 , 𝒔𝒊 𝒏𝒐𝒏

9

ème

XXXVII

Adresse : 2, Rue Ibn Arafa, av de l'environnement Manouba Centre, Manouba

PROMOTION 2017/ Thème❶