Postila Bètran

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Ale nan: Navigasyon, Fouye

Postila Bètran[edite | modifier le wikicode]

ant yon antye n ki pi gran ke 1 e doub antye sila a , toujou gen yon nonm premye. se yon teyorèm ki te konjektire pa Bertrand e ki demontre an 1850 pou lapremyè fwa pa matematisyen rus Tchebycheff Gen plizyè varyant pou demontre teyorèm oubyen postila a . Men de ladan yo ki privilejye apròch elemantè.

premye demonstrasyon[edite | modifier le wikicode]

premye lèm[edite | modifier le wikicode]

premye demach la ap kondwi nou redemontre tankou yon lèm ke

\frac{2^{2n}}{2 \sqrt n}<N<\frac{2^{2n}}{\sqrt{2n}} pou tout n ki siperyè ou egal ak 2 avèk

 N = C_{2n}^n={{2n} \choose n}

 N =\frac{\left(\left(2n\right)!\right)}{\left(n!\right)\times\left(2n-n\right)!}

 N =\frac{\left(\left(2n\right)!\right)}{\left(n!\right)\times\left(n\right)!}

 N =\frac{\left(\left(2n\right)!\right)}{\left(n!\right)^2}

Nou pale de redemontre paske se yon demonstrasyon ki te la deja ....

An nou konsidere pwodwi tout antye empè yo divize pa pwodwi tout antye pè yo ki enferyè oubyen egal ak 2n

Konfòméman ak Arnel Mercier ak Jean Marie de Koninck an desiye pwodwi sa pa P

P = \frac{1\times\times3\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{2\times4\times...\times\left(2n-2\right)\times 2n}

An nou fè parèt anlè ya pwodwi ki anba a

P = \frac{1\times\times3\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{2\times4\times...\times\left(2n-2\right)\times 2n}\times\frac{2\times4\times6\times ...\times\left(2n-2\right)\times2n}{2\times4\times6\times ...\times\left(2n-2\right)\times2n}


P = \frac{\left(1\times\times3\times ...\times\left(2n-3\right)\times\left(2n-1\right)\right)\times\left(2\times4\times6\times ...\times\left(2n-2\right)\times2n\right)}{2\times4\times ...\times\left(2n-2\right)\times 2n\times\left(2\times4\times6\times ...\times\left(2n-2\right)\times2n\right)}

P=\frac{\left(2n\right)!}{\left(2\times4\times6\times ...\times\left(2n-2\right)\times2n\right)^2}

P=\frac{\left(2n\right)!}{{\left(2\times1\times2\times2\times2\times3\times ... \times2\times\left(n-1\right)\times2\times n\right)}^2}


P=\frac{\left(2n\right)!}{{\left(2^n\times n!\right)}^2}



P=\frac{\left(2n\right)!}{{\left(2^n\right)}^2\times{\left(n!\right)}^2}


P=\frac{\left(2n\right)!}{2^{2n}\times{\left(n!\right)}^2}


P\times{2^{2n}}=\frac{\left(2n\right)!}{2^{2n}\times{\left(n!\right)}^2}\times{2^{2n}}


P\times{2^{2n}}=\frac{\left(2n\right)!}{{\left(n!\right)}^2}


P\times{2^{2n}}=N


An nou konsidere



\prod_{i = 1}^n{\left(1-\frac{1}{{\left(2i\right)}^2}\right)}{}


si nou devlope prodwi nap genyen

\prod_{i = 1}^n{\left(1-\frac{1}{{\left(2i\right)}^2}\right)}=\left(1-\frac{1}{2^2}\right)\times\left(1-\frac{1}{4^2}\right)\times\left(1-\frac{1}{6^2}\right)\times ... \times \left(1-\frac{1}{{\left(2n-2\right)}^2}\right)\times\left(1-\frac{1}{{\left(2n\right)}^2}\right)


chak faktè yo estrikteman pozitif epi yo enferyè ak 1 , kidonk pwodwi li menm tou enferyè ak 1.

\prod_{i = 1}^n{\left(1-\frac{1}{{\left(2i\right)}^2}\right)}{<1}


\prod_{i = 1}^n{\left(\frac{{\left(4i\right)}^2-1}{{\left(2i\right)}^2}\right)}{<1}


\prod_{i = 1}^n{\left(\frac{\left(2i-1\right)\left(2i+1\right)}{{\left(2i\right)}^2}\right)} {<1}

sa ki bay

 \left(\frac{1\times3}{2^2}\right)\times\left(\frac{3\times5}{4^2}\right)\times\left(\frac{5\times7}{6^2}\right)\times ... \times\left(\frac{\left(2n-3\right)\times\left(2n-1\right)}{{\left(2n-2\right)}^2}\right)\times\left(\frac{\left(2n-1\right)\times\left(2n+1\right)}{{\left(2n\right)}^2}\right) {<1}

lè nou diseke numeratè yo ak denominatè yo :


 \left(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times ... \times\frac{2n-3}{2n-2}\times\frac{2n-1}{2n}\right)\times\left(\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times ... \times\frac{2n-1}{2n-2}\times\frac{2n+1}{2n}\right){<1}


\frac{1\times3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{2\times4\times6\times. ..\times\left(2n-2\right)\times2n}

\times\frac{3\times5\times ...\times\left(2n-1\right)\times\left(2n+1\right)}{2\times4\times6\times ...\times\left(2n-2\right)\times2n} {<1}


\frac{1\times3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{2\times4\times6\times  ...\times\left(2n-2\right)\times2n}

\times\frac{1\times3\times5\times ... \times\left(2n-3\right)\times\left(2n-1\right)}{2\times4\times6\times ... \times\left(2n-2\right)\times2n}\times\left(2n+1\right) {<1}

P\times P \times\left(2n+1\right){<1}



P^2\times\left(2n+1\right){<1}


P^2\times2n{<}P^2\times\left(2n+1\right){<1}

kidonk

{P^2\times 2n}{<}1

Nou te wè ke

2^{2n}P = N


{\left(2^{2n}\right)}^2P^2 = N^2

{\left(2^{2n}\right)}^2P^2 \times2n {<} {\left(2 ^ {2n}\right)} ^2

N^2\times{2n} < {\left(2^{2n}\right)}^2


N^2 < \frac{{\left(2^{2n}\right)}^2}{2n}

Fonksyon rasin kare kwasan, nou genyen :

N {<} \sqrt{\frac{{\left(2^{2n}\right)}^2}{2n}}


N < \frac{2^{2n}}{\sqrt{2n}}

Inegalite dwat la demontre

li pa fè answa apèl a oken nosyon nonm premye jiska prezan sinon ke factoryèl la se nosyon de predileksyon nonm premye


Kounye an nou redemontre dezyèm pati inegalite , inegalite goch la.

An nou konsidere pwodwi sa :

\prod_{i=2}^n{\left(1-\frac{1}{{\left(2i-1\right)} ^2}\right)}

chak faktè yo pozitif e enferyè ak 1, kidonk pwodwi ya globalman enferyè ak 1 .


\prod_{i=2}^n{\left(1-\frac{1}{{\left(2i-1\right)} ^2}\right)}{<1}




\prod_{i=2}^n{\left( \frac{{\left(2i-1\right)}^2-1}{{\left(2i-1\right)}^2}\right)}{<1}


\prod_{i=2}^n{\left(\frac{\left(2i-1-1\right)\left(2i-1+1\right)}{{\left(2i-1\right)}^2}\right)}{<1}

\prod_{i=2}^n{\left(\frac{\left(2i-2\right)\times 2i}{{\left(2i-1\right)}^2}\right)}<1


{\prod_{i=2}^n{\left(\frac{2i-2}{2i-1}\right)}}\times{\prod_{i=2}^n{\left(\frac{2i}{2i-1}\right)}}{<1}

sa ki bay

\left(\frac{2}{3}\times\frac{4}{5}\times ... \times\frac{2n-4}{2n-3}\times\frac{2n-2}{2n-1}\right)\times\left(\frac{4}{3}\times\frac{6}{5}\times...\times\frac{2n-2}{2n-3}\times\frac{2n}{2n-1}\right)<1


 \frac{2\times4\times...\times\left(2n-4\right)\times\left(2n-2\right)}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}\times\frac{4\times6\times ...\times\left(2n-2\right)\times2n}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{<1}


 \frac{2\times4\times...\times\left(2n-4\right)\times\left(2n-2\right)}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}\times2n\times2\times\frac{4\times6\times ...\times\left(2n-2\right)\times2n}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{<2\times2n}


 \frac{2\times4\times...\times\left(2n-2\right)\times2n}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}\times\frac{2\times4\times ...\times\left(2n-2\right)\times2n}{3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}{<4n}


 \left(\frac{2\times4\times...\times\left(2n-2\right)\times2n}{1\times3\times5\times...\times\left(2n-3\right)\times\left(2n-1\right)}\right)^2{<4n}


{\left({\frac{1}{ \frac{{1\times3\times5\times ...\times\left(2n-3\right)\times\left(2n-1\right)}}{{2\times4\times ...\times\left(2n-2\right)\times2n}}}}\right)}^2 {<4n}

{\left(\frac{1}{P}\right)}^2 {< 4n}


 \frac{1}{P^2} {< 4n}

 \frac{1}{\frac{1}{P^2}} {>} {\frac{1}{4n}}

p^2 > \frac{1}{4n}

nou genyen

{ 2^{2n}\times P} = N

P ={\frac{N}{2^{2n}}}



 {\left(\frac{N}{2^{2n}}\right)}^2 {>}{\frac{1}{4n}}


 \left(\frac{N}{2^{2n}}\right) {>} \sqrt{{\frac{1}{4n}}}


 \left(\frac{N}{2^{2n}}\right) {>} {\frac{1}{\sqrt{4n}}}


 \left(\frac{N}{2^{2n}}\right) {>} {\frac{1}{2\sqrt{n}}}


 {\left(\frac{N}{2^{2n}}\right)}\times{2^{2n}} {>} {\frac{1}{2\sqrt{n}}}\times{2^{2n}}

 N {>} \frac {2^{2n}}{2\sqrt{n}}

\frac {2^{2n}}{2\sqrt{n}}  {<} N

Inegalite goch la demontre

kidonk lè nou konbine inegalite goch la ak inegalite dwat la nou gen premye lèm


\frac{2^{2n}}{2 \sqrt n}< N <\frac{2^{2n}}{ \sqrt{2n}}

dezyèm lèm[edite | modifier le wikicode]

An nou demontre yon dezyèm pwopozisyon sa ki va sèvi kòm lèm

\theta\left(n\right) {<} 2nLn2

An soulye ke fonksyon \theta Tchebychèf la defini konsa

\theta\left(x\right) = \sum_{p\le x}^{}{Ln p} avèk x \le 0 ...

Nan kad sa a nou sipoze ke n \in \mathbb{N}^*

 \theta\left(1\right) = 0 < 2\times1\times Ln2

 \theta\left(2\right) = Ln2 {<} 2\times2\times Ln2


Inegalite a verifye pou n = 1 ak n = 2 .

An nou sipoze li vrè pou yon sèten  n>2


kounye a an nou konsidere \frac{N}{2}

 \frac{N}{2}=\frac{1}{2} N = \frac{1}{2}{{2n} \choose n} =\frac{\left(2n\right)!}{2\times {\left(n!\right)}\times {n!}}=\frac{\left(\left(2n-1\right)!\right)\times2n}{2\times\left( n!\right) \times n!} =\frac{\left(\left(2n-1\right)!\right)\times n}{\left(n!\right)\times{\left(\left(n-1\right)!\right)}\times n} =\frac{\left(2n-1\right)!}{\left(\left(n-1\right)!\right)\times n!} = \frac{\left(2n-1\right)}{\left(\left(n-1\right)!\right)\times\left(\left(2n-1\right)-\left(n-1\right)\right)!} = C^{{n-1}}_{{2n-1}} = { {2n-1} \choose{n-1}}



pou tou n siperyè ak 1, ekspresyon sa gen yon reyalite konkrè e se yon antye : se pa egzanp kantite souzansanm de n-1 eleman diferan ke nou ka fòme a pati de yon ansanm de 2n-1 eleman.

An nou gade byen nimeratè a ak denominatè a

\frac{\left(2n-1\right)!}{\left(n!\right)\times\left(\left(n-1\right)!\right)} = \frac{1\times2\times3\times ... \times\left(2n-2\right)\times\left(2n-1\right)}{\left(1\times2\times3\times ... \times {\left(n-1\right)}\times n\right)\times{\left(1\times2\times3\times ... \times \left(n-2\right)\times\left(n-1\right)\right)}}


Tout nonm antye non nil ki enferyè ak 2n-1 se yon divizè de nimeratè a , an patikilye tout nonm premye ki enferyè ou egal ak 2n-1 se yon divizè nimeratè a .

Menm konsiderasyon an ka fèt pou denominatè a jiska n .

An nou konsidere nonm premye p ki siperyè ak n e ki enferyè oubyen egal ak 2n-1.


 n{< p}\le {2n-1}

 p\nmid {\left(\left(n!\right)\times\left(\left(n-1\right)!\right) \right)}


 p\mid \left(2n-1\right)!

kidonk yon faktè premye p ki tèl ke  n <p\leq {2n-1} pa senplifye .


 \frac{N}{2} se donk miltipl non nil chak nonm premye tèl ke  n <p\leq {2n-1} . Se donk yon miltipl non nil pwodwi tout nonm premye sila yo .

 \frac{N}{2} siperyè oubyen egal ak pwodwi nonm sa yo


 \frac{N}{2}\ge {\prod_{n<p\le{2n-1}}{p}}


Fonksyon Ln lan kwasant, nou genyen


 {Ln{\frac{N}{2}}}\ge {Ln{\prod_{n<p\le{2n-1}}{p}}}


tout fonksyon logaritm transforme prodwi an sòm logaritm


 {Ln{\frac{N}{2}}}\ge {\sum_{n<p\le{2n-1}}{Lnp}}


 {Ln{\frac{N}{2}}}\ge {\sum_{1\le{p}\le{n}}{Lnp}} - {\sum_{1\le{p}\le{n}}{Lnp}} + {\sum_{n{<p}\le{2n-1}}{Lnp}}


 {Ln{\frac{N}{2}}}\ge \left({\sum_{1\le{p}\le{n}}{Lnp}}  + {\sum_{n{<p}\le{2n-1}}{Lnp}}\right)- {\sum_{1\le{p}\le{n}}{Lnp}}



 {Ln{\frac{N}{2}}}\ge {\sum_{1\le{p}\le{2n - 1}}{Lnp}} - {\sum_{1\le{p}\le{n}}{Lnp}}


 {Ln{\frac{N}{2}}}\ge \theta\left(2n-1\right) - \theta\left(n\right)


 \theta\left(2n-1\right) - \theta\left(n\right)\le {Ln{\frac{N}{2}}}


An nou konsidere inegalite dwat premye lèn lan oubyen dezyèm inegalite premye lèm lan

 {N} {<} {{\frac{2^{2n}}{\sqrt{2n}}}}

nou kapab ekri li 

{N} < {\frac{2^{2n}}{{\left(2n\right)}^{\frac{1}{2}}}}


An nou pran logaritm Neperyen de manm yon.

Kòm se yon fonksyon kwasant, nap genyen

Ln N < {{\frac{2^{2n}}{{\left(2n\right)}^{\frac{1}{2}}}}}


Ln N < {Ln2^{2n}} - Ln{\left(2n\right)}^{\frac{1}{2}}


Ln N {<} 2nLn2 - \frac{1}{2}Ln2n


An nou konbine



\{_{\theta\left(2n - 1\right) - \theta \left(n\right) \le Ln {\frac{N}{2}}}^{Ln N < 2nLn2 - \frac{1}{2}Ln{2n}}


\theta\left(2n - 1\right) - \theta\left(n\right) + Ln N{<}{Ln{\frac{N}{2}} + 2nLn N  - \frac{1}{2}Ln{2n}}


\theta\left(2n - 1\right) - \theta\left(n\right) < Ln{\frac{N}{2}} - LnN + 2nLn2 - \frac{1}{2}Ln{2n}


\theta\left(2n - 1\right) - \theta\left(n\right) {<} Ln{\frac{\frac{N}{2}}{N}} + 2nLog2 - \frac{1}{2}Ln{2n}


\theta\left(2n - 1\right) - \theta\left(n\right) < Ln{\frac{1}{2}} + 2nLn2 - \frac{1}{2}Ln{2n}

\theta\left(2n - 1\right) - \theta\left(n\right) < - Ln2 + 2nLn2 - \frac{1}{2}Ln{2n}

\theta\left(2n - 1\right) - \theta\left(n\right) <  \left(2n - 1\right)Ln2 - \frac{1}{2}Ln{2n}


An nou pa bliye ke nou vle demontre kòm dezyèm lèm

\theta\left(n\right) {<} 2nLn2


An nou konsidere kòm Ipotèz e an montre ke li vre pou 2n - 1 ak 2n pou tout n > 2

 \theta\left(2n - 1\right) - \theta\left(n\right)+ \theta\left(n\right) {<} \left(2n - 1\right)Ln2 - \frac{1}{2}Ln2n + 2nLn2


 \theta\left(2n - 1\right) {<} \left(4n - 1\right)Ln2 - \frac{1}{2}Ln2n

....

kòm  n>2

 2n {>} {2\times2}

 2n  {>} { 2^2}

 Ln2n {>} {Ln{2^2}}


 Ln2n {>} 2Ln2


 \frac{1}{2}Ln2n {>} Ln2

 - \frac{1}{2}Ln2n {<} - Ln2

 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {<} \left(4n - 1\right)Ln2 - Ln2

 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {<} \left(4n - 2\right)Ln2


 \left(4n-1\right)Ln2 - \frac{1}{2}Ln2n {<} 2\left(2n-1\right)Ln2


 \theta\left(2n -1 \right) {<} \left(4n - 1\right)Ln2 - \frac{1}{2}Ln2n < 2\left(2n - 1\right) Ln 2

 {\theta\left(2n - 1\right)} < {2\left(2n - 1\right)Ln2}


si nou gen  n \ge 2 alò  2n \ge 4

 2n \ne 2

 2n \notin \mathbb{P}

{\{_ {n \in \mathbb{N}}^{n\ge2}}\Rightarrow {2n \notin \mathbb{P}}\Leftrightarrow \not\exists p \in \mathbb{P} {/} 2n-1{<}p\le 2n


\theta\left(2n\right) = \sum_{1\le p \le 2n}Lnp = \sum_{1\le p \le 2n - 1}Lnp + \sum_{2n - 1\le p \le 2n}Lnp = \sum_{1\le p \le 2n - 1}{Lnp}  + 0  = \sum_{1\le p \le 2n - 1}{Lnp} = \theta\left(2n\right)


kidonk


 n \ge 2 \Rightarrow \theta\left(2n-1\right) = \theta\left(2n\right)


pou tout  n\ge 2

si lèm lan vre pou n li vre pou 2n - 1

an nou pwouve rapidman ke inegalite vre pou 2n tou .

\theta\left(2n-1\right) {<} 2\left(2n-1\right)Ln2

\theta\left(2n\right) {<} 2\left(2n - 1\right)Ln2

\theta\left(2n\right){<}4nLn2 - 2Ln2

\theta\left(2n\right) {<} 4nLn2 - 2Ln2 {<} 4nLn2

\theta\left(2n\right) {<} 4nLn2

\theta\left(2n\right) < 2\times 2nLn2

si inegalite a vre pou  n\ge 2 li vre alafwa pou 2n - 1 ak 2n .

An nou montre kounye a si fòmil la vrè pou entèval ]{2^{r-1}} , 2^r] alò li vrè pou entèval ]2^r , 2^{r+1}]

Avan nou montre sa an nou verifye ke ingalite a vrè pou entèval  ]2,4]

Inegalite a vre pou n = 2 selon enplikasyon avan an li vre pou  2\times2 - 1 et pou 2\times2 ki donk li vre e pou 3 e pou 4. Pwopriyete a vre pou  [2 , 4 ] oubyen ankò li vre pou  ]2 ,4] Pwopriyete a vre pou  ]2^1 , 2^2]


Ipotèz rekirans

An nou konsidere ke inegalite sou tout entèval ]2^{r-1} , 2^r] la e an nou montre ke li vre pou tout entèval

]2^r , 2^{r+1}] la tou .

An nou pran yon antye m kèlkon sou entèval

]2^r , 2^{r+1}]

sa vle di  m\in ]2^r , 2^{r+1}]

 m\in ]2^r , 2^{r+1}] \Leftrightarrow {2^r<m\le{2^{r+1}}}


Nan de bagay ki ekskli yo youn lòt , nou gen youn fòseman .

m se yon nonm pè ou m se yon nonm enpè

1)

m pè   \Leftrightarrow \exists n\in \mathbb{N} {/}m = 2n 


\{_{m={2n}}^{2^r {<} m \le {2^{r+1}}}\Rightarrow {2^r {<} {2n} \le {2^{r+1}}}\Rightarrow {{2\times{2^{r - 1}} {<} {2n} \le {2\times2^{r}}}}\Rightarrow {2^{r - 1} {<} n \le {2^r}} \Rightarrow n\in ]2^{r - 1} , 2^r]

ò pa ipotèz rekirans , inegalite a vre pou tout n ki nan entèval presedan an . Inegalite a vre pou \theta\left(n\right) Kidonk selon premye dediksyon an li vre pou \theta\left(2n\right) . Li vre donk pou tout m pè ki nan entèval  ]{2^r} , {2^{r + 1}}]

2)

m enpè  \Leftrightarrow \exists n\in \mathbb{N}{/}m = {2n - 1}

\{_{m={2n-1}}^{2^r {<} m \le {2^{r+1}}}\Rightarrow {2^r {<} {2n - 1} \le {2^{r+1}}} \Rightarrow{{2^r + 1} < {2n} \le {2^{r +1} +1}}\Rightarrow{\{_{2^r < {2n} < 2^{r + 1} + 1}^{{2n}\ne 2^{r + 1} + 1}}\Rightarrow{2^{r} {<}{2n}\le{2^{r + 1}}}\Rightarrow \frac{2^r}{2}  {<}n\le \frac{2^{r + 1}}{2}\Rightarrow {2^{r - 1} {<} n \le{2^r}}\Rightarrow n\in ]2^{r - 1} , 2^r]


selon ipotèz rekirans

\theta\left( n\right) {<} {2n}Ln2

inegalite a vre pou tout n ki nan entèval presedan an . Inegalite a vre pou \theta\left(n\right) 

Kidonk selon premye dediksyon an li vre pou \theta\left(2n - 1\right) . Li vre donk pou tout m enpè ki nan entèval  ]{2^r} , {2^{r + 1}}]



An fendkont pwopriyete a vre pou tout {n}\ge{2}


\forall n\ge 2, \theta\left( n\right) {<} {2n}Ln2 

Teyorèm Legendre[edite | modifier le wikicode]

Si nou gen yon nonm premye fiks, pi gwo ekspozan k tèl ke {p^k}\mid n! alò

 k = \sum_{i=1}^{\infty}{\left[\frac{n}{p^i}\right]}

p^k \|{n!}

An nou demontre sa pa rekirans

fòmil la vre pou n = 1

 {\sum_{i=1}^{\infty}{\left[\frac{1}{p^i}\right]}} = 0

 p^i > 1

 {\frac{1}{p^i}} < 1

an sipoze fòmil la bon pou  n epi an nou montre li bon pou n + 1


swa p^\alpha\|{n+1}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left[\frac{n + 1}{p^i}\right]} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n + 1}{p^i}\right]}}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left(\left[\frac{n }{p^i}\right] + 1\right)} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i = 1}^{\alpha}{\left[\frac{n }{p^i}\right]} + \sum_{i = 1}^{\alpha}{1} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} ={ \sum_{i = 1}^{\alpha}{\left[\frac{n }{p^i}\right]} + \alpha + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]}}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {{\sum_{i = 1}^{\alpha}{\left[\frac{n }{p^i}\right]} + \sum_{i=\alpha + 1}^{\infty}{\left[\frac{n}{p^i}\right]} + \alpha }}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\sum_{i=1}^{\infty}{\left[\frac{n}{p^i}\right]}} + {\alpha}


{\sum_{i=1}^{\infty}{\left[\frac{n+1}{p^i}\right]}} = {\alpha + k}


selon ipotèz rekirans lan

ò  {\left(n + 1\right)!} = {\left(n + 1 \right)\times{n!}}

 {p^k} \| {n!}

ak

 {p^{\alpha}} \| { n + 1}

enplike

{ {p^{\alpha}}\times {p^k}}\| {{ \left(n + 1 \right)}\times {n!}}

Sa ki enplike

 {p^{\alpha + k}} \| {{\left( n + 1\right)}!}


kidonk si fòmil la vre pou  n li vre pou  n + 1 kidonk li vre pou tout antye n ki pa egal ak zero.


premye demonstrasyon an answa[edite | modifier le wikicode]

An nou montre an premye lye ke  \theta\left(2n\right) - \theta\left(n\right) > 0

pou chak  n> 2^6 sa ki va pwouve ekzistans de yon nonm premye pou pi piti nan entèval  ]{n} , {2n}[ pou chak  n > {2^6}

An nou konsidere yon lòt fwa nonm

 N =  \dbinom{2n}{n}

 N =  \dbinom{2n}{n} = \frac{{\left(2n\right)}!}{{n!}\times\left(2n - n\right)!} = \frac{{\left(2n\right)}!}{{\left(n!\right)}^2}

Oken nonm premye ki siperyè ak 2n pa yon divizè de N .

Tandiske tout nonm premye ki konpri ant n ekstrikteman ak 2n lajman divize N yon sèl fwa .

An nou konsidere n ak 2n pou nou wè kisa ki rive .

 N = \frac{\left(2n\right)!}{\left(n!\right)\times\left(n!\right)}

si n premye, li parèt de fwa anlè a sou fòm n ak  2\times n

menm jan an li parèt 2 fwa anba a sèlman. Kidonk, n pa yon faktè de N .

Pou 2n , si  n>1 2n pa yon nonm premye.

Nan entèval  ]n, 2n[ oubyen ankò  ]2 , 2n] tout nonm premye se divizè N yon sèl fwa, paske nonm premye sa a prezante yon sèl fwa nan nimeratè a e li pa prezante nan denominatè a ,

De tout fason nou ap gen pou nou retounen sou konsiderasyon sa a .

Si nou konsidere teyorèm Legendre lan nou kapab ekri :

 N = \frac{{2n}!}{{n!}\times\left(n!\right)} = \frac{\prod_{p\le {2n}}{\left(p^{\sum_{i = 1}^{\infty}{\left[\frac{2n}{p^i}\right]}}\right)}}{{\prod_{p\le n}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}}

An nou konsidere  \prod_{n < p \le {2n}}{p^{2\times\sum_{i=1}^{\infty}{\left(\left[\frac{n}{p^i}\right]\right)}}}

Pwodwi sa vo 1 paske ekspozan toujou vo 0 .

pou tout antye n > 1 , 2n pa premye kidon

 \forall n > 1  , {n < p \le {2n}} \Rightarrow {n < p < {2n}} \Rightarrow {{\frac{1}{2n}} < {\frac{1}{p}} <{\frac{1}{n}}}  \Rightarrow {{\frac{n}{2n}} < {\frac{n}{p}} <{\frac{n}{n}}} \Rightarrow {{\frac{1}{2}} < {\frac{n}{p}} < 1} \Rightarrow{\left[\frac{n}{p}\right] = 0}

pou tout  i \ge 1

 \frac{n}{p^i} < \frac{n}{p} kidonk  \left[\frac{n}{p^i}\right]= 0 pou tout  i \ge 1

\prod_{n<p\le{2n}}{p^{2\times \sum_{i=1}^{\infty}{\left(\left[\frac{n}{p^i}\right]\right)}}} =\prod_{n<p\le{2n}}{p^{2\times 0}} =1


Nou ka ekri

 N = \frac{{2n}!}{{n!}\times\left(n!\right)} = \frac{\prod_{p\le {2n}}{\left(p^{\sum_{i = 1}^{\infty}{\left[\frac{2n}{p^i}\right]}}\right)}}{{\prod_{p\le n}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}}


 N = \frac{{2n}!}{{n!}\times\left(n!\right)} = \frac{\prod_{p\le {2n}}{\left(p^{\sum_{i = 1}^{\infty}{\left[\frac{2n}{p^i}\right]}}\right)}}{{{\prod_{p\le n}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}}\times{\prod_{n < p \le {2n}}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}}


 N = \frac{{2n}!}{{n!}\times\left(n!\right)} = \frac{\prod_{p\le {2n}}{\left(p^{\sum_{i = 1}^{\infty}{\left[\frac{2n}{p^i}\right]}}\right)}}{{\prod_{p\le {2n}}{\left(p^{2\times\sum_{i = 1}^{\infty}{\left[\frac{n}{p^i}\right]}}\right)}}}


 N = \frac{{2n}!}{{n!}\times\left(n!\right)} = \prod_{p\le {2n}}{\left(p^{\left(\sum_{i=1}^{\infty}{\left[\frac{2n}{p^i}\right]} - 2\sum_{i=1}^{\infty}{\left[\frac{n}{p^i}\right]} \right)}\right)}


 N =  \prod_{p\le {2n}}{\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)}

An nou pran logarithm Neperyen de manmb egalite ya .


 Ln N = Ln\left( \prod_{p\le {2n}}{\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)}\right)

 Ln N =\sum_{p\le{2n}}{\left(Ln\left({\left(p^{\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\right)}\right)\right)}


 Ln N =\sum_{p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}

An nou diseke sòm sa an 4 tèrm .


 Ln N =\sum_{n < p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}    +   \sum_{{\frac{2n}{3}} < p\le{n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}                        +    \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}                                     +   \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}

pou premye tèrm lan nou genyen :

 \sum_{n < p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} = \sum_{n < p\le{2n}}{\left(Ln p\right)}

....

kidonk

 \sum_{n < p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} = \sum_{1 \le {p} \le{2n}}{\left(Ln p\right)} - \sum_{1 \le {p} \le{n}}{\left(Ln p\right)} = \theta\left(2n\right) - \theta\left(n\right)


 n < p \le {2n} \Rightarrow {\frac{1}{2n} \le{\frac{1}{p}}<{\frac{1}{n}}}    \Rightarrow {\frac{n}{2n} \le{\frac{n}{p}}<{\frac{n}{n}}}    \Rightarrow {\frac{1}{2} \le{\frac{n}{p}}< 1}     \Rightarrow \left[\frac{n}{p}\right] = 0



 \forall i > 0 , \frac{n}{p^i}

 \left[\frac{n}{p^i}\right] = 0


sou yon lòt ang

 n < p \le {2n} \Rightarrow {\frac{1}{2n} \le{\frac{1}{p}}<{\frac{1}{n}}}    \Rightarrow {\frac{2n}{2n} \le{\frac{2n}{p}}<{\frac{2n}{n}}}    \Rightarrow {1 \le{\frac{2n}{p}}< 2}     \Rightarrow \left[\frac{2n}{p}\right] = 1


 \forall p , p\le 2

 p\le 2 \Rightarrow \frac{1}{p} \le \frac{1}{2}

nou genyen sou baz ipotèz la

 \frac{1}{p} \le \frac{2n}{p^2} < \frac{2}{p} \le 1

 \frac{2n}{p^2} < 1

pou tout  i > 2 , \left[ \frac{2n}{p^i}\right]=0





An nou konsidere dezyèm sòm lan


\sum_{{\frac{2n}{3}} < p\le{n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}


 \frac{2n}{3} < p \le n  \Rightarrow {\frac{1}{n}\le\frac{1}{p}<\frac{3}{2n}} \Rightarrow {\frac{n}{n}\le\frac{n}{p}<\frac{3n}{2n}} \Rightarrow 1\le\frac{n}{p}<\frac{3}{2} \Rightarrow \left[\frac{n}{p}\right] = 1


 \frac{2n}{3} < p \le n  \Rightarrow {\frac{1}{n}\le\frac{1}{p}<\frac{3}{2n}}\Rightarrow {\frac{2n}{n}\le\frac{2n}{p}<3}\Rightarrow {2\le\frac{2n}{p}<3}\Rightarrow \left[\frac{2n}{p}\right] = 2

 \left[\frac{2n}{p}\right] - 2\left[\frac{n}{p}\right] = 2 - 2\times 1 = 0

sou yon lòt ang ,

 \frac{1}{p} \le \frac{n}{p^2} < \frac{3}{2p} < \frac{3}{2\times2}


 \left[\frac{n}{p^2}\right] = 0 


 \frac{2}{p}\le\frac{2n}{p^2} < \frac{3\times2}{2\times p}


 \frac{2}{p}\le\frac{2n}{p^2} < \frac{3}{p}


si p = 2

 1\le\frac{2n}{2^2} < \frac{3}{2}

nap genyen lè sa a :

\left[\frac{2n}{p^2}\right] = 1


 1\le\frac{n}{2} < \frac{3}{2}

kidonk tou nou tap genyen

 n < 3


si  n \le 3 

An nou gade ki sa nou ap genyen

 2n \ge 3\times 2

 2n \ge 6  \frac{2n}{3}\ge 2

 2\le \frac{2n}{3}<p\le n

kidonk  p> 2

 p > 3  \frac{2n}{p^2} < \frac{3}{p}


 \frac{1}{p} < \frac{1}{3}

 \frac{3}{p} < 1

 \frac{2n}{p^2} < \frac{3}{p} < 1

\frac{2n}{p^2} < 1

 \left[\frac{2n}{p^2}\right] = 0

pou  n\le 3 e  i \ge 2


 \left[\frac{2n}{p^i}\right] = 0

 \left[\frac{n}{p^i}\right] = 0


 \left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] = 0


kidonk pou tout  n\ge 3


\sum_{{\frac{2n}{3}} < p\le{n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} = 0


An nou konsidere twazyèm sòm lan


\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}

An konsidere nan ki kondisyon inegalite ki anba somasyon ak gen sans :

 {\sqrt{2n}} \le \frac{2n}{3}

2n\le \frac{4n^2}{9}

18n\le4n^2

18n - 4n^2 \le 0

2n\left(9 -2n\right) \le 0

 9 - 2n \le 0

 9\le 2n

 \frac{9}{2}\le n

n\ge\frac{9}{2}

n\ge 5




An nou endike dabò ke  \left[2\times\frac{n}{p^r}\right] - 2\left[\frac{n}{p^r}\right]  toujou egal oubyen ak zero oubyen ak 1 .


anfèt pou tout reyèl x nou genyen

 \left[2x\right] - 2\left[x\right] egal oubyen a 0 oubyen ak 1 .

 x = \left[x\right]+\{x\}

 2x = 2\left[x\right]+ 2\{x\}

 \left[2x\right] = \left[2\left[x\right]+ 2\{x\}\right]

\left[2x\right] = 2\left[x\right] + \left[2\{x\}\right]

\left[2x\right] - 2\left[x\right] = \left[2\{x\}\right]


 0\le\{x\}<1

0\le 2\{x\}<2

\left[2x\right] - 2\left[x\right] = \left[2\{x\}\right] egal swa zero soit 1 .


\left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] vo 0 oubyen 1 si  i = 1

An nou montre ke li vo 0 nan tout lòt ka .

\sqrt{2n}< p \le \frac{2n}{3}\Rightarrow 2n < p^2 \le \frac{4n^2}{9} \Rightarrow \frac{9}{4n^2}\le\frac{1}{p^2}<\frac{1}{2n}\Rightarrow \frac{9n}{4n^2}\le\frac{n}{p^2}<\frac{n}{2n} \Rightarrow \frac{9n}{4n^2}\le\frac{n}{p^2}<\frac{1}{2} \Rightarrow \left[\frac{n}{p^2}\right] = 0

\sqrt{2n}< p \le \frac{2n}{3}\Rightarrow 2n < p^2 \le \frac{4n^2}{9} \Rightarrow \frac{9}{4n^2}\le\frac{1}{p^2}<\frac{1}{2n}\Rightarrow \frac{9n}{4n^2}\le\frac{n}{p^2}<\frac{n}{2n} \Rightarrow \frac{9n}{2n^2}\le\frac{2n}{p^2}<1 \Rightarrow \frac{2n}{p^{2+j}} < 1 ,  \forall j \in \mathbb{N}

 \forall i\ge 2 , \frac{2n}{p^i} < 1

 \forall i\ge 2 , \left[\frac{2n}{p^i}\right] = 0


 \forall n\ge 5

\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =   \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right) }     +  \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=2}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =  \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right) }    + 0 = \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right) }

ò

 \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] oubyen egal ak 1 oubyen egal ak 0

Kidonk  \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \le 1


{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right)}\le Ln p

{\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right) }}\le {\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{Ln p }} 

 {\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{Ln p }} = \theta\left(\frac{2n}{3}\right) - \theta\left(\sqrt{2n}\right)

{\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left(\left( \left[  \frac{2n}{p} \right] - 2\left[\frac{n}{p}  \right] \right)\times Ln P\right) }}\le{\theta\left(\frac{2n}{3}\right) - \theta\left(\sqrt{2n}\right)}


\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}\le{\theta\left(\frac{2n}{3}\right) - \theta\left(\sqrt{2n}\right)}


 \theta\left(\sqrt{2n}\right) = Ln2 + Ln3+ ... + Ln p_{\left(\pi\left(\sqrt{2n}\right)\right)}\ge{\pi\left(\sqrt{2n}\right)\times{Ln2}}

 - \theta\left(\sqrt{2n}\right) \le {- \pi\left(\sqrt{2n}\right)\times{Ln2}}


\theta\left( \frac{2n}{3}\right) - \theta\left(\sqrt{2n}\right) \le {\theta\left( \frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)\times{Ln2}}

Pou tout  n \ge 5

\sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le  {\theta\left( \frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)\times{Ln2}}



An nou konsidere katryèm sòm lan





\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}


An nou raple ke \left[\frac{2n}{p^i}\right] - 2\left[\frac{n}{p^i}\right] swa egal ak 0 swa egal ak 1 .

yon lòt kote  \left[\frac{2n}{p^i}\right] \ge {  2\left[\frac{n}{p^i}\right]}


An nou gade apati de kilè   \left[ \frac{2n}{p^i}\right] toujou vo zero pou yon p fikse


  \left[ \frac{2n}{p^i}\right] = 0 \Leftrightarrow  \frac{2n}{p^i} < 1 \Leftrightarrow {2n} < {p^i} \Leftrightarrow {p^i} > {2n}\Leftrightarrow Ln{p^i} > Ln{2n} \Leftrightarrow i Ln{p} > Ln{2n} \Leftrightarrow i > \frac{Ln2n}{Ln p} \Leftrightarrow i > \left[ \frac{Ln2n}{Ln p}\right] \Leftrightarrow i \ge \left[ \frac{Ln2n}{Ln p}\right] + 1 , etandone ke i se yon antye natirèl ki pa egal ak 0 .


 i \ge \left[ \frac{Ln2n}{Ln p}\right] + 1 \Rightarrow \left[ \frac{2n}{p^i}\right] -2\left[ \frac{n}{p^i}\right] = 0


sou ipotèz  n\ge 5

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}   +   \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{\left[ \frac{Ln2n}{Ln p}\right] + 1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}


\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}  + 0


\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} =    \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le{\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\left[ \frac{Ln2n}{Ln p}\right]}{1}\right)}\times Ln p\right)}}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[ \frac{Ln2n}{Ln p}\right]\times Ln P\right)}


\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\frac{Ln 2n}{Ln p}\times Ln p\right)}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{Ln 2n}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \sum_{p\le{\sqrt{2n}}}{\left(\left[\frac{Ln 2n}{Ln p}\right]\times Ln p\right)} \le Ln 2n\sum_{p\le{\sqrt{2n}}}{1}

\sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)} \le \pi\left(\sqrt{2n}\right)Ln 2n


An nou raple ke  Ln N =\sum_{n < p\le{2n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}    +   \sum_{{\frac{2n}{3}} < p\le{n}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}                        +    \sum_{{\sqrt{2n}}<p\le{{\frac{2n}{3}}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}                                     +   \sum_{p\le{\sqrt{2n}}}{\left({\left(\sum_{i=1}^{\infty}{\left(\left[\frac{2n}{p^i}\right]- 2\left[\frac{n}{p^i}\right]\right)}\right)}\times Ln p\right)}


lè nou anvizaje tout estimasyon yo oswa tout estime yo nou kapab ekri :

pou tout  n\le 5

 Ln N \le \left(\theta\left(2n\right) - \theta\left(n\right)\right)  + 0 + \left( \theta\left(\frac{2n}{3}\right)-\pi\left(\sqrt{2n}\right)\times  Ln 2\right) + \left( \pi\left(\sqrt{2n}\right)\times  Ln 2n\right)


\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) + \pi\left(\sqrt{2n}\right)\left(Ln2 - Ln 2n\right)


\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)\left(Ln\left(\frac{2n}{2}\right)\right)


\theta\left(2n\right) -\theta\left(n\right) \ge Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)Ln2


Li rete pou nou pwouve ke Ln N - \theta\left(\frac{2n}{3}\right) - \pi\left(\sqrt{2n}\right)Ln estriteman pozitif .

Nou te genyen  N > \frac{2^n}{2\sqrt{n}}

 Ln N > Ln \left( \frac{2^n}{2\sqrt{n}}\right)

Ln N > Ln 2^{2n} - ln{2\sqrt{n}}

 Ln N > 2n ln2 - ln\left(2\sqrt{n}\right)

Ln N > 2nLn2 - Ln2 -Ln{\sqrt{n}}

 Ln N > 2n Ln2 -Ln2 - Ln{n^{\frac{1}{2}}}


  Ln N >2n Ln2 -Ln2 - \frac{1}{2}Ln n

Anplis lè  n \le 2 , nou genyen

 \theta\left(\frac{2n}{3}\right) = \theta\left(\left[\frac{2n}{3}\right]\right) < 2\times\left[\frac{2n}{3}\right]Ln 2 < \frac{4n}{3}Ln 2


 - \theta\left(\frac{2n}{3}\right) > - \frac{4n}{3}Ln2

yon lòt kote,  \pi\left(n\right)\le\frac{n}{2}, \forall n\ge 8 , etandone ke tout nonm pè pi gran ke 2 pa premye ou di mwens pa konpoze .

Nou vin genyen :


\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 - Ln\left(2\sqrt{n}\right)-\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n


\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 -Ln2 -Ln{\sqrt{n}} -\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n


\theta\left(2n\right) -  \theta\left(n\right) > 2nLn2 -Ln2 -\frac{1}{2}Lnn -\frac{4n}{3}Ln2 - \frac{\sqrt{2n}}{2}Ln n





 \theta\left(2n\right) -  \theta\left(n\right) > \left(2n - \frac{4n}{3} - 1\right)Ln2 - \left(\frac{1}{2} + \frac{\sqrt{2n}}{2}\right)Lnn


 \theta\left(2n\right) -  \theta\left(n\right) > \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n

Li rete pou nou pwouve ke manm dwat la pi gran ke zero e konsa ant n ak 2n nap va asire ke gen o mwen yon nonm premye .

 \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n  > 0 \Leftrightarrow \sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}} > 0



An nou pwouve ekivalans lan


 \sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}}>0 \Leftrightarrow \frac{2\sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} }{2\sqrt{4n}Ln 2} > 0 \Leftrightarrow \sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 


\sqrt{4n}Ln2\times{\sqrt{2n}}- 3\sqrt{4n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 \Leftrightarrow 2\sqrt{8n^2}Ln2 -6\sqrt{n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n - 6\sqrt{2}Ln{\sqrt{4n}} > 0 \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln{\sqrt{4}}+Ln{\sqrt{n}}\right)        \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln2 + \frac{1}{2}Ln2\right)>0


   \Leftrightarrow 4n\sqrt{2}Ln2 - 6\sqrt{n}Ln n -6\sqrt{2}\left(Ln2 + \frac{1}{2}Ln2\right)>0 \Leftrightarrow \frac{4n}{6}\sqrt{2}Ln2 -\sqrt{n}Ln n - \sqrt{2}Ln2 - \frac{\sqrt{2}}{2}Ln n > 0  \Leftrightarrow   \frac{2n}{3}\sqrt{2}Ln2 -\sqrt{n}Ln n - \sqrt{2}Ln2 - \frac{\sqrt{2}}{2}Ln n > 0  \Leftrightarrow \sqrt{2}\times \left( \frac{2n}{3}Ln 2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n -Ln 2 -Ln \sqrt{n}\right) > 0

\Leftrightarrow \sqrt{2}\times \left( \frac{2n}{3}Ln 2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n -Ln 2 -Ln \sqrt{n}\right) > 0 \Leftrightarrow \left(\frac{2n}{3} - 1\right)Ln2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n - Ln {\sqrt{n}} > 0  \Leftrightarrow \left(\frac{2n}{3} - 1\right)Ln2 -\frac{\sqrt{2}}{2}\sqrt{n}Ln n - \frac{1}{2}Ln n > 0 \Leftrightarrow \left(\frac{2n}{3} - 1 \right)Ln2  - \left( \frac{1 + \sqrt{2n}}{2}\right)Ln n  > 0 

An nou deziye pa  f\left(n\right)

 \sqrt{2n} -\frac{3}{2}\frac{Ln n}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4n}}}{\sqrt{4n}}

An nou montre ke li pozitif pou tout  n \ge 2^6


An etann fonksyon an sou tout  \mathbb{R}

 f\left(x\right) =  \sqrt{2x} -\frac{3}{2}\frac{Ln x}{Ln 2} -\frac{3\sqrt{2}}{Ln2}\frac{Ln{\sqrt{4x}}}{\sqrt{4x}}


an nou pwouve ke  f\left(2^6\right) > 0


 f\left(2^6\right) = 2^3\times\sqrt{2} - \frac{3}{2}\times\frac{6Ln2}{Ln2} - \frac{3\sqrt{2}}{Ln2}\times\frac{Ln\sqrt{4\times2^6}}{\sqrt{4\times2^6}} = 8\sqrt{2} - 9 - \frac{3\sqrt{2}}{Ln2}\times\frac{4Ln2}{2^4}  =  8\sqrt{2} - 9 - \frac{12\sqrt{2}}{2^4} =   8\sqrt{2} - 9 - \frac{3\sqrt{2}}{4} = \frac{32\sqrt{2} - 36 - 3 \sqrt{2}}{4} = \frac{29\sqrt{2} -36}{4}


 f\left(2^6\right) = \frac{29\sqrt{2} -36}{4} = \frac{\left({29\sqrt{2} -36}\right)\times\left({29\sqrt{2} +36}\right)}{4}\times\frac{1}{{29\sqrt{2} +36}} = \frac{\left(29\times29\times2 - 36\times36\right)}{4}\times\frac{1}{{29\sqrt{2} +36}} = \frac{386}{4}\times\frac{1}{{29\sqrt{2} +36}} > 0

an nou pwouve tou ke  f'\left(x\right) > 0 pou tout  n \ge 2^6


kidonk fonksyon f la estriteman kwasant


   \left(\sqrt{2x}\right)' = \left( \left(2x\right)^{\frac{1}{2}}\right) = \frac{1}{2}\left(2x\right)^{-\frac{1}{2}}\times2 = \left(2x\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x}} = \frac{\sqrt{2x}}{\sqrt{2x}\times \sqrt{2x}} = \frac{\sqrt{2x}}{2x}

 \left(-\frac{3}{2}\frac{Lnx}{Ln2}\right)' = - \frac{3}{2Ln2}\times\frac{1}{x} = \frac{-3}{2xLn2}


 \left(\sqrt{4x}\right)'  = \left( \left(4x\right)^{\frac{1}{2}}\right)' = \frac{1}{2}\times\left(4x\right)^{\frac{1}{2}}\times4 = 2\times\left(4x\right)^{\frac{1}{2}} = \frac{2}{\sqrt{4x}}= \frac{2\sqrt{4x}}{4x} = \frac{\sqrt{4x}}{2x} = \frac{\sqrt{x}}{x}


 \left(Ln\sqrt{4x}\right)' = \frac{1}{\sqrt{4x}}\times\frac{\sqrt{4x}}{2x} = \frac{1}{2x}

oubyen

 \left(Ln\sqrt{4x}\right)' = \left(Ln\left( 4x\right)^{\frac{1}{2}}\right)' = \left( \frac{1}{2}Ln{4x}\right)' = \frac{1}{2}\times\frac{1}{4x}\times 4 = \frac{4}{8x} = \frac{1}{2x}

 \left(\frac{Ln\sqrt{4x}}{\sqrt{4x}}\right)' = \frac{\frac{1}{2x}\times\sqrt{4x} - \frac{\sqrt{4x}}{2x}\times Ln{\sqrt{4x}}}{\left({\sqrt{4x}}\right)^2}  =  \frac{\frac{\sqrt{4x}-\sqrt{4x}\times Ln{\sqrt{4x}}}{2x}}{4x}  = \frac{\sqrt{4x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2}



 \left(\frac{-3\sqrt{2}}{Ln2}\times\frac{Ln\sqrt{4x}}{\sqrt{4x}}\right)' = \frac{\frac{\sqrt{4x}-\sqrt{4x}\times Ln{\sqrt{4x}}}{2x}}{4x}  = \frac{-3\sqrt{2}}{Ln2}\times\frac{\sqrt{4x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2}


 f'\left(x\right) = \frac{\sqrt{2x}}{2x} - \frac{3}{2xln2}  - \frac{3\sqrt{2}}{Ln2}\times \frac{\sqrt{4x}\times\left( 1- Ln{\sqrt{4x}} \right)}{8x^2}

 f'\left(x\right) =  \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x}\left(1 - Ln{\sqrt{4x}}\right)}{8x^2Ln2}


 f'\left(x\right) =  \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}ln{\sqrt{4x}}}{8x^2Ln2}

f'\left(x\right)>0\Leftrightarrow \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}Ln{\sqrt{4x}}}{8x^2Ln2}> 0 \Leftrightarrow 4x\times Ln2\times\sqrt{2x} + 3\sqrt{8x}\times Ln{\sqrt{4x}} > 12x +3\sqrt{8x}


si nou miltiplye pa  \sqrt{2x} \ne 0


f'\left(x\right)>0\Leftrightarrow \frac{4x\times Ln2\times\sqrt{2x} - 12x - 3\sqrt{8x} + 3\sqrt{8x}Ln{\sqrt{4x}}}{8x^2Ln2}> 0 \Leftrightarrow 4x\times Ln2\times\sqrt{2x} + 3\sqrt{8x}\times Ln{\sqrt{4x}} > 12x +3\sqrt{8x} \Leftrightarrow 4xLn2\times2x + 3\times4x\times Ln{\sqrt{4x}} > 12x \times\sqrt{2x} + 3\times 4x \Leftrightarrow 8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12


sou ipotèz  x > 2^6 an nou pwouve ke  8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12


 x > 2^6 \Rightarrow x> 18 \Rightarrow 16x > 288 \Rightarrow 16x \times Ln4 \times Ln4 > 288\times Lne \times Lne \Rightarrow 2\times 2\times Ln2 \times Ln2 \times 16x \Rightarrow 64x\times \left(ln2\right)^2>288 \Rightarrow 64x^2\times \left(ln2\right)^2>288x \Rightarrow 64x^2\times \left(ln2\right)^2>144\times 2x \Rightarrow 8xln2 > 12\sqrt{2x}

yon lòt kote nou genyen

 12 Ln\sqrt{4x} > 12 \Leftrightarrow Ln\sqrt{4x} > 1  \Leftrightarrow sqrt{4x} > e \Leftrightarrow 4x > e^2 \Leftrightarrow x > \frac{e^2}{4}

non genyen tou :

 e < 3

 0 < e < 3 \Rightarrow e^2 < 9 \Rightarrow \frac{e^2}{4} < \frac{9}{4}

 2^6 > \frac{9}{4} > \frac{e^2}{4}

 x > 2^6 \Rightarrow x > \frac{e^2}{4} \Rightarrow 4x > e^2 \Rightarrow \sqrt{4x} > e \Rightarrow Ln\sqrt{4x} > 1 \Rightarrow 12 Ln\sqrt{4x} > 12


donk


 x > 2^6 \Rightarrow 8xLn2 + 12Ln{\sqrt{4x}} > 12\sqrt{2x} + 12

list nonm premye superyè ou egal ak 3 epi ki enferyè ak 64[edite | modifier le wikicode]

3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61

dezyèm demonstrasyon[edite | modifier le wikicode]

Majorasyon apati postila Bètran[edite | modifier le wikicode]

P_n \le {2^n}

 P_1 = 2 = 2^1  2^2 =4

 P_2 = 3 < 2^2

 P_3 = 5

 3^2 P_3 < 3^2



an sipoze majorasyon an vrè pou  n > 3 e an montre ke li vrè pou n+1

 P_n \le 2^n \Rightarrow \pi\left(P_n\right) \le \pi\left(2^n\right) \Rightarrow n\le \pi\left(2^n\right)\ge n

kidonk nan entèval \left[1, 2^n\right] gen o mwen n nonm premye selon ipotèz rekirans lan .


selon ipotèz Bètran an egziste pou pi piti yon nonm premye nan entèval \left]2^n , 2^{n+1}\right[


nou kapab di tou ke nan entèval \left]2^n , 2^{n+1}\right] gen pou piti yon nonm premye

 \pi\left(2^{n+1}\right) - \pi\left(2^n\right)  \ge 1

 \pi\left(2^{n+1}\right) \ge 1+\pi\left(2^n\right)


 \pi\left(2^{n+1}\right) \ge 1+ n


gen pou pi piti n+1 nonm premye nan entèval


 \left[1, 2^{n+1}\right]

 P_{n+1} < 2^{n+1}

Anfendkont si majorasyon an vrè pou n, li vrè pou n + 1 .


 \forall n \in \mathbb{N} , P_{n}\le 2^n

.....


P_k\left(n\right) \le {n\times{2^k}}


sa nou rele P_k\left(n\right) se k yèm nonm premye ki siperyè oubyen egal ak n .


Avèk , sipoze fikse an nou montre ke majorasyon an vre pou  k+1 .

P_k\left(n\right) \le n\times 2^k

Anvan nou fè sipozisyon an , an nou verifye ke majorasyon vre pou k = 1


 P_1\left( n\right) \le 2n  paske nan entèval \left] n, 2n \right[   li egziste pou pi piti yon nonm premye selon postila Bètran  .

Anfèt si n premye,  P_1\left(n\right) = n

si n pa premye  P_1\left(n\right) \in \left]n, 2n\right]


kidonk  P_1{\left(n\right)} \in \left[n, 2n\right]

 P_1\left(n\right) \le 2n

Postila Bètran avèk de twa fòmil matematisyen Ayisyen Lainé Jean Lhermite Junior konsènan nonm premye[edite | modifier le wikicode]

Ekspresyon nonm premye ran n selon modèl flèch Jonatan anakò avèk postila Bètran an[edite | modifier le wikicode]

  P_n = \sum_{m=1}^{2^n}{\left(    \left[  \frac  {1+\sum_{m=1}^{i}{\left( -\left[ \frac{\left(m!\right)^2}{m^3} \right] -\left[ -  \frac{\left(m!\right)^2}{m^3}   \right]     \right)}  }       {n+1} \right]   \times      \left[ \frac{n+1}  {1+\sum_{m=1}^{i}{\left( -\left[ \frac{\left(m!\right)^2}{m^3} \right] -\left[ -  \frac{\left(m!\right)^2}{m^3}   \right]     \right)}  }    \right] \times i \times \left(  -\left[\frac{\left(i!\right)^2}{i^3}\right] -   \left[-\frac{\left(i!\right)^2}{i^3}\right] \right)       \right)}

K yèm nonm premye ki siperyè oubyen egal ak n[edite | modifier le wikicode]

premye apròch[edite | modifier le wikicode]

P_K\left(n\right) = \left(\sum_{i=1}^{2^k\times n}{\left(\left[\frac{1+\sum_{m=1}^{i}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{k + 1 + \sum_{m=1}^n{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times \left[\frac{k + 1 + \sum_{m=1}^{n}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{ 1 + \sum_{m=1}^i{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times i \times \left(1 - \left[\frac{\left[\frac{{\left(i!\right)}^2}{i^3}\right]}{\frac{\left(i!\right)^2}{i^3}}\right]\right)\right)}\right)\times\left[\frac{\left[\frac{{\left(n!\right)}^2}{n^3}\right]}{\frac{{\left(n!\right)}^2}{n^3}}\right] + {\left(\sum_{i=1}^{2^k\times n}{\left(\left[\frac{1+\sum_{m=1}^{i}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{k + \sum_{m=1}^n{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times \left[\frac{k + \sum_{m=1}^{n}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{ 1 + \sum_{m=1}^i{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times i \times \left(1 - \left[\frac{\left[\frac{{\left(i!\right)}^2}{i^3}\right]}{\frac{\left(i!\right)^2}{i^3}}\right]\right)\right)}\right)\times \left(1 - \left[\frac{\left[\frac{{\left(n!\right)}^2}{n^3}\right]}{\frac{{\left(n!\right)}^2}{n^3}}\right] \right)}

Demonstrasyon...[edite | modifier le wikicode]

nan de evenman kontrè nou gen fòseman youn . Nou kapab di ke oubyen :

 i = P_K\left(n\right)  oubyen i\ne P_K\left(n\right)

De presizyon dabò

nou gen asirans ke  P_K\left(n\right) egziste e nou gen asirans tou ke egziste yon i nan entèval ... kote i = P_K\left(n\right) akoz majorasyon an .

dezyèm apròch[edite | modifier le wikicode]

P_K\left(n\right) = \left(\sum_{i=1}^{2^k\times n}{\left(\left[\frac{1+\sum_{m=1}^{i}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{k + 1 + \sum_{m=1}^{n - 1}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times \left[\frac{k + 1 + \sum_{m=1}^{n - 1}{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}}{ 1 + \sum_{m=1}^i{\left(1 - \left[\frac{\left[\frac{{\left(m!\right)}^2}{m^3}\right]}{\frac{{\left(m!\right)}^2}{m^3}}\right]\right)}} \right]\times i \times \left(1 - \left[\frac{\left[\frac{{\left(i!\right)}^2}{i^3}\right]}{\frac{\left(i!\right)^2}{i^3}}\right]\right)\right)}\right)

twazyèm apròch[edite | modifier le wikicode]

 P_k\left(n\right) = \sum_{i=n}^{2^k\times n}{\left( \left[\frac{1+\sum_{m=1}^i{\left(1-\left[\frac{\left[\frac{\left(m!\right)^2}{m^3}\right]}{\frac{\left(m!\right)^2}{m^3}}\right]\right)}}{k+1}\right]\times\left[\frac{k+1}{1+\sum_{m=1}^i{\left(1-\left[\frac{\left[\frac{\left(m!\right)^2}{m^3}\right]}{\frac{\left(m!\right)^2}{m^3}}\right]\right)}}\right]\times i \times\left(1 - \left[\frac{\left[\frac{\left(i!\right)^2}{i^3}\right]}{\frac{\left(i!\right)^2}{i^3}}\right] \right)\right)}

wè tou[edite | modifier le wikicode]

Referans[edite | modifier le wikicode]

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