Postila Bètran

Postila Bètran

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 ris Tchebycheff Gen plizyè varyant pou demontre teyorèm oubyen postila a. Men de ladan yo ki privilejye apwòch elemantè.

Premyè demonstrasyon

Premye lèm

Premyè demach la ap mennen 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

$N=C_{2n}^n=(\genfrac{}{}{0ex}{}{2n}{n})$

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

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

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

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

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

Konfòmeman ak Arnel Mercier ak Jean Marie de Koninck an desiye pwodui sa pa P

$P=\frac{1\times \times 3\times ...\times (2n-3)\times (2n-1)}{2\times 4\times ...\times (2n-2)\times 2n}$

An nou fè parèt anlè ya pwodui ki anba a

$P=\frac{1\times \times 3\times ...\times (2n-3)\times (2n-1)}{2\times 4\times ...\times (2n-2)\times 2n}\times \frac{2\times 4\times 6\times ...\times (2n-2)\times 2n}{2\times 4\times 6\times ...\times (2n-2)\times 2n}$

$P=\frac{(1\times \times 3\times ...\times (2n-3)\times (2n-1))\times (2\times 4\times 6\times ...\times (2n-2)\times 2n)}{2\times 4\times ...\times (2n-2)\times 2n\times (2\times 4\times 6\times ...\times (2n-2)\times 2n)}$

$P=\frac{\left(2n\right)!}{{(2\times 4\times 6\times ...\times (2n-2)\times 2n)}^2}$

$P=\frac{\left(2n\right)!}{{(2\times 1\times 2\times 2\times 2\times 3\times ...\times 2\times (n-1)\times 2\times n)}^2}$

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

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

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

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

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

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

An nou konsidere

$\prod _{i=1}^n(1-\frac{1}{{\left(2i\right)}^2})$

si nou devlope pwodui nap genyen

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

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

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

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

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

sa ki bay

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

lè nou diseke nimeratè yo ak denominatè yo :

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

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

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

$P\times P\times (2n+1)<1$

$P^2\times (2n+1)<1$

$P^2\times 2n<P^2\times (2n+1)<1$

kidonk

$P^2\times 2n<1$

Nou te wè ke

$2^{2n}P=N$

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

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

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

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

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 faktoryèl la se nosyon de predileksyon nonm premye

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

An nou konsidere pwodui sa :

$\prod _{i=2}^n(1-\frac{1}{{(2i-1)}^2})$

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

$\prod _{i=2}^n(1-\frac{1}{{(2i-1)}^2})<1$

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

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

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

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

sa ki bay

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

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

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

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

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

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

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

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

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

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

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

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

$\theta \left(x\right)=\sum _{p\le x}^{}Lnp$ avèk $x\le 0$ ...

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

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

$\theta \left(2\right)=Ln2<2\times 2\times Ln2$

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

An nou sipoze li vrè pou yon sèten $n>2$

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

$\frac{N}{2}=\frac{1}{2}N=\frac{1}{2}(\genfrac{}{}{0ex}{}{2n}{n})=\frac{\left(2n\right)!}{2\times (n!)\times n!}=\frac{((2n-1)!)\times 2n}{2\times (n!)\times n!}=\frac{((2n-1)!)\times n}{(n!)\times ((n-1)!)\times n}=\frac{(2n-1)!}{((n-1)!)\times n!}=\frac{(2n-1)}{((n-1)!)\times ((2n-1)-(n-1))!}=C_{2n-1}^{n-1}=(\genfrac{}{}{0ex}{}{2n-1}{n-1})$

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

$\frac{(2n-1)!}{(n!)\times ((n-1)!)}=\frac{1\times 2\times 3\times ...\times (2n-2)\times (2n-1)}{(1\times 2\times 3\times ...\times (n-1)\times n)\times (1\times 2\times 3\times ...\times (n-2)\times (n-1))}$

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.

$n<p\le 2n-1$

$p\nmid ((n!)\times ((n-1)!))$

$p\mid (2n-1)!$

kidonk yon faktè premye p ki tèl ke $n<p\le 2n-1$ pa senplifye.

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

$\frac{N}{2}$ siperyè oubyen egal ak pwodui 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 pwodui an sòm logaritm

$Ln\frac{N}{2}\ge \sum _{n<p\le 2n-1}Lnp$

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

$Ln\frac{N}{2}\ge (\sum _{1\le p\le n}Lnp+\sum _{n<p\le 2n-1}Lnp)-\sum _{1\le p\le n}Lnp$

$Ln\frac{N}{2}\ge \sum _{1\le p\le 2n-1}Lnp-\sum _{1\le p\le n}Lnp$

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

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

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

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

$LnN<Ln{2}^{2n}-Ln{\left(2n\right)}^{\frac{1}{2}}$

$LnN<2nLn2-\frac{1}{2}Ln2n$

An nou konbine

${\{}_{\theta (2n-1)-\theta \left(n\right)\le Ln\frac{N}{2}}^{LnN<2nLn2-\frac{1}{2}Ln2n}$

$\theta (2n-1)-\theta \left(n\right)+LnN<Ln\frac{N}{2}+2nLnN-\frac{1}{2}Ln2n$

$\theta (2n-1)-\theta \left(n\right)<Ln\frac{N}{2}-LnN+2nLn2-\frac{1}{2}Ln2n$

$\theta (2n-1)-\theta \left(n\right)<Ln\frac{\frac{N}{2}}{N}+2nLog2-\frac{1}{2}Ln2n$

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

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

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

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

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

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

....

kòm $n>2$

$2n>2\times 2$

$2n>2^2$

$Ln2n>Ln{2}^2$

$Ln2n>2Ln2$

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

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

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

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

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

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

$\theta (2n-1)<2(2n-1)Ln2$

si nou gen $n\ge 2$ alò $2n\ge 4$

$2n\ne 2$

$2n\notin \mathbb{P}$

${\{}_{n\in \mathbb{N}}^{n\ge 2}\Rightarrow 2n\notin \mathbb{P}\iff \mathrm{\exists }p\in \mathbb{P}/2n-1<p\le 2n$

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

kidonk

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

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.

$\theta (2n-1)<2(2n-1)Ln2$

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

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

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

$\theta \left(2n\right)<4nLn2$

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

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 vrè pou n = 2 selon enplikasyon avan an li vrè pou $2\times 2-1etpou2\times 2$ ki donk li vre e pou 3 e pou 4. Pwopriyete a vrè pou $[2,4]$ oubyen ankò li vrè pou $]2,4]$ Pwopriyete a vrè 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 vrè 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 $m\in ]2^r,2^{r+1}]$

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

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 $\iff \mathrm{\exists }n\in \mathbb{N}/m=2n$

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

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

2)

m enpè $\iff \mathrm{\exists }n\in \mathbb{N}/m=2n-1$

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

selon ipotèz rekirans

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

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

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

An fendkont pwopriyete a vrè pou tout $n\ge 2$

$\mathrm{\forall }n\ge 2,\theta \left(n\right)<2nLn2$

Teyorèm Legendre

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

$k=\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]$

$p^k\Vert n!$

An nou demontre sa pa rekirans

fòmil la vrè pou $n=1$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{1}{p^i}\right]=0$

$p^i>1$

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

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

swa $p^{\alpha }\Vert n+1$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\alpha }\left[\frac{n+1}{p^i}\right]+\sum _{i=\alpha +1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\alpha }(\left[\frac{n}{p^i}\right]+1)+\sum _{i=\alpha +1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\alpha }\left[\frac{n}{p^i}\right]+\sum _{i=1}^{\alpha }1+\sum _{i=\alpha +1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\alpha }\left[\frac{n}{p^i}\right]+\alpha +\sum _{i=\alpha +1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\alpha }\left[\frac{n}{p^i}\right]+\sum _{i=\alpha +1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]+\alpha$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]+\alpha$

$\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n+1}{p^i}\right]=\alpha +k$

selon ipotèz rekirans lan

ò $(n+1)!=(n+1)\times n!$

$p^k\Vert n!$

ak

$p^{\alpha }\Vert n+1$

enplike

$p^{\alpha }\times p^k\Vert (n+1)\times n!$

Sa ki enplike

$p^{\alpha +k}\Vert (n+1)!$

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

Premyè demonstrasyon an answa

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

$N=(\genfrac{}{}{0ex}{}{2n}{n})$

$N=(\genfrac{}{}{0ex}{}{2n}{n})=\frac{\left(2n\right)!}{n!\times (2n-n)!}=\frac{\left(2n\right)!}{{(n!)}^2}$

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.

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

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 :

$N=\frac{2n!}{n!\times (n!)}=\frac{\prod _{p\le 2n}\left(p^{\sum _{i=1}^{\mathrm{\infty }}\left[\frac{2n}{p^i}\right]}\right)}{\prod _{p\le n}\left(p^{2\times \sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]}\right)}$

An nou konsidere $\prod _{n<p\le 2n}{p}^{2\times \sum _{i=1}^{\mathrm{\infty }}\left(\left[\frac{n}{p^i}\right]\right)}$

Pwodui sa vo 1 paske ekspozan toujou vo 0.

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

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

pou tout $i\ge 1$

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

$\prod _{n<p\le 2n}{p}^{2\times \sum _{i=1}^{\mathrm{\infty }}\left(\left[\frac{n}{p^i}\right]\right)}=\prod _{n<p\le 2n}{p}^{2\times 0}=1$

Nou ka ekri

$N=\frac{2n!}{n!\times (n!)}=\frac{\prod _{p\le 2n}\left(p^{\sum _{i=1}^{\mathrm{\infty }}\left[\frac{2n}{p^i}\right]}\right)}{\prod _{p\le n}\left(p^{2\times \sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]}\right)}$

$N=\frac{2n!}{n!\times (n!)}=\frac{\prod _{p\le 2n}\left(p^{\sum _{i=1}^{\mathrm{\infty }}\left[\frac{2n}{p^i}\right]}\right)}{\prod _{p\le n}\left(p^{2\times \sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]}\right)\times \prod _{n<p\le 2n}\left(p^{2\times \sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]}\right)}$

$N=\frac{2n!}{n!\times (n!)}=\frac{\prod _{p\le 2n}\left(p^{\sum _{i=1}^{\mathrm{\infty }}\left[\frac{2n}{p^i}\right]}\right)}{\prod _{p\le 2n}\left(p^{2\times \sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right]}\right)}$

$N=\frac{2n!}{n!\times (n!)}=\prod _{p\le 2n}\left(p^{(\sum _{i=1}^{\mathrm{\infty }}\left[\frac{2n}{p^i}\right]-2\sum _{i=1}^{\mathrm{\infty }}\left[\frac{n}{p^i}\right])}\right)$

$N=\prod _{p\le 2n}\left(p^{\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)}\right)$

An nou pran logarithm Neperyen de manm egalite ya.

$LnN=Ln\left(\prod _{p\le 2n}\left(p^{\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)}\right)\right)$

$LnN=\sum _{p\le 2n}\left(Ln\left(\left(p^{\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)}\right)\right)\right)$

$LnN=\sum _{p\le 2n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

An nou diseke sòm sa an 4 tèrm .

$LnN=\sum _{n<p\le 2n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{\frac{2n}{3}<p\le n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

pou premye tèm lan nou genyen :

$\sum _{n<p\le 2n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{n<p\le 2n}\left(Lnp\right)$

....

kidonk

$\sum _{n<p\le 2n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{1\le p\le 2n}\left(Lnp\right)-\sum _{1\le p\le n}\left(Lnp\right)=\theta \left(2n\right)-\theta \left(n\right)$

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

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

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

sou yon lòt ang

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

$\mathrm{\forall }p,p\le 2$

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

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

$\sum _{\frac{2n}{3}<p\le n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

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

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

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

sou yon lòt ang ,

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

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

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

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

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 $n\ge 3$

$\sum _{\frac{2n}{3}<p\le n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=0$

An nou konsidere twazyèm sòm lan

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

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

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

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

$18n\le 4n^2$

$18n-4n^2\le 0$

$2n(9-2n)\le 0$

$9-2n\le 0$

$9\le 2n$

$\frac{9}{2}\le n$

$n\ge \frac{9}{2}$

$n\ge 5$

An nou endike dabò ke $[2\times \frac{n}{p^r}]-2\left[\frac{n}{p^r}\right]$ 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 .

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

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

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

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

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

$0\le \{x\}<1$

$0\le 2\{x\}<2$

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

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

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

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

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

$\mathrm{\forall }n\ge 5$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{\sqrt{2n}<p\le \frac{2n}{3}}((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)+\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=2}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{\sqrt{2n}<p\le \frac{2n}{3}}((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)+0=\sum _{\sqrt{2n}<p\le \frac{2n}{3}}((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)$

ò

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

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

$((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)\le Lnp$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)\le \sum _{\sqrt{2n}<p\le \frac{2n}{3}}Lnp$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}Lnp=\theta \left(\frac{2n}{3}\right)-\theta \left(\sqrt{2n}\right)$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}((\left[\frac{2n}{p}\right]-2\left[\frac{n}{p}\right])\times LnP)\le \theta \left(\frac{2n}{3}\right)-\theta \left(\sqrt{2n}\right)$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \theta \left(\frac{2n}{3}\right)-\theta \left(\sqrt{2n}\right)$

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

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

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

Pou tout $n\ge 5$

$\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \theta \left(\frac{2n}{3}\right)-\pi \left(\sqrt{2n}\right)\times Ln2$

An nou konsidere katryèm sòm lan

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

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

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

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

sou ipotèz $n\ge 5$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\left[\frac{Ln2n}{Lnp}\right]}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{p\le \sqrt{2n}}(\left(\sum _{\left[\frac{Ln2n}{Lnp}\right]+1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\left[\frac{Ln2n}{Lnp}\right]}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+0$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)=\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\left[\frac{Ln2n}{Lnp}\right]}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\left[\frac{Ln2n}{Lnp}\right]}1\right)\times Lnp)$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \sum _{p\le \sqrt{2n}}(\left[\frac{Ln2n}{Lnp}\right]\times LnP)$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \sum _{p\le \sqrt{2n}}(\left[\frac{Ln2n}{Lnp}\right]\times Lnp)\le \sum _{p\le \sqrt{2n}}(\frac{Ln2n}{Lnp}\times Lnp)$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \sum _{p\le \sqrt{2n}}(\left[\frac{Ln2n}{Lnp}\right]\times Lnp)\le \sum _{p\le \sqrt{2n}}Ln2n$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \sum _{p\le \sqrt{2n}}(\left[\frac{Ln2n}{Lnp}\right]\times Lnp)\le Ln2n\sum _{p\le \sqrt{2n}}1$

$\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)\le \pi \left(\sqrt{2n}\right)Ln2n$

An nou raple ke $LnN=\sum _{n<p\le 2n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{\frac{2n}{3}<p\le n}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{\sqrt{2n}<p\le \frac{2n}{3}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)+\sum _{p\le \sqrt{2n}}(\left(\sum _{i=1}^{\mathrm{\infty }}(\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right])\right)\times Lnp)$

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

pou tout $n\le 5$

$LnN\le (\theta \left(2n\right)-\theta \left(n\right))+0+(\theta \left(\frac{2n}{3}\right)-\pi \left(\sqrt{2n}\right)\times Ln2)+(\pi \left(\sqrt{2n}\right)\times Ln2n)$

$\theta \left(2n\right)-\theta \left(n\right)\ge LnN-\theta \left(\frac{2n}{3}\right)+\pi \left(\sqrt{2n}\right)(Ln2-Ln2n)$

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

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

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

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

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

$LnN>Ln{2}^{2n}-ln2\sqrt{n}$

$LnN>2nln2-ln\left(2\sqrt{n}\right)$

$LnN>2nLn2-Ln2-Ln\sqrt{n}$

$LnN>2nLn2-Ln2-Ln{n}^{\frac{1}{2}}$

$LnN>2nLn2-Ln2-\frac{1}{2}Lnn$

Anplis lè $n\le 2$ , nou genyen

$\theta \left(\frac{2n}{3}\right)=\theta \left(\left[\frac{2n}{3}\right]\right)<2\times \left[\frac{2n}{3}\right]Ln2<\frac{4n}{3}Ln2$

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

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

Nou vin genyen :

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

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

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

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

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

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 .

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

An nou pwouve ekivalans lan

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

$\sqrt{4n}Ln2\times \sqrt{2n}-3\sqrt{4n}Lnn-6\sqrt{2}Ln\sqrt{4n}>0\iff 2\sqrt{8n^2}Ln2-6\sqrt{n}Lnn-6\sqrt{2}Ln\sqrt{4n}\iff 4n\sqrt{2}Ln2-6\sqrt{n}Lnn-6\sqrt{2}Ln\sqrt{4n}>0\iff 4n\sqrt{2}Ln2-6\sqrt{n}Lnn-6\sqrt{2}(Ln\sqrt{4}+Ln\sqrt{n})\iff 4n\sqrt{2}Ln2-6\sqrt{n}Lnn-6\sqrt{2}(Ln2+\frac{1}{2}Ln2)>0$

$\iff 4n\sqrt{2}Ln2-6\sqrt{n}Lnn-6\sqrt{2}(Ln2+\frac{1}{2}Ln2)>0\iff \frac{4n}{6}\sqrt{2}Ln2-\sqrt{n}Lnn-\sqrt{2}Ln2-\frac{\sqrt{2}}{2}Lnn>0\iff \frac{2n}{3}\sqrt{2}Ln2-\sqrt{n}Lnn-\sqrt{2}Ln2-\frac{\sqrt{2}}{2}Lnn>0\iff \sqrt{2}\times (\frac{2n}{3}Ln2-\frac{\sqrt{2}}{2}\sqrt{n}Lnn-Ln2-Ln\sqrt{n})>0$

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

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

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

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

An etann fonksyon an sou tout $ℝ$

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

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

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

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

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

kidonk fonksyon f la estriteman kwasant

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

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

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

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

oubyen

$\left(Ln\sqrt{4x}\right)=\left(Ln{\left(4x\right)}^{\frac{1}{2}}\right)=\left(\frac{1}{2}Ln4x\right)=\frac{1}{2}\times \frac{1}{4x}\times 4=\frac{4}{8x}=\frac{1}{2x}$

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

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

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

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

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

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

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

$f^{\prime }\left(x\right)>0\iff \frac{4x\times Ln2\times \sqrt{2x}-12x-3\sqrt{8x}+3\sqrt{8x}Ln\sqrt{4x}}{8x^{2}Ln2}>0\iff 4x\times Ln2\times \sqrt{2x}+3\sqrt{8x}\times Ln\sqrt{4x}>12x+3\sqrt{8x}\iff 4xLn2\times 2x+3\times 4x\times Ln\sqrt{4x}>12x\times \sqrt{2x}+3\times 4x\iff 8xLn2+12Ln\sqrt{4x}>12\sqrt{2x}+12$

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

$12Ln\sqrt{4x}>12\iff Ln\sqrt{4x}>1\iff sqrt4x>e\iff 4x>e^2\iff x>\frac{e^2}{4}$

non genyen tou :

$e<3$

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

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

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

donk

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

Lis nonm premye superyè ou egal ak 3 epi ki enferyè ak 64

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

Majorasyon apati postila Bètran

$P_n\le 2^n$

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

$P_2=3<2^2$

$P_3=5$

$3^2$ $P_3<3^2$

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 $]2^n,2^{n+1}[$

nou kapab di tou ke nan entèval $]2^n,2^{n+1}]$ 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

$[1,2^{n+1}]$

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

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

$\mathrm{\forall }n\in \mathbb{N},P_n\le 2^n$

.....

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

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 $]n,2n[$  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 $P_1\left(n\right)\in ]n,2n]$

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

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

Postila Bètran avèk de twa fòmil matematisyen Ayisyen Lainé Jean Lhermite Junior konsènan nonm premye

Ekspresyon nonm premye ran n selon modèl flèch Jonatan anakò avèk postila Bètran an

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

K yèm nonm premye ki siperyè oubyen egal ak n

Premye apwòch

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

Demonstrasyon

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 apwòch

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

Twazyèm apwòch

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

Gade tou

• Fonksyon ki konte kantite divizè n
• Fonksyon ki konte nonm premye ki sou fòm an+b lè a ak b premye entre yo
• Fonksyon lanmbda Liouville
• Fonksyon pi
• Fonksyon Omega ki konte kantite faktè premye ki rantre nan dekonpozisyon yon nonm antye n
• Fonksyon omega ki konte kantite nonm premye ki diferan ki divize yon nonm n
• Fonksyon phi Euler
• Fonksyon pi, ki konte kantite nonm premye
• Fonksyon psi Tchebycheff
• Fonksyon tèta Tchebycheff
• Fonksyon Von Mangolt
• Fòmil pou nonm konpoze
• Modèl boul wouj ak boul ble
• Modèl flèch Jonatan
• Modèl Lhermite yo
• Modèl mouton lòt patiraj
• Modèl pitit prodig ou modèl ti mouton pèdi
• Modèl predesesè
• Modèl siksesè
• Modèl ti triyang ak gwo triyang
• Modèl triyang ak kwadrilatè
• Nonm pafè
• Nonm premye
• Nonm premye ak modèl boul rouj ak boul ble
• Nonm premye Ezau
• Nonm premye Fermat
• Nonm premye Izaak
• Nonm premye Jakòb
• Nonm premye Mersenne
• Nonm zanmi
• Primoryel
• Pwisansyèl
• Fonksyon ki konte nonm premye ki sou fòm an+b lè a ak b premye entre yo
• Rezolisyon ekasyon premye degre ak sentèz