A proof for prime number theorem!

From here we get:

ln(N)1\ln(N)-1 ~ PNln(P)P\sum_{P≤N}\frac{\ln(P)}{P}

Transform this into following using function p(x)p(x) defined as the xthx_{th} prime, and let kk be a variable

0kln(p(x))p(x)dx∫_{0}^{k}\frac{\ln(p(x))}{p(x)}dx ~ ln(p(k)1)\ln(p(k)-1)

Take derivative with respect to kk on both sides

ln(p(k))p(k)\frac{\ln(p(k))}{\cancel{p(k)}} ~ dp(k)d(k)×1p(k)\frac{dp(k)}{d(k)} \times \frac{1}{\cancel{p(k)}}

ln(p(k))\ln(p(k)) ~ dp(k)dk\frac{dp(k)}{dk}

Now consider instead of taking derivative of p(k)p(k) we take derivative of p(kln(k))p(\frac{k}{\ln(k)})

dp(kln(k))dk\frac{dp(\frac{k}{\ln(k)})}{dk} ~ ln(p(kln(k)))×(1ln(k)1(ln(k))2)\ln(p(\frac{k}{\ln(k)})) \times (\frac{1}{\ln(k)}-\frac{1}{(\ln(k))^2}) ~ ln(p(kln(k)))ln(k)(11ln(k))\frac{\ln(p(\frac{k}{\ln(k)}))}{\ln(k)}(1-\frac{1}{\ln(k)}) ~ ln(p(kln(k)))ln(k)\frac{\ln(p(\frac{k}{\ln(k)}))}{\ln(k)}

Let p(kln(k))=f(k)p(\frac{k}{\ln(k)})=f(k)

f(k)f'(k) ~ ln(f(k))ln(k)\frac{\ln(f(k))}{\ln(k)}

We can observe that if f(k)=kf(k)=k then f(k)=ln(f(k))ln(k)f'(k)=\frac{\ln(f(k))}{\ln(k)}

So p(kln(k))p(\frac{k}{\ln(k)}) ~ kk

π(k)\pi(k) ~ kln(k)\frac{k}{\ln(k)}

Note by Zakir Husain
1 year, 1 month ago

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You can use the above result to find the Euler constant 0.577....

Aruna Yumlembam - 1 year, 1 month ago

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@Aruna Yumlembam - Can you tell me how? I have tried a lot.

Zakir Husain - 1 year, 1 month ago

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The following is the argument,since ln(N)-1 is approximately ln(2)/2+ln(3)/3+ln(5)/5+....ln(p)/p ,this implies -ln(N)+1 is approx. -(ln(2)/2+ln(3)/3+ln(5)/5+.....ln(p)/p),then add 1/2+1/3+1/4+1/5+....+1/N to both sides and take the limit of N approaches infinity. Then limit as N approaches infinity (1/2+1/3+1/4+1/5+....1/N)-(ln(2)/2+ln(3)/3+ln(5)/5+....+ln(p)/p) must approximately be 0.577..,where p<N. I also did a test ,the result came out correctly up to the second decimal place.

Aruna Yumlembam - 1 year, 1 month ago

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Very cool! Also, what is the Euler constant? @Zakir Husain @Aruna Yumlembam

A Former Brilliant Member - 1 year, 1 month ago

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@Yajat Shamji See this and this

Zakir Husain - 1 year, 1 month ago

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