Log of factorial

n=1xln(n)=ln(x!)\sum_{n=1}^{x}ln(n)=ln(x!)

We can approximate it's derivative by taking 2 secants equidistant (in the x-axis) from the point for which we have to approximate the derivative.

let f(x)f(x) be derivative of ln(x!)ln(x!)

f(x)>ln(x)f(x)>ln(x) ...(because secant from point of coordinates (x1,ln((x1)!))(x-1,ln((x-1)!)))

f(x)<ln(x+1)f(x)<ln(x+1) ...(because secant from point of coordinates (x+1,ln((x+1)!))(x+1,ln((x+1)!)))

As the difference between ln(x+1)ln(x+1) and ln(x)ln(x) decreases then we can more accurately approximate the value of f(x)f(x)

ln(x+1)ln(x)=ln(x+1x)=ln(1+1x)ln(x+1)-ln(x)=ln(\frac{x+1}{x})=ln(1+\frac{1}{x})

We know that as xx approaches infinity, 1x\frac{1}{x} approaches 0

=> As x approaches infinity ln(1+1x)ln(1+\frac{1}{x}) approaches ln(1)=0ln(1)=0

=> f(x)f(x) ~ ln(x)+ln(x+1)2\frac{ln(x)+ln(x+1)}{2}

ln(x)+ln(x+1)2=ln(x(x+1))2=21ln(x(x+1))=ln(x(x+1))\frac{ln(x)+ln(x+1)}{2}=\frac{ln(x(x+1))}{2}=2^{-1}ln(x(x+1))=ln(\sqrt{x(x+1)})

So as x(x+1)\sqrt{x(x+1)} ~ x+12x+\frac{1}{2} ...(1)

Proof of above statement (1) :

let say as xx approaches infinity x(x+1)\sqrt{x(x+1)} approaches x+kx+k for any constant k

Therefore, x(x+1)=x+k\sqrt{x(x+1)}=x+k must led to contradictions

let's simplify it

x(x+1)=(x+k)2x(x+1)=(x+k)^{2}

x2+x=x2+2xk+k2x^{2}+x=x^{2}+2xk+k^{2}

x=2kx+k2x=2kx+k^{2}

if k=12k=\frac{1}{2} then x=x+k2x=x+k^{2} => k2=0k^{2}=0

this led to contradiction

this mean that it only leads to contradiction only if k=12k=\frac{1}{2}

this mean that kk must equal 12\frac{1}{2}

Hence x(x+1)\sqrt{x(x+1)} ~ x+12x+\frac{1}{2}

=> f(x)f(x) ~ ln(x+12)ln(x+\frac{1}{2})

integrating both sides

ln(x!)ln(x!) ~ ln(x+12)dx∫ln(x+\frac{1}{2})dx

ln(x+12)=(x+12)(ln(x+12)1)+c∫ln(x+\frac{1}{2})=(x+\frac{1}{2})(ln(x+\frac{1}{2})-1)+c

ln(x!)ln(x!) ~ (x+12)(ln(x+12)1)+c(x+\frac{1}{2})(ln(x+\frac{1}{2})-1)+c ...(A)

PNPfloor(NP)=N!∏_{P≤N}P^{floor(\frac{N}{P})}=N! ...(here P is prime)

floor(NP)floor(\frac{N}{P}) ~ NP\frac{N}{P}

ln(N!)ln(N!) ~ PN(NP)ln(p)\sum_{P≤N}(\frac{N}{P})ln(p) ...(B)

Combining (A) and (B)

(N+12)(ln(N+12)1)+c(N+\frac{1}{2})(ln(N+\frac{1}{2})-1)+c ~ NPNln(P)PN\sum_{P≤N}\frac{ln(P)}{P}

(N+12)(ln(N+12)1)+cN\frac{(N+\frac{1}{2})(ln(N+\frac{1}{2})-1)+c}{N} ~ PNln(P)P\sum_{P≤N}\frac{ln(P)}{P}

(1+12N)(ln(N+12)1)+cN(1+\frac{1}{2N})(ln(N+\frac{1}{2})-1)+\frac{c}{N} ~ PNln(P)P\sum_{P≤N}\frac{ln(P)}{P}

ln(n+12)1ln(n+\frac{1}{2})-1 ~ PNln(P)P\sum_{P≤N}\frac{ln(P)}{P}

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

Note:

  • Here, floor() is the floor function I used this notation because I was unable to find it's symbol.

  • If you find any problem or difficulty in mathematics used here please comment.

Note by Zakir Husain
1 week, 1 day ago

No vote yet
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