0

I have come across a general BBP type formula to calculate the natural log of any integer greater than one. How do I prove that it is true? $$\ln(n) = \sum_{k=1}^\infty \left( n^{-k} \cdot\frac{(n-1)^k}{k} \right) \text{when } n \gt 0$$

The digit-extraction formula is: $$s(x,n) = \sum_{k=1}^{n} \left( \frac{(x-1)^k \cdot x^{n-k} \mod k}{k} \right) + \sum_{k=n+1}^\infty \left( \frac{(x-1)^k \cdot x^{n-k}}{k} \right)$$ $$ts(x,n) = \left\lfloor (s(x,n) \mod 1 ) \cdot x \right\rfloor$$ when $\ln(x)$, and $n + 1$ is the the digit you are trying to find. Generates digits in base $x$.

JacobTDC
  • 140
  • 1
    Isn't the BBP formula the Bailey–Borwein–Plouffe formula for computing $\pi$? Where did you come across this; do you have a link? – Dietrich Burde Feb 02 '18 at 16:14
  • Sorry, let me change that to BBP type formula, and I discovered it on my own, @DietrichBurde – JacobTDC Feb 02 '18 at 16:17
  • \begin{eqnarray} \frac{1}{n+1} = \int_{0}^{1} x^n dx. \end{eqnarray} part of the summand will be of the form $\frac{1}{n+1}$, substitute the above, interchange the order of the integration and summation, perform the geometric sum, be bamboozled by the resulting integral. – Donald Splutterwit Feb 02 '18 at 16:24
  • @Somos, I have tested it, and it appears accurate, but how do I prove that it's true? – JacobTDC Feb 02 '18 at 19:29
  • 1
    After you correct the typo, this is a very simple consequence of the Taylor series $\ln(1-x) = -\sum_{n>0} x^n/n$. – Erick Wong Feb 02 '18 at 20:27
  • Well, I may not have discovered it after all, @ErickWong, but I have successfully implemented it as a digit-extraction algorithm, so I'm VERY happy regardless. – JacobTDC Feb 02 '18 at 20:52
  • Re: Numerical calculation of $\log x$ when $2<x$ : (i). We have $\log 2=\log (1+1/3)-\log (1-1/3)=(2/3) \sum_{n=0}^{\infty} 9^{-n}/(2n+1).$....(ii). If $2<x$ and $x$ is not an integral power of $2,$ let $k\in \Bbb N$ with $2^k<x<2^{k+1}.$ Now if $y=x2^{-k}\leq 3/2$ let $z=(y+1)/(y-1) .$ Then $z\geq 5$ and $(2z)\sum_{n=0}^{\infty}z^{2-2n}/(2n+1)=\log (1+z)-\log (1-z)=\log y.$.. And an analogous "manoevre" can be employed when $1/2<y<1,$ using $y'=x^{-1}2^{1+k}.$ – DanielWainfleet Feb 02 '18 at 21:34
  • @JacobTDC Well done on the independent discovery of this spigot algorithm. If I recall correctly, the series for $\ln 2$ was one of inspirations that Plouffe (the P in BBP) drew on to search for the BBP formula. – Erick Wong Feb 03 '18 at 07:16

1 Answers1

2

For $|x|<1$ we have $\ln (1-x)=-\sum_{k=1}^{\infty}x^k/k.$

Let $x=1-1/n.$

JacobTDC
  • 140
DanielWainfleet
  • 57,985