0
$\begingroup$

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 $x$.

$\endgroup$
8
  • 1
    $\begingroup$ Isn't the BBP formula the Bailey–Borwein–Plouffe formula for computing $\pi$? Where did you come across this; do you have a link? $\endgroup$
    – Dietrich Burde
    Feb 2, 2018 at 16:14
  • $\begingroup$ Sorry, let me change that to BBP type formula, and I discovered it on my own, @DietrichBurde $\endgroup$
    – JacobTDC
    Feb 2, 2018 at 16:17
  • $\begingroup$ \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. $\endgroup$
    – Donald Splutterwit
    Feb 2, 2018 at 16:24
  • $\begingroup$ @Somos, I have tested it, and it appears accurate, but how do I prove that it's true? $\endgroup$
    – JacobTDC
    Feb 2, 2018 at 19:29
  • 1
    $\begingroup$ After you correct the typo, this is a very simple consequence of the Taylor series $\ln(1-x) = -\sum_{n>0} x^n/n$. $\endgroup$
    – Erick Wong
    Feb 2, 2018 at 20:27

1 Answer 1

Reset to default
2
$\begingroup$

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

Let $x=1-1/n.$

$\endgroup$
2
  • $\begingroup$ I just changed the formula a little. $\endgroup$
    – JacobTDC
    Feb 2, 2018 at 20:01
  • $\begingroup$ Thanks for the typo notice. $\endgroup$
    – JacobTDC
    Feb 2, 2018 at 20:42

Not the answer you're looking for? Browse other questions tagged .