Ln(x) using repeated square root extraction
03-21-2016, 05:09 AM (This post was last modified: 03-21-2016 05:12 AM by Gerson W. Barbosa.)
Post: #2
 Gerson W. Barbosa Senior Member Posts: 1,473 Joined: Dec 2013
RE: Ln(x) using repeated square root extraction
For the sake of completeness here is the recursive limit definition of ln(x), even though this is not quite necessary for our purpose:

$\ln (x)=\lim_{n\rightarrow \infty} \left [ n\cdot \left ( x^{\frac{1}{n}}-1 \right )-\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!\cdot n^{k-1}} \right ]$

If the limit is removed and n is set to 1 then we'll have the following recursive series representation:

$\ln (x)= \ x-1 -\sum_{k=2}^{\infty }\frac{\ln ^{k}(x)}{k!}$
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 Messages In This Thread Ln(x) using repeated square root extraction - Gerson W. Barbosa - 03-21-2016, 12:03 AM RE: Ln(x) using repeated square root extraction - Gerson W. Barbosa - 03-21-2016 05:09 AM RE: Ln(x) using repeated square root extraction - Paul Dale - 03-21-2016, 07:11 AM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-03-2022, 11:20 PM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-06-2022, 12:04 AM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 05:32 PM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 05:53 PM RE: Ln(x) using repeated square root extraction - Dan C - 03-05-2022, 08:02 PM RE: Ln(x) using repeated square root extraction - Namir - 03-05-2022, 08:29 PM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-05-2022, 08:39 PM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-06-2022, 12:50 AM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-06-2022, 09:33 AM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-09-2022, 07:47 PM RE: Ln(x) using repeated square root extraction - Thomas Klemm - 03-10-2022, 05:18 AM RE: Ln(x) using repeated square root extraction - Albert Chan - 03-10-2022, 03:17 PM

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