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Hi I've been playing with ln(Natural logarithm) since I first learn it in physic class and my goal is to calculate it without using the calculator. I done $$\ln(12,14,15,16,18,\ldots)$$

just fine but for

$$\ln(17,13,11)$$

I can't seen to divide them into anything other than $\ln(17)=(17\cdot1).$ I was wondering if there's a method to do this? Thank you :D

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    $\begingroup$ How did you compute $\ln{12}$ without a calculator? $\endgroup$ – cws Dec 12 '17 at 19:14
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    $\begingroup$ Surely, 11, 13, 17 are prime and cannot be written as a product of smaller numbers, but I wonder what you did once you managed to break up e.g. 15. For example, $\ln 15=\ln 3 + \ln 5$, but how did you calculate $\ln 3$ and $\ln 5$ ... and why cannot you do the same for $\ln 17$. (3 and 5 are prime too!) $\endgroup$ – user491874 Dec 12 '17 at 19:14
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    $\begingroup$ I memorize Ln from 1-10 $\endgroup$ – SkullboyBZ Dec 12 '17 at 19:16
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    $\begingroup$ Write $17=16+1=2^4+1$ we know $\ln 2$ then you Taylor expansion of $\ln(x+1)$ $\endgroup$ – Guy Fsone Dec 12 '17 at 19:17
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    $\begingroup$ And how many digits do you memorize? I know at least 50 digits of $\ln 1$, but hardly more than 3 or for of $\ln 2$ or $\ln 10$. $\endgroup$ – Professor Vector Dec 12 '17 at 19:28
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You might try $\ln(17)=\ln(16)+\int_{16}^{17} \frac1x dx$.

You can then estimate the integral as a rectangular area with width $1$ and height $\frac{1}{16.5}$.

This gives $\ln(17)\approx \ln(16)+\frac{2}{33}$

According to my calculator, $\ln(17)\approx 2.83321334$, and $\ln(16)+\frac{2}{33}\approx 2.833194783$.

For better accuracy, you could partition $[16,17]$ into two (or more) subintervals.

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    $\begingroup$ Thank you and I'll try this method :D $\endgroup$ – SkullboyBZ Dec 12 '17 at 19:37
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You can first calculate : $$ \ln(50) = \ln2+2\ln5 $$ and then $$ \ln(52) = \ln13+2\ln2 $$ After this you can get with precision to 3E-8 a value for ln(17) : $$ \ln(17) \approx { {\ln50 + \ln52} \over 2} + ({ 1 \over 50}-{1 \over 52})/4 - \ln3 $$ If you calculate pretty exactly you can get a difference of only 0.000000037 with ln(17)
greatings, Daniel

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  • $\begingroup$ Are you using the spline interpolation formula: en.wikipedia.org/wiki/Cubic_Hermite_spline ? $\endgroup$ – user491874 Dec 12 '17 at 20:38
  • $\begingroup$ I don't use this interpolation and also never hear about it :) , but you can imagine this like a middle point (between ln(50) and ln(52)) in which we also do a kind of middle point of his derivative. May be you will expect 1/2, but not, using 1/4 it is very much close. I have even more extensive formulas for ln(n), but even more for sqrt(n). I dont know if this exist already in Internet. Daniel $\endgroup$ – Daniel Pol Dec 13 '17 at 1:01

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