ln(x) is an analytic function. This means you can find its Taylor expansion at any point of its domain.
Write x = a 10^n, where a \in [1, 10), Then we get that
ln x = ln a + n ln 10, which roughly equals ln a + 2.3025850929940457 * n.
So 2.3 n for rather large n should be quite good as an approximation.
Thank you for your answer. now, i need an approximation for ln(1-y) and 0
You can do it in several way, by Taylor expansion , or by linear interpolation :
f(x) is approximated by f'(a)(x-a)+f(a)
Dear Abdelhay,
I think you received some very interesting solutions (or advice) but more of them are connected with Taylor expansions. Only Peter said about continued fraction expansion. I specialised in Pade approximation. This approximation is close to continued fraction expansion and connected with Taylor expansions (it is fractional approximation).
I recommend you this type of approximation. You can find example for Log(1+x) and similar functions here (and in a lot of papers):
http://www.nezumi.demon.co.uk/consult/logx.htm
This approximation is very accurate for some kinds of function.If you want to get to know much more about Pade approximation you can look at some papers in my profile.
Please refer to that
Morse P, Feshbach H. Method of Theoretical Physics. New York: McGraw-Hill; 1953
ln(a-b)=ln(a)-Summing((1/n)*(b/a)^n)), where |b/a|
for any positive x
ln(x)=a (x)^(1/a) - a
where a is any large constant
The larger a, the better approximation.
Please refer to research paper at
http://mathematical-forum.org/approximation-of-logarithm-factorial-and-euler-mascheroni-constant-using-odd-harmonic-series/
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