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Hi bincybn,Hii everyone,
Can anyone tell me a decent upper bound of Ln[x](which can mimic Ln[x]) where x is in [0,1]
regards,
Bincy
Although not well known, it exist a series expansion of the function $\ln x$ ...Hii everyone,
Can anyone tell me a decent upper bound of Ln[x](which can mimic Ln[x]) where x is in [0,1]
regards,
Bincy
You start from the well known series expansion...Thanks. May I know the source of it?
Hi bincybn,Hii,
Thanks for your spontaneous reply.
But what i want is a function f(x), such that Ln(x)<=f(x) which can mimic the variations of Ln(x).
For eg (1-x)/x is a lower bound of Ln[x].
Hi chisigma,Although not well known, it exist a series expansion of the function $\ln x$ ...
$\displaystyle \ln x= 2\ \{(\frac{x-1}{x+1}) + \frac{1}{3}\ (\frac{x-1}{x+1})^{3} + \frac{1}{5}\ (\frac{x-1}{x+1})^{5}+...\}$ (1)
It is easy to see that any 'truncation' of the series (1) to a finite power of $\displaystyle \frac{x-1}{x+1}$ is an upper bound of $\ln x$ in $0<x<1$...
Kind regards
$\chi$ $\sigma$