MHB Finding an Upper Bound for ln(x) in [0,1]

[bincy]

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

 

[Sudharaka]

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

Hi bincybn, :)

The natural logarithm function is non-positive when \(x\in(0,1]\) and \(\ln\,x=0\) when \(x=1\). Therefore,

\[\mbox{sup}\{\ln\,x\,:\,x\in(0,1]\}=0\]

Hence any non-negative real number would be an upper bound of, \(f(x)=\ln\,x\mbox{ where }x\in[0,1]\).

Kind Regards,
Sudharaka.

 

[bincy]

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].

 

[chisigma]

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 $\ln x$ ...

$\displaystyle \ln x= 2\ \{(\frac{x-1}{x+1}) + \frac{1}{3}\ (\frac{x-1}{x+1})^{3} + \frac{1}{5}\ (\frac{x-1}{x+1})^{5}+...\}$ (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

$\chi$ $\sigma$

 

[bincy]

Thanks. May I know the source of it?

 

[chisigma]

Thanks. May I know the source of it?

You start from the well known series expansion...

$\displaystyle \ln \frac{1+t}{1-t}= 2\ (t + \frac{t^{3}}{3} + \frac{t^{5}}{5}+...)$ (1)

... and now setting $\displaystyle x=\frac{1+t}{1-t} \implies t=\frac{x-1}{x+1}$ You obtain...

$\displaystyle \ln x = 2\ \{(\frac{x-1}{x+1}) + \frac{1}{3}\ (\frac{x-1}{x+1})^{3} + \frac{1}{5}\ (\frac{x-1}{x+1})^{5}+...\}$ (2)

Kind regards

$\chi$ $\sigma$

 

[Sudharaka]

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 bincybn, :)

The lower bound function that you have given as the example should be, \(\dfrac{x-1}{x}\).

It can be shown without much difficulty that the function, \(\ln(\sqrt{x})\) is an upper bound of \(\ln(x)\) where \(x\in(0,1]\). Note that \(\ln(\sqrt{x})\) has the same behavior as \(\ln(x)\) in the interval \((0,1]\).

Generalizing this you can show that, any function of the form, \(\displaystyle\ln(x^{\frac{1}{n}})\mbox{ where }n>1\) can be taken as an upper bound of \(\ln(x)\) and they exhibit the same behavior as \(\ln(x)\) in the interval \((0,1]\).

Kind Regards,
Sudharaka.

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

$\chi$ $\sigma$

Hi chisigma, :)

I think bincybn wants a function that exhibits the same behaviour of \(\ln(x)\) in the interval \((0,1]\). Now, \(\ln(x)\rightarrow -\infty\mbox{ as }x\rightarrow 0^+\). However, if we truncate the series of \(\ln(x)\) this limiting behavior as \(x\rightarrow 0^+\) is not satisfied.

Kind Regards,
Sudharaka.