# Upperbound

#### bincybn

##### Member
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

##### Well-known member
MHB Math Helper
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.

#### bincybn

##### Member
Hii,

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

##### Well-known member
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

$\chi$ $\sigma$

Last edited:
• bincybn

#### bincybn

##### Member
Thanks. May I know the source of it?

#### chisigma

##### Well-known member
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

$\chi$ $\sigma$

• bincybn

#### Sudharaka

##### Well-known member
MHB Math Helper
Hii,

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.