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For $1\leq x < \infty$, we know $\ln x$ can be bounded as following: $\ln x \leq \frac{x-1}{\sqrt{x}}$.

Then what is the upper bound of $\ln x$ for following condition?

$2\leq x <\infty$

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    $\begingroup$ The tightest upper bound is $+\infty$. What are you actually trying to ask? $\endgroup$ – dfeuer Jul 1 '13 at 7:19
  • $\begingroup$ Are you searching for a function $g(x)$ such that $\ln x\leq g(x)~~\forall x>1$ ? $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 7:47
  • $\begingroup$ Well, for $1\leq x < \infty$ we get $\ln x \leq \frac{x-1}{\sqrt{x}}$, then what for $2\leq x <\infty$? $\endgroup$ – mash Jul 1 '13 at 7:50
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    $\begingroup$ Maybe you should give some restrictions on the type of function for the upper bound. For example in your $\ln x \le (x-1)/\sqrt{x}$ the bounding function is a simple form involving polynomials and radicals, whereas the function $\ln x$ is not of that type. After all, the "tightest bound" with no restrictions at all would just be $\ln x$ itself. $\endgroup$ – coffeemath Jul 1 '13 at 7:54
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    $\begingroup$ Sure! Let $g = \ln$ and you're done. Should I convert this comment to an answer so you can accept it, or do you actually want something else? (Hint: I think you want something else, but I don't think you have figured out what it is that you want just yet.) $\endgroup$ – dfeuer Jul 1 '13 at 8:38
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Given $a>0$, if $a \le u < \infty$ then also $1 \le u/a < \infty$, and you can apply your inequality taking $x=u/a$ to get $$\ln(u/a)\le \frac{u/a-1}{\sqrt{u/a}}.$$ Then cleaning it up you have $$\ln(u)\le \ln(a)+\frac{u-a}{\sqrt{au}}.$$ This is a bit less appealing than the $a=1$ case wherein the logarithm doesn't appear in the bounding function, but actually it only appears in a constant.

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  • $\begingroup$ @mash Welcome. The idea is really just re-scaling the original inequality to other intervals. There may be better inequalities as suggested in the other answers. $\endgroup$ – coffeemath Jul 1 '13 at 10:09
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$\ln(x)=-\ln\left(1-(1-\frac{1}{x})\right)$

Since $x\geq 2$ then $\left|1-\frac{1}{x}\right|<1$

Using the Taylor series of the logarithm: $$\ln(1-t)=-\sum\limits_{k=1}^{\infty}\frac{t^n}{n}$$ with $t=1-\frac{1}{x}$ in this case, we obtain: $$\ln x\leq 1-\frac{1}{x}$$ $$\ln x\leq \left(1-\frac{1}{x}\right)+\frac{\left(1-\frac{1}{x}\right)^2}{2}+\frac{\left(1-\frac{1}{x}\right)^3}{3}$$ You can stop at any add degree. In general: $$\ln x\leq\sum\limits_{k=1}^{2n+1}\frac{\left(1-\frac{1}{x}\right)^k}{k}$$

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  • $\begingroup$ The taylor series doesn't converge for all $x>0$... $\endgroup$ – coffeemath Jul 1 '13 at 8:52
  • $\begingroup$ @coffeemath I corrected it. $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 9:31
  • $\begingroup$ Yes, +1 for seeing how to use series of log to attack this, via the use of $\ln(x)=-\ln(1/x)$ [so the series for right side converges for large $x$]. $\endgroup$ – coffeemath Jul 1 '13 at 9:47
  • $\begingroup$ @RaymondManzoni Sorry about that. $\endgroup$ – llllllllllllllllllllllllllllll Jul 1 '13 at 16:36
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To get your first upper bound we may start with the expansion : $$\ln(1+x)=x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 + O(x^5)$$ with the upper bound : $$\frac x{\sqrt{1+x}}=x - \frac{x^2}2 + \frac{3x^3}8 - \frac{5x^4}{16} + O(x^5)$$

To stay in the same spirit you may use the expansion : $$\ln\left(1+\frac x2\right)=\frac x2 - \frac{x^2}8 + \frac{x^3}{24} - \frac{x^4}{64} + O(x^5)$$ and more generally (for $a$ positive) : $$\ln\left(1+\frac xa\right)=\frac xa - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + O(x^5)$$ we could use the same formula for the upper bound (so that the initial proof remains valid) : $$\frac {\frac xa}{\sqrt{1+\frac xa}}= \frac xa - \frac{x^2}{2a^2} + O(x^3)$$ Let's use $\ln\left(1+\frac xa\right)=\ln(a+x)-\ln(a)$ and $z:=a+x$ to get : $$\ln(z)\le\ln(a)+\frac {\frac za-1}{\sqrt{1+\frac za-1}}$$ or for $x\ge a$ and $a>0$ : $$\ln(x)\le\ln(a)+\frac{x-a}{\sqrt{ax}}$$ (as obtained by coffeemath...)

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  • $\begingroup$ @mash: yes of course you only need to replace $2$ by $a$ in my answer (I'll update it). $\endgroup$ – Raymond Manzoni Jul 1 '13 at 8:54
  • $\begingroup$ You are welcome @mash (and coffeemath complete answer was earlier). $\endgroup$ – Raymond Manzoni Jul 1 '13 at 9:02
  • $\begingroup$ I'm a bit confused about the use of the series for $\ln(1+x)$, which as it is converges only for $|x|<1$. It seems somewhere one should invert things, e.g use $1/x$ for $x$, in order to get convergence for all $x>1$. I see how the two series for $\ln x$ and $x/\sqrt{1+x}$ may be compared termwise, where they are convergent. But somehow one wants an inequality for all $x>1$. Other than that (likely something I didn't see) I like the use of comparing series. +1 $\endgroup$ – coffeemath Jul 1 '13 at 9:27
  • $\begingroup$ You are of course right @coffeemath this is not a proof since the expansions are divergent for $|x|>a$ (this is why I indicated that the initial proof for the upper bound should remain valid...) but rather an attempt to find out the origin of the initial inequality and to expend it. You went directly at the essential so +1 too. Cheers $\endgroup$ – Raymond Manzoni Jul 1 '13 at 9:40
  • $\begingroup$ If you check out metacompactness' answer above, the trick involving a pre-manipulation so as to make the log series converge is there. I'd bet with that thrown in your methods would yield a collection of upper bounds, progressively better than that given in the OP, but being progressively more involved. $\endgroup$ – coffeemath Jul 1 '13 at 9:45

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