3
$\begingroup$

Let $x\geq 0$ be fixed. By Bernoulli's inequality we know that for all $n \in \mathbb{N}_0$, $$ (1+x)^n \geq 1+nx $$ Now, let $\alpha \geq 0$ be fixed as well. By a limit argument, we know that $$ (1+x)^n \geq 1+\alpha nx $$ for large $n$. But how large does $n$ have to be? More specifically, can we find some expression for $N \in \mathbb{N}$ depending on $x$ and $\alpha$ such that the inequality holds for all $n \geq N$?

If it helps, $N$ does not necessarily have to be optimal. I am specifically interested in the case where $\alpha \in \mathbb{N}$, and even more specifically in the case $\alpha=2$.

$\endgroup$
4
$\begingroup$

If $n\ge 2$ then $$(1+x)^n\ge 1+nx+\frac{n(n-1)}{2}x^2.$$So you just need $\frac{(n-1)}{2}x>\alpha-1.$

$\endgroup$
1
$\begingroup$

Well, of course. Equations in $x$ of the form $p^x=ax+b$ have solutions given by the Lambert-$W$ function. In your specific case, we have

$$(1+x)^n=(\alpha x)n+1$$

And the corresponding solution in $n^*$ is

$$n^*=-\frac{W_{-1}\left(-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)}{\ln(1+x)}-\frac1{\alpha x}$$

Hence, for all $n\geq n^*$ we have $(1+x)^n\geq 1+n\alpha x$.


There's the bound

$$W_{-1}(-e^{-u-1})\geq -1-\sqrt{2u}-u.$$

Setting $-e^{-u-1}=-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}$, we get

\begin{align} u&=-1-\ln\left(\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)\\ &=-1-\left(\ln\left(\frac{\ln(1+x)}{\alpha x}\right)+\ln\left((1+x)^{-\frac1{\alpha x}}\right)\right)\\ &=-1-\left(\ln(\ln(1+x))-\ln(\alpha x)-\frac1{\alpha x}\ln(1+x)\right)\\ &=\ln(\alpha x)+\frac{\ln(1+x)}{\alpha x}-1-\ln\Big(\ln(1+x)\Big). \end{align}

So, depending on how precise you wish to be, you may use

$$n^*\geq \frac{1+\sqrt{2u}+u}{\ln(1+x)}-\frac1{\alpha x},$$

where $u$ is as above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.