1
$\begingroup$

Is the following inequality true? $$1-\alpha x\le (1-x)^{\alpha}\le 1-\alpha x+\frac{\alpha(\alpha-1)}{2}x^2$$ for real numbers $x,\alpha.$ We may assume $0\le x\le 1$ and put some requirements about $\alpha.$ And $(1-x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}x^k,$ where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$
I think I might meet with this inequality in Stanley's book Enumerative Combinatorics 1 http://www-math.mit.edu/~rstan/ec/ec1.pdf or somewhere else, saying that $(1-x)^\alpha$ greater the sum of its expansion terms up to a negative sign, and less than sum up to a positive sign. But I cannot remember it now and I am not sure. And it seems natural if we view it as a Taylor expansion. Any proof, comments or reference are very welcome!

$\endgroup$
  • 1
    $\begingroup$ Have you looked at Bernoulli's inequality? $\endgroup$ – Jacky Chong Apr 26 '17 at 2:30
  • 1
    $\begingroup$ It seems Bernoulli's inequality en.m.wikipedia.org/wiki/Bernoulli%27s_inequality gives the lower bound. How about the upper bound? $\endgroup$ – Connor Apr 26 '17 at 4:31
  • $\begingroup$ If $\alpha \in (0,1)$ and $x\in (0,1)$ then the left inequality is a strict inequality in the OPPOSITE direction. For example if $\alpha =1/2$ and $x\in (0,1)$ then $1-x/2>0$ so $1-x/2>(1-x)^{1/2}\iff (1-x/2)^2>1-x\iff 1-x+x^2/4>1-x.$ $\endgroup$ – DanielWainfleet Apr 26 '17 at 9:22
0
$\begingroup$

For the upper bound, use \begin{align} (1-x)^r\leq e^{-rx}= 1-rx+ \frac{r^2x^2}{2!}-\frac{r^3x^3}{3!}+\ldots \end{align} for all $x<1$ and $r>0$.

Next, consider the following fact: for $x<0$, we have \begin{align} e^x \leq T_{2n}(x) \end{align} for all $n$ where \begin{align} T_n(x) = 1+x+\frac{x}{2!}+\ldots +\frac{x^n}{n!}. \end{align}

Combining both result yields the desired inequality for all $r>0$ and $0<x<1$.

$\endgroup$

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.