Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\alpha$ be a real number in $[1/2,5/6]$.

How do I easily prove that $$AGM(1,\sqrt{1-\alpha})\leq AGM(1,\sqrt{\alpha}) \leq 3AGM(1,\sqrt{1-\alpha})?$$

AGM denotes the arithmetic geometric mean.

share|improve this question

1 Answer 1

Arithmetic geometric mean $M(1,x)$ for $x>0$ is related to complete elliptic integral: $$ M(1,x) = \frac{\pi (x+1)}{4 K\left(\frac{(1-x)^2}{(1+x)^2}\right)} $$ From the definition of $M(1,x)$ is clear that for $0<x<1$, $0<M(1,x)<1$ and that $M(1,x)$ is increasing.

Thus to verify your inequality it suffices to check end-points. The left inequality is saturated for $\alpha=\frac{1}{2}$, and so the right inequality follows since $M(1,x)>0$ for $x = 2^{-1/2}$. Let $m_1 = M\left(1,\frac{1}{\sqrt{6}}\right)$ and $m_2 = M\left(1,\sqrt{\frac{5}{6}}\right)$, but $$ m_1 = 0.6711 \qquad m_2 = 0.9559 \qquad \implies \qquad m_1 < m_2 < 3 m_1 $$ From this analysis it follows that the inequality can be strengthened to : $$ \operatorname{AGM}(1,\sqrt{1-\alpha})\leq \operatorname{AGM}(1,\sqrt{\alpha}) \leq \frac{3}{2} \operatorname{AGM}(1,\sqrt{1-\alpha}) $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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