5
$\begingroup$

We know the Arithmetic-Geometric mean:

$$a_0=a, \qquad b_0=b$$

$$a_n=\frac{a_{n-1}+b_{n-1}}{2}, \qquad b_n=\sqrt{a_{n-1} b_{n-1}}$$

$$\text{agm} (a,b)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n$$


Now we introduce a family of modified 'iterated means':

$$a_n=\frac{a_{n-1}+b_{n-1}}{2}\left(1-\frac{(a_{n-1}-b_{n-1})^2}{(a_{n-1}+b_{n-1})^2} \right)^p$$

$$b_n=\sqrt{a_{n-1} b_{n-1}}\left(1-\frac{(a_{n-1}-b_{n-1})^2}{(a_{n-1}+b_{n-1})^2} \right)^q$$

$$M_{pq} (a,b)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n$$


I obtained the following numerical results:

$$M_{10} (a,b)=\frac{ab}{\text{agm}(a,b)}$$

(This is actually the Harmonic-Geometric mean, and this result is known and easy to prove).

$$M_{11} (a,b)=\frac{a^2b^2}{\text{agm}^3(a,b)}$$

$$M_{21} (a,b)=\frac{a^3b^3}{\text{agm}^5(a,b)}$$

$$M_{22} (a,b)=\frac{a^4b^4}{\text{agm}^7(a,b)}$$

$$M_{32} (a,b)=\frac{a^5b^5}{\text{agm}^9(a,b)}$$

I think the pattern is easy to see here. We increase $p$ by one and leave $q$ the same. Then we increase $q$ by one as well.

For negative $p,q$ we have a corresponding situation:

$$M_{0-1} (a,b)=\frac{\text{agm}^3(a,b)}{ab}$$

$$M_{-1-1} (a,b)=\frac{\text{agm}^5(a,b)}{a^2b^2}$$

And so on.

How can we prove this observation? Is there some way to prove it for the general case, because I find it unlikely that we can directly prove something like:

$$M_{100,100} (a,b)=\frac{a^{200}b^{200}}{\text{agm}^{399}(a,b)}$$


In particular cases, the 'modification' introduced here turns one mean into another.

$$\frac{a+b}{2}\left(1-\frac{(a-b)^2}{(a+b)^2} \right)=\frac{a+b}{2} \frac{4 ab}{(a+b)^2}=\frac{2 ab}{a+b}$$

$$\frac{a+b}{2}\left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{1/2}=\frac{a+b}{2} \frac{2 \sqrt{ab}}{a+b}=\sqrt{ab}$$

$$\sqrt{ab} \left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{1/2}=\frac{2a b }{a+b}$$

$$\sqrt{ab} \left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{-1/2}=\frac{a+b}{2}$$

etc. In the general case it probably doesn't always result in a mean.

And the 'iterated means' above certainly aren't means in the general case, since the limits for large positive or negative $p,q$ can go to zero or infinity.

$\endgroup$
2
+150
$\begingroup$

I'll show (skipping some parts of the proof) $$M_p(a,b):=M_{p,p}(a,b)=(ab)^{2p} \mathrm{agm}(a,b)^{1-4p},$$ in the same way one can show $$M_{p,p-1}(a,b)=(ab)^{2p-1} \mathrm{agm}(a,b)^{3-4p}.$$ Let us write $S(a,b)=\frac12 (a+b)$ and $P(a,b) = \sqrt{ab}$. One can rewrite the "$p$ iterated means" as follows: \begin{align} a_n & = Q_1(a_{n-1},b_{n-1}) := S(a_{n-1},b_{n-1})^{1-2p} P(a_{n-1},b_{n-1})^{2p},\\ b_n & = Q_2(a_{n-1},b_{n-1}) := S(a_{n-1},b_{n-1})^{-2p} P(a_{n-1},b_{n-1})^{2p+1}. \end{align} Let us also define $F_p(a,b) = (ab)^{2p} \mathrm{agm}(a,b)^{1-4p}$, we want to prove $M_p(a,b) = F_p(a,b)$. The key observation is that we have (this follows immediately from the definition), $$\mathrm{agm}(a,b) = \mathrm{agm}(S(a,b),P(a,b)) \label{1}\tag{1}.$$ Suppose we can prove $$ F_p(a,b) = F_p(Q_1(a,b),Q_2(a,b)) = F_p(a_1,b_1) \label{2} \tag{2}.$$ Iterating the same equality (and using the continuity of $F_p$) we find \begin{align} F_p(a,b) & = F_p(a_1,b_1) = F_p(a_2,b_2) = \dots = F_p(a_n,b_n) = \dots = F_p(M_p(a,b),M_p(a,b))\\ & = (M_p(a,b) \cdot M_p(a,b))^{2p} \cdot \mathrm{agm} (M_p(a,b),M_p(a,b))^{1-4p} = M_p(a,b), \end{align} where we used $\mathrm{agm}(\lambda, \lambda) = \lambda$.

To prove $\eqref{2}$ we use property $\eqref{1}$. We have \begin{align} F_p(a_1,b_1) & = F_p(Q_1(a,b),Q_2(a,b)) = (Q_1(a,b) \cdot Q_2(a,b))^{2p} \cdot \mathrm{agm}(Q_1(a,b), Q_2(a,b))^{1-4p}\\ & = \left( S(a,b)^{1-2p} P(a,b)^{2p} \cdot S(a,b)^{-2p} P(a,b)^{2p+1} \right)^{2p} \cdot \\ & \quad \mathrm{agm}\left( S(a,b)^{1-2p} P(a,b)^{2p} , S(a,b)^{-2p} P(a,b)^{2p+1} \right)^{1-4p}\\ & = S(a,b)^{2p(1-4p)} P(a,b)^{2p(4p+1)} \cdot \left( S(a,b)^{-2p} P(a,b)^{2p} \mathrm{agm}\left( S(a,b), P(a,b) \right)\right)^{1-4p}\\ & = P(a,b)^{4p} \mathrm{agm}\left( a, b \right)^{1-4p}\\ & = F_p(a,b). \end{align}

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's a nice approach. It makes sense that since we have the product of some powers of $ab$ and $(a+b)$ this should be related to the agm. I'll study your proof some more for the next day $\endgroup$ – Yuriy S Aug 27 '16 at 20:49

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.