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Define the following family of one parameter sequences:

$$a_0=x,~~~b_0=y$$

$$a_{n+1}=\sqrt{a_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}},~~~b_{n+1}=\sqrt{b_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}}$$

I conjecture that this family of sequences has the limit:

$$L_p(x,y)=\left(\frac{x^{2p}-y^{2p}}{2p(\ln x-\ln y)} \right)^{\dfrac{1}{2p}}$$

The proof:

$$\ln a_n-\ln b_n=\frac{\ln x-\ln y}{2^n}=\frac{\delta_1}{2^n} \tag{1}$$

$$a_n^{2p}-b_n^{2p}=(a_{n-1}^p-b_{n-1}^p)\frac{a_{n-1}^p+b_{n-1}^p}{2}=\frac{x^{2p}-y^{2p}}{2^n}=\frac{\delta_2}{2^n} \tag{2}$$

From $(2)$:

$$2p \ln a_n=2p \ln b_n+\ln \left(1+\frac{\delta_2}{2^nb_n^{2p}} \right)$$

From $(1)$:

$$2p \ln a_n=2p \ln b_n+\frac{2p\delta_1}{2^n}$$

$$\ln \left(1+\frac{\delta_2}{2^nb_n^{2p}} \right)=\frac{2p\delta_1}{2^n} \tag{3}$$

Assuming $2^nb_n^{2p} \to \infty$ at $n \to \infty$ we expand the logarithm, considering only the first term:

$$\frac{\delta_2}{2^nb_n^{2p}}+O\left(\frac{1}{2^{2n}} \right)=\frac{2p\delta_1}{2^n}$$

$$\lim_{n \to \infty} b_n=\left(\frac{\delta_2}{2p \delta_1} \right)^{\dfrac{1}{2p}}=\left(\frac{x^{2p}-y^{2p}}{2p(\ln x-\ln y)} \right)^{\dfrac{1}{2p}} \tag{4}$$

Is this proof correct for any $p \in \mathbb{R}$?

How would I know that $2^nb_n^{2p} \to \infty$, i.e. $b_n^{2p}$ doesn't vanish?

Does the mean defined by $(4)$ has a special name?

Also, do you know other such parametric families of sequences, related to means?

The case $p=\frac{1}{2}$ is familiar and appeared in my recent question.

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Yuriy S
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  • I have corrected my hint, your considerations are right. – user90369 Aug 23 '16 at 11:08
  • Yes, indeed. You have answered your question (if your proof is correct for any $p$ and if $b_n^{2p}$ vanishes) yourself. – user90369 Aug 23 '16 at 11:20
  • @user90369, no, I did not answer my question. Notice the tag 'proof verification'. Which means I ask someone to verify my proof is correct, not merely repeat it verbatim. And there are additional questions as well – Yuriy S Aug 23 '16 at 11:22
  • O.k., I have deleted my post. I think you have written everything which is needed. E.g. to show, why $b_n^{2p}$ doesn't vanish and that $2^n b_n^{2p}\to\infty$, I think you can use equation (3) multiplicated with $2^n$ and $n\to\infty$. – user90369 Aug 23 '16 at 11:49
  • @user90369, thank you for the hint – Yuriy S Aug 23 '16 at 11:51
  • You are welcome. I hope it helps. Other such parametric families of sequences I don't know, sorry. – user90369 Aug 23 '16 at 11:55

1 Answers1

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do you know other such parametric families of sequences, related to means?

I just figured out a family of sequences with the exact same limit, but with much faster convergence.

Instead of taking:

$$a_{n+1}=\sqrt{a_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}},~~~b_{n+1}=\sqrt{b_n \sqrt[p]{\frac{a_n^p+b_n^p}{2}}}$$

We take:

$$A_{n+1}=\sqrt[q]{A_n \frac{A_n^q-B_n^q}{q(A_n-B_n)}},~~~B_{n+1}=\sqrt[q]{B_n \frac{A_n^q-B_n^q}{q(A_n-B_n)}}$$

This gives us:

$$\ln A_n-\ln B_n=\frac{\ln x-\ln y}{q^n}=\frac{\Delta_1}{q^n} \tag{3}$$

$$A_n^{q}-B_n^{q}=(A_{n-1}-B_{n-1})\frac{A_{n-1}^q-B_{n-1}^q}{q(A_{n-1}-B_{n-1})}=\frac{x^{q}-y^{q}}{q^n}=\frac{\Delta_2}{q^n} \tag{4}$$

Thus, the same proof as in the OP can be used to obtain the limit:

$$L_q(x,y)=\left(\frac{x^{q}-y^{q}}{q(\ln x-\ln y)} \right)^{\dfrac{1}{q}}$$

Obviously, if we compare $(3)$ with $(1)$ we can see that for larger $q$ the convergence is much better than for the original sequence in the OP.

Comparing directly for $q=6,p=3,x=1,y=2$ we have:

$$\begin{array}( n & a_n & b_n & A_n & B_n & L \\ 3 & \color{blue}{1.5}03487040 & \color{blue}{1}.639564243 & \color{blue}{1.57}0468969 & \color{blue}{1.57}5516722 & \color{green}{1.572994870} \\ 7 & \color{blue}{1.5}68730080 & \color{blue}{1.57}7248129 & \color{blue}{1.57299}2923 & \color{blue}{1.57299}6818 & \color{green}{1.572994870} \\ 12 & \color{blue}{1.572}861769 & \color{blue}{1.57}3127960 & \color{blue}{1.572994870} & \color{blue}{1.57299487}1 & \color{green}{1.572994870} \end{array}$$

The correct digits are highlighted. The limit in this case is: $$L_6(1,2)=\left(\frac{63}{6\ln 2} \right)^{1/6}$$

Note, that we really don't need to subtract any numbers during iterations, if we expand $\frac{A_{n-1}^q-B_{n-1}^q}{A_{n-1}-B_{n-1}}$

Yuriy S
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