Let $\{x_n\}$ and $\{y_n\}$ be defined iteratively, $x_0:=\beta >1, \ y_0:= 1$ and $x_{n+1}= \frac{x_n+y_n}{2}$, $y_{n+1} = (x_n.y_n)^{\frac{1}{2}}$; i.e. they are respectively the arithmetic and geometric mean of the previous terms. We know that their limit is called the arithmetic-geometric mean of $\beta$ and $1$ (denoted by $AGM(\beta,1)$).

Now, let's define $\xi_0:= \beta^2$, $\eta_0:= 1$ and $\zeta_0:=0$. Then let's define iteratively, $\xi_{n+1}:= \frac{\xi_n + \eta_n}{2}$, $\eta_{n+1}:= \zeta_n + ((\xi_n-\zeta_n)(\eta_n-\zeta_n))^{\frac{1}{2}}$ and finally $\zeta_{n+1}:= \zeta_n - ((\xi_n-\zeta_n)(\eta_n-\zeta_n))^{\frac{1}{2}}$. The common limit of $\xi_n$ and $\eta_n$ is called the modified arithmetic-geometric mean of $\beta^2$ and $1$ (denoted by $MAGM(\beta^2, 1)$).

My question is wheter there is an easy way to prove the equality $\xi_n = \beta^2-\sum_{m=0}^{n-1}2^m\frac{x_m^2-y_m^2}{2}$ and if there is, how? Thank you very for any help.

Note: This is from an article in Notices of the AMS, Volume 59, Number 8.

This is my first input into this forum, so sorry for the bad font etc...

The basic proof that you want is in this article: http://www.maths.lancs.ac.uk/~jameson/ellagm.pdf

after one recognizes that the arithmetic averages of Adlaj's MAGM (a_n) are just the same as the one minus the partial sums of jameson's (Gauss's) series.

The first iterations being equal is easy to show, one just can see it in arithmetic... but how to prove this is true for any n?

Well, I took the MAGM formulae, and using d_2,d_3,d_4,d_5 (*) as d_n,d_(n+1),d(n+2),d(n+3) I get the following:

AGM                                  MAGM

a b                                d1 e1 f1

a2 b2                              d2 e2 f2

a3 b3                              d3 e3 f3

a4 b4                              d4 e4 f4

a5 b5                              d5 e5 f5

the following apply to MAGM (I use REDUCE to help myself through these), del^2 and dsq being two ways to calculate (f3-f2)^2...

del:=(D3-D2)^2/4/(D4-D3)+D4-(D4-D3)^2/4/(D5-D4)-D5;
dsq:=(E2-F2)*(D2-F2);
f2:=(D3-D2)^2/4/(D4-D3)+D4;
e2:=2*d3-D2;
bigd:=dsq-del^2;
bign:=bigd*(d5-d4)^2*(d4-d3)^2;

for the MAGM to be obeyed we need dsq=del^2 or equivalently bign=0

now, from the Jamieson article

d1=1-c0^2/2
d2=d1-c1^2
d3=d2-c2^2*2
etc...

which I write as:

d3:=d2-s3;
d4:=d3-s4;
d5:=d4-s5;

reduce shows me that bign is quite a bit simpler now:

bign/s4;
on factor;
bign/s4;

now finally, from the AGM...:

s3:=(a^2-b^2)/2;
s4:=(a/2-b/2)^2;
s5:=(2*sqrt(a*b)-a-b)^2/8;

there after, I ask REDUCE to calculate...

bign/s4;

and it returns zero, QED...

To make a nice formatted version of this proof, I would copy and paste the output from REDUCE... but I guess anybody can do that for themselves. The REDUCE script you need is all the lines ending with a semicolumn above..

So let me know if you want clarifications or if you are interested that I develop this into a well-written mini-article. I would rather let someone else do that as my day job is not mathematician...! but eventually I could do it myself too...

Francois

(*) I used d_n instead of a_n from Adlaj's article because I wanted to use Jameson's a_n for the plain AGM...

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