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.


1 Answer 1


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...


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

now, from the Jamieson article


which I write as:


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

on factor;

now finally, from the AGM...:


there after, I ask REDUCE to calculate...


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...


(*) 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...

Post-scriptum, December 2019:

It is easy to see the first iterations of Jamieson/Gauss match that of Semjon Adlaj. Now, as far as generalizing this to all the iterations, I have built an easier "proof by contradiction". Here it is: Suppose there is a breaking point of the series of equalities, i.e. a number B where R_B ≠ A_B, while R_{B-1} = A_{B-1}. One remarkable property of the MAGM series is that one can add a constant K to all the A, S, and D numbers, and it will still be a MAGM, as "K" comes out added on each row via D (for S and D) and via the average (for A), the differences not changing. This means that we can build a MAGM where we add b to each and all the numbers.

Apart from a needed scale factor, the second row (the N=1 row) now looks exactly like a first row (with a different value of b), because the D value is zero, which is the characteristic of the first row of a MAGM.

The different value of b is b*=(2b/(1/2+b^2/2+b)^(1/2)=2b^(1/2)/(1+b)

Another remarkable property of the MAGM is that we can apply any scale factor to a MAGM sequence, and still get a MAGM, because when calculating a new row, if all the values are a factor 𝑆_MAGM times as large as before, all the values of the new row will simply be a factor 𝑆_MAGM times as large as before. The scale factor that we need to make the A of the second row a "1", like the A of the first row was, is easy to figure out:

S_MAGM = 1/(1/2+b^2/2+b) = 2/(1+b)^2

After these two operations, the second row has become a first row (for 𝑏∗ instead of 𝑏, but it holds for any value of either in the range ]0,1]).

The increments in the R column are scaled by the square of that factor. If we rescale by 𝑆_AGM=1/((1+b)/2), the second row of the AGM looks like a first row, with the "new b value" 𝑏∗ being 2√𝑏/(1 + 𝑏), as in the MAGM above, and the running sum increments will be scaled by the square of 𝑆_AGM, 1/((1+b)/2)/((1+b)/2)= 4/(1+b)^2. This is twice the scaling of MAGM. In order to use the second row of the R column as a first row, we need to make one further change: divide all the numbers by 2, because 2^{P-2} is half of 2^{N-2} when P=N-1 is the row number counting from the second row, which differs by one from the row number N counting from the first row. After setting the new first row R value 𝑅_1 to 1, we can now see that we are getting a set of values R matching (at least initially) the MAGM's set of values A, since we have set the first values the same, and the increments are the same due to identical scaling factors. The breaking point of the equality being at B means that as far as the newly built table is concerned, the breaking point of the equality will be at B-1. The breaking point of the equality cannot depend on the value of b, and yet the breaking point of the equality is B when considering b, and B-1, when considering b*. We arrive at the impossible conclusion that B=B-1. The conclusion being impossible, the hypothesis of a finite B where equality of the R and A series breaks is itself impossible, so 𝐴_n = 𝑅_n for any n, and thus the limit is the same, and MAGM(1,b)=1-S, where S is the scaled sum of the differences between the square of the arithmetic and geometric means of the plain AGM. QED.


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