# Thread: Deriving fourth value from three other values and AGM

1. ## Deriving fourth value from three other values and AGM

I have three known values: a, b, c, and a fourth unknown value: D.

I know that (a, b) has the same arithmetic-geometric mean as (c, D).

I.e.:

$\displaystyle M(a, b)=M(c, D)$

What is the quickest way to derive the unknown value D from this info? Do I just have to "eyeball" it iteratively, or is there a quicker more effective method?

Thanks.

2. Originally Posted by rainer
I have three known values: a, b, c, and a fourth unknown value: D.

I know that (a, b) has the same arithmetic-geometric mean as (c, D).

What do you mean here? The pair of elements $\displaystyle a,b$ have the same arithmetic AND geometric mean as the pair of elements $\displaystyle c,D$ , or what? And if this is so, why do you write $\displaystyle (a,b)$ and not $\displaystyle \{a,b\}$ ?

Tonio

I.e.:

$\displaystyle M(a, b)=M(c, D)$

What is the quickest way to derive the unknown value D from this info? Do I just have to "eyeball" it iteratively, or is there a quicker more effective method?

Thanks.
.

3. No, I mean "arithmetic-geometric mean" as defined here:

Arithmetic-geometric mean - Wikipedia, the free encyclopedia

According to this webpage, my use of brackets is correct.

The verbal definition is included below for convenience:

"In mathematics, the arithmetic-geometric mean (AGM) of two positive real numbers x and y is defined as follows:
First compute the arithmetic mean of x and y and call it a1. Next compute the geometric mean of x and y and call it g1; this is the square root of the product xy.

Then iterate this operation with a1 taking the place of x and g1 taking the place of y. In this way, two sequences (an) and (gn) are defined.

These two sequences converge to the same number, which is the arithmetic-geometric mean of x and y; it is denoted by M(x, y), or sometimes by agm(x, y)."

4. Originally Posted by rainer
I have three known values: a, b, c, and a fourth unknown value: D.

I know that (a, b) has the same arithmetic-geometric mean as (c, D).

I.e.:

$\displaystyle M(a, b)=M(c, D)$

What is the quickest way to derive the unknown value D from this info? Do I just have to "eyeball" it iteratively, or is there a quicker more effective method?

Thanks.

Since agm is a homogeneous function of degree 1

$\displaystyle M(\alpha \, a, \alpha \, b) = \alpha \, M(a,b)$

You could backtrack from the end to determine the value of D.

The number of iterations would be no more than the number of iterations required to calculate the agm initially.

5. Yeah, that property has me confused.

From that property I thought it was safe to conclude that c and D absoltely must stand in the same ratio to each other as a and b stand to each other. And that solving for D was thus a simple matter of:

$\displaystyle \frac{a}{b}=\frac{c}{D}$

But I have now seen that there may exist two pairs of values that converge to the same AGM and which do not stand in the same ratio. Follow up question: Doesn't this violate the property?