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Arithmetic-geometric mean

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In mathematics, the arithmetic-geometric mean

M(x, y)
of two positive real numbers x and y is defined as follows: we first form the arithmetic mean of x and y and call it a1, i.e.

a1 = (x + y) / 2.
We then form the geometric mean of x and y and call it g1, i.e. g1 is the square root of xy. Now we can 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:

[a_ = \frac]
and

[g_ = \sqrt.]
These two sequences converge to the same number, which we call the arithmetic-geometric mean M(x, y) of x and y.

M(x, y) is a number between the geometric and arithmetic mean of x and y; in particular it is between x and y. If r > 0, then M(rx, ry) = r M(x, y).

M(x, y) is sometimes denoted agm(x, y).

Implementation

The following example code in the Scheme programming language computes the arithmetic-geometric mean of two positive real numbers:
(define agmean
(lambda (a b epsilon)
(letrec ((ratio-diff       ; determine whether two numbers
(lambda (a b) ; are already very close together (abs (/ (- a b) b)))) (loop ; actually do the computation (lambda (a b) ;; if they're already really close together, ;; just return the arithmetic mean (if (< (ratio-diff a b) epsilon) (/ (+ a b) 2) ;; otherwise, do another step (loop (sqrt (* a b)) (/ (+ a b) 2))))))
;; error checking
(if (or (not (real? a))
(not (real? b)) (<= a 0) (<= b 0)) (error 'agmean "~s and ~s must both be positive real numbers" a b) (loop a b)))))

One can show that

[\Mu(x,y) = \frac \cdot \frac \right) }]
where K(x) is the complete elliptic integral of the first kind.

The reciprocal of the arithmetic-geometric mean of 1 and the square root of 2 is called Gauss's constant.

[ \frac)} = G ]
named after Carl Friedrich Gauss.

The geometric-harmonic mean can be calculated by an analogous method, using sequences of geometric and harmonic means. The arithmetic-harmonic mean is none other than the geometric mean.

See also

 


From Wikipedia, the Free Encyclopedia. Original article here. Support Wikipedia by contributing or donating.
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