Mathematical function of two positive real arguments
This article is about the particular type of mean. For the similarly named inequality, see
Inequality of arithmetic and geometric means
.
Plot of the arithmetic–geometric mean
agm
(
1
,
x
)
{\displaystyle \operatorname {agm} (1,x)}
generalized means . In
computing π
.
The AGM is defined as the limit of the interdependent sequences
a
i
{\displaystyle a_{i}}
g
i
{\displaystyle g_{i}}
a
0
=
x
,
g
0
=
y
a
n
+
1
=
1
2
(
a
n
+
g
n
)
,
g
n
+
1
=
a
n
g
n
.
{\displaystyle {\begin{aligned}a_{0}&=x,\\g_{0}&=y\\a_{n+1}&={\tfrac {1}{2}}(a_{n}+g_{n}),\\g_{n+1}&={\sqrt {a_{n}g_{n}}}\,.\end{aligned}}}
These two sequences
converge to the same number, the arithmetic–geometric mean of
x and
y ; it is denoted by
M (x , y ), or sometimes by
agm(x , y ) or
AGM(x , y ) .
The arithmetic–geometric mean can be extended to complex numbers and when the branches of the square root are allowed to be taken inconsistently, it is, in general, a multivalued function .[1]
Example To find the arithmetic–geometric mean of a 0 = 24g 0 = 6
a
1
=
1
2
(
24
+
6
)
=
15
g
1
=
24
⋅
6
=
12
a
2
=
1
2
(
15
+
12
)
=
13.5
g
2
=
15
⋅
12
=
13.416
407
8649
…
⋮
{\displaystyle {\begin{array}{rcccl}a_{1}&=&{\tfrac {1}{2}}(24+6)&=&15\\g_{1}&=&{\sqrt {24\cdot 6}}&=&12\\a_{2}&=&{\tfrac {1}{2}}(15+12)&=&13.5\\g_{2}&=&{\sqrt {15\cdot 12}}&=&13.416\ 407\ 8649\dots \\&&\vdots &&\end{array}}}
The first five iterations give the following values:
The number of digits in which a n g n 171 481 725 615 420 766 813 156 974 399 243 053 838 8544 [2]
History The first algorithm based on this sequence pair appeared in the works of
Properties The geometric mean of two positive numbers is
never greater than the arithmetic mean.
[3] So
(gn ) is an increasing sequence,
(an ) is a decreasing sequence, and
gn ≤ M (x , y ) ≤ an . These are strict inequalities if
x ≠ y .
M (x , y )x y x y
If r ≥ 0M (rx ,ry ) = r M (x ,y )
There is an integral-form expression for M (x ,y )[4]
M
(
x
,
y
)
=
π
2
(
∫
0
π
2
d
θ
x
2
cos
2
θ
+
y
2
sin
2
θ
)
−
1
=
π
(
∫
0
∞
d
t
t
(
t
+
x
2
)
(
t
+
y
2
)
)
−
1
=
π
4
⋅
x
+
y
K
(
x
−
y
x
+
y
)
{\displaystyle {\begin{aligned}M(x,y)&={\frac {\pi }{2}}\left(\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {x^{2}\cos ^{2}\theta +y^{2}\sin ^{2}\theta }}}\right)^{-1}\\&=\pi \left(\int _{0}^{\infty }{\frac {dt}{\sqrt {t(t+x^{2})(t+y^{2})}}}\right)^{-1}\\&={\frac {\pi }{4}}\cdot {\frac {x+y}{K\left({\frac {x-y}{x+y}}\right)}}\end{aligned}}}
where
K (k ) is the
complete elliptic integral of the first kind :
K
(
k
)
=
∫
0
π
2
d
θ
1
−
k
2
sin
2
(
θ
)
{\displaystyle K(k)=\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}(\theta )}}}}
Since the arithmetic–geometric process converges so quickly, it provides an efficient way to compute elliptic integrals, which are used, for example, in
elliptic filter design.
[5] Jacobi theta function
θ
3
{\displaystyle \theta _{3}}
[6]
M
(
1
,
x
)
=
θ
3
−
2
(
exp
(
−
π
M
(
1
,
x
)
M
(
1
,
1
−
x
2
)
)
)
=
(
∑
n
∈
Z
exp
(
−
n
2
π
M
(
1
,
x
)
M
(
1
,
1
−
x
2
)
)
)
−
2
,
{\displaystyle M(1,x)=\theta _{3}^{-2}\left(\exp \left(-\pi {\frac {M(1,x)}{M\left(1,{\sqrt {1-x^{2}}}\right)}}\right)\right)=\left(\sum _{n\in \mathbb {Z} }\exp \left(-n^{2}\pi {\frac {M(1,x)}{M\left(1,{\sqrt {1-x^{2}}}\right)}}\right)\right)^{-2},}
which upon setting
x
=
1
/
2
{\displaystyle x=1/{\sqrt {2}}}
gives
M
(
1
,
1
/
2
)
=
(
∑
n
∈
Z
e
−
n
2
π
)
−
2
.
{\displaystyle M(1,1/{\sqrt {2}})=\left(\sum _{n\in \mathbb {Z} }e^{-n^{2}\pi }\right)^{-2}.}
Related concepts The reciprocal of the arithmetic–geometric mean of 1 and the
Gauss's constant
.
1
M
(
1
,
2
)
=
G
=
0.8346268
…
{\displaystyle {\frac {1}{M(1,{\sqrt {2}})}}=G=0.8346268\dots }
In 1799, Gauss proved
[note 1] that
M
(
1
,
2
)
=
π
ϖ
{\displaystyle M(1,{\sqrt {2}})={\frac {\pi }{\varpi }}}
where
ϖ
{\displaystyle \varpi }
is the
lemniscate constant .
M
(
1
,
2
)
{\displaystyle M(1,{\sqrt {2}})}
G
{\displaystyle G}
transcendental by Theodor Schneider .[note 2] [7] [8]
{
π
,
M
(
1
,
1
/
2
)
}
{\displaystyle \{\pi ,M(1,1/{\sqrt {2}})\}}
algebraically independent over
Q
{\displaystyle \mathbb {Q} }
[9] [10]
{
π
,
M
(
1
,
1
/
2
)
,
M
′
(
1
,
1
/
2
)
}
{\displaystyle \{\pi ,M(1,1/{\sqrt {2}}),M'(1,1/{\sqrt {2}})\}}
derivative with respect to the second variable) is not algebraically independent over
Q
{\displaystyle \mathbb {Q} }
[11]
π
=
2
2
M
3
(
1
,
1
/
2
)
M
′
(
1
,
1
/
2
)
.
{\displaystyle \pi =2{\sqrt {2}}{\frac {M^{3}(1,1/{\sqrt {2}})}{M'(1,1/{\sqrt {2}})}}.}
The
geometric–harmonic mean GH can be calculated using analogous sequences of geometric and
harmonic means, and in fact
GH(x,y ) = 1/M(1/x , 1/y ) = xy /M(x,y ) .
[12]
The arithmetic–harmonic mean
is equivalent to the
geometric mean .
The arithmetic–geometric mean can be used to compute – among others – logarithms , complete and incomplete elliptic integrals of the first and second kind ,[13] Jacobi elliptic functions .[14]
Proof of existence The
inequality of arithmetic and geometric means
implies that
g
n
≤
a
n
{\displaystyle g_{n}\leq a_{n}}
and thus
g
n
+
1
=
g
n
⋅
a
n
≥
g
n
⋅
g
n
=
g
n
{\displaystyle g_{n+1}={\sqrt {g_{n}\cdot a_{n}}}\geq {\sqrt {g_{n}\cdot g_{n}}}=g_{n}}
that is, the sequence
gn is nondecreasing and bounded above by the larger of
x and
y . By the
monotone convergence theorem , the sequence is convergent, so there exists a
g such that:
lim
n
→
∞
g
n
=
g
{\displaystyle \lim _{n\to \infty }g_{n}=g}
However, we can also see that:
a
n
=
g
n
+
1
2
g
n
{\displaystyle a_{n}={\frac {g_{n+1}^{2}}{g_{n}}}}
and so:
lim
n
→
∞
a
n
=
lim
n
→
∞
g
n
+
1
2
g
n
=
g
2
g
=
g
{\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\frac {g_{n+1}^{2}}{g_{n}}}={\frac {g^{2}}{g}}=g}
Q.E.D.
Proof of the integral-form expression This proof is given by Gauss.[1]
I
(
x
,
y
)
=
∫
0
π
/
2
d
θ
x
2
cos
2
θ
+
y
2
sin
2
θ
,
{\displaystyle I(x,y)=\int _{0}^{\pi /2}{\frac {d\theta }{\sqrt {x^{2}\cos ^{2}\theta +y^{2}\sin ^{2}\theta }}},}
Changing the variable of integration to
θ
′
{\displaystyle \theta '}
sin
θ
=
2
x
sin
θ
′
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
,
{\displaystyle \sin \theta ={\frac {2x\sin \theta '}{(x+y)+(x-y)\sin ^{2}\theta '}},}
cos
θ
=
(
x
+
y
)
2
−
2
(
x
2
+
y
2
)
sin
2
θ
′
+
(
x
−
y
)
2
sin
4
θ
′
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
,
{\displaystyle \cos \theta ={\frac {\sqrt {(x+y)^{2}-2(x^{2}+y^{2})\sin ^{2}\theta '+(x-y)^{2}\sin ^{4}\theta '}}{(x+y)+(x-y)\sin ^{2}\theta '}},}
cos
θ
d
θ
=
2
x
(
x
+
y
)
−
(
x
−
y
)
sin
2
θ
′
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
2
cos
θ
′
d
θ
′
,
{\displaystyle \cos \theta \ d\theta =2x{\frac {(x+y)-(x-y)\sin ^{2}\theta '}{((x+y)+(x-y)\sin ^{2}\theta ')^{2}}}\ \cos \theta 'd\theta '\ ,}
d
θ
=
2
x
cos
θ
′
(
(
x
+
y
)
−
(
x
−
y
)
sin
2
θ
′
)
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
(
x
+
y
)
2
−
2
(
x
2
+
y
2
)
sin
2
θ
′
+
(
x
−
y
)
2
sin
4
θ
′
d
θ
′
,
{\displaystyle d\theta ={\frac {2x\cos \theta '((x+y)-(x-y)\sin ^{2}\theta ')}{((x+y)+(x-y)\sin ^{2}\theta '){\sqrt {(x+y)^{2}-2(x^{2}+y^{2})\sin ^{2}\theta '+(x-y)^{2}\sin ^{4}\theta '}}}}d\theta '\ ,}
x
2
cos
2
θ
+
y
2
sin
2
θ
=
x
2
(
(
x
+
y
)
2
−
2
(
x
2
+
y
2
)
sin
2
θ
′
+
(
x
−
y
)
2
sin
4
θ
′
)
+
4
x
2
y
2
sin
2
θ
′
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
2
=
x
2
(
(
x
+
y
)
−
(
x
−
y
)
sin
2
θ
′
)
2
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
2
{\displaystyle x^{2}\cos ^{2}\theta +y^{2}\sin ^{2}\theta ={\frac {x^{2}((x+y)^{2}-2(x^{2}+y^{2})\sin ^{2}\theta '+(x-y)^{2}\sin ^{4}\theta ')+4x^{2}y^{2}\sin ^{2}\theta '}{((x+y)+(x-y)\sin ^{2}\theta ')^{2}}}={\frac {x^{2}((x+y)-(x-y)\sin ^{2}\theta ')^{2}}{((x+y)+(x-y)\sin ^{2}\theta ')^{2}}}}
This yields
d
θ
x
2
cos
2
θ
+
y
2
sin
2
θ
=
2
x
cos
θ
′
(
(
x
+
y
)
−
(
x
−
y
)
sin
2
θ
′
)
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
(
x
+
y
)
2
−
2
(
x
2
+
y
2
)
sin
2
θ
′
+
(
x
−
y
)
2
sin
4
θ
′
(
(
x
+
y
)
+
(
x
−
y
)
sin
2
θ
′
)
x
(
(
x
+
y
)
−
(
x
−
y
)
sin
2
θ
′
)
=
2
cos
θ
′
d
θ
′
(
x
+
y
)
2
−
2
(
x
2
+
y
2
)
sin
2
θ
′
+
(
x
−
y
)
2
sin
4
θ
′
,
{\displaystyle {\frac {d\theta }{\sqrt {x^{2}\cos ^{2}\theta +y^{2}\sin ^{2}\theta }}}={\frac {2x\cos \theta '((x+y)-(x-y)\sin ^{2}\theta ')}{((x+y)+(x-y)\sin ^{2}\theta '){\sqrt {(x+y)^{2}-2(x^{2}+y^{2})\sin ^{2}\theta '+(x-y)^{2}\sin ^{4}\theta '}}}}{\frac {((x+y)+(x-y)\sin ^{2}\theta ')}{x((x+y)-(x-y)\sin ^{2}\theta ')}}={\frac {2\cos \theta 'd\theta '}{\sqrt {(x+y)^{2}-2(x^{2}+y^{2})\sin ^{2}\theta '+(x-y)^{2}\sin ^{4}\theta '}}},}
gives
I
(
x
,
y
)
=
∫
0
π
/
2
d
θ
′
(
1
2
(
x
+
y
)
)
2
cos
2
θ
′
+
(
x
y
)
2
sin
2
θ
′
=
I
(
1
2
(
x
+
y
)
,
x
y
)
.
{\displaystyle {\begin{aligned}I(x,y)&=\int _{0}^{\pi /2}{\frac {d\theta '}{\sqrt {{\bigl (}{\frac {1}{2}}(x+y){\bigr )}^{2}\cos ^{2}\theta '+{\bigl (}{\sqrt {xy}}{\bigr )}^{2}\sin ^{2}\theta '}}}\\&=I{\bigl (}{\tfrac {1}{2}}(x+y),{\sqrt {xy}}{\bigr )}.\end{aligned}}}
Thus, we have
I
(
x
,
y
)
=
I
(
a
1
,
g
1
)
=
I
(
a
2
,
g
2
)
=
⋯
=
I
(
M
(
x
,
y
)
,
M
(
x
,
y
)
)
=
π
/
(
2
M
(
x
,
y
)
)
.
{\displaystyle {\begin{aligned}I(x,y)&=I(a_{1},g_{1})=I(a_{2},g_{2})=\cdots \\&=I{\bigl (}M(x,y),M(x,y){\bigr )}=\pi /{\bigr (}2M(x,y){\bigl )}.\end{aligned}}}
The last equality comes from observing that
I
(
z
,
z
)
=
π
/
(
2
z
)
{\displaystyle I(z,z)=\pi /(2z)}
.
Finally, we obtain the desired result
M
(
x
,
y
)
=
π
/
(
2
I
(
x
,
y
)
)
.
{\displaystyle M(x,y)=\pi /{\bigl (}2I(x,y){\bigr )}.}
Applications The number π According to the Gauss–Legendre algorithm ,[15]
π
=
4
M
(
1
,
1
/
2
)
2
1
−
∑
j
=
1
∞
2
j
+
1
c
j
2
,
{\displaystyle \pi ={\frac {4\,M(1,1/{\sqrt {2}})^{2}}{1-\displaystyle \sum _{j=1}^{\infty }2^{j+1}c_{j}^{2}}},}
where
c
j
=
1
2
(
a
j
−
1
−
g
j
−
1
)
,
{\displaystyle c_{j}={\frac {1}{2}}\left(a_{j-1}-g_{j-1}\right),}
with
a
0
=
1
{\displaystyle a_{0}=1}
g
0
=
1
/
2
{\displaystyle g_{0}=1/{\sqrt {2}}}
c
j
=
c
j
−
1
2
4
a
j
.
{\displaystyle c_{j}={\frac {c_{j-1}^{2}}{4a_{j}}}.}
Complete elliptic integral K (sinα ) Taking
a
0
=
1
{\displaystyle a_{0}=1}
g
0
=
cos
α
{\displaystyle g_{0}=\cos \alpha }
M
(
1
,
cos
α
)
=
π
2
K
(
sin
α
)
,
{\displaystyle M(1,\cos \alpha )={\frac {\pi }{2K(\sin \alpha )}},}
where K (k )elliptic integral of the first kind :
K
(
k
)
=
∫
0
π
/
2
(
1
−
k
2
sin
2
θ
)
−
1
/
2
d
θ
.
{\displaystyle K(k)=\int _{0}^{\pi /2}(1-k^{2}\sin ^{2}\theta )^{-1/2}\,d\theta .}
That is to say that this quarter period may be efficiently computed through the AGM,
K
(
k
)
=
π
2
M
(
1
,
1
−
k
2
)
.
{\displaystyle K(k)={\frac {\pi }{2M(1,{\sqrt {1-k^{2}}})}}.}
Other applications Using this property of the AGM along with the ascending transformations of John Landen ,[16] Richard P. Brent [17] transcendental functions (e x cos x , sin x ). Subsequently, many authors went on to study the use of the AGM algorithms.[18]
See also References Notes
^ By 1799, Gauss had two proofs of the theorem, but neither of them was rigorous from the modern point of view.
^ In particular, he proved that the beta function
B
(
a
,
b
)
{\displaystyle \mathrm {B} (a,b)}
a
,
b
∈
Q
∖
Z
{\displaystyle a,b\in \mathbb {Q} \setminus \mathbb {Z} }
a
+
b
∉
Z
0
−
{\displaystyle a+b\notin \mathbb {Z} _{0}^{-}}
M
(
1
,
2
)
{\displaystyle M(1,{\sqrt {2}})}
M
(
1
,
2
)
=
1
2
B
(
1
2
,
3
4
)
.
{\displaystyle M(1,{\sqrt {2}})={\tfrac {1}{2}}\mathrm {B} \left({\tfrac {1}{2}},{\tfrac {3}{4}}\right).}
Citations Sources
