FEATURE I ARTICLE
The Arithmetico-geometric Mean of Gauss
How to find the Perimeter of an Ellipse
B Sury
Elementary Calculus
As all of us learn very early in school, the length of a
circle is 7r • diameter. Somewhat later, one learns t h a t
elementary calculus gives the length of an arc of the
unit circle to be the value of the integral s = s ( x ) =
The author enjoys
f~ ~
= sin-l(x).
writing about such great
masters as:
A person who does arouse
much admiration a n d a
million wows!
Such a mathematicians'
prince
has not been born since.
I talk of Carl FHedrich
Gauss!
72
Before going further, it is worthwhile to recall how length
of (part of) a curve is defined. Suppose first t h a t t h e
curve is given on the plane and is described by the pair
of equations x = x ( t ) , y = y ( t ) in terms of a p a r a m e t e r
t. Intuitively, one would think of the length of any part
of the curve to be t h a t of parts of a polygon which increasingly approximates t h a t part of the curve. Thus,
one divides the interval [a, b] where t varies into finitely
m a n y parts a = to < tl < " " < tn = b. The points
P~ = ( x ( t , ) , y ( t i ) ) can be joined by line segments and
the sum PoP1 + PIP2 + "'" + P n - l P n of the lengths of
these segments is an approximation to the length of the
part of the curve which corresponds to the p a r a m e t e r
t varying from a to b. This approximation might be a
crude one but it is at once clear (see Figure 1) t h a t one
can take more points to obtain a b e t t e r approximation.
At this point, calculus comes to the rescue. It is heuristically clear t h a t the best possible notion of length is obtained as the limiting case as one increases the n u m b e r
n of points Pi indefinitely while simultaneously letting
the lengths of all the line segments PiPi+l diminish indefinitely. In other words, if the limit exists, it can be
defined as the length of the curve. A curve for which
this limit exists is called rectifiable for evident reasons.
Clearly, circles, ellipses and hyperbolae are rectifiable
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FEATURE I ARTICLE
curves.
E x e r c i s e S h o w that the curve defined by
y = x2sinl/x
, O < x < 1 , y(O) = O
is rectifiable while t h a t defined by
y = xsinl/x
is not.
, O < x < 1 , y(O) = O
W h a t is the length o f the f i r s t c u r v e ?
Is there a nice condition which is sufficient to ensure
t h a t a given curve is rectifiable? Let us look at a curve
w i t h a p a r a m e t r i s a t i o n x = x ( t ) , y -- y ( t ) for a < t < b
where these functions are assumed to have continuous
first derivatives. This condition is indeed enough to enforce rectifiability as we show now. Let a = to < tl <
9 .. < tn = b be a subdivision; then the approximating
perimeter
n--1
Sn =
PiPi+l =
i=0
n-1
-
x(t,))2 + {y(t,+,)
-
y(t,)}2].
i=0
B y the mean-value theorem, there are hi, bi E (ti, ti+l)
and y(t,+l)such t h a t x ( t i + l ) - x ( t i )
= dz
y(ti) = ~ ( b i ) ( t , + l - ti). W h e n we let the n u m b e r of
intervals (ti, ti+l) t e n d to infinity while simultaneously
letting their lengths t e n d to 0, the limit is, by definition, just the integral fb ~ / ( ~ ) 2 + (~)2dt" We note t h a t
the above limit exists by uniform continuity of d~
~ _
dt ~ dt
consequence of the assumption t h a t the parametrising
functions x = x ( t ) , y = y ( t ) for a < t < b are functions
which have continuous first derivatives. Actually, one
can even allow the functions d~ and ~ to have finitely
m a n y discontinuities and still the integral makes sense.
It is clear from the change of variable formula for integrals t h a t the above notion of length is independent of
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Figure 1.
FEATURE I ARTICLE
which parametrisation we use. For this reason, if one is
given the curve in nonparametric form as y = f(x), then
one can write its length from x = a to x = b to be the integral ~b 9/1 +
If the curve is given in polar coordinates as x = r(0)cosO, y = r(O)sinO, the length can,
[dy~2dXkdz/
.
therefore, be written as the integral fo~1 ~ / r 2 + (~)2d0
where r(Oo)COSOo = a, r(O1)cos01 = b.
Let us look at the circle of radius to; its polar equation
is r = ro (constant); the integral gives the length (01 -
Oo)ro.
Already, it might be clear t h a t even if a sequence of
functions fn defined on some interval (a, b) converges in
the interval to a function f , the corresponding integrals
f : fn(x)dx need not converge to f : f ( x ) d x . The same
problem persists with the integrals of their derivatives
etc.
This, for instance, provides the correct explanation of
one of the 'Think-it-over' problems: 'Where is the missing string?' posed in Resonance, Vol. 3, No. 1, 1998.
The area under the string in its various positions can
get arbitrarily close to that under the diagonal string
without the arc lengths tending to the diagonal length.
Elliptic Integrals
After the brief digression to recall the fundamental concept of length of a curve, let us come back a whole circle
viz., look at x 2 + y2 = I. T h e length s = s(xo) from
(x,y) = (0, 1) to (xo, Yo)is given as f~o : ~ = sin-lxo.
We express this as x = sin(s) and, viewed this way, the
addition formula for the sine function amounts to the
formula
foX
where z = x ~
dt
+ foU ~ dt
= fo z
+ y ~ .
W h a t happens when we try to calculate the length of an
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FEATURE I ARTICLE
arc in an ellipse
Lemniscate can be
X2
2
~+~=i?
described as the
Such a length is given by an integral of the form
product of whose
distances from two
fixed points is
constant.
locus of points, the
s(xo) = ~0=~ l+(d'~y)2dxax
a 2 _ e2x 2
-_ f o
.
.
.
.
dx
where R is a rational function (i.e. a quotient of two
polynomials) and p is a polynomial of degree 4 in x.
Here e = ~
is the eccentricity of the ellipse.
Similarly, let us look at a lemniscate which can be described as the locus of points, the product of whose distances from two fixed points is constant. It is given in
polar co-ordinates by the equation r 2 = cos(20), the arc
length is given by the integral
s(00)
=
J~0O~ (r2"t-(dr)2)d0d0
= fOO~cos(20)-l/2dO
=
1=o(1
--
Jo
_
x4)-l/2dx.
The analogy to the circular length f~~
x2)-l/2dx
is evident. For the circle, one evaluates the integral
by rationalising the integrand i.e., by making a change
of variables which makes the integrand into a rational function i.e., into a quotient of two polynomials.
One can rationalise the integrand by the substitution
2t (How?). Guided by this analogy in 1718,
x = 1--$~
Fagnano tried the substitution x = 2t2 for the lemniscate. Although the integrand does not rationalise, he
noticed that f~~ - x 4 ) - l / 2 d x = V~f~~ + t 4 ) - l / 2 d t
-
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Using a ruler and
compass, one can
double the arc
length of a
lemniscate.
if 0 _< x _< 1.
Following this up with the substitution t - 1-=4,
2=2 one obtains f0x~ - x4)-l/2dx =
2j'~~ - u4)-l/2du. Let us note the interesting consequence of this t h a t using a ruler and compass, one can
double the arc length of a lemniscate - see [1] for a m o r e
general discussion of this aspect.
A fundamental idea in the general problem of evaluating integrals of the form f~o R(x, ~ ) d x
is due to
Abel and Jacobi. This is simply to invert the problem!
So, one considers x0 as a function of s in the formula
s(xo) = f~o R(x, ~ - ~ ) d x . Note t h a t the integrand is a
multi-valued function while however their inverse functions arise as single-valued functions as Abel and Jacobi
realised in the 1820's. This yields addition formulae for
these functions akin to those for the trigonometric functions. These integrals and, in general, the integrals of
the form s(xo) = f~o R(x, x/~)dx are called elliptic integrals where R is a rational function and p is a polynomial
of degree 3 or 4 in x (why did we leave out degrees 1 a n d
2?). The inverse functions x = x(s) are called elliptic
functions. T h e y have properties analogous to those for
trigonometric functions; in place of the usual periodicity,
they are doubly 1 periodic!
Although we can't prove it here, the f u n d a m e n t a l fact
involved here is t h a t any cubic curve over the complex
numbers which is smooth i.e., has no singularities, is
biholomorphic to a complex torus. We digress now to
discuss a notion first introduced by Gauss and studied
in relation with elliptic integrals to get some beautiful
results.
The agM
Let a > b > 0 be two real numbers. Construct the
following sequence of arithmetic and geometric means:
al---- a+b2 ' bl = V f ~ '
a=+b.,
bn+l = ~/anbn,
2
76
a2 =
al+b12 , b2 ----
av/-~lbl," ' " , a n + l
=
9 9 9 etc.
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FEATURE I ARTICLE
For example, look at a - 2, b -- 1. We have the following
sequence of values correct to eight decimal places:
n
0
1
2
3
4
an
2
1.5
1.45710678
1.45679105
1.45679103
bn
1
1.41421356
1.45647531
1.45679101
1.45679103
The fundamental
fact involved here
is that any cubic
curve over the
complex numbers
which is smooth is
biholomorphic to a
complex torus.
It is natural to guess t h a t the two sequences approach a
c o m m o n number. This is indeed true in general and is
seen quite easily as follows. From the familiar inequality
which asserts t h a t the arithmetic m e a n is at least as
big as the geometric mean, one has an _> bn. In fact,
a >_ al >_ a2 >_ " " >_ an >_ bn >_ bn-1 >_ "'" >_ bl >_ b.
C a n one say more?
Indeed, it is easy to see t h a t any of the a's is at least as
big as any of the b's (why?) i.e.,
a >__al >_a2>_"" >_an>_"" >_ bm >_ bin-1 >_""
>_bl >_b.
Moreover, by m a t h e m a t i c a l induction, it follows t h a t
an - bn < ~ for all n > 1. W h a t do these two observations give us? T h e first implies t h a t the limits limn--.~an
and limn--.o~bn exist. T h e second implies t h a t these two
limits are actually equal! This c o m m o n limit, denoted
by M (a, b) is called t h e arithmetico-geometric mean and
d e n o t e d by agM, a notation introduced by Gauss himself. Some of the evident properties of the agM are:
M(a,a)=a
M ( a , b ) --- M(an, bn) V n
M(ta, tb)=tM(a,b)
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agM and the Perimeter
Gauss proved the following beautiful result on the perimeter of an ellipse:
Theorem
If a > b > 0, then
fo'~/2( a 2cos2 0 + b2sin20)-t/2dO = 2M ~a,
~r b).
Gauss's proof is ingenious and goes as follows. If I(a, b)
denotes the integral above, the key step is to show t h a t
I(a, b) = I(al, bl) for then, repeating this, and going to
the limit, one will get I(a,b) = I(M(a,b),M(a,b)) =
M(a, b)-l~, which is the assertion. The proof of the key
step is achieved in the following m a n n e r by making the
substitution
sin0 =
2asinr
a + b + (a -- b)sin2r
This is now called Gauss's transformation.
The change of variables gives easily t h a t
a2cos2 0 + b2sin2 0 = a2 (a + b - (a - b)sin2r 2
(a + b + (a - b)sin r 2
Let us differentiate both sides of
(a2cos20 -]- b2sin20) 1/2 ~- a
a + b - (a - b)sin2r
... (A)
a + b + (a - b)sin2r
~:o compute the integral in terms of the new variable r
Derivative of the left hand side of (A) is
(a2cos20 -I" b2sin20 ) - l / 2 ( b 2 - a2)sinOcosOdO.
Differentiating the right hand side of (A), one gets the
expression
4a(b 2 - a2)sinr162
(a + b + (a - b)sin2r 2dr
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FEATURE J ARTICLE
Substituting for the denominator, this further simplifies
to 4a(b 2 - a 2 ) s i n r 1 6 2 1 6 2
i.e., to the expression
(b 2 - a2)cosCsin28sinr162
asinr
On the other hand, using a + b = 2al and ab = b21, one
has
(al2cos2r + bl2sin2r162
(al2-b2)sin4r =
= a 2 _ (2b 2 - a~)sin2r
(a + b) 2 - 2(a 2 + b2)sin2r + (a - b)2sin4r
4
(a + b + (a - b)sin2r 2
4
Thus,
(a2cos2r + b2sin2r
4a2sin2r
a2cos28sin2r
sin28
_ sinScosr .
acos~sinr
Mutiplying this with (b 2 - a2)sinScos6dr yields just the
derivative of the right hand side of (A). In other words,
we have verified that
(a2cos28 + b2sin28)-l/2d~ =
(al2cos2r +
b21sin2r162
From this, the key step follows immediately and so does
Gauss's theorem.
The agM M (a, b) can be expressed in terms of one parameter, viz., the eccentricity e = ~ / a
of the
ellipse. Indeed, M(1 + e, 1 - e) = M(1, x/T-L-'~) =
M(1, b) = M ( a , b ) / a .
Therefore, the theorem can be
restated as
7F
M(l+e,l-e)-
2K(e)
where K ( e ) = fo~/2(1 - e2sin2•)-l/2d•.
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There is a transformation similar to the one used by
Gauss in the proof above. It is due to a little-known
English mathematician J Landen and has turned out to
be very fruitful in evaluating elliptic integrals. Landen's
transformation is the substitution esint9 = sin(2r -/9).
Exercise. Verify that this transformation can also be
used to prove the above theorem of Gauss.
Yet another result that can be proved
usage of Landeri's transformation is the
a > b > 0, denote by J(a, b), the integral
b2sin2O)W2dS. Note that J(a, b) = aE(e)
f~/2(1 - e2sin20)l/2dO. Then, we have:
by a repeated
following. For
f~/2(a2cos20 +
where E(e) =
Proposition
oo
J(a,b) = (a 2 - Z 2n-l(a2 - b2))I(a'b)"
n=O
We shall return to this proposition in the next section.
For some other applications of Landen's transformation,
the interested reader may see [2].
In the case of a lemniscate, the total perimeter is
f01r cos(2O)-l/2d/9 = 4 f0~r/2(2COS2r + sin2~)-l/2d~
from the change of variables
cos(2O) = cos2a.
So, in this case, the constants a and b in Gauss's theorem
are V~ and 1, respectively. Therefore, one has
M ( v ~ , 1 ) = 2w where w =
-
fo (1-z4)-l/2dz.
(0)
Gauss's 92nd entry made in 1798 in his famous mathematical diary reads "I have obtained most elegant results concerning the lemniscate, which surpasses all expectation - indeed, by methods which open an entirely
80
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FEATURE
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new field to us". T h e 98th entry m a d e in May 1799
says that"demonstration of this fact (the identity ~ )
will open an entirely new field of analysis". What is
this 'new field of analysis' that is being talked about?
W h e n a and b are complex numbers, the square-root
causes trouble i.e., at each stage bn has two possible
choices. Thus, M(a, b) is a multi-valued function and
t h e various different values are related. Determining the
relations between t h e m requires the so-called theory of
modular forms of weight 1 for congruence subgroups of
SL(2, Z). This is what Gauss refers to as 'the new field
of analysis.'
Ways to Approximate
The fascination of
calculating
approximations to
is said to have
existed among
mathematicians for
as long as 4000
years - even
laymen have had
their fingers in the
7r
T h e fascination of calculating approximations to 7r is
said to have existed among mathematicians for as long
as 4000 years - even laymen have had their fingers in the
~r! T h e study of zr has led to deep questions - an example is the ancient Greek problem of squaring the circle see the discussion of this problem and other related ones
in [3]. It was only at the end of the 19th century that
L i n d e m a n n (1852-1939) finally proved t h a t 7r is transcendental i.e., that there is no nonconstant polynomial
with rational coefficients that has ~r as a root.
Several great mathematicians haste given'formulae' or
approximations. Perhaps, the earliest was Archimedes
(287-212 B.C.) who gave us the approximation ~13 <
7r < -~. The notation 7r is also due to h i m . Aryabhatta
2832
(476-550 A.D.) gave the approximation 7r ,,~ 62000o
-_ _
3.1416. Leibniz (1646-1716) and Euler (1707-1783) gave
the respective formulae:
lr
1
1
~=1-~+g-7+""
~2
1
1
1
1
--=1+
+
+
+...
8
Legendre gave the following formula for lr in terms of the
elliptic integrals ([4] has a new and rather easy proof):
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FEATURE
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Legendre's Formula
Let O < e < 1 and e' = ~ .
Then
7r
K ( e ) E ( e ' ) + K ( e t ) E ( e ) - K ( e ) K ( e ' ) = 7"
Let us use this formula with the special value e = e~ =
1/x/~ to give an approximation for 7r which involves the
agM. This will be seen to be reminiscent of Archimedes's
method. His method was to approximate the unit circle by inscribed and circumscribed regular polygons.
Then, one has Pn < 21r < Qn where Pn, Qn are, respectively, the perimeters of the inscribed and the circumscribed regular polygons of n sides. In fact, the
numbers an := 1/Q2, and b, = 1/P2~ satisfy the recursion an+l = (an + bn)/2, bn+l = x/a,+lbn. This is
very similar to the defining recursion for M(1/2, 1/4).
However, Archimedes's sequences converge much more
slowly.
Now, Legendre's formula gives
2K(1/V~)E(1/v~)-
K(1/V~)2 = 7"
On the other hand, the proposition above implies (with
a = 1, b - l/v/2) that
E(1/v/2) = (1 - E 2n-l( a2 - b 2 ) ) K ( 1 / V ~ ) 9
n=0
Combining these two with Gauss's theorem, we have
~r(1 - E 2i+1( a2 - b2)) = 4M(1, l/V/'2) 2"
i=1
Thus, one has the approximations
2
n
^2
~rn = 4an+
f f ~'~i-=1
2 i + 1 (~i
- b2) which converge quadratically to ~r. In other words, the number of correct digits
doubles at each step.
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Finally, we end with a formula due to Ramanujan (see
[5]). R a m a n u j a n used the theory of modular functions
(this was alluded to at the end of the last section) to give
some amazing formulae for ~r as well as for the perimeters of ellipses. One of these is:
1
oo 1103 + 26390n
2~rv~ = n=0
~
994~+2
(4n)!
44~(n!) 4"
Using just the first term, one gets the approximation
0.112539536... to ~
= 0.112539539..:
Suggested Reading
[1]
[2]
[3]
[4]
IS]
[6]
B Sury, Resonance, Vol.4, No.12, p.48, 1999.
B Berndt, Ramanujan'$Notebooks,Part III, Springer-Verlag, 1991.
Shashidhar Jagadeesan, Resonance, VoL 4, No. 3, 1999.
G Almkvist and B Berndt, Amer. Math. Monthly,Vol.94 585-608, 1987.
J H Borwein and P B Borwein, P/and theAGM, Wiley, Canada, 1987.
D Cox, L'EnseignonentMatlwmatique, Vol.30, 275-330,1984.
Address for Correspondence
B Sury
Statistics & Mathematics Unit
Indian Statistical Institute
Bangalore 560050, India.
Email: [email protected]
The Square Root by Infinite Descent
One of the first instances of a proof by contradiction that one learns
in school is that of the irrationality of x/~. Here is a proof due to
T a n n e n b a u m by 'infinite descent' - the m e t h o d employed by Fermat
a2
and by B r a h m a g u p t a even earlier in other contexts. Suppose 2 = ~where a, b are n a t u r a l numbers. In the figure, ABCD is a square of
side a, P B Q V and STRD are b o t h squares of side b. As a 2 = 2b 2,
one has Area(TUVW) = Area(PWSA) + Area(QCRU). This means
( 2 b - a ) 2 = 2 ( a - b) 2 i.e., 2 = ~ .
On the other hand, it is clear from
the figure t h a t 0 < 2b - a < a which produces a smaller numerator t h a n
the original one setting in a process of infinite descent. The resultant
contradiction forced on us proves the irrationality of v~.
A
S
T
W
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Kanakku Puly
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