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Ramanujan JDOI 10.1007/s11139-013-9498-7
Ramanujan’s inverse elliptic arc approximation
Mark B. Villarino
Received: 17 October 2012 / Accepted: 21 May 2013© Springer Science+Business Media New York 2013
Abstract We suggest a continued fraction origin to Ramanujan’s approximation to( a−ba+b
)2 in terms of the arc length of an ellipse with semiaxes a and b.
Keywords Ramanujan · Arc length of an ellipse · Approximation
Mathematics Subject Classification 33E05 · 41A25
Entry 14 of Chapter 38 (page 541) in Berndt’s edition of Volume 5 of Ramanujan’sNotebooks [2] states (in part):
“Let the perimeter of ellipse = π(a + b)(1 + h), then
(a − b
a + b
)2
= 4h − 3h2
2 + √1 − 3h
(1)
very nearly . . .”
Berndt’s subsequent discussion of this entry merits comment on two points:
1. If one writes
λ2(h) ≈ 4h − 3h2
2 + √1 − 3h
, (2)
then Berndt discusses the (asymptotic) accuracy of the inverse approximation
h ≈ h(λ2), (3)
Support from the Vicerrectoría de Investigación of the University of Costa Rica is acknowledged.
M.B. Villarino (�)Depto. de Matemática, Universidad de Costa Rica, 2060 San José, Costa Ricae-mail: [email protected]
M.B. Villarino
i.e., how well the function of λ2 approximates h, instead of the stated directapproximation, i.e., how well the approximative function of h in (1) approxi-mates λ2.
2. Berndt provides no suggestion as to how Ramanujan might have proved (or dis-covered) the approximation (1).
We therefore offer the following treatment of Entry 14 in which we deal with bothpoints. Our starting point is Ivory’s well-known series representation for the perimeterof an ellipse with semiaxes a and b ([1], p. 146):
Theorem 1 If
λ := a − b
a + b, (4)
and if L denotes the perimeter of the ellipse
x2
a2+ y2
b2= 1, (5)
then
L = π(a + b)
{1 +
(1
2
)2
λ2 +(
1 · 1
2 · 4
)2
λ4 +(
1 · 1 · 3
2 · 4 · 6
)2
λ6 + · · ·}. (6)
We sketch his clever proof. In 1796, J. Ivory [3] published the following identity(in somewhat different notation):
Theorem 2 (Ivory’s Identity) If 0 � x � 1 then the following formula for B(x) isvalid:
B(x) ≡ 1
π
∫ π
0
√1 + 2
√x cos(2φ) + x dφ =
∞∑n=0
{1
2n − 1
1
4n
(2n
n
)}2
xn. (7)
Proof We sketch his elegant proof.
1
π
∫ π
0
√1 + 2
√x cos(2φ) + x dφ
= 1
π
∫ π
0
√1 + √
xe2iφ
√1 + √
xe−2iφ dφ
= 1
π
∫ π
0
∞∑m=0
{1
2m − 1
1
4m
(2m
m
)(√
x)me2πimφ
}
×∞∑
n=0
{1
2n − 1
1
4n
(2n
n
)(√
x)ne−2πinφ
}dφ
= 1
π
∞∑m=0
{1
2m − 1
1
4m
(2m
m
)(√
x)m}
Ramanujan’s inverse elliptic arc approximation
×∞∑
n=0
{1
2n − 1
1
4n
(2n
n
)(√
x)n}∫ π
0e2πi(m−n)φ dφ
=∞∑
n=0
{1
2n − 1
1
4n
(2n
n
)}2
xn.�
Now we substitute x := ( a−ba+b
)2. Then the integral becomes
1
π
∫ π
0
√√√√1 + 2
√(a − b
a + b
)2
cos(2φ) +(
a − b
a + b
)2
dφ
= 4
π(a + b)
∫ π2
0
(a2 sin2 φ + b2 cos2 φ
)dφ,
and therefore
B{(
a − b
a + b
)2}= 4
π(a + b)
∫ π2
0
(a2 sin2 φ + b2 cos2 φ
)dφ.
But, it is well known (Berndt [2]) that the perimeter, p, of an ellipse with semiaxes a
and b is given by
p = 4∫ π
2
0
(a2 sin2 φ + b2 cos2 φ
)dφ,
and thus
p = π(a + b) · B{(
a − b
a + b
)2}. (8)
and this is Ivory’s formula for the perimeter of an ellipse.We deal with the first point.Writing
L := π(a + b)(1 + h) (9)
it follows from (6) that
h ≡ h(λ) = 1
4λ2 + 1
43λ4 + 1
44λ6 + 25
47 λ8 + 49
48λ10 + 441
410λ12 + 1089
411λ14
+ 184041
415λ16 + 511225
416λ18 + · · · . (10)
If we invert this series, we obtain the true expansion:
λ2 ≡(
a − b
a + b
)2
= 4h − h2 − 1
2h3 − 5
8h4 − 17
16h5 − 273
128h6
− 609
128h7 − 23391
2048h8 + · · · . (11)
M.B. Villarino
On the other hand, if we develop the RHS of Ramanujan’s approximation (1) intopowers of h, we obtain the approximate expansion:
λ2 ≡(
a − b
a + b
)2
≈ 4h − h2 − 1
2h3 − 5
8h4 − 17
16h5 − 269
128h6
− 1163
256h7 − 10657
1024h8 − · · · . (12)
Therefore, we conclude that
(Approximate λ2) − (
True λ2) = h6
32+ 55
256h7 + 2077
2048h8 + · · · ,
and thus, asymptotically, the following error estimate is valid:
Theorem 3 Ramanujan’s approximation overestimates the true value by about 132h6.
The accuracy is quite impressive.Now we deal with the second point.If we expand the RHS of the true value (11) into a continued fraction, we obtain
(a − b
a + b
)2
= 4h − h2
1 − 12 h
1−34 h
1−34 h
1−3136 h
1−···
. (13)
We suggest that starting with the third numerator on, Ramanujan took all of thenumerators = 3
4h. Then
(a − b
a + b
)2
≈ 4h − h2
1 − 12 h
1−34 h
1−34 h
1−34 h
1−···
. (14)
Let us define B by
B := 1 −34h
1 − 34 h
1−34 h
1−···
. (15)
To evaluate B , we observe that B satisfies the equation
B = 1 −34h
B⇒ B2 − B + 3
4h = 0
⇒ B = 1 + √1 − 3h
2
Ramanujan’s inverse elliptic arc approximation
since B = 1 when h = 0. Therefore, (13) becomes
(a − b
a + b
)2
≈ 4h − h2
1 − 12 h
1+√1−3h2
= 4h − 3h2
2 + √1 − 3h
,
and we have recovered (or discovered!) Ramanujan’s approximation (1).This derivation also explains why Ramanujan’s approximation is so accurate, since
the continued fraction of the true value coincides with that of the approximative valuefor the first four convergents, so that one knows that the error is that of the fifthconvergent.
Ramanujan’s mastery of continued fractions is justly famous, and we suggest thathe discovered the approximation (1) by a method that follows the march of our pre-sentation.
Acknowledgements We thank the reviewer for his suggestions. Assistance from the Vicerectoria deInvestigación of the University of Costa Rica is acknowledged.
References
1. Berndt, B.: Ramanujan’s Notebooks, vol. 3. Springer, New York (1991)2. Berndt, B.: Ramanujan’s Notebooks, vol. 5. Springer, New York (1998)3. Ivory, J.: A new series for the rectification of the ellipsis; together with some observations on the
evolution of the formula (a2 + b2 − 2ab cosφ)n . Trans. R. Soc. Edinb. 4(2), 177–190 (1796)