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Ramanujan J DOI 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 ( ab a+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’s Notebooks [2] states (in part): Let the perimeter of ellipse = π(a + b)(1 + h), then a b a + b 2 = 4h 3h 2 2 + 1 3h (1) very nearly ...Berndt’s subsequent discussion of this entry merits comment on two points: 1. If one writes λ 2 (h) 4h 3h 2 2 + 1 3h , (2) then Berndt discusses the (asymptotic) accuracy of the inverse approximation h h ( λ 2 ) , (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 Rica e-mail: [email protected]

Ramanujan’s inverse elliptic arc approximation

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Page 1: Ramanujan’s inverse elliptic arc approximation

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]

Page 2: Ramanujan’s inverse elliptic arc approximation

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}

Page 3: Ramanujan’s inverse elliptic arc approximation

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

= 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)

Page 4: Ramanujan’s inverse elliptic arc approximation

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

Page 5: Ramanujan’s inverse elliptic arc approximation

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)