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Eisenstein series and approximations to pi - Mathematics

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8 BRUCE C. BERNDT AND HENG HUAT CHAN<br />

Now, by rearranging the second equation in (4.4), with the help of (4.1)<br />

<strong>and</strong> (4.2), we find that<br />

(4.5) P(q) = R(q) 12q dz<br />

+<br />

Q(q) z dq .<br />

From the chain rule <strong>and</strong> (4.5), it follows that, for any positive integer n,<br />

P(q n ) = R(qn )<br />

Q(qn 12q<br />

+<br />

) nz(qn )<br />

Subtracting (4.5) from the last equality <strong>and</strong> setting<br />

we find that<br />

(4.6)<br />

nP(q n ) − P(q) =n R(qn )<br />

Q(qn R(q)<br />

−<br />

) Q(q)<br />

m := z(q)<br />

z(q n ) ,<br />

q<br />

+ 12<br />

z(qn )<br />

=n R(qn )<br />

Q(qn R(q) q dm<br />

− − 12<br />

) Q(q) m dq .<br />

dz(qn )<br />

.<br />

dq<br />

dz(q n )<br />

dq<br />

q dz(q)<br />

− 12<br />

z(q) dq<br />

Our next aim is <strong>to</strong> replace dm<br />

dm<br />

in (4.6) by<br />

dq dJ (J(q), J(qn )). From (2.1),<br />

the definition of J, (4.4), (4.1), <strong>and</strong> (4.2), upon differentiation, we find that<br />

(4.7)<br />

q dJ<br />

dq =(3Q2 Q ′ − 2RR ′ )Q3 − 3Q2Q ′ (Q3 − R2 )<br />

Q6 = {Q2 (PQ − R) − R(PR − Q 2 )}Q 3 − Q 2 (PQ − R)(Q 3 − R 2 )<br />

Q 6<br />

= RQ3 − R 3<br />

Q 4<br />

which implies that<br />

= R∆<br />

Q 4 = z6√ 1 − J J<br />

Q = z2 J √ 1 − J,<br />

(4.8) z 2 1<br />

(q) =<br />

J(q) dJ(q)<br />

q<br />

1 − J(q) dq .<br />

Replacing q by q n in (4.8) <strong>and</strong> simplifying, we deduce that<br />

(4.9) z 2 (q n ) =<br />

1<br />

nJ(qn ) 1 − J(qn ) q dJ(qn )<br />

.<br />

dq

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