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270 D. Peña Peña and F. Sommen CMFT<br />

An <strong>in</strong>terest<strong>in</strong>g particular case of the toroidal expansion is the case where the<br />

coefficients do not depend on the variable x 0 and satisfy the symmetric relation<br />

A k,l (x) = A l,k (x). With this assumption we can calculate explicitly us<strong>in</strong>g (3) the<br />

coefficients of series (2). Indeed, they are determ<strong>in</strong>ed by the <strong>for</strong>mula<br />

A k,l (x) = (−1)k+l<br />

2 k+l k! l! ∂k+l x A 0,0 (x).<br />

Substitut<strong>in</strong>g the above <strong>in</strong> (2) gives<br />

∞∑ k∑<br />

f(x 0 , x) = Z k−l Z l A k−l,l (x)<br />

=<br />

=<br />

=<br />

k=0<br />

∞∑<br />

k=0<br />

k=0<br />

l=0<br />

(−1) k<br />

2 k k!<br />

k∑<br />

l=0<br />

k!<br />

(k − l)! l! Zk−l Z l ∂ k xA 0,0 (x)<br />

∞∑ (−1) k<br />

2 k k! (Z + Z)k ∂xA k 0,0 (x)<br />

∞∑ (−x 0 ) k<br />

∂ k<br />

k!<br />

xA 0,0 (x)<br />

k=0<br />

which is the classical Cauchy-Kowalewski extension (1).<br />

We can also solve the recurrence <strong>for</strong>mula (3) if the coefficients satisfy the relation<br />

A k,l (x 0 , x) = (−1) k+l A l,k (x 0 , x). In this case we obta<strong>in</strong> that<br />

⎧<br />

(−1)<br />

⎪⎨<br />

l<br />

2<br />

A k,l (x 0 , x) =<br />

k+l k! l! D x(P x D x ) (k+l−1)/2 A 0,0 (x 0 , x) if k + l is odd,<br />

⎪⎩ (−1) l<br />

2 k+l k! l! (P xD x ) (k+l)/2 A 0,0 (x 0 , x) if k + l is even,<br />

where the differential operator P x is def<strong>in</strong>ed by<br />

P x g = ∂ x0 g + ω∂ r g + 1 r Γ x(ωg).<br />

Then we obta<strong>in</strong> the follow<strong>in</strong>g expansion<br />

∞∑ 1<br />

f(x 0 , x) =<br />

2 2k (2k)! (Z − Z)2k (P x D x ) k A 0,0 (x 0 , x)<br />

k=0<br />

∞∑ 1<br />

+<br />

2 2k+1 (2k + 1)! (Z − Z)2k+1 D x (P x D x ) k A 0,0 (x 0 , x)<br />

k=0<br />

∞∑<br />

k (r − 1)2k<br />

= (−1) (P x D x ) k A 0,0 (x 0 , x)<br />

(2k)!<br />

k=0<br />

∞∑<br />

+ (−1) k (r − 1)2k+1 ω<br />

D x (P x D x ) k A 0,0 (x 0 , x).<br />

(2k + 1)!<br />

k=0

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