27.07.2014 Views

Lectures on String Theory

Lectures on String Theory

Lectures on String Theory

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

– 127 –<br />

where 1 F 1 is a degenerate hypergeometric functi<strong>on</strong>:<br />

C. Useful formulae<br />

1F 1 (α, β; x) =<br />

∞∑<br />

k=0<br />

(α) k<br />

(β) k<br />

x k<br />

k! . (B.7)<br />

In discussi<strong>on</strong> of the central charge of the Virasoro algebra we encounter the sum<br />

S n =<br />

n∑<br />

q 2 .<br />

q=1<br />

Here we present a general method of computing this and similar sums. The method<br />

is based <strong>on</strong> c<strong>on</strong>sidering the generating functi<strong>on</strong><br />

so that<br />

For x < 1 we have<br />

℘(x) =<br />

∞∑<br />

S n x n =<br />

n=1<br />

℘(x) =<br />

S n = 1 n!<br />

∞∑<br />

n=1 q=1<br />

∞∑<br />

S n x n ,<br />

n=1<br />

( ∂ n ℘<br />

∂x n )<br />

| x=0 .<br />

n∑<br />

q 2 x n =<br />

∞∑<br />

q=1<br />

q 2<br />

∞<br />

∑<br />

n=q<br />

x n =<br />

∞∑<br />

q=1<br />

q 2 x q<br />

1 − x .<br />

We further notice that<br />

∂ 2<br />

∂x 2<br />

∞<br />

∑<br />

q=1<br />

x q =<br />

∞∑<br />

q(q − 1)x q−2 = 1 x 2<br />

q=2<br />

∞<br />

∑<br />

q=2<br />

q 2 x q − 1 x 2<br />

∞<br />

∑<br />

q=2<br />

qx q = 1 x 2<br />

∞<br />

∑<br />

q=1<br />

q 2 x q − 1 x 2<br />

∞<br />

∑<br />

q=1<br />

qx q .<br />

Thus,<br />

∞∑ ( ∂<br />

q 2 x q = x 2 2<br />

x<br />

∂x 2 1 − x + 1 ∂<br />

x 2 ∂x<br />

q=1<br />

and, therefore, we obtain the generating functi<strong>on</strong><br />

Finally we compute<br />

1<br />

( ∂ n ℘<br />

)<br />

= 1 n! ∂x n n!<br />

x<br />

)<br />

1 − x<br />

= x2 + x<br />

(1 − x) 3<br />

℘(x) = x2 + x<br />

(1 − x) 4 = 1<br />

(1 − x) 2 − 3<br />

(1 − x) 3 + 2<br />

(1 − x) 4 .<br />

( )<br />

2 · 3 · · · (n + 1) 3 · 4 · · · (n + 2) 4 · 5 · · · (n + 3)<br />

− 3 + 2 .<br />

(1 − x) n+2 (1 − x) n+3 (1 − x) n+4<br />

The last formula results into<br />

S n = (n + 1)<br />

(1 − 3 2 (n + 2) + 1 )<br />

3 (n + 2)(n + 3)<br />

= 1 n(n + 1)(2n + 1) .<br />

6

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!