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Concrete mathematics : a foundation for computer science

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338 GENERATING FUNCTIONS<br />

Table 337 is the database we need. The identities in this table are not<br />

difficult to prove, so we needn’t dwell on them; this table is primarily <strong>for</strong><br />

reference when we meet a new problem. But there’s a nice proof of the first<br />

<strong>for</strong>mula, (7.43), that deserves mention: We start with the identity<br />

1<br />

= t (xy)zn<br />

(1 -2)x+’ n<br />

and differentiate it with respect to x. On the left, (1 - z)-~-’ is equal to<br />

elx+l~ln~llll-rll so d/dx contributes a factor of ln(l/( 1 - 2)). On the right,<br />

the numerator df (“‘-,“) is (x +n) . . . (x + 1 ), and d/dx splits this into n terms<br />

whose sum is equivalent to :multiplying (“‘,“) by<br />

-+...+-<br />

1 1<br />

= H x+Tl - H, .<br />

x+n x+1<br />

Replacing x by m gives (7.43). Notice that H,+n - H, is meaningful even<br />

when x is not an integer.<br />

By the way, this method of differentiating a complicated product - leaving<br />

it as a product-is usually better than expressing the derivative as a sum.<br />

For example the right side of<br />

$(i x+n)“...(x+l)‘)<br />

= (x+n)n...(x+l)’<br />

( *+...+A ><br />

would be a lot messier written out as a sum.<br />

The general identities in Table 337 include many important special cases.<br />

For example, (7.43) simplifies to the generating function <strong>for</strong> H, when m = 0:<br />

&ln& = tH,z”. (7.57)<br />

n<br />

This equation can also be derived in other ways; <strong>for</strong> example, we can take the<br />

power series <strong>for</strong> ln(l/( 1 - z)) and divide it by 1 - z to get cumulative sums.<br />

Identities (7.51) and (7.52) involve the respective ratios {,~,}/(“~‘)<br />

and [,“‘J /(“c’), which have the undefined <strong>for</strong>m O/O when n 3 m. However,<br />

there is a way to give them a proper meaning using the Stirling polynomials<br />

of (6.45), because we have<br />

{mmn}/(m~l) .= (-l)n+‘n!mo,(n-m);<br />

[m~n]/(m~l) = n!mo,(m). (7.59)

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