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“mcs” — 2017/3/3 — 11:21 — page 691 — #699<br />

16.5. Formal Power Series 691<br />

used above to derive generating functions <strong>for</strong> the Fibonacci and Tower of Hanoi<br />

examples carries over to yield a quotient of polynomials that defines the generating<br />

function f .0/ C f .1/x C f .2/x 2 C . Then partial fractions can be used to find<br />

a <strong>for</strong>mula <strong>for</strong> f .n/ that is a linear combination of terms of the <strong>for</strong>m n k˛n where k<br />

is a nonnegative integer d and ˛ is the reciprocal of a root of the denominator<br />

polynomial. For example, see Problems 16.14, 16.15, 16.18, and 16.19.<br />

16.5 Formal Power Series<br />

16.5.1 Divergent Generating Functions<br />

Let F .x/ be the generating function <strong>for</strong> nŠ, that is,<br />

F .x/ WWD 1 C 1x C 2x 2 C C nŠx n C :<br />

Because x n D o.nŠ/ <strong>for</strong> all x ¤ 0, this generating function converges only at<br />

x D 0. 3<br />

Next, let H.x/ be the generating function <strong>for</strong> n nŠ, that is,<br />

H.x/ WWD 0 C 1x C 4x 2 C C n nŠx n C :<br />

Again, H.x/ converges only <strong>for</strong> x D 0, so H.x/ and F .x/ describe the same,<br />

trivial, partial function on the reals.<br />

On the other hand, F .x/ and H.x/ have different coefficients <strong>for</strong> all powers of<br />

x greater than 1, and we can usefully distinguish them as <strong>for</strong>mal, symbolic objects.<br />

To illustrate this, note than by subtracting 1 from F .x/ and then dividing each<br />

of the remaining terms by x, we get a series where the coefficient if x n is .n C 1/Š.<br />

That is<br />

F .x/ 1<br />

Œx n <br />

D .n C 1/Š : (16.16)<br />

x<br />

Now a little further <strong>for</strong>mal reasoning about F .x/ and H.x/ will allow us to<br />

deduce the following identity <strong>for</strong> nŠ: 4<br />

nŠ D 1 C<br />

nX<br />

.i 1/ .i 1/Š (16.17)<br />

iD1<br />

3 This section is based on an example from “Use of everywhere divergent generating function,”<br />

mathoverflow, response 8,147 by Aaron Meyerowitz, Nov. 12, 2010.<br />

4 A combinatorial proof of (16.17) is given in Problem 15.70

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