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A PRIMER OF ANALYTIC NUMBER THEORY: From Pythagoras to ...

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I2.4 Geometric Series 125<br />

can be detected by considering the residue of f ′ (z)/f (z). When residues can<br />

be computed by means of integrals, this provides a way of counting the <strong>to</strong>tal<br />

number of zeros of f (z) in any given region.<br />

Exercise I2.3.1. Compute the Laurent expansion of 2/(x 2 − 1) at x =−1.<br />

Do the same for x = 0.<br />

Exercise I2.3.2. What value should you assign <strong>to</strong> the function x 2 /(cos(x) −<br />

1) at x = 0?<br />

Exercise I2.3.3. Suppose a function f (z) has a zero of order N at z = a,<br />

that is,<br />

f (z) = aN (z − a) N + O((z − a) N+1 ), with aN �= 0.<br />

Compute the residue of f ′ (z)/f (z)atz = a.<br />

I2.4. Geometric Series<br />

The ancient Greeks created a lot of beautiful mathematics, but they were not<br />

very comfortable with the idea of infinity. Zeno’s paradox is an example.<br />

Zeno of Elea (circa 450 b.c.) is known for the following s<strong>to</strong>ry about the<br />

Tor<strong>to</strong>ise and Achilles. (Achilles was the hero of Homer’s Iliad.) Achilles and<br />

the Tor<strong>to</strong>ise were <strong>to</strong> have a race, and Achilles gave the Tor<strong>to</strong>ise a head start<br />

of one kilometer. The Tor<strong>to</strong>ise argued that Achilles could never catch up,<br />

because in the time it <strong>to</strong>ok Achilles <strong>to</strong> cover the first kilometer, the Tor<strong>to</strong>ise<br />

would go some further small distance, say one tenth of a kilometer. And in<br />

the time it <strong>to</strong>ok Achilles <strong>to</strong> cover that distance, the Tor<strong>to</strong>ise would have gone<br />

still further, and so forth. Achilles gets closer and closer but is never in the<br />

same place (asserts Zeno).<br />

Actually, the mathematicians of the ancient world had the techniques<br />

needed <strong>to</strong> resolve the paradox. Archimedes knew the Geometric series<br />

1 + x + x 2 +···+x n n+1 1 − x<br />

= . (I2.9)<br />

1 − x<br />

Exercise I2.4.1. The formula (I2.9) was derived in Exercise 1.2.10, using<br />

finite calculus. But there is a shorter proof, which you should find. Start by<br />

multiplying<br />

(1 − x)(1 + x + x 2 ···+x n ).

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