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Statistical Mechanics - Physics at Oregon State University

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5.1. FERMIONS IN A BOX. 97<br />

fα(z) =<br />

∞<br />

n=1<br />

z n<br />

n α (−1)n+1 , |z| < 1 (5.34)<br />

This defines the function for a disc around the origin. The values for arbitrary<br />

values of z are obtained by analytical continu<strong>at</strong>ion. For example:<br />

∞<br />

f0(z) = z n (−1) n+1 ∞<br />

= 1 − z n (−1) n = 1 − 1 z<br />

=<br />

1 + z 1 + z<br />

n=1<br />

n=0<br />

and this defines a function everywhere, except <strong>at</strong> the pole z = −1.<br />

Why is this important?<br />

(5.35)<br />

One might argue th<strong>at</strong> now we have computers available and we do not need<br />

all this analytical detail anymore. Since one use of power series is to be able<br />

to evalu<strong>at</strong>e functions over a reasonable domain of arguments, we can simply let<br />

computers take over and do the integr<strong>at</strong>ions for all values. This is not really<br />

correct, however. The analysis above tells us where we have poles and other<br />

singularities. These are places where computer calcul<strong>at</strong>ions will fail! Relying on<br />

computers also assumes th<strong>at</strong> there are no errors in the computer calcul<strong>at</strong>ions,<br />

and th<strong>at</strong> is not always true. Therefore, we need to be able to check our computer<br />

calcul<strong>at</strong>ions against the analytical solutions we can obtain via power series and<br />

other means. The analytical solutions can tell us th<strong>at</strong> near a limiting point<br />

or singularity an energy will behave like |T − T0| −0.3 , for example. This is<br />

very difficult to extract precisely from a computer calcul<strong>at</strong>ion. But, of course,<br />

comput<strong>at</strong>ional physics also has its place, because it can extend the analytical<br />

solutions to domains of the parameters where we cannot get analytical solutions.<br />

Simple example.<br />

As a simple example, study the differential equ<strong>at</strong>ion<br />

¨x + x = 0 (5.36)<br />

and try a power series of the form x(t) = ∞<br />

0 cnt n . This gives:<br />

∞<br />

cnn(n − 1)t n−2 ∞<br />

+ cnt n = 0 (5.37)<br />

0<br />

or after combining terms with equal powers of t, and defining c−2 = c−1 = 0 we<br />

get:<br />

∞<br />

0<br />

0<br />

(cn+2(n + 2)(n + 1) + cn)t n = 0 (5.38)

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