# Combinatorial Physics by Ted Bastin

By Ted Bastin

An essay within the conceptual foundations of physics. Its objective is to introduce what's known as a combinatorial method.

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Extra info for Combinatorial Physics

Sample text

1)r a n! 2r , r! 2n−1 (1) 38 Some irrational numbers where for r > 0 the denominator r! contains the prime factor 2 at most r − 1 times, while n! contains it exactly n − 1 times. ) And since n is even (we assume that n = 2m ), the series that we get for r ≥ n + 1 are 2b 4 8 2 + + + ... n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3) resp. 2a − 2 4 8 + − ± ... n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3) 4a These series will for large n be roughly 4b n resp. − n , as one sees again by comparison with geometric series.

The lattice points observing (3) are precisely the points in the upper strip 0 < py − qx < p2 , and those of (4) the points in the lower strip 0 < qx − py < 2q . Hence the number of lattice points in the two strips is s + t. 3. The outer regions R : py − qx > p2 and S : qx − py > q2 contain the same number of points. To see this consider the map ϕ : R → S which q+1 maps (x, y) to ( p+1 2 − x, 2 − y) and check that ϕ is an involution. Since the total number of lattice points in the rectangle is q−1 infer that s + t and p−1 2 · 2 have the same parity, and so q p p q = (−1)s+t = (−1) p−1 q−1 2 2 .

Clearly the even 1 1 terms 212 + 412 + 612 + . . = k≥1 (2k) 2 sum to 4 ζ(2), so the odd terms 1 1 1 1 k≥0 (2k+1)2 make up three quarters of the total 12 + 32 + 52 + . . = sum ζ(2). Thus Euler’s series is equivalent to The Substitution Formula To compute a double integral Z f (x, y) dx dy. I = S we may perform a substitution of variables x = x(u, v) k≥0 π2 1 = . (2k + 1)2 8 Proof. As above, we may express this as a double integral, namely 1 1 J = 0 0 if the correspondence of (u, v) ∈ T to (x, y) ∈ S is bijective and continuously differentiable.