# Elementary Differential Topology by James R. Munkres

By James R. Munkres

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Extra resources for Elementary Differential Topology

Example text

G. 1, Sect. 3) (1 − u2 ) d2 G dG + l(l + 1)G = 0 − 2u du2 du where E= 1 + k 2 = −l(l + 1) 4 and 1 l = − − ik . 2 As for the plane case the required solution of the above equation should grow as eikd when d → ∞ and should behave like ln d/2π when d → 0. 1, Sect. 3 it follows that (0) GE (x, x ) = − 1 Q 1 (cosh d(x, x )) . 2π − 2 −ik Here Q− 12 −ik (cosh d) is the Legendre function of the second kind with the integral representation [32], Vol. 4) 1 Q− 12 −ik (cosh d) = √ 2 ∞ d √ eikr dr cosh r − cosh d and the following asymptotics d→0 Q− 12 −ik (cosh d) −→ − log d and Quantum and Arithmetical Chaos d→∞ Q− 12 −ik (cosh d) −→ 23 π ei(kd−π/4) .

But pi = − ∂S , ∂yi pf = ∂S . ∂yf Therefore δpi = − ∂2S ∂2S ∂2S ∂2S δy − δy , δp = δy + δyf . i f f i ∂yi2 ∂yi ∂yf ∂yi ∂yf ∂yf2 From comparison of these two expression one obtains the expressions of the second derivatives of the action through monodromy matrix elements y periodic orbit classical orbit Fig. 7. A periodic orbit and a closed classical orbit in its vicinity Quantum and Arithmetical Chaos 41 1 ∂2S m11 ∂2S m22 ∂2S =− , = , = . 2 ∂yi ∂yf m12 ∂yi m12 ∂yf2 m12 Substituting these expressions to the contribution to the trace formula from one periodic orbit one gets (in two dimensions) (E) = d(osc) p 1 i(2πi )3/2 i m11 + m22 − 2 2 dx |m12 |−1/2 exp( Sp + i y )dy 2 m12 k(x) where x and y are respectively coordinates parallel and perpendicular to the trajectory.

32], Vol. 1, Sect. 1). Therefore if Re s > 1 ∞ Γ (s/2)ζ(s) = xs/2−1 Ψ (x)dx π s/2 0 where Ψ (x) is given by the following series ∞ Ψ (x) = e−πn 2 x . n=1 Using the Poisson summation formula (5) one obtains ∞ e−πn 2 n=−∞ x ∞ 2 1 =√ e−πn /x x n=−∞ which leads to the identity 1 2Ψ (x) + 1 = √ x 1 2Ψ ( ) + 1 x . Hence 1 ξ(s) ≡ π −s/2 Γ ( s)ζ(s) = 2 1 = 0 s/2 x 1 0 xs/2 Ψ (x)dx + 1 1 1 1 √ Ψ( ) + √ − x x 2 x 2 ∞ 1 xs/2 Ψ (x)dx = ∞ dx + 1 xs/2 Ψ (x)dx 1 ∞ 1 1 1 − + = xs/2−3/2 Ψ ( )dx + xs/2 Ψ (x)dx = s−1 s x 0 1 ∞ 1 = + x−s/2−1/2 + xs/2−1 Ψ (x)dx .