 | Krystian Pilarczyk - 2000 - 944 páginas
...tension t can be computed from the solution k of the equation 2[K(k)-E(k)]Pbot=l (12) where K(k) and E(k) are the complete elliptic integrals of the first and second kind, respectively. In this formulation, for a solution to exist, it is required that 0 < k2 < 1, so that / must be less... | |
 | Mohammed Ameen - 2001 - 288 páginas
...y-\ (2.60) and «,/ = H0." = UQ,' = u:g = 0. In the above, 2Rr 2 i+r The quantities K(m) and £(w) are the complete elliptic integrals of the first and second kind respectively. Also, m is called the parameter of the elliptic integral and k is the modulus. From the above displacement... | |
 | A.M. Alexandrov, D.A. Pozharskii - 2001 - 432 páginas
...25, /2 c(/ 13 -/i 2 ) c 2 5 10 - /$ '00 c 2 5, '01 where K, = /C(\/l - c 2 ), £ — £(\/\ - c 2 ) are the complete elliptic integrals of the first and second kind, respectively, we finally obtain that (4 - 145) B 25 00 / 19 (1 - ni 4 - i 2 - l ( . , , , : )(/ll/16 + /18)/A 3 '... | |
 | Dieter W. Langbein - 2002 - 388 páginas
...— C2 COS2 </9 O^? C 1 + -S(c2) 1-c2 K(c2 cc TT TT o STT (5.14) (5.15) (5.16) and K(c2) and E(c2) are the complete elliptic integrals of the first and second kind, respectively. The volume and surface area may be integrated and expressed in terms of these complete elliptic integrals... | |
 | I. David Abrahams, Paul A Martin, Michael J. Simon - 2002 - 376 páginas
...and a = 2mß2. (3.5) Here cn(x) is the Jacobian elliptic function of modulus m, while K(m) and E(m) are the complete elliptic integrals of the first and second kind respectively. The mean value of A over one period is d, while the spatial period is 1K(m}/ß. As m — » 1, cn2(x)... | |
 | Rodney Vaughan, J Bach Andersen - 2003 - 785 páginas
...(p2(*)) - (i - p2(*)) K (p2(r))] lp2(r)j, (7.6.10) where F is the Gaussian hypergeometric function, K and E are the complete elliptic integrals of the first and second kind respectively, r/2 ( i \-'/2 I•7t/2 / i \1/2 K(m)= fi-mSm2# d0, E(m) = I l-msin20 d0; (7.6.11) Aspects of simulation... | |
 | Enrico Lipparini - 2003 - 452 páginas
...This potential is analytic: RE(r/R) r<R r(E(R/r} - [1 - (R/r)2)K(R/r}} r>R. (5.62) In this equation, K and E are the complete elliptic integrals of the first and second kind respectively. In the second model, a harmonic oscillator potential was employed with LJ0 = 2.78 meV. For the N —... | |
 | Muthusamy Lakshmanan, Shanmuganathan Rajaseekar - 2002 - 644 páginas
...¿ = i—)¿ (1.38) Further defining the new function Z(u) = e(u) - ¿u, (1.39) where K(k) and E(k) are the complete elliptic integrals of the first and second kind respectively, one can show that Z(u) is a periodic function with period 4K. In terms of this new function, the addition... | |
 | William Coffey, Yu. P. Kalmykov, J. T. Waldron - 2004 - 706 páginas
...variables »{£,w) and recall that [14] msn («|m) = lE(m) K(m) nq -cos nxu K(m) <C6) where K(m) and E(m) are the complete elliptic integrals of the first and second kind, respectively [12], and q = e.."K( '""'. Thus, we have from Eqs. (C3)-(C6) == , 2K(m) I cos 6dw 2K(m) m K(m K(m)... | |
 | A.I. Lurie - 2010 - 1036 páginas
...(5.11.1) oo -I i 00 (¿1 — A2A(A) e2 dA (A2-e2)A(A) e)-E(e}\, E(e)-(l-e2}K(e) e2(l-e2) (5.11.2) where /C and E are the complete elliptic integrals of the first and second kind respectively, so that E 1-e2 (5.11.3) and e = 0 yields formulae (5.10.2) for the case of the circular slot. The solution... | |
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F and E are the complete elliptic integrals of the first and second kind, respectively, to modulus k. 
