By Lucia M., Magtone R., Zhou H.-S.
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Additional info for A dirichlet problem with asymptotically linear and changing sign nonlinearity
1, u < v . 0, u ≥ v The function σ is needed to count the simple zeros of the polynomial h(x) and its derivatives. Let #h(k) denote the number of simple zeros that the polynomial equation (k) h (x) = 0 has. Clearly, #h = r = ∑ri=1 σ (0, mi ). By the classical Rolle’s theorem, Proof. 38], define the function σ (u, v) := r r r i=1 i=1 i=1 #h = ∑ σ (1, mi ) + (#h) − 1 = ∑ σ (1, mi ) + ∑ σ (0, mi ) − 1. Continuing this process, we find that r r r r i=1 i=1 i=1 i=1 #h = ∑ σ (2, mi ) + (#h ) − 1 = ∑ σ (2, mi ) + ∑ σ (1, mi ) + ∑ σ (0, mi ) − 2, 18 1 Modular Number Systems and more generally that r r k #h(k) = ∑ σ (k, mi ) + (#h(k−1)) − 1 = ∑ ∑ σ ( j, mi ) − k i=1 i=1 j=0 for all integers k ≥ 0.
8) w− 1 w2 = (x1 − uy1 )(x2 + uy2 ) = (x1 x2 − y1 y2 ) + u(x1 y2 − x2 y1 ). 9) and The real and imaginary parts of the conjugate product z1 z2 , denoted by z1 z2 0 and z1 z2 i , respectively, are called the inner and outer products of the complex numbers z1 and z2 . Likewise, the real and unipotent parts of the conjugate product w− 1 w2 , − denoted by w− w and w w , respectively, are called the hyperbolic inner and u 2 0 2 1 1 outer products of the hyperbolic numbers w1 and w2 . The vectors z1 and z2 in the complex number plane and w1 and w2 in the hyperbolic number plane are said to be, respectively, Euclidean orthogonal or hyperbolic orthogonal if their respective inner products are zero.
5 The hyperbolic angle φ is defined by φ ≡ tanh−1 (y/x) in the quadrants H-I and H-III or φ ≡ tanh−1 (x/y) in H-II and H-IV, respectively. Just as the area of the sector of the unit circle with central angle θ is 12 θ , the area of the unit hyperbolic sector determined by the ray from the origin to the point exp uφ = cosh φ + u sinh φ (shaded in Fig. 2b) is 12 φ . Let us consider a simple example. 693147. ˙ Alternatively, we can calculate the hyperbolic angle φ by calculating the hyperbolic arc length along the hyperbola x2 − y2 = 1, getting dsh = where dy dx |dx2 − dy2| = = x/y = √ x2 x −1 5 4 1 |1 − dy dx 2 |dx = log 2 .