Báo cáo hóa học: "ON THE DIFFERENCE EQUATION xn+1 = axn − bxn /(cxn − dxn−1 )" pptx

... − h h − k  . (5.3) Now, it follows from (5.1)that x 2n−1 = x 2n−2 − x 2n−2 x 2n−2 − x 2n−3 = h +(n − 1)  h − k − (n − 1) − h h − k  − h+(n−1)  h−k−(n−1)−h/(h−k)   h+(n−1)  h−k−(n−1)−h/(h−k)  −  k+(n−1)  h−k−(n−2)−h/(h−k)  = h ... by (h − k)weget x 2n−1 = k +(n − 1)  h − k − (n − 1) − h h −...

Ngày tải lên: 22/06/2014, 22:20

... X, the mapping g x : ℝ ® Y given by g x (t) := f(tx)(t Î ℝ) be continuous. Then the mapping f satisfies the functional equa- tion (1.4). 3. Stability of the quadratic equation (1.4) For r = 1, the ... satis- fies the functional equation f (x + y)+f (x − y )=2 f (x)+2f (y). (1:1) Hence, the above equation i s called the quadratic functional equation. In part...

Ngày tải lên: 20/06/2014, 22:20

... 1) (cb−1)ae b = b a 2 e(ad − 1)(cb − 1) for n = k, assume that x 2k−1 = x 2k−3 y 2k−2 x 2k−3 − 1 = d (ad − 1) k , x 2k = x 2k−2 y 2k−1 x 2k−2 − 1 = c(cb − 1) k , y 2k−1 = y 2k−3 x 2k−2 y 2k−3 − 1 = b (cb − ... 1 , y 1 = y −1 x 0 y −1 − 1 = b cb − 1 , z 1 = 1 y 0 z 0 = 1 ae , x 2 = x 0 y 1 x 0...

Ngày tải lên: 20/06/2014, 22:20

... 8 n−1 t))) , T ∞ k=1 (T(ρ 2 n+k−1 x,2 n+k−1 x (8 n−1 t), ρ 2 n+k x,2 n+k−1 x (4 · 8 n−1 t))) ≥ T(T ∞ k=1 (T(ρ 2 n+k−1 x,2 n+k−1 x (2 n−3 t), ρ 2 n+k x,2 n+k−1 x ))), T ∞ k=1 (T(ρ 2 n+k−1 x , 2 n+k−1 x (2 n−3 t), ... 16 n−1 t))) , T ∞ k=1 (T(ρ 2 n+k−1 x,2 n+k−1 x (2 · 16 n−1 t), ρ 2 n+k x,2 n+k−1 x (8 · 16 n−1 t))) ≥ T(T ∞ k=1 (T(ρ 2 n+k−1 x,2 n+k−1 x (2 · 4 n−2 t), ρ 2 n+k x,2 n+k−1 x (2 · 4...

Ngày tải lên: 21/06/2014, 01:20

... that A i ◦ σ = −A i for i =1 ,2and, for all x Î V 1 ,   2f 1 (x) − A 1 (x) − A 2 (x) − v(x) − q(x) − 2f 1 (0)    1 | 2 | δ , (3:3)   2f 2 (x) − A 1 (x)+A 2 (x)+v(x) − q(x) − 2f 2 (0)    1 |2| δ , (3:4)   f 3 (x) ... in Equation 2.4 with x − σ (x) 2 , we get     f  x − σ(x)  + f (0) − 4f  x − σ(x) 2          ...

Ngày tải lên: 21/06/2014, 02:20

... variables 2f  n  i=1 x i  +  1≤i,j≤n i=j f(x i − x j ) = (n + 1) n  i=1 f(x i ) + (n − 1) n  i=1 f(−x i ) as the equation for the spaces of generalized functions. Using the fundamental solution of the ... ◦ A +  1≤i,j≤n, i=j u ◦ B ij = (n + 1) n  i=1 u ◦ P i + (n − 1) n  i=1 u ◦ Q i ,(1.2)    2u ◦ A +  1≤i,j≤n, i=j u ◦ B ij − (n + 1) n  i=1 u...

Ngày tải lên: 21/06/2014, 20:20

... +1}, g(M + ε) <y l s − j <g(M − ε), if j ∈{k +1, ,2k +1}. (2.17) From (1.2)and(2.17), we have y l s −k = f  y l s −k−1 ,x l s −2 k−1  <f  g(M − ε),M − ε  = g(M − ε). (2.18) Also (1.2), ... Consider equation x n+1 = 1+ x n y n−k , y n+1 = 1+ y n x n−k , n = 0,1, , (3.4) where k ∈{1,2, } and the initial conditions x −k ,x −k+1 , ,x 0 ,...

Ngày tải lên: 22/06/2014, 22:20

... = − (a − 1)(b − 1) + c − d +   (a − 1)(b − 1) + d − c  2 +4c(a − 1)(b − 1) 2(a − 1) , M 2 = m 2 = y = − (a − 1)(b − 1) + d − c +   (a − 1)(b − 1) + c − d  2 +4d(a − 1)(b − 1) 2(b − 1) . (3.4) It ... that lim k→∞ x l k (s+1)+R−q−t−2 = α,lim k→∞ y l k (s+1)+R−q−p−2 = β. (2.10) By (1.3)w...

Ngày tải lên: 22/06/2014, 22:20

... C 8  C λ,K   m i =1 χ 2(−i) χ 2  m i =1 iχ −i . (4.32) Taking limit as m →∞,weobtain |u| L ∞ (Ω) ≤ C 8  C λ,K  σ 1 χ σ 2 , (4.33) where σ 1 =  ∞ i=1 χ 2(−i) and σ 2 = 2  ∞ i=1 iχ −i . In order ... of the changes in the length of the strings during the vibra- tions. The parameters in (1.3) have the following meanings: L is the length of the string, h is...

Ngày tải lên: 22/06/2014, 22:20

... importantly, superiority of one type of therapy over another could not be distinguished, and therefore that the specific choice of therapy was diffi- cult to justify over another type of treatment. This ... of the 'magic' in the hands of the individual therapist might be the ability to engage the patient and direct appropriate attention to task. It is apparent then, that...

Ngày tải lên: 19/06/2014, 08:20