Báo cáo toán học: "Some Results on the Properties D3 (f ) and D4 (f )" ppsx

... ∈ D 4 (f) then the sequence splits. Proof. Since E ∈ D 4 (f) andL f (α, ) ∈ D 3 (f) (see Proposition 2.11 in [1 ]) then (E,L f (α, )) ∈ S (see Proposition 2.9 in [1 ]). Then by Lemma 1.7 in [1] there ... ν( 2) = max((k, j),m( 2)) . Next we apply ( 4) to k =2,j = ν( 2) and choose ν( 3) = max((k, j),m( 3)) . Continuing this way and by putting a nk = a...

Ngày tải lên: 06/08/2014, 05:20

... holomorphic on ∆ satisfying condition h(z)=g(z, 0) on {|z| < 1 3 }∪{ 2 3 < |z| < 1}. Consider the function on B n given by f(z,w)=  g(z, w)ifw =0 h(z)ifw =0. Then f is continuous on every ... ×f −1 (V ) → V × f −1 (V ) given by θ(z,w)=(p(z),w). Then we have θ(Z ∩ (U × f −1 (V )) ) = {(υ, w) ∈ C × f −1 (V ): υ = f(w)}. It follows that the graph of f |...

Ngày tải lên: 06/08/2014, 05:20

... − 1 2  k−i q q i(x− 1 2 ) = ⎛ ⎝ q x− 1 2 2 E(q)+  x − 1 2  q ⎞ ⎠ k , (2:1 4) where the symbol B k (q)andE k (q) are interpreted to mean that (B(q )) k and (E(q )) k must be replaced by B k (q )andE k (q) when we expanded the one on the ... the limit q ® 1, i.e., I 1 (f ) = lim N→∞ 1 p N p N −1  a=0 f (a)=  p f (z)dμ 1 (z). (2: 3) From (2. 1), we have...

Ngày tải lên: 20/06/2014, 21:20

... function. If s( f ) <1, we get Q(z) is a constant. Then Theorem 1.3 holds. Next, we s uppose that 1 < s(f) <2andl(f) < s(f)=s.SetF(z)=f(z) -P(z), then s(F)=s(f). Substituting F(z)=f(z) ... as follows. Theorem 1.1. Let f(z) be a non-constant entire function, s(f) <1 or 1 < s(f) <2 and l (f) < s(f)=s. Set L 1 (f) =a n (z) f(z + n)+a n-1 (z) f(z + n - 1) +...

Ngày tải lên: 20/06/2014, 21:20

... dim ( R ( A ) ∩ N ( C )) . Hence, (r(B − A) − r(B)+r( A )) − (r(CB − CA) − r(CB)+r(CA )) =dim ( R ( B − A ) ∩ N ( C )) +dim ( R ( A ) ∩ N ( C )) − dim ( R ( B ) ∩ N ( C )) . On account of (1. 6) this theorem can ... ∩ N(C )) - dim (R(A) ∩ N(C )) . Proof. Applying Lemma 1, we have r(CB − CA)=r(C(B − A )) = r(B − A) − dim (R(B − A) ∩ N(C )) , r(CB)=...

Ngày tải lên: 21/06/2014, 00:20

Ngày tải lên: 05/08/2014, 09:46

Ngày tải lên: 05/08/2014, 10:20

... set T j .Then P (T 1 )+ P (T 2 ) ≥ (n − s  ) − (a − s) and consequently at least one of the numbers P (T 1 )andP (T 2 ) is not smaller than (1/ 2)[ (n − s  ) − (a − s)]. Some Results on Mid-Point ... Numbers 87 Therefore by the definition of T 1 and T 2 and also recalling A(n)=  a∈A,a≤n 1 we arrive at the inequality A(n) ≤ n − (1/ 2)( (n − s  ) −...

Ngày tải lên: 06/08/2014, 05:20

... := Z/uZ and v (j) ∈ Z/qZ with j ∈ J and v (j) ≤ v (k) for j<k. Then  j∈J d L (v (j) ,v (j+m) ) ≤ mq ( 8) and equality holds in estimation ( 8) iff d L (v (j) ,v (j+m) )=  v (j+m) − v (j) if j<u− ... 1 4 . Hence, P L q (u) ≥  {j,k}⊆J d L (v (j) ,v (k) )= a(u + b) q 2 − 1 8 + P L q (b) is valid. ✷ One might conjecture equality in (1 0). The combination of the...

Ngày tải lên: 07/08/2014, 07:21

... of the symmetry classes of lines from the [6, 4] configuration (30q)#(10q) (6q) (12q) (10q) (9q) (3q) (see Section 5. 2). Through each point in the configuration (30q)#(10q) (6q) (12q) (10q) (9q) (3q) * pass ... is the farthest symmetry class from the center of the configuration. (For an example of such a configuration, see Figure 13 .) To classify these configurations, the noti...

Ngày tải lên: 07/08/2014, 13:21