A bivariate extension of Bleimann-Butzer-Hahn operator by Khan R.A.

By Khan R.A.

Show description

Read or Download A bivariate extension of Bleimann-Butzer-Hahn operator PDF

Best mathematics books

Field Theory and Its Classical Problems (Carus Mathematical Monographs, Volume 19)

Put up 12 months word: First released January 1st 1978
------------------------

Field thought and its Classical difficulties shall we Galois idea spread in a usual means, starting with the geometric building difficulties of antiquity, carrying on with in the course of the development of standard n-gons and the houses of roots of harmony, after which directly to the solvability of polynomial equations by way of radicals and past. The logical pathway is old, however the terminology is in line with glossy remedies.

No prior wisdom of algebra is believed. awesome issues taken care of alongside this path comprise the transcendence of e and p, cyclotomic polynomials, polynomials over the integers, Hilbert's irreducibility theorem, and lots of different gem stones in classical arithmetic. old and bibliographical notes supplement the textual content, and whole recommendations are supplied to all difficulties.

Combinatorial mathematics; proceedings of the second Australian conference

A few shelf put on. half" skinny scrape to backbone. Pages are fresh and binding is tight.

Additional resources for A bivariate extension of Bleimann-Butzer-Hahn operator

Example text

6. There exist e > 0 and an analytic matrix-valued function R(X,Y ) on Be (0) ⇥ Be (0) such that ⇣ ⌘ 1 exp X expY = exp X +Y + [X,Y ] + R(X,Y ) 2 24 1 Lie Groups and Algebraic Groups when X,Y 2 Be (0). Furthermore, kR(X,Y )k  C(kXk + kY k)3 for some constant C and all X,Y 2 Be (0). 6 we now obtain the fundamental identities relating the Lie algebra structure of gl(n, R) to the group structure of GL(n, R). 7. 14) ⇣ 1 ⌘ X exp k 1 ⌘⌘k2 Y . 15) k Proof. 6 implies that exp ⇣1 ⌘ ⇣1 ⌘ ⇣1 ⌘ X exp Y = exp (X +Y ) + O(1/k2 ) , k k k where O(r) denotes a matrix function of r whose norm is bounded by Cr for some constant C (depending only on kXk + kY k) and all small r.

Identify Hn with C2n as a vector space over C by the map a + jb 7! ab , where a, b 2 Cn . Let T = A + jB 2 Mn (H) with A, B 2 Mn (C). (a) Show left multiplication by T on Hn corresponds to multiplication by the ⇥ Athat ¯⇤ B matrix B A¯ 2 M2n (C) on C2n . (b) Show that multiplication by i on Mn (H) becomes the transformation   A B¯ iA iB¯ 7! B A¯ iB iA¯ 11. 6) as an H-valued function on C2n ⇥ C2n . 2 The Classical Lie Algebras 13 (a) Show that B(x, y) = B0 (x, y) + jB1 (x, y), where B0 is a C-Hermitian form on C2n of signature (2p, 2q) and B1 is a nondegenerate skew-symmetric C-bilinear form on C2n .

Then H is a closed subgroup of GL(n, R), and Lie(H) = {X 2 Lie(G) : exp(tX) 2 H for all t 2 R} . Proof. It is obvious that H is a closed subgroup of GL(n, R). If X 2 Lie(H) then exp(tX) 2 H ⇢ G for all t 2 R. Thus X 2 Lie(G). t u We consider Lie(Sp(n, C)). Since Sp(n, C) ⇢ GL(2n, C) ⇢ GL(2n, R), we can look upon Lie(Sp(n, C)) as the set of X 2 M2n (C) such that exptX 2 Sp(n, C) for all t 2 R. This condition can be expressed as exp(tX t )J exp(tX) = J for all t 2 R . 19) Differentiating this equation at t = 0, we find that X t J + JX = 0 for all X 2 Lie(Sp(n, C)).

Download PDF sample

Rated 4.65 of 5 – based on 5 votes