# Download An Integrated Approach to Image Watermarking and JPEG-2000 by Su Wang Kuo PDF

April 5, 2017 | | By admin |

By Su Wang Kuo

Similar mathematics books

Algebra II (Cliffs Quick Review)

In terms of pinpointing the things you really want to understand, not anyone does it higher than CliffsNotes. This speedy, powerful educational is helping you grasp center algebraic innovations -- from linear equations, relatives and capabilities, and rational expressions to radicals, quadratic structures, and factoring polynomials -- and get the very best grade.

Bitopological Spaces: Theory, Relations with Generalized Algebraic Structures, and Applications

This monograph is the 1st and an preliminary creation to the idea of bitopological areas and its purposes. particularly, diverse households of subsets of bitopological areas are brought and numerous kinfolk among topologies are analyzed on one and an analogous set; the speculation of measurement of bitopological areas and the speculation of Baire bitopological areas are built, and diverse periods of mappings of bitopological areas are studied.

Lectures on Lie Groups (University Mathematics , Vol 2)

A concise and systematic advent to the idea of compact attached Lie teams and their representations, in addition to an entire presentation of the constitution and type concept. It makes use of a non-traditional procedure and association. there's a stability among, and a common mix of, the algebraic and geometric features of Lie conception, not just in technical proofs but additionally in conceptual viewpoints.

Extra info for An Integrated Approach to Image Watermarking and JPEG-2000 Compression

Example text

So suppose N x + y − 2i x y ≡ 2 αk (x + i y)k . 2 k=0 Setting y = 0, we obtain N x2 ≡ αk x k k=0 or α0 + α1 x + (α2 − 1)x 2 + · · · + α N x N ≡ 0. Setting x = 0 gives α0 = 0; dividing out by x and again setting x = 0 shows α1 = 0, etc. We conclude that α1 = α3 = α4 = · · · = α N = 0 α2 = 1, and so our assumption that N x 2 + y 2 − 2i x y ≡ αk (x + i y)k k=0 has led us to x 2 + y 2 − 2i x y ≡ (x + i y)2 = x 2 − y 2 + 2i x y, which is simply false! A bit of experimentation, using the method described above (setting y = 0 and “comparing coefﬁcients”) will show how rare the analytic polynomials are.

Sin2 z + cos2 z = 1, c. (sin z) = cos z. * Show that a. sin( π2 + iy) = 12 (e y + e−y ) = cosh y b. | sin z| ≥ 1 at all points on the square with vertices ±(N + 12 )π ± (N + 12 )π i, for any positive integer N. c. | sin z| → ∞, as Imz = y → ±∞. Exercises 43 17. Find (cos z) . 18. Find sin−1 (2)– that is, ﬁnd the solutions of sin z = 2. * Find all solutions of the equation: z ee = 1. 20. Show that sin(x + iy) = sin x cosh y + i cos x sinh y. 21. Show that the power series f (z) = 1 + z + z2 + ...

Power Series Expansion about z = α All of the previous results on power series are easily adapted to power series of the form Cn (z − α)n . By the simple substitution w = z − α, we see, for example, that series of the above form converge in a disc of radius R about z = α and are differentiable throughout |z − α| < R where R = 1/lim|Cn |1/n . ) Exercises 1. 3 by showing that for an analytic polynomial P, Py = i Px . * a. Suppose f (z) is real-valued and differentiable for all real z. Show that f (z) is also real-valued for real z.