1. Subdivision Analysis via JSR We already know the z-transform formulation of schemes: To check if the scheme generates a continuous limit curve ( the scheme is ), we have to check if the difference scheme is contractive. We hereby present the ‘joint spectral radius’ analysis (JSR) of schemes, and examine the relation between the two methods. Let us consider a univariate scheme with support [0,s]. 2005 Subdivision Summer School

2. Since the support of the basic limit function is also [0,s], and the limit function is, the values of on [0,1] are fully determined by

3. Since the support of the basic limit function is also [0,s], and the limit function is, the values of on [0,1] are fully determined by The values on [0,1/2] are determined by The values on [1/2,1] are determined by

4. Let us recall the matrix representation of subdivision schemes, and use this representation to compute limit values of the subdivision process, and to analyze the convergence and the smoothness of the limit function. We present everything for a mask with support [0,4], i.e. the scheme coefficients are and the symbol of the scheme is. The scheme is: The key model is the scheme generating the cubic B-spline:

5. The subdivision matrix S: Infinite – 2 slanted

7. The values of on [0,1] are determined by The values on [0,1/2] are determined by The values on [1/2,1] are determined by

8. The sub-matrix :

10. The vector of initial values determines the limit values at [0,1]. determines the values at [0,1/2] while determines the values at [1/2,1]. And so on, while determines the limit function on the interval, determine the limit function on the left half and the right half of respectively. It thus follows that at any point in [0,1], the limit value of a convergent subdivision, starting with initial values, is given by

11. First, a scheme must reproduce constants, therefore, Having that, we should make sure that the components in the subspace complementary to the constant eigenvector tend to 0. Let us show how to check it, and let us also try to understand the relation to the difference scheme analysis. The idea is to represent the operators and in another basis, the vectors comprising this basis are (e.g.) the columns of the following matrix:

12. In the new basis, the two operators are: Since is an eigenvector of and it follows that Note that determine the behavior of the last 3 components in the new basis.

13. Recalling the products, which in the new basis appear as The last three coordinates of this product will converge to zero if the joint spectral radius is less than 1. Each matrix has its own spectral radius. The joint spectral radius is defined as: If then the last three components of any vector of the form (#) will tend to zero.

14. We already know that a scheme is convergent to a limit if the differences of the generated values tend to zero. In the new basis, namely, the columns of, it is clear that differences within the first column are zero. The differences within the other columns tend to zero simply because the corresponding coefficients tend to zero. The number tells us how fast the differences are decaying to zero, and from it we can derive the Holder exponent of the limit function:

15. The choice of and the relation to the difference scheme. As an exercise, show that with the specific choice of as: the two matrices are just the two matrices generating the limit values on [0,1] for the difference scheme related with the generating polynomial In fact, any other completion of the constant vector into a basis may be used here.

16. The analysis of higher smoothness is very similar. First we have to show that if a scheme is, it must reproduce polynomials up to degree k. That is, To show it we consider eigenvectors of the infinite matrix S

17. The non-zero eigenvalues of the infinite matrix S, due to its two-slanted structure, are just the non-zero eigenvalues of the finite matrix marked below.

18. Let be an eigen-vector of S, and let denote the limit function of the subdivision process, starting with. It thus follows that Exercise: Show that if then Also, the entries of the eigen-vector are of the form It can further be shown that if the scheme is non-degenerate, then all the monomials up to degree k are eigen-functions of the subdivision scheme.

19. In particular, To verify that a given scheme is, we first check that it satisfies the necessary conditions. I.e., its symbol has the factor, or, in the matrix language, it has the polynomial eigenvectors up to degree k. Then, we have to show the decay to zero of all the differences of the k-th order divided differences One way is by showing that the scheme is contractive. The other way..., why do we need the other way?

20. Here again we represent the operators and in another basis, where the first k+1 vectors in the basis are the polynomial vectors (for a scheme with support [0,s]) : The differences of the k-th order divided differences within these vectors are all zero. And, since these vectors span the polynomial eigenvectors of and, the operators now take the form:

21. To show that the scheme is it remains to verify that Holder exponent of the k-th derivative of the limit function:

22. An example : The 4-point scheme for w=1/16

23. It can be shown here that

24. Limit values at grid points To represent the curve or surface at the m-th refinement level, we prefer to use the limit values and derivatives. Can we? We would like to use only values at the m-th level. It is enough to show how to do it for the 0-th level : Using Therefore, it is enough to know the values of the basic limit function and its derivatives at the integers ! 2005 Subdivision Summer School

25. Here we use the refinement equation: The values are left eigenvectors of the subdivision matrix S

26. Example 1 : The limit values of the cubic B-spline scheme. Normalization : Using the polynomial reproduction property

27. Example 2 : The limit derivatives of the 4-point scheme.

28. Approximation order and Quasi-interpolation A local method has approximation order n+1 if it is exact on 2005 Subdivision Summer School

29. If the symbol of a convergent scheme has the factor then To see this let us start with an initial polynomial data There exists a scheme transforming into

30. If the symbol of a convergent scheme has the factor then Interpolatory subdivision schemes reproduce polynomials. Non-interpolatory scheme usually do not.

31. Example: For the cubic B-spline scheme We would like to find a local operator Q such that I.e., Q is the inverse of The ‘method’ will be local and exact on

32. Constructing Q

33. Example : For the cubic B-spline scheme ( n=3 )

34. Approximation order: Given values This suggests ‘lifting’ the given data points, or control points, by the quasi-interpolation operator Q before applying subdivision. That works if the control points are taken on a ‘uniform grid’ on a smooth curve (surface)

35. 2005 Subdivision Summer School

36. Refinement on regular rectangular meshes

37. Refinement on regular triangular meshes

38. Can we use a univariate scheme on each of the three directions?

39. Tensor Product Refinement

40. A univariate refinement along one direction

41. A univariate refinement along the second direction

43. The tensor product bi-cubic spline scheme with the symbol :

44. The butterfly scheme on a grid :

45. The averaged butterfly scheme (w=1/16) :

46. Analysis of bivariate schemes on regular grids Let us consider a biivariate scheme with the symbol The generating functions satisfy

47. We have four subdivision rules, according to the parity of and the symbol should satisfy Using the Laurent polynomial formulation starting with we have Plot coefficients’ array ?

48. An example:. This scheme satisfies the necessary conditions for convergence. Does it converge to a continuous limit? How to check it? Let us try to find a difference schemes: If the symbol has the factor, we have a difference scheme transforming differences in one direction from one level to the next. To verify convergence to a continuous limit we have to show that differences in two independent directions tend to zero. In the example above, we do have a nice difference scheme for differences in the diagonal direction. Let us check it!

49. Hence, is the symbol of the scheme taking differences into differences of the same kind in the next level. Therefore, differences in the diagonal direction tend to zero.

50. The symbol of the butterfly scheme :

51. The symbol of the the averaged butterfly scheme :

52. Theorem 1. Theorem 2.

53. For the butterfly scheme on a regular mesh The scheme related to the following symbol should be contractive

55. What if the symbol is not factorizable? 1. We may use the JSR approach. Here one has to check the joint spectral radius of four local matrices. 2. If the symbol satisfies the necessary conditions for convergence there still exists a difference scheme, but in general it is a matrix scheme. I.e., a scheme of the form where For convergence, this matrix scheme has to be contractive.

56. We want to show the existence of a difference scheme: The existence of a matrix difference scheme

57. An example of a matrix difference scheme Together with the difference scheme for the y-direction (using the symmetry) we have contraction for two independent directional differences, hence, the scheme is convergent.

58. Thank you!