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Published byAgatha Hancock Modified over 9 years ago
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Anupam Saxena Associate Professor Indian Institute of Technology KANPUR 208016
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(tj t)m1+(tj t)m1+ tjtj t f(t j ; t) = (t j t) m 1 + is continuous at t = t j and so are its derivatives up to m 2 (t j t) m 1 + is a spline of order m An nth order (n 1 degree) spline is C n-2 continuous
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Since each truncated power function is a spline, their weighted linear combination will be a spline of order m treat t as constant and compute the m th divided difference using s = t i m, t i m+1, …, t i, mth divided difference of the truncated function f(s; t) = (s t) m 1 + B spline basis function w(t) = (t t i m )(t t i m+1 )… (t t i ) and w ’ (t) = dw/dt
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t > tit > ti (t) = 0 (t i+r m t) m 1 +, r = 0, …, m are all zero t t i-m (t) = 0 (t) is the m th divided difference of a pure (m 1) degree polynomial (t) is standardized Peano’s theorem g(t) = t m, g m (t) = (m)! mth divided difference of t m is 1 An nth order (n 1 degree) spline is C n-2 continuous
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divided differences of the product of two functions, h(t) = f(t)g(t) h[t 0, t 1,…,t k ] = f[t 0, t 1, …,t r ] g[t r, t r+1, …,t k ] = f[t 0 ]g[t 0, t 1, …,t k ] + f[t 0, t 1 ]g[t 1, …,t k ] + … + f[t 0, t 1, …,t k 1 ]g[t k 1,t k ] + f[t 0, t 1, …,t k ] g[t k ] h k (t j ; t) = (t j t) k 1 + = (t j t) k 2 + (t j t) + = h k 1 (t j ; t) (t j t) +
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h k [t i k,..., t i ;t ] = h k 1 [t i k,…, t i 1 ;t] + h k 1 [t i k,…, t i ;t](t i t) k th divided difference of (t j t) k 1 + : B-spline M k, i (t) M k-1,i-1 (t)
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h k [t i k,..., t i ;t ] = h k 1 [t i k,…, t i 1 ;t] + h k 1 [t i k,…, t i ;t](t i t) M k, i (t)M k 1,i 1 (t) + {h k 1 [t i k+1,…, t i ;t ] h k 1 [t i k,…, t i 1 ;t]} = {M k 1, i (t) M k 1, i 1 (t)} M k, i (t) + = M k 1,i 1 (t) M k, i (t) = M k 1,i 1 (t) + M k 1, i (t) similar to the de Casteljau’s algorithm repeated linear interpolation is performed between two consecutive splines a table to construct splines may also be generated
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