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o u t p u t ******
iflag.....integer indicating success (=1) or failure (=2) .
if iflag = 1, then
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t.....array of length | | |
n+k containing the optimal knots ready for
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use in optimal recovery. specifically, | | |
t(1) = ... = t(k) =
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tau(1) |
and t(n+1) = ... = t(n+k) = tau(n) , while the n-k
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interior knots | | |
t(k+1), ..., t(n) are calculated as described
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below under | | |
*method* .
if iflag = 2, then
k .lt. 3, or n .lt. k, or a certain linear system was found to
be singular.
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and nowhere else in (tau(1),tau(n)) is orthogon-
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al to the spline space | | |
spline( k , tau ) on that interval .
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let |
xi(j) be the current guess for t(k+j), j=1,...,n-k. then
the next newton iterate is of the form
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xi(j) |
+ (-)**(n-k-j)*x(j) , j=1,...,n-k,
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with |
x the solution of the linear system
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here, |
c(i,j) = b(i)(xi(j)), all j, with b(i) the i-th b-spline of
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order |
k for the knot sequence tau , all i, and d is the vector
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given by | | |
d(i) = sum( -a(j) , j=i,...,n )*(tau(i+k)-tau(i))/k, all i,
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with |
a(i) = sum ( (-)**(n-k-j)*b(i,k+1,tau)(xi(j)) , j=1,...,n-k )
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for i=1,...,n-1, and | | |
a(n) = -.5 .
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(see chapter | | |
xiii of text and references there for a derivation)
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the first guess for | | |
t(k+j) is (tau(j+1)+...+tau(j+k-1))/(k-1) .
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iteration terminates if | | |
max(abs(x(j))) .lt. t o l , with
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t o l |
= t o l r t e *(tau(n)-tau(1))/(n-k) ,
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or else after | | |
n e w t m x iterations , currently,
newtmx, tolrte / 10, .000001
dimension scrtch((n-k)*(2*k+3)+5*k+3), t(n+k)
current fortran standard makes it impossible to specify the precise dim-
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ensions of | | |
scrtch and t without the introduction of otherwise
superfluous additional arguments .
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