function [ynew,a,b,c,d] = NatSpline(x,y,xnew)
% ynew = NatSpline(x,y)
% Natural spline interpolation.
% x,y are column n-vectors.
% Assume that n >= 4 and x(1) < ... x(n).
% xnew -- is a set of points in the interval x(1) to x(n) for which
% the natural spline needs to be evaluated.
% On the i-th interval [x(i),x(i+1)] the spline has the form:
% S(z) = a(i) + b(i)(z-x(i)) + c(i)(z-x(i))^2 + d(i)(z-x(i))^2(z-x(i+1).
%
% Outputs:
% ynew -- the values of the natural spline evaluated at the points xnew.
% a,b,c,d,-- the vector of coefficients of the splines
% Usage:
% [ynew,a,b,c,d] = NatSpline(x,y,xnew)

% First, set up all but the first and last equations that
% define the vector of interior slopes s(2:n-1).

n = length(x);
Dx = diff(x);
yp = diff(y) ./ Dx;
T = zeros(n-2,n-2);
r = zeros(n-2,1);
for i=2:n-3
T(i,i) = 2*(Dx(i) + Dx(i+1));
T(i,i-1) = Dx(i+1);
T(i,i+1) = Dx(i);
r(i) = 3*(Dx(i+1)*yp(i) + Dx(i)*yp(i+1));
end

T(1,1) = 2*Dx(1) + 1.5*Dx(2);
T(1,2) = Dx(1);
r(1) = 1.5*Dx(2)*yp(1) + 3*Dx(1)*yp(2);
T(n-2,n-2) = 1.5*Dx(n-2)+2*Dx(n-1);
T(n-2,n-3) = Dx(n-1);
r(n-2) = 3*Dx(n-1)*yp(n-2) + 1.5*Dx(n-2)*yp(n-1);
%we now need to solve
%stilde = T\r;
A=diag(T);
B=zeros(size(A));
for i=2:n-2
B(i)=T(i,i-1);
end
C=zeros(size(A));
for i=1:n-3
C(i)=T(i,i+1);
end

%call the tri diagonal solver: (below)
stilde=tridiag( A, B, C, r );

s1 = (3*yp(1) - stilde(1));
sn = (3*yp(n-1) - stilde(n-2));
s = [s1;stilde;sn];

% Compute the a,b,c,d vector of coefficients for the splines.

a = y(1:n-1);
b = s(1:n-1);
c = (yp - s(1:n-1)) ./ Dx;
d = (s(2:n) + s(1:n-1) - 2*yp) ./ (Dx.* Dx);

%next evaluate at the points xnew to obtain the values ynew

for j=1:length(xnew)
x_val=xnew(j);

(1) Use bisection method to find a positive root of x2 x 0 such t...