First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 3. INTERPOLATION 128
end
for i in 2:n
u[i]=3*(a[i+1]-a[i])/h[i]-3*(a[i]-a[i-1])/h[i-1]
end
s=Array{Float64}(undef,m)
z=Array{Float64}(undef,m)
t=Array{Float64}(undef,n)
s[1]=1
z[1]=0
t[1]=0
for i in 2:n
s[i]=2*(x[i+1]-x[i-1])-h[i-1]*t[i-1]
t[i]=h[i]/s[i]
z[i]=(u[i]-h[i-1]*z[i-1])/s[i]
end
s[m]=1
z[m]=0
c[m]=0
for i in reverse(1:n)
c[i]=z[i]-t[i]*c[i+1]
b[i]=(a[i+1]-a[i])/h[i]-h[i]*(c[i+1]+2*c[i])/3
d[i]=(c[i+1]-c[i])/(3*h[i])
end
Out[2]: CubicNatural (generic function with 1 method)
Once the matrix equation is solved, and the coefficients of the cubic polynomials are
computed by CubicNatural, the next step is to evaluate the spline at a given value.
This is done by the following function CubicNaturalEval. The inputs are the value
at which the spline is evaluated, w, and the x-coordinates of the data. The function
first finds the interval [x i ,x i+1 ],i =1, ..., m − 1, wbelongs to, and then evaluates the
spline at w using the corresponding cubic polynomial. Note that in the previous section
the data and the intervals are counted starting at 0, so the formulas used in the Julia
code do not exactly match the formulas given earlier.
In [3]: function CubicNaturalEval(w,x::Array)