In the Week 2 Practical, we explored the application of scattered data interpolation. We suppose that we have sampled the value of a function f:R2R at a set of input vectors - let's call them points - x(i), for i=1,,n. Note that we use the superscript notation x(i) to mean the i th input point. We fit a function g delined by g(x)=i=1nci(xx(i)), where is the thin plate spline RBF (r)=r2log(r). If we demand that g(x(i))=f(x(i)) for i=1,,n,1.e, that g interpolates the data exactly, we tound that this amounts to the coefficient vector c satisfying the linear system: (x(1)x(1))(x(2)x(1))(x(n)x(1))(x(1)x(2))(x(2)x(2))(x(n)x(2))(x(1)x(n))(x(2)x(n))(x(n)x(n))c1c2cn =f1f2fn or compactly, Ae=f. In the Week 2 Practical, we observed that demanding an exact fit to the data resulted in an overly "wiggly" interpolant, which was ill-suited to an application where we wished to minimise f over the unit square. We overcame this by imposing a "wiggliness penalty" to our formulation. Rather than solve Ac =f, we found c by minimising an expression involving the interpolation error fAc2 and a wiggliness penalty term. An alternative, which we will explore in this question, is to treat this as a least squares problem instead of an interpolant. To this end, we introduce a second set of points: x~(i), for i=1,.k, where k is less than n. We now define our function g by reterence to these points instead: g(x)=i=1kci(xx~(i)) But (and here's why it becomes a least squares problem) we still impose that g interpolates the data at the original points x(i) (no tildes): g(x(i))=f(x(i))fori=1,n. So this is n equations, defined by (3), but with only k unknowns, the coefficients ci in (2), with k.

1. In the Week 2 Practical, we explored the application of scattered data interpolation. We suppose
that we have sampled the value of a function f:R2R at a set of input vectors - let's call them
points - x(i), for i=1,,n. Note that we use the superscript notation x(i) to mean the i th input point.
We fit a function g delined by g(x)=i=1nci(xx(i)), where is the thin plate spline RBF
(r)=r2log(r). If we demand that g(x(i))=f(x(i)) for i=1,,n,1.e, that g interpolates the data exactly,
we tound that this amounts to the coefficient vector c satisfying the linear system:
(x(1)x(1))(x(2)x(1))(x(n)x(1))(x(1)x(2))(x(2)x(2))(x(n)x(2))(x(1)x(n))(x(2)x(n))(x(n)x(n))c1c2cn
=f1f2fn or compactly, Ae=f. In the Week 2 Practical, we observed that demanding an exact fit to
the data resulted in an overly "wiggly" interpolant, which was ill-suited to an application where
we wished to minimise f over the unit square. We overcame this by imposing a "wiggliness
penalty" to our formulation. Rather than solve Ac =f, we found c by minimising an expression
involving the interpolation error fAc2 and a wiggliness penalty term. An alternative, which we
will explore in this question, is to treat this as a least squares problem instead of an interpolant.
To this end, we introduce a second set of points: x~(i), for i=1,.k, where k is less than n. We now
define our function g by reterence to these points instead: g(x)=i=1kci(xx~(i)) But (and here's
why it becomes a least squares problem) we still impose that g interpolates the data at the
original points x(i) (no tildes): g(x(i))=f(x(i))fori=1,n. So this is n equations, defined by (3), but
with only k unknowns, the coefficients ci in (2), with k