Question

Given the nodes x0 < x1 < · · · < xn, let V be the vector space of functions that are twice continuously differentiable at each node xi and cubic polynomial on (−∞, x₀), (x_n, ∞), and each of the intervals (xi , xi+1). Show that any function s(x) ∈ V can be uniquely represented as

s(x) = a0 + a₁x + a₂x^2 + a₃x^3 + $summation$ (i=0 to n) c_i (x − xi) ^3 +

where (x − xi)^3₊ is 0 for x ≤ xi and (x − xi)^3 otherwise. Conclude that this vector space has dimension n + 5.

Answer 1

Function $s(x)$ can be twice differentiated:

$s'(x)=a_1+2a_2x+3a_3x^2+3\displaystyle\sum^n_{i=0}c_i(x-x_i)^2_+$

$s''(x)=2a_2+6a_3x+9\displaystyle\sum^n_{i=0}c_i(x-x_i)_+$

Also, by the theorem of Existence and uniqueness of the polynomial interpolant:

Given the data pairs $\{x_j,f_j\}^n_{j=0}$ , where the points $\{x_j\}^n_{j=0}$ are distinct, there exists a unique polynomial $p_n(x)$ that satisfies the interpolation conditions $p_n(x_i)=f_i,\ i=0,1,2,...,n$

So, any function s(x) ∈ V can be uniquely represented as

$s(x)=a_0+a_1x+a_2x^2+a_3x^3+\displaystyle\sum^n_{i=0}c_i(x-x_i)^3_+$