# Given a truncated power basis function show that it represents a cubic spline for one knot

Given the truncated power basis function $$h_1(x)=1, h_2(x)=x, h_3(x)=x^2, h_4(x)=x^3, h_5(x)=(x-\epsilon)^3_+$$ Show that a function of the form $$f(x)=\beta_0+\beta_1x+\beta_2x^2+\beta_3x^3+\beta(x-\epsilon)^3_+$$ is a cubic regression spline by finding $$f_1(x)=a_1+b_1x+c_1x^2+d_1x^3$$ s.t $$f(x)=f_1(x)$$

There's quite a few things I'm confused about here. I was under the impression that a truncated power basis function is of the form: some polynomial function for a range and 0 otherwise. So I don't understand what the $$h_i()$$ functions above mean because they don't give a range and I don't know how to interpret $$h_5(x)$$. Also how does showing that you can represent $$f(x)$$ equal to $$f_1(x)$$ prove it's a cubic regression spline?

What you've been asked to show is false as stated: $$f$$ is a cubic spline with break point $$\epsilon$$, and cannot necessarily be represented by a cubic polynomial $$f_1$$. To see this, note that $$f_1$$ is a cubic polynomial, so it must be a smooth function, i.e. it is continuous and has continuous derivatives of all orders, i.e. $$f_1 \in \mathcal{C}^\infty$$. However, because \begin{align} h_5 (x) & = (x - \epsilon)_+^3 = \begin{cases} (x - \epsilon)^3 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases} \\ h_5' (x) & = \begin{cases} 3 (x - \epsilon)^2 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}\\ h_5'' (x) & = \begin{cases} 6 (x - \epsilon) , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}\\ h_5''' (x) & = \begin{cases} 6 , \text{ if } x - \epsilon > 0 \\ \text{undefined} , \text{ if } x = \epsilon \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}, \end{align}

we have that $$f$$ is only a $$\mathcal{C}^2$$ function (if the coefficient $$\beta \ne 0$$) and can have a discontinuous third derivative, so $$f$$ is not a $$\mathcal{C}^\infty$$ function.

To clear up the confusion on terminology, a truncated polynomial in this context is any function of the form: $$g(x) = p(x) I_A (x)$$ where $$p(x)$$ is a polynomial $$I_A (x)$$ is the indicator function of a set $$A$$, given by $$I_A (x) = \begin{cases} 1 , & \text{ if } x \in A \\ 0 , & \text{ if } x \notin A \end{cases}.$$ This means that all polynomials are also truncated polynomials by taking the set $$A = \mathbb{R} = (-\infty, \infty)$$. So $$h_1$$, $$h_2$$, $$h_3$$, and $$h_4$$ are all truncated polynomials. And to reiterate the meaning of $$h_5$$, $$h_5 (x) = (x - \epsilon)_+^3 = \left(\max(0, x - \epsilon)\right)^3 = \begin{cases} (x - \epsilon)^3 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}.$$

What you actually want to prove is that the functions $$h_i$$ form a basis for all $$\mathcal{C}^2$$ piecewise-polynomial functions of degree 3 with a break point $$\epsilon$$. Let's call this space of functions $$\mathcal{P}^2_3$$. First, let's describe the functions in this space: each one is a piecewise-polynomial of degree 3 with break point $$\epsilon$$, which means that any $$f \in \mathcal{P}_3^2$$ can be written as $$f(x) = \begin{cases} a_0x^3 + b_0x^2 + c_0x + d_0 , & \text{ if } x \le \epsilon \\ a_1x^3 + b_1x^2 + c_1x + d_1 , & \text{ if } x > \epsilon \\ \end{cases}.$$ However, the constraint that this function be continuous and have continuous derivatives up to order two, i.e. that $$f \in \mathcal{C}^2$$, means that the coefficients have to satisfy the constraints: \begin{align} a_0\epsilon^3 + b_0\epsilon^2 + c_0\epsilon + d_0 & = a_1\epsilon^3 + b_1\epsilon^2 + c_1\epsilon + d_1 \\ 3a_0\epsilon^2 + 2b_0\epsilon + c_0 & = 3a_1\epsilon^2 + 2b_1\epsilon + c_1 \\ 6a_0\epsilon + 2b_0 & = 6a_1\epsilon + 2b_1 . \end{align}

You have to show that every function in $$\mathcal{P}^2_3$$, i.e. all the piece-wise polynomials of degree 3 that satisfy these constraints, can be written as a unique linear combination of the functions $$h_i$$.

• your truncated power basis functions look off, shouldn't it be $(x-\epsilon)^3$ if $x-\epsilon>0$ not greater than or equal to? Aug 25 '20 at 5:02
• Because $h_5$ is continuous you can write it either way, because the two are equal. One of the two inequalities in the expression has to be "or equal to," otherwise $h_5$ is undefined at the point $x = \epsilon$. It just doesn't matter which one has the "or equal to." Aug 25 '20 at 5:12
• The only inequality here where it actually makes a difference is the one in the definition of $\mathcal{P}_3^2$, where I've implicitly assumed that the piece-wise polynomials are left-continuous. However, since we assume in the next part that these functions are continuous, ultimately this is the same either way as well. Aug 25 '20 at 5:21