In using a basis model to represent a cubic spline, what is the meaning of the truncated power basis function and its math in lay terms? I am going over “Introduction to Statistical Learning” (James, et al). While I understand the concept of a spline (piecewise polynomial with continuous 1st and 2nd derivative at each knot), I’ve been having a lot of trouble understanding this “power basis” function, “$\,(x−ξ)^3_+\,$” and how it was derived. I understand that its purpose is to represent a spline with that specific smoothness restriction, but I don't really understand hy mathematically. I'm having trouble understanding why this power basis function looks the way it does and how it helps define the spline.  
 A: A cubic spline is a piece-wise cubic that is continuous and has continuous first and second derivatives. 
Consider a single knot, $\xi$. Start with a cubic representing the curve before the knot and add in a multiple of a basis function giving $\alpha_3x^3+\alpha_2x^2+\alpha_1x+\alpha_0+\beta_\xi \phi_\xi(x)$ and let's try to find a $\phi$ that will give us a second cubic piece after the knot that joins up at the knot in the way that a cubic spline does.
Consider the following requirements on the basis function, $\phi_\xi(x)$:


*

*when added to an existing fit, $\beta_\xi \phi_\xi(x)$  must leave alone the cubic to the left of $\xi$. (That is, $\phi_\xi(x)$ must be zero for $x\leq\xi$.)

*to the right of $\xi$ you must end up with another cubic. Consequently $\phi_\xi(x)$ must be cubic when $x>\xi$

*continuity conditions: we need this cubic to be 0 and have 0 first and second derivative at $\xi$. (These conditions on the right fix the cubic part to be $a(x-\xi)^3$. Nothing else will do. So we have that to the right and 0 to the left.)
Consequently, you're left with $\phi_\xi(x)=a(x-\xi)_+^3$ for $a\neq 0$ as the only possibility satisfying all those conditions.
However, you don't need to have $a$ be anything but $1$, since any necessary scaling of   $(x-\xi)_+^3$ is what $\beta$ is for; any $a$ other than $1$ is redundant. 
Consequently we're left with $\phi_\xi(x)=(x-\xi)_+^3$.
