I am trying to determine whether or not a function is piecewise polynomial. The function is as below:
Let $\ X$ be a continuous random variable with support on $\ \Omega_x$, and with corresponding cdf $\ F$. Is the function $g_r$, $r\geq 1$, piecewise polynomial?
$\ g_r(x) = x\sum_{k=0}^{r-1} \left( (-1)^{r-1-k} {r-1 \choose k}{r-1+k \choose k}(F(x)^k)\right)$
If it helps (though I doubt it makes a difference) then $\ \Omega_x$ could be restricted to the positive reals.
I am not exactly sure of what I would need to do to prove this is (or isn't) the case. I have spent some time reading on the subject but it is not at all clear to me. My initial guess is yes, since the function is smooth, but I would be grateful for any thoughts.
The definition I have found for a function to be piecewise-polynomial is:
"A piecewise polynomial function is a continuous function $\ f : A \rightarrow \mathbb{R}$ for which there exist finitely many polynomials $\ p_1, \ldots , p_k$ such that for every $\ a \in A$, $\ f(a) = p_i(a)$ for some $i$."
Many thanks for all the help.
According to the definition that I am familiar with, a function on $\mathbb R$ is piecewise $P$ (for some property $P$, e.g. continuous, differentiable, constant, linear, polynomial, ...) if its domain can be divided into a union of non-degenerate intervals such that the function is $P$ over each interval.
Fix $r = 2$, so that $g_r(x) = x(-1 + 2F(x))$. Suppose $X$ follows an exponential distribution with parameter $\lambda = 1$. Then $F(x) = 1 - e^{-x}$, and $g_2(x) = x - 2xe^{-x}$. The function $g_2$ has an infinite Taylor expansion about any $x$, so it cannot be polynomial in any neighbourhood.
Of course, if the distribution of $X$ is such that $F(x)$ itself is piecewise polynomial, then so is $g_r$ for any $r$. However, I'm not sure if there are any $F$ and $r$ such that $F$ is not piecewise polynomial but $g_r$ is.