Polynomials for which $f''$ divides $f$

Question

Let $n \geq 2$ and let $a < b$ be real numbers. Then it is easy to see that there is a unique up to scale polynomial $f(x)$ of degree $n$ such that $$f(x) = \frac{(x-a)(x-b)}{n(n-1)} f''(x).$$

1. Have these polynomials been studied? Do they have a standard name?

2. Is it true that all the roots of $f(x)$ are real numbers in $[a,b]$?

Here are the polynomials $f$, with $a=-1$ and $b=1$, for $2 \leq n \leq 10$:

-1 + x^2, 
-x + x^3, 
1/5 - (6 x^2)/5 + x^4, 
(3 x)/7 - (10 x^3)/7 + x^5, 
-(1/21) + (5 x^2)/7 - (5 x^4)/3 + x^6, 
-((5 x)/33) + (35 x^3)/33 - (21 x^5)/11 + x^7, 
5/429 - (140 x^2)/429 + (210 x^4)/143 - (28 x^6)/13 + x^8, 
(7 x)/143 - (84 x^3)/143 + (126 x^5)/65 - (12 x^7)/5 + x^9, 
-(7/2431) + (315 x^2)/2431 - (210 x^4)/221 + (42 x^6)/17 - (45 x^8)/17 + x^10

Answer 1

Recording a CW-answer to take this off the unanswered list. The Gegenbauer polynomials are defined by the differential equation $$(1-x^2) g'' - (2 \alpha+1) x g' +n(n+2 \alpha) g =0.$$ Putting $\alpha = -1/2$, we get $$g=\frac{(x+1)(x-1)}{n(n-1)} g''.$$ If we want some other values for $a$ and $b$, we can put $f(x) = g(\ell(x))$ where $\ell$ is the affine linear function with $\ell(a) = -1$ and $\ell(b)=1$.

The Gegenbauer polynomials are orthogonal with respect to the weight $(1-x^2)^{\alpha - 1/2}$, so $(1-x^2)^{-1}$ in our case, so standard results on orthogonal polynomials tell us that the roots are in $[-1,1]$. If you are worried about the poles at $x = \pm 1$, then put $g_n(x) = (1-x^2) h_n(x)$, and the $h$'s are orthogonal with respect to $(1-x^2)$.