# Determine the leading coefficient of polynomials given by a recurrence relation

Suppose that we have the expression $$xr_n(x)=b_nr_{n+1}(x)+a_nr_n(x)+b_{n-1}r_{n-1}(x),\quad n\geq 0$$ where $$a_n\in \mathbb{R}$$, $$b_n>0$$ for $$n\geq 0$$ and $$b_{-1}=1$$, with the initial conditions $$r_0(x)=1$$ and $$r_{-1}(x)=0$$.

From here, I can see that $$r_n$$ is a (real) polynomial of degree $$n$$ (by induction). Then, we can write $$r_n(x)=\sum_{k=0}^nc_{k,n}x^k$$ for some coefficients $$c_{k,n}$$. How can one deduce from the expression above that $$c_{n,n}=1/(b_0b_1\dots b_{n-1})?$$

Use induction. Notice that $$\deg a_n r_n(x) \leq n$$ and $$\deg b_{n-1}r_{n-1}(x) \leq n-1$$ (building on your result that $$r_n(x)$$ has degree $$n$$). So when looking at coefficient of $$x^{n+1}$$ in the recurrence expression, the only terms left are $$xr_n(x)$$ and $$b_nr_{n+1}(x)$$. The coefficient of $$x^{n+1}$$ in $$xr_n(x)$$ is the coefficient of $$x^n$$ in $$r_n(x)$$, which by induction hypothesis is $$1/(b_0b_1\dots b_{n-1})$$. This must be equal to coefficient of $$x^{n+1}$$ in $$b_nr_{n+1}(x)$$, and hence coefficient of $$x^{n+1}$$ in $$r_{n+1}(x)$$ is $$(1/(b_0b_1\dots b_{n-1}))/b_n=1/(b_0b_1\dots b_{n})$$, which was to be proven.
• Thanks for your answer. Took me time to understand your process (wordings aren't my strong point). One can do a slightly different way: by comparing the coefficients $xr_n$ and $b_nr_{n+1}$, we get $c_{n,n}=b_nc_{n+1,n+1}$, so $c_{n+1,n+1}=\frac{1}{b_n}c_{n,n}$. Then use induction to get the conclusion, where $c_{0,0}=r_0=1$. Commented Aug 23, 2020 at 21:36