# Why is the sign of terms in orthogonal polynomials always alternating?

I checked some expressions of orthogonal polynomials, e.g. Laguerre Polymonial, Legendre Polynomial, Hermite Polynomials, etc. And the sign of terms in them are always alternating. For example, $$H_8(x) = 256 x^8-3584 x^6+13440 x^4-13440 x^2+1680$$ The sign of each term is $(+~-~+~-~+)$. This seems to be true for every polynomials. Is there an explanation for this?

• Simply being an orthognal family is of course not sufficient for such a property. For example, we might replace $H_6$ and $H_8$ with $uH_6+vH_8$ and $vH_6-uH_8$ and break the sign rule by suitably choosing $u$ and $v$. Aug 15, 2017 at 19:05

I think this follows from the fact that all these families satisfy a three-term recurrence relation $$P_n(x) = (x-\alpha_{n-1}) P_{n-1}(x)-\beta_{n-2} P_{n-2}(x)$$
• If $\alpha$ and $\beta$ are both positive, then there seems to be a competition between $P_{n-1}(x)$ and $P_{n-2}(x)$, the second term may inverse the sign? Aug 15, 2017 at 20:37
Once we have a Rodrigues-like formula (encoding a particular choice of an orthogonal base) the alternating signs are a pretty straightforward consequence. For instance, in the Legendre case $$P_n(x) = \frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n.\tag{1}$$ The binomial expansion of $(x^2-1)^n$ has alternating signs: trivial. The operator $\frac{d^n}{dx^n}$ does not change that, neither it does the multiplication by $\frac{1}{2^n n!}$.