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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?

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    $\begingroup$ 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 $v$. $\endgroup$ Aug 15, 2017 at 19:05

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I think this follows from the fact that all these families satisfy a three-term recurrence relation $$ P_n(x) = (x-\alpha_{n-1}) P_{n-1}(x)-\beta_{n-2} P_{n-2}(x) $$

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  • $\begingroup$ See also doi.org/10.1016/j.aml.2005.10.023. $\endgroup$
    – lhf
    Aug 15, 2017 at 19:10
  • $\begingroup$ 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? $\endgroup$
    – Doris
    Aug 15, 2017 at 20:37
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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 $\frac{1}{2^n n!}$.

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