Editing Gaussian quadrature (section)

==== Recurrence relation ====
Orthogonal polynomials <math>p_r</math> with <math>(p_r, p_s) = 0</math> for <math>r \ne s</math> for a scalar product <math>(\cdot , \cdot)</math>, degree <math>(p_r) = r</math> and leading coefficient one (i.e. [[monic polynomial|monic]] orthogonal polynomials) satisfy the recurrence relation
<math display="block">p_{r+1}(x) = (x - a_{r,r}) p_r(x) - a_{r,r-1} p_{r-1}(x) \cdots - a_{r,0}p_0(x)</math>

and scalar product defined
<math display="block">(f(x),g(x))=\int_a^b\omega(x)f(x)g(x)dx</math>

for <math>r = 0, 1, \ldots, n - 1</math> where {{mvar|n}} is the maximal degree which can be taken to be infinity, and where <math display="inline">a_{r,s} = \frac{\left(xp_r, p_s\right)}{\left(p_s, p_s\right)}</math>. First of all, the polynomials defined by the recurrence relation starting with <math>p_0(x) = 1</math> have leading coefficient one and correct degree. Given the starting point by <math>p_0</math>, the orthogonality of <math>p_r</math> can be shown by induction. For <math>r = s = 0</math> one has
<math display="block">(p_1,p_0) = (x-a_{0,0}) (p_0,p_0) = (xp_0,p_0) - a_{0,0}(p_0,p_0) = (xp_0,p_0) - (xp_0,p_0) = 0.</math>

Now if <math>p_0, p_1, \ldots, p_r</math> are orthogonal, then also <math>p_{r+1}</math>, because in
<math display="block">(p_{r+1}, p_s) = (xp_r, p_s) - a_{r,r}(p_r, p_s) - a_{r,r-1}(p_{r-1}, p_s)\cdots - a_{r,0}(p_0, p_s)</math>
all scalar products vanish except for the first one and the one where <math>p_s</math> meets the same orthogonal polynomial. Therefore,
<math display="block">(p_{r+1},p_s) = (xp_r,p_s) - a_{r,s}(p_s,p_s) = (xp_r,p_s)-(xp_r,p_s) = 0.</math>

However, if the scalar product satisfies <math>(xf, g) = (f,xg)</math> (which is the case for Gaussian quadrature), the recurrence relation reduces to a three-term recurrence relation: For <math>s < r - 1, xp_s</math> is a polynomial of degree less than or equal to {{math|''r'' − 1}}. On the other hand, <math>p_r</math> is orthogonal to every polynomial of degree less than or equal to {{math|''r'' − 1}}. Therefore, one has <math>(xp_r, p_s) = (p_r, xp_s) = 0</math> and <math>a_{r,s} = 0</math> for {{math|''s'' < ''r'' − 1}}. The recurrence relation then simplifies to
<math display="block">p_{r+1}(x) = (x-a_{r,r}) p_r(x) - a_{r,r-1} p_{r-1}(x)</math>

or
<math display="block">p_{r+1}(x) = (x-a_r) p_r(x) - b_r p_{r-1}(x)</math>

(with the convention <math>p_{-1}(x) \equiv 0</math>) where
<math display="block">a_r := \frac{(xp_r,p_r)}{(p_r,p_r)}, \qquad b_r := \frac{(xp_r,p_{r-1})}{(p_{r-1},p_{r-1})} = \frac{(p_r,p_r)}{(p_{r-1},p_{r-1})}</math>

(the last because of <math>(xp_r, p_{r-1}) = (p_r, xp_{r-1}) = (p_r, p_r)</math>, since <math>xp_{r-1}</math> differs from <math>p_r</math> by a degree less than {{mvar|r}}).