ORTHOGONAL POLYNOMIALS/PROOFS

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Contents

Proof of the recurrence relation

Proof of the existence of real roots

Proof of interlacing of roots

Proof of the recurrence relation

Any orthogonal series has a 'recurrence formula' relating any three consecutive polynomials
in the series.
:$p\_\{n+1\}\; =\; (a\_nx+b\_n)\; p\_n\; -\; c\_n\; p\_\{n-1\}$
The coefficients ''a'', ''b'', and ''c'' depend on ''n''. They also depend on the standardization, obviously.
Proof:
We will prove this for fixed ''n'', and omit the subscripts on ''a'', ''b'', and ''c''.
First, choose ''a'' so that the $x^\{n+1\}$ terms match, so we have
:$a\; x\; p\_n\; -\; p\_\{n+1\}\; =$ a polynomial of degree ''n''.
Next, choose ''b'' so that the $x^n$ terms match, so we have
:$(ax+b)\; p\_n\; -\; p\_\{n+1\}\; =$ a polynomial of degree ''n'' − 1
Expand the right-hand-side in terms of polynomials in the sequence
:$(ax+b)\; p\_n\; -\; p\_\{n+1\}\; =\; sum\_\{i=0\}^\{n-1\}\; \{lambda\}\_i\; p\_i$
Now if $jle\; n-2$, then
:$langle\; (ax+b)\; p\_n,\; p\_j$
angle - langle p_{n+1}, p_j
angle = sum_{i=0}^{n-1} {lambda}_i langle p_i, p_j
angle = {lambda}_j langle p_j, p_j
angle.
But
:$langle\; p\_n,\; p\_j$
angle = 0 and $langle\; p\_\{n+1\},\; p\_j$
angle = 0,
so
:$a\; langle\; x\; p\_n,\; p\_j$
angle = {lambda}_j langle p_j, p_j
angle.
Since the inner product is just an integral involving the product:
:$langle\; x\; p\_n,\; p\_j$
angle = langle p_n, x p_j
angle
we have
:$a\; langle\; p\_n,\; x\; p\_j$
angle = {lambda}_j langle p_j, p_j
angle
If , then $x\; p\_j$ has degree $le\; n-1$, so it is orthogonal to $p\_n$;
hence $\{lambda\}\_j\; langle\; p\_j,\; p\_j$
angle = 0, which implies $\{lambda\}\_j\; =\; 0$ for .
Therefore, only $\{lambda\}\_\{n-1\}$ can be nonzero, so
:$(ax+b)\; p\_n\; -\; p\_\{n+1\}\; =\{lambda\}\_\{n-1\}\; p\_\{n-1\}$
Letting $c\; =\; \{lambda\}\_\{n-1\}$, we have
:$p\_\{n+1\}\; =\; (ax+b)\; p\_n\; -\; c\; p\_\{n-1\}.$

Proof of the existence of real roots

Each polynomial in an orthogonal sequence has all ''n'' of its roots real, distinct,
and strictly inside the interval of orthogonality.
Proof:
Let ''m'' be the number of places where the sign of ''P''_''n'' changes inside the
interval of orthogonality, and let $x\_1\; dots\; x\_m$ be those points.
Each of those points is a root of ''P''_''n''. By the
fundamental theorem of algebra, ''m'' ≤ ''n''.
Now ''m'' might be strictly less than ''n'' if some roots of ''P''_''n'' are complex, or not inside
the interval of orthogonality, or not distinct. We will show that ''m'' = ''n''.
Let $S(x)\; =\; prod\_\{j=1\}^m\; (x\; -\; x\_j)$
This is an ''m''^th-degree polynomial that changes sign at each of the
''x''_''j'', the same way that ''P''_''n''(''x'') does. ''S''(''x'')''P''_''n''(''x'')
is therefore strictly positive, or strictly negative, everywhere except at
the ''x''_''j''. ''S''(''x'')''P''_''n''(''x'')''W''(''x'') is also strictly positive
or strictly negative except at the ''x''_''j'' and possibly the end points.
Therefore, $langle\; S,\; P\_n$
angle, the integral of this, is nonzero.
But, by Lemma 2, ''P''_''n'' is orthogonal to any polynomial of lower degree,
so the degree of ''S'' must be ''n''.

Proof of interlacing of roots

The roots of each polynomial lie strictly between those of the next higher polynomial
in the sequence.
Proof:
First, standardize all of the polynomials so that their leading terms are
positive. This will not affect the roots.
Next, a lemma: For all ''n'' and all ''x'',
:$P\_\{n+1\}\text{'}(x)\; P\_n(x)\; >\; P\_\{n+1\}(x)\; P\_n\text{'}(x),$
Proof by induction. For ''n'' = 0, $P\_1\text{'}(x)\; >\; 0,$, $P\_0(x)\; >\; 0,$,
and $P\_0\text{'}(x)\; =\; 0,$.
Otherwise, the recurrence formula has
:$P\_\{n+1\}(x)\; =\; (ax\; +\; b)\; P\_n(x)\; -\; c\; P\_\{n-1\}(x),$
with and .
So
:$P\_\{n+1\}\text{'}\; =\; a\; P\_n\; +\; (ax\; +\; b)\; P\_n\text{'}\; -\; c\; P\_\{n-1\}\text{'},$.
So
:$P\_\{n+1\}\text{'}\; P\_n\; -\; P\_\{n+1\}\; P\_n\text{'}\; =\; [a\; P\_n\; +\; (ax\; +\; b)\; P\_n\text{'}\; -\; c\; P\_\{n-1\}\text{'}]\; P\_n\; -\; [(ax\; +\; b)\; P\_n\; -\; c\; P\_\{n-1\}]\; P\_n\text{'},$
:$=\; [a\; P\_n\; -\; c\; P\_\{n-1\}\text{'}]\; P\_n\; +\; c\; P\_\{n-1\}\; P\_n\text{'},$
:$=\; a\; P\_n^\{\; 2\}\; +\; c\; (P\_n\text{'}\; P\_\{n-1\}\; -\; P\_n\; P\_\{n-1\}\text{'}),$
:$ge\; c\; (P\_n\text{'}\; P\_\{n-1\}\; -\; P\_n\; P\_\{n-1\}\text{'}),$
But $P\_n\text{'}\; P\_\{n-1\}\; -\; P\_n\; P\_\{n-1\}\text{'}\; >\; 0,$ by the induction step.
Now if ''x'' is a root of ''P''_''n+1'', the lemma tells us that
:$P\_\{n+1\}\text{'}(x)\; P\_n(x)\; >\; P\_\{n+1\}(x)\; P\_n\text{'}(x)\; =\; 0,$
So $P\_\{n+1\}\text{'}(x),$ and $P\_n(x),$ have the same sign.
But $P\_\{n+1\}\text{'}(x),$ must change sign from any root of ''P''_''n+1''
to the next. Therefore, ''P''_''n'' must change sign also, so ''P''_''n''
must have a root in that interval.