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When I was experimenting with orthogonalization of polynomials

$$p_n(x)=\begin{cases} 1-x^n&\text{if }n\equiv0\; (\operatorname{mod}2),\\ x-x^n&\text{otherwise}, \end{cases}$$

i.e. simplest binomials vanishing at $x=-1$ and $x=1$, with respect to inner product of

$$\langle p_n,p_m\rangle=\int\limits_{-1}^1 p_n(x)p_m(x)\,dx,$$

i.e. the one used to generate Legendre polynomials, I found that the resulting orthogonal polynomials are (up to normalization) nothing other than associated Legendre polynomials $P_l^m$ with $m=2$.

Similarly, I can get a series of orthogonal polynomials vanishing at the endpoints with dot product the same as that of Chebyshev polynomials. The first several such polynomials are:

$$\begin{align} c_2(x)=&\sqrt{\frac8{3\pi}}(x^2-1),\\ c_3(x)=&\sqrt{\frac{16}\pi}(x^3-x),\\ c_4(x)=&\sqrt{\frac{32}{15\pi}}(6x^4-7x^2+1),\\ c_5(x)=&\sqrt{\frac{16}{3\pi}}(8x^5-11x^3+3x),\\ c_6(x)=&\sqrt{\frac8{35\pi}}(80x^6-128x^4+51x^2-3). \end{align}$$

After some more experiments I've found that these polynomials solve a modified version of Chebyeshev equation, with modification highly resembling that of associated vs usual Legendre polynomials:

$$(1-x^2)y''(x)-xy'(x)+\left(n^2-\frac2{1-x^2}\right)y(x)=0,$$

which, if we omit the $\frac2{1-x^2}$ term, becomes the usual equation for Chebyshev polynomials.

Actually, they can be generated from the Chebyshev polynomials themselves by differentiation (again, inspiration comes from associated Legendre polynomials):

$$c_n(x)=\sqrt{\frac2\pi}\frac{x^2-1}{n\sqrt{n^2-1}}\frac{\mathrm d^2}{\mathrm dx^2}T_n(x).$$

Is this a known series of polynomials? I failed to find anything resembling "associated Chebyshev polynomials", but maybe they have another name?

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  • $\begingroup$ No hits on OEIS (which has various triangles in addition to sequences), although I may have entered it incorrectly (I think the coefficients have to be listed in increasing order of exponents). $\endgroup$ – Qiaochu Yuan Jan 15 '18 at 22:11
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    $\begingroup$ Wouldn't it make more sense do divide your polynomials by $x^2-1$ and consider: $$\{1,~x,~6x^2-1,~8x^3-3x,~80x^4-48x^2+3 \}$$? $\endgroup$ – Yuriy S Jan 16 '18 at 0:31
  • $\begingroup$ @YuriyS How would it help? I expect that if there's a known extension of the family of Chebyshev polynomials, then they'll have to be divided by something dependent on additional parameter $m$ to be brought to order $0$ for $n=0$ — like we have $(1-x^2)^{m/2}$ for associated Legendre functions. $\endgroup$ – Ruslan Jan 16 '18 at 10:56
  • $\begingroup$ @Ruslan, if you had no success in searching for the original forms of the polynomials, you might have more luck with reduced forms? $\endgroup$ – Yuriy S Jan 16 '18 at 13:26

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