Editing Chebyshev polynomials (section)

==Relations between the two kinds of Chebyshev polynomials==
The Chebyshev polynomials of the first and second kinds correspond to a complementary pair of [[Lucas sequence]]s {{math|''Ṽ<sub>n</sub>''(''P'', ''Q'')}} and {{math|''Ũ<sub>n</sub>''(''P'', ''Q'')}} with parameters {{math|1=''P'' = 2''x''}} and {{math|1=''Q'' = 1}}:
<math display="block">\begin{align}
{\tilde U}_n(2x,1) &= U_{n-1}(x), \\
{\tilde V}_n(2x,1) &= 2\, T_n(x).
\end{align}</math>
It follows that they also satisfy a pair of mutual recurrence equations:{{sfn|Bateman|Bateman Manuscript Project|1953|loc=[https://archive.org/details/highertranscende02bate/page/184/ {{pgs|184}}, eqs. 3–4]}}
<math display="block">\begin{align}
T_{n+1}(x) &= x\,T_n(x) - (1 - x^2)\,U_{n-1}(x), \\
U_{n+1}(x) &= x\,U_n(x) + T_{n+1}(x).
\end{align}</math>

The second of these may be rearranged using the [[#Recurrence definition|recurrence definition]] for the Chebyshev polynomials of the second kind to give:
<math display="block">T_n(x) = \frac{1}{2} \big(U_n(x) - U_{n-2}(x)\big).</math>

Using this formula iteratively gives the sum formula:
<math display="block">
U_n(x) = \begin{cases}
2\sum_{\text{ odd }j>0}^n T_j(x)      & \text{ for odd }n.\\
2\sum_{\text{ even }j\ge 0}^n T_j(x) - 1      & \text{ for even }n,
\end{cases}
</math>
while replacing <math>U_n(x)</math> and <math>U_{n-2}(x)</math> using the [[#Differentiation and integration|derivative formula]] for <math>T_n(x)</math> gives the recurrence relationship for the derivative of <math>T_n</math>:

<math display="block">2\,T_n(x) = \frac{1}{n+1}\, \frac{\mathrm{d}}{\mathrm{d}x}\, T_{n+1}(x) - \frac{1}{n-1}\,\frac{\mathrm{d}}{\mathrm{d}x}\, T_{n-1}(x), \qquad n=2,3,\ldots</math>

This relationship is used in the [[Chebyshev spectral method]] of solving differential equations.

[[Turán's inequalities]] for the Chebyshev polynomials are:<ref>{{citation|mr=0040487 | last1=Beckenbach | first1= E. F.| last2= Seidel|first2= W.| last3= Szász|first3= Otto | title=Recurrent determinants of Legendre and of ultraspherical polynomials | journal=Duke Math. J. | volume= 18 | year=1951 | pages= 1–10 | doi=10.1215/S0012-7094-51-01801-7}}</ref>
<math display="block">\begin{align}
T_n(x)^2 - T_{n-1}(x)\,T_{n+1}(x)&= 1-x^2 > 0 &&\text{ for } -1<x<1 &&\text{ and }\\
U_n(x)^2 - U_{n-1}(x)\,U_{n+1}(x)&= 1 > 0~.
\end{align}</math>

The [[integral]] relations are{{sfn|Bateman|Bateman Manuscript Project|1953|loc=[https://archive.org/details/highertranscende02bate/page/187/ {{pgs| 187}}, eqs. 47–48]}}{{sfn|Mason|Handscomb|2002}}
<math display="block">\begin{align}
\int_{-1}^1 \frac{T_n(y)}{y-x} \, \frac{\mathrm{d}y}{\sqrt{1 - y^2}} &= \pi\,U_{n-1}(x)~, \\[1.5ex]
\int_{-1}^1\frac{U_{n-1}(y)}{y-x}\, \sqrt{1 - y^2}\mathrm{d}y &= -\pi\,T_n(x)
\end{align}</math>
where integrals are considered as principal value.