How do you prove following relation between Chebyshev poly of first kind and Chebyshev poly of second kind: $$dT_n(x)/dx=nU_{n-1}(x)$$
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Since both sequences of polynomials have nice recursions, one idea is to do this by induction. Have you tried? – Mariano Suárez-Álvarez Oct 04 '15 at 05:15
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I am not sure what you mean...you mean indicator function? – jfcjohn Oct 04 '15 at 05:17
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Since $T_n(x) =\cos(n\arccos(x)) $ and $U_n(x) =\frac{\sin((n+1)\arccos(x))}{\sin(\arccos(x))} $,
$\begin{array}\\ T_n'(x) &=-(n\arccos(x)))'\sin(n\arccos(x))\\ &=-n\frac{-1}{\sqrt{1-x^2}}\sin(n\arccos(x))\\ &=\frac{n}{\sqrt{1-(\cos(\arccos(x))^2}}\sin(n\arccos(x)) \qquad\text{(this is a sort of sneaky part)}\\ &=\frac{n\sin(n\arccos(x))}{\sin(\arccos(x))}\\ &=nU_{n-1}(x) \end{array} $
marty cohen
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I see...is there a proof for conditions Tn(x)=cos(narccos(x)) and Un(x)=sin((n+1)arccos(x))sin(arccos(x)) you used? Or it is theorem that people can directly use it without proving? – jfcjohn Oct 04 '15 at 14:44
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