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$T_n(x)$ is a polynomial of degree $n$ such that $T_n(x)=\cos(n\cos^{-1}(x))$

(i) Show that $T_2(x)=2x^2-1$ and $T_3(x)=4x^3-3x$

(ii) Show that $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$

$U_n(x)$ is an $n$ degree polynomial defined by $$U_n(x)=\frac{\sin((n+1)\theta)}{\sin\theta}$$ where $x=\cos\theta$.

(iii) Show that $$\frac{d}{dx}T_n(x)=nU_{n-1}(x)$$

(iv) Show that $$\frac{d}{dx}U_{n-1}(x)=\frac{xU_{n-1}(x)-nT_n(x)}{1-x^2}$$

I'm having a lot of problems with part (iii). I typed out part (ii) because I thought perhaps the result from part (ii) maybe be cleverly used to solve part (iii). Differentiating $T_n(x)$ would yield an algebraic expression in $x$. On the other hand, $nU_{n-1}(x)$ is in $\theta$ unless we explicitly express $U_n(x)$ in $x$ which would be very troublesome because there is no simple way of expressing $\sin n\theta$ in terms of $\cos\theta$ and therefore $x$. Some help would be of great help thank you!

[I don't need help with part (i) and (ii)]

WilliamKin
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  • You need to use the chain rule for iii. the result should really be not that hard using some trigonometric identities – JohnColtraneisJC Aug 16 '17 at 11:53
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    https://math.stackexchange.com/questions/1463422/relation-between-first-kind-chebyshev-poly-and-second-kind-chebyshev-poly – Math-fun Aug 16 '17 at 11:54

1 Answers1

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Hints:

Since $$ T_n(x)=\cos(n\arccos(x)), $$ we can differentiate $T_n(x)$ with respect to $x$ to get $$ \frac{d}{dx}T_n(x)=-\sin(n\arccos(x))\cdot n\cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)=\frac{n\sin(n\arccos(x))}{\sqrt{1-x^2}} $$

On the other hand, $$ nU_{n-1}(x)=n\cdot\frac{\sin(n\arccos(x))}{\sin(\arccos x)}. $$

Therefore, all you need to do is to show that

$\sin(\arccos x)=\sqrt{1-x^2}.$

This can be done by

drawing an appropriate right triangle (one leg has length $x$ while the hypotenuse has length $1$).

Michael Burr
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