Relationship between the First and Second Kind Chebyshev Polynomials
Theorem
The first kind Chebyshev polynomials $T_{n} (x) = \cos \left( n \cos^{-1} x \right)$ and second kind Chebyshev polynomials $\displaystyle U_{n} (x) = {{1} \over {n+1} } T_{n+1} ’ (X)$ have the following relationship:
- [1]: $$U_{n} (x) - U_{n-2} (x) = 2 T_{n} (X)$$
- [2]: $$T_{n} (x) - T_{n-2} (x) = 2( x^2 - 1 ) U_{n-2} (x)$$
- Typically, for $0 \le \theta \le \pi$, it is set to $\theta := \cos^{-1} x $.
Proof
The following fact is essential for proving the above equations.
Another expression of the second kind Chebyshev polynomials: $$U_{n} (x) = {{\sin \left( ( n +1 ) \theta \right)} \over { \sin \theta }}$$
[1]
By the sum and difference formulas of trigonometric functions, $$ \begin{align*} & U_{n} (x) - U_{n-2} (x) \\ =& {{1} \over { \sin \theta }} \left[ \sin (n+1) \theta - \sin (n-1) \theta \right] \\ =& {{1} \over { \sin \theta }} 2 \cos \left( {{ (n+1) + (n-1) } \over {2}} \theta \right) \sin \left( {{ (n+1) - (n-1) } \over {2}} \theta \right) \\ =& {{2} \over { \sin \theta }} \cos n \theta \sin \theta \\ =& 2 \cos n \theta \\ =& 2 T_{n} (x) \end{align*} $$
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[2]
By the addition theorem of trigonometric functions, $$ T_{n \pm 1} (x) = \cos (n \pm 1) \theta = \cos (n \theta ) \cos \theta \mp \sin ( n \theta ) \sin \theta $$ Therefore, $$ \begin{align*} & T_{n-1} (x) - T_{n+1} (x) \\ =& 2 \sin (n \theta ) \sin \theta \\ =& 2 \sin^2 \theta {{ \sin (n \theta ) } \over { \sin \theta}} \\ =& 2 \sin^2 \theta U_{n-1} (x) \\ =& 2 (1-x^2) U_{n-1} (x) \end{align*} $$
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