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Proof of a Chebyshev Identity

Reference: Wiki on Chebyshev Polynomials, Proof problem 6.

$$\huge T_n(x) = \frac{ ( x - \sqrt{ x^2 - 1} )^n + ( x + \sqrt{ x^2 - 1 } ) ^ n } { 2} .$$

Solution: $$\text{From the above identity, we get}$$ $$\large{ T }_{ n }(\cos { \theta )= } \frac { { \left( \cos { \theta -\sqrt { \cos ^{ 2 }{\theta}- 1 } } \right) }^{ n }+{ \left( \cos { \theta +\sqrt { \cos ^{ 2 }{ \theta } -1} } \right) }^{ n } }{ 2 } =\frac { { \left( \cos { \theta -i\sin { \theta } } \right) }^{ n }+{ \left( \cos { \theta +i\sin { \theta } } \right) }^{ n } }{ 2 }$$.

$$\text{Using}$$ De Moivre's Theorem, $$\text{we get}$$ $$\large\frac { { \left( \cos { \theta -i\sin { \theta } } \right) }^{ n }+{ \left( \cos { \theta +i\sin { \theta } } \right) }^{ n } }{ 2 } =\frac { \cos { n\theta -i\sin { \theta +\cos { n\theta +i\sin { n\theta } } } } }{ 2 } =\cos { n\theta }$$.

$$\text{This turns out to be the definition of the}$$ Chebyshev Polynomial of the first kind.

$$\therefore\text{ Proved}$$

Note by Swapnil Das
1 year, 2 months ago

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Nice work. I have a doubt $$\sqrt{1-cos^2(\theta)}=\sin(\theta)$$. Can you please explain how you got $$i\sin(\theta)$$ and how $$\sqrt{1+cos^2(\theta)}=i\sin(\theta)$$. · 1 year, 2 months ago

Correct. That's a typo, thanks for pointing out! · 1 year, 2 months ago

Yup, I don't understand how Swapnil wrote $$\sqrt{1-\cos^2 \theta}$$ and $$\sqrt{1+\cos^2 \theta}$$ · 1 year, 2 months ago

Bro, I am Swapnil. Is it fine now? (It was a typo) · 1 year, 2 months ago