Proof of a Chebyshev Identity

Reference: Wiki on Chebyshev Polynomials, Proof problem 6.

Tn(x)=(xx21)n+(x+x21)n2.\huge T_n(x) = \frac{ ( x - \sqrt{ x^2 - 1} )^n + ( x + \sqrt{ x^2 - 1 } ) ^ n } { 2} .


Solution: From the above identity, we get\text{From the above identity, we get} Tn(cosθ)=(cosθcos2θ1)n+(cosθ+cos2θ1)n2=(cosθisinθ)n+(cosθ+isinθ)n2\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 }.

Using\text{Using} De Moivre's Theorem, we get\text{we get} (cosθisinθ)n+(cosθ+isinθ)n2=cosnθisinθ+cosnθ+isinnθ2=cosnθ\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 }.

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

 Proved\therefore\text{ Proved}

Note by Swapnil Das
3 years, 4 months ago

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Nice work. I have a doubt 1cos2(θ)=sin(θ)\sqrt{1-cos^2(\theta)}=\sin(\theta). Can you please explain how you got isin(θ)i\sin(\theta) and how 1+cos2(θ)=isin(θ)\sqrt{1+cos^2(\theta)}=i\sin(\theta).

Brilliant Member - 3 years, 4 months ago

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Yup, I don't understand how Swapnil wrote 1cos2θ\sqrt{1-\cos^2 \theta} and 1+cos2θ\sqrt{1+\cos^2 \theta}

Nihar Mahajan - 3 years, 4 months ago

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Bro, I am Swapnil. Is it fine now? (It was a typo)

Swapnil Das - 3 years, 4 months ago

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@Swapnil Das Yep, it is correct now. Sorry, Swaplin was a typo, I meant to type Swapnil :P :P :P

Nihar Mahajan - 3 years, 4 months ago

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Correct. That's a typo, thanks for pointing out!

Swapnil Das - 3 years, 4 months ago

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