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: From the above identity, we get\text{From the above identity, we get} Tn(cos⁡θ)=(cos⁡θ−cos⁡2θ−1)n+(cos⁡θ+cos⁡2θ−1)n2=(cos⁡θ−isin⁡θ)n+(cos⁡θ+isin⁡θ)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} (cos⁡θ−isin⁡θ)n+(cos⁡θ+isin⁡θ)n2=cos⁡nθ−isin⁡θ+cos⁡nθ+isin⁡nθ2=cos⁡nθ\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}

#Algebra

Note by Swapnil Das
5 years, 2 months ago

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Comments

Nice work. I have a doubt 1−cos2(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).

A Former Brilliant Member - 5 years, 2 months ago

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

Nihar Mahajan - 5 years, 2 months ago

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

Swapnil Das - 5 years, 2 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 - 5 years, 2 months ago

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

Swapnil Das - 5 years, 2 months ago
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