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Chebyshev roots

Authors
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    Name
    Vu Hung
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Problem Statement

The first-kind Chebyshev polynomial of degree 55 is

T5(x)=16x5−20x3+5x.T_5(x)=16x^5-20x^3+5x.

Use the identity T5(cos⁡θ)=cos⁡(5θ)T_5(\cos\theta)=\cos(5\theta) to find all roots of T5(x)T_5(x) in the interval [−1,1][-1,1].

Hints

Set cos⁡(5θ)=0\cos(5\theta)=0, then convert the resulting angles back to x=cos⁡θx=\cos\theta.

Solutions

Let x=cos⁡θx=\cos\theta. We need

cos⁡(5θ)=0.\cos(5\theta)=0.

Thus

5θ=(2k+1)π2,k=0,1,2,3,4.5\theta=\frac{(2k+1)\pi}{2},\qquad k=0,1,2,3,4.

So

θ=(2k+1)π10,\theta=\frac{(2k+1)\pi}{10},

and the five roots are

x=cos⁡π10, cos⁡3π10, cos⁡5π10, cos⁡7π10, cos⁡9π10.x=\cos\frac{\pi}{10},\ \cos\frac{3\pi}{10},\ \cos\frac{5\pi}{10},\ \cos\frac{7\pi}{10},\ \cos\frac{9\pi}{10}.

Takeaways

  • Roots of Tn(x)T_n(x) on [−1,1][-1,1] come from zeros of cos⁡(nθ)\cos(n\theta).
  • The substitution x=cos⁡θx=\cos\theta is natural because cos⁡θ\cos\theta covers [−1,1][-1,1].
  • Degree 55 gives five roots, which is a useful check.

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