Chebyshev Polynomials - Application to Polynomial Interpolation

Recall that the Chebyshev polynomials are defined by

$$T_n (x) = \cos ( n \arccos x ),\quad T_n ( \cos \theta) = \cos n \theta .$$

The initial values of $T_{n} (x)$ are

$$\begin{aligned} T_0(x) &= 1 \\ T_1(x) &= x \\ T_2(x) &= 2x^2 - 1 \\ T_3(x) &= 4x^3 - 3x \\ T_4(x) &= 8x^4 - 8x^2 + 1 \\ T_5(x) &= 16x^5 - 20x^3 + 5x \\ T_6(x) &= 32x^6 - 48x^4 + 18x^2 - 1 \\ T_7(x) &= 64x^7 - 112x^5 + 56x^3 - 7x \\ T_8(x) &= 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1 \\ T_9(x) &= 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x. \\ T_{10}(x) &= 512x^{10} - 1280x^8 + 1120x^6 - 400x^4 + 50x^2-1. \\ \end{aligned}$$

For a given value $y$ between -1 and 1, the solutions to $T_n (x) = y$ are $\cos \frac{ \theta + 2 \pi k } { n }$, where $k$ ranges from 1 to $n$ and $\cos \theta = y$.

For each of these values, we have

$$T_n \left( \cos \frac{ \theta + 2 \pi k } { n }\right) = \cos\left( n \times \frac{ \theta + 2 \pi k } { n } \right ) = \cos \theta = y.$$

Since we have $n$ solutions to a degree $n$ polynomial, these are all of the roots. $_\square$

In terms of $\cos A$, what are the roots of $T_3 ( x) = 0?$

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In terms of $\cos A$, what are the roots of $2 T_4 ( x) - 1 = 0?$

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In terms of $\cos A$, what are the roots of $4 T_5 ^2 ( x) - 3 = 0?$

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The converse of the above theorem is as follows:

The polynomial whose roots are $\cos \frac{ \theta + 2\pi k } { n }$, where $k$ ranges from 1 to $n$, is

$$T_n ( x) = \cos \theta .$$

In terms of $T_n$, find a polynomial whose roots are $\cos 10 ^ \circ , \cos 100^\circ, \cos 190^ \circ, \cos 180 ^ \circ$.

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In terms of $T_n$, find a polynomial whose roots are $\cos \left( \theta + \frac{ 2 n } { 5} \pi \right)$ for $i = 1$ to $5.$

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To answer problems in this section, you should be familiar with Vieta's formula to help you find the sum and product of roots.

Evaluate

$$\cos 10 ^ \circ \times \cos 50 ^ \circ \times \cos 70 ^ \circ.$$

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Evaluate

$$\frac{ 1}{ \cos 25 ^ \circ } + \frac{ 1}{ \cos 115 ^ \circ } + \frac{ 1}{ \cos 205 ^ \circ } + \frac{ 1}{ \cos 295 ^ \circ }.$$

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In terms of $T_n$, find a polynomial whose roots are exactly

$$\cos^2 1 ^ \circ, \cos^2 3 ^ \circ, \cos^2 5 ^ \circ, \ldots , \cos^2 89 ^ \circ .$$

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