Chebyshev Series

• By rearranging the Chebyshev polynomials,
• we can express powers of $x$ in terms of them:

  $$\begin{array}{l} 1=T_0 x = T_1 x^2 = \frac{1}{2}(T_0 + T_... ... x^9=\frac{1}{256}(126T_1 + 84T_3+36T_5+9T_7+T_9) \end{array}$$ (6.2)

  
  

• By substituting these identities into an infinite Taylor series
• and collecting terms in $T_i(x)$, we create a Chebyshev series.

• For example, we can get the first four terms of a Chebyshev series
  $$e^x=1.2661T_0+1.1302T_1+0.2715T_2+0.0443T_3$$
  by starting with the Maclaurin expansion for $e^x$.
• Such a series converges more rapidly than does a Taylor series on $[-1, 1]$;
  $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\ldots$$

• Replacing terms by Eqn. 6.2, but omitting polynomials beyond $T_3(x)$ because we want only four terms, we have;
  $$e^x=1.2661T_0+1.1302T_1+0.2715T_2+0.0443T_3+\ldots$$

• The number of terms that are employed determines the accuracy of the computed values.
• To compare the Chebyshev expansion with the Maclaurin series, we convert back to powers of $x$, using Eqn. 6.1:

  $$e^x=0.9946+0.9973x+0.5430x^2+0.1772x^3+\ldots$$ (6.3)

  
  

Table 6.2: Comparison of Chebyshev series for $e^x$ with Maclaurin series.

• Table 6.2 and Figure 6.2 compare the error of the Chebyshev expansion ( $0.9946+0.9973x+0.5430x^2+0.1772x^3$) with the Maclaurin series $(1+x+ 0.5x^2 +0.1667x^3)$.
  • Chebyshev expansion, the errors can be considered to be distributed more or less uniformly throughout the interval.
  • Maclaurin expansion, which gives very small errors near the origin, allows the error to bunch up at the ends of the interval.

Figure 6.2: Comparison of the error of Chebyshev series for $e^x$ with the error of Maclaurin series.