An important example of such a class of polynomial functions is the Cheby- v polynomials denoted by the symbol T(x). They are defined by the re- rence formula (;) Tk+2 – xTk+1 + T = 0, (7.59) ere |æ| < 1, and To = 2, T1 = x. In this section, we will investigate certain perties of these functions. Using the above equation and the given values for To and T1, the first eral Chebyshev polynomials can be easily calculated; they are 1 T2(x) = x² – 2' 3x T2(x) = x³ 4' (7.60) 1 T4(x) = xª – x² + 8' oceeding in this way, we can obtain T(x) for any finite integer k. How- r, this procedure is very laborious and it would be much better to have Difference Equations ompact expression giving Tx(x) explicitly in terms of x and k. This can easily done because equation (7.59) is a second-order, linear difference ation, and its general solution can be determined by the techniques of apter 4. The characteristic equation corresponding to equation (7.59) is 1 p2 - xr + = 0. (7.61) 4 _s equation has the two solutions T'1,2 = (x± / Va2 - - 1]. (7.62) erefore, the kth Chebyshev polynomial takes the form [A(r1)* + B(r2)*], (7.63) ere A and B are constants that can be determined by requiring To = 2 and = x; doing this gives A+B = 2, (7.64) rịA+ r2B = 2x, A = B = 1. (7.65) ostitution of these values for A and B into equation (7.63) gives an explicit -ression for T(æ): (ne) - g2 – 1)*). TR(x) = ) [(x + /x² – 1)* + (x – (7.66) check of equation (7.66) for k = 2,3, and 4 shows that it reproduces the ults previously given by equation (7.60).
An important example of such a class of polynomial functions is the Cheby- v polynomials denoted by the symbol T(x). They are defined by the re- rence formula (;) Tk+2 – xTk+1 + T = 0, (7.59) ere |æ| < 1, and To = 2, T1 = x. In this section, we will investigate certain perties of these functions. Using the above equation and the given values for To and T1, the first eral Chebyshev polynomials can be easily calculated; they are 1 T2(x) = x² – 2' 3x T2(x) = x³ 4' (7.60) 1 T4(x) = xª – x² + 8' oceeding in this way, we can obtain T(x) for any finite integer k. How- r, this procedure is very laborious and it would be much better to have Difference Equations ompact expression giving Tx(x) explicitly in terms of x and k. This can easily done because equation (7.59) is a second-order, linear difference ation, and its general solution can be determined by the techniques of apter 4. The characteristic equation corresponding to equation (7.59) is 1 p2 - xr + = 0. (7.61) 4 _s equation has the two solutions T'1,2 = (x± / Va2 - - 1]. (7.62) erefore, the kth Chebyshev polynomial takes the form [A(r1)* + B(r2)*], (7.63) ere A and B are constants that can be determined by requiring To = 2 and = x; doing this gives A+B = 2, (7.64) rịA+ r2B = 2x, A = B = 1. (7.65) ostitution of these values for A and B into equation (7.63) gives an explicit -ression for T(æ): (ne) - g2 – 1)*). TR(x) = ) [(x + /x² – 1)* + (x – (7.66) check of equation (7.66) for k = 2,3, and 4 shows that it reproduces the ults previously given by equation (7.60).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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