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interpolation and the Lagrange polynomial 5. Polynomial Irrte rpolation. {a} Use the Lagrange interpolation process to obtain a polynomial of least degree that assumes these

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interpolation and the Lagrange polynomial

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5. Polynomial Irrte rpolation. {a} Use the Lagrange interpolation process to obtain a polynomial of least degree that assumes these values. 1 I] 2 3 4 y i" 11 28 63 {b} Rearrange the points o'fthe problem in part {a} in a table and nd the Newton form ofthe interpolating polynomial. Show that the polynomials obtained are identical, although their forms may differ. {c} Assume that the data set in part [a] to be the values of some function y = ffx]. i. Approximate f[2}, the rst derivative of f at x = 2, by using the {a} forward difference [bi backward difference and {c} centered difference approximations. ii. Another approximation of 2] would be P3i2] [the derivative of the cubic interpolant you constructed above]. Find the value. iii. Construct the quadratic polynomial Fax] interpolating the rst 3 poirrts. P32] is yet another approximaon of ll. Does it agree with any of the above?I listing of your code. (c) If one attempts to fit a much higher-degree interpolation polynomial to the Runge function using MATLAB's polyfit routine, MATLAB will issue a warning that the problem is ill conditioned and answers may not be accurate. Still, the barycentric interpolation formula (8.6) with the special weights (8.16) for the values of a polynomial that interpolates f at the Chebyshev points may be used. Use this formula with 21 scaled Chebyshev interpolation points, x; = 5 cos 5 0. .... 20, to compute a more accurate approximation to this function. Turn in a plot of the interpolation polynomial pro(x) and the errors If(xx) - Pzo(x)| over the interval [-5, 5]. (d) Download the chebfun package from the website Solving this equation for (x) and substituting into (8.4) leads to the second form or what is simply called the barycentric interpolation formula: Wi Wi p(x) = Vi (8.6) x - Xi =0& ( ho athand ) hmot An - 2 Interpolation and the Lagrange Polynomial: Problem : Let f (x ) = e , for 0 ex = 2. a) Approximate f (o, 25) using linear interpolation with x, =0 and x, =0.5 bj Approximate f (o. 75 ) using linear interpolation with * = 0.5 and * , =1. () Approximate f(0. 25) and f (0. 75) using the second interpolating polynomial with 7,=0, 2, = 1 and 2 = 2. d) which approximations are better and why!close to LI nciency in the chebfun is to convert between different representations of this polynomial ant using the Fast Fourier Transform (FFT), an algorithm that will ibed in section 14.5.1. be shown that when xo, ..., In are the Chebyshev points, the factor - x;) in the error formula (8.11) satisfies maxxel-1,1] | II"=0(x - x;)Is hich is near optimal [88]. Moreover, the weights in the barycentric tion formula (8.6) take on a very simple form: 27-1 [(-1)1/2 if j = 0 or j = n, ((-1) otherwise, (8.16) kes this formula especially convenient for evaluating p(x) at points x interpolation points. The factor 2"-1 in each w; can be dropped pears as a factor in both the numerator and the denominator of (8.6). llowing theorem [9], which we state without proof, shows that omial interpolant at the Chebyshev points is close to the best al approximation in the co-norm; that is, its maximum deviation unction on the interval [-1, 1] is at most a moderate size multiple of Anloo, where the minimum is taken over all nth-degree polynomialsProblem : Let fix) = x'cos ( x) + 3x and consider the values of fix) at the points x = 0. 1, 0.4, 0.5 , 1. a) find the natural cubic polynomial spline function . by Find the clamped cubic polynomial spline function . c) Use both splines to approximate fro.25). d) use both splines to approximate f'(o.25)

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