The Chebyshev polynomials can be determined from Tn (2) = cos(n cos-1.). (c) Show that n!...
2. The Chebyshev polynomials can be determined from T.(x) = cos(n cos-?x). (a) Find T-(x) from above formula. (b) Find Tn+1(x) + Tn-1(x) in terms of T,(x). (c) Show that In/2] n! Tn () = KO (2k)! (n – 2k);?"*2*(x2 – 1)". (Note: You need to prove it in detail. To do it, you may need to consider two cases: n=2p-1 (odd) and n=2p (even). )
2. (Chebyshev Polynomials). Below is a guideline for finding the coefficients in T,l(x) = cos(n cos-1 x), Chebyshev polynomials equivalently or T,,(cosa.) = cos(na). For example, To(x)-1, T1(x)=x, T2(x)= 2x2-1 (b) Calculate T3(x) using T2(x) and T1(x) (c) Keep iterating and calculate T(x) and T( 2. (Chebyshev Polynomials). Below is a guideline for finding the coefficients in T,l(x) = cos(n cos-1 x), Chebyshev polynomials equivalently or T,,(cosa.) = cos(na). For example, To(x)-1, T1(x)=x, T2(x)= 2x2-1 (b) Calculate T3(x) using T2(x)...
2. Exercise 2. Consider the sequence (xn)n≥1 defined by xn = Xn k=1 cos(k) k + n2 = cos(1) 1 + n2 + cos(2) 2 + n2 + · · · + cos(n) n + n2 . (a) Use the triangle inequality to prove that |xn| ≤ n 1 + n2 for all n ≥ 1. (b) Use (a) and the -definition of limit to show that limn→∞ xn = 0. Exercise 2. Consider the sequence (In)n> defined by cos(k)...
1 point Prove the following statement: If n2 is even, then n is even. Order each of the following sentences so that they form a logical proof. Proof by Contrapositive: Choose from these sentences: Your Proof: Suppose n is odd. Then by definitionn 2k +1 for some integer k Required to show if n is not even (odd), then n is not even (odd). Thus n2(2k1)2. n24k2 4k1. 22(22+2k) +1 Thus n2 (an integer) +1 and by definition is odd....
Problem 1: Recall that the Chebyshev nodes 20, 21, ...,.are determined on the interval (-1,1) as the zeros of Tn+1(x) cos((n + 1) arccos(x)) and are given by 2; +17 Tj = COS , j = 0,1,...n. n+1 2 Consider now interpolating the function f(x) = 1/(1 + x2) on the interval (-5,5). We have seen in lecture that if equispaced nodes are used, the error grows unbound- edly as more points are used. The purpose of this problem is...
numerical methods 2+17), j = 0,1...... Problem 1: Recall that the Chebyshev nodes x0, 71,..., are determined on the interval (-1,1) as the zeros of Tn+1(x) = cos((n +1) arccos(x)) and are given by 2j +17 X; = cos in +12 Consider now interpolating the function f(x) = 1/(1+22) on the interval (-5,5). We have seen in lecture that if equispaced nodes are used, the error grows unbound- edly as more points are used. The purpose of this problem is...
2. For n . define functions T inductivelv such that 0, 1, 2, . . . (cosx) = cos(nx), with Folz) 1. (a) Prove that Tn is a polynomial for every n and compute its degree. b) Prove the recursion formula (c) Compute the integral dr 山 for every n, m E N 2. For n . define functions T inductivelv such that 0, 1, 2, . . . (cosx) = cos(nx), with Folz) 1. (a) Prove that Tn is...
1. (Taylor Polynomial for cos(ax)) For f(x)cos(ar) do the following. (a) Find the Taylor polynomials T(x) about 0 for f(x) for n 1,2,3,4,5 (b) Based on the pattern in part (a), if n is an even number what is the relation between Tn (x) and TR+1()? (c) You might want to approximate cos(az) for all in 0 xS /2 by a Taylor polynomial about 0. Use the Taylor polynomial of order 3 to approximate f(0.25) when a -2, i.e. f(x)...
b) Using the binomial theorem show that Σ (-1)"/2 (n) cos" k(z) sink(z), Σ (-1)(k-1)/2C) cox"-"(x) sink(z). cos(nx) = sin(nx) = COS k-odd 6 marks]
2. Consider the polynomials 0-k (z) := (1 + z) for k-0,..., 10 and let B-bo,b1bo) can be shown that B is a basis for Pio the vector space of polynomials of degree at most 10. (You do not need to prove this.) Let Pk (z)-rk for k = 0, 1, . . . , 10, so that S = {po, pi, . . . , pio) is the standard basis for P10. Use Mathematica to: (a) Compute the change...