Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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(2) Practice using Algorithm 8.3.12 by working through the Fibonacci sequence to find a closed formula. Algorithm 8.3.12 Algorithm for Solving Homogeneous Finite Order Linear Relations. (a) Write out the characteristic equation of the relation S(k)+C₁S(k − 1) + ...+ C₂S(kn) = 0, which is a" +C₁a-1 ++C₁-1a+C₁ = 0. (b) Find all roots of the characteristic equation, the characteristic roots. (c) If there are n distinct characteristic roots, a, a,...an, then the general solution of the recurrence relation is S(k) = b₁a₁k + b₂a2k +...+ b₂ªn*. If there are fewer than n characteristic roots, then at least one root is a multiple root. If a, is a double root, then the bjajk term is replaced with (bjo+bjak) a. In general, if a, is a root of multiplicity p, then the bajk term is replaced with (bo+bj.1k+·· + bj.(p-1) k³−¹) at. (d) If n initial conditions are given, we get n linear equations in n unknowns (the by's from Step 3) by substitution. If possible, solve these equations to determine a final form for S(k). Although this algorithm is valid for all values of n, there are limits to the size of n for which the algorithm is feasible. Using just a pencil and paper, we can always solve second-order equations. The quadratic formula for the roots of ar²+bx+c=0 is -b± √b²-4ac 2a The solutions of a² + C₁a + C₂ = 0 are then x= (-C₁+√C₁²-40₂) and (-C₁-√C₁² - 40₂) Although cubic and quartic formulas exist, they are too lengthy to introduce here. For this reason, the only higher-order relations (n ≥ 3) that you could be expected to solve by hand are ones for which there is an easy factorization of the characteristic polynomial.
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Transcribed Image Text:(2) Practice using Algorithm 8.3.12 by working through the Fibonacci sequence to find a closed formula. Algorithm 8.3.12 Algorithm for Solving Homogeneous Finite Order Linear Relations. (a) Write out the characteristic equation of the relation S(k)+C₁S(k − 1) + ...+ C₂S(kn) = 0, which is a" +C₁a-1 ++C₁-1a+C₁ = 0. (b) Find all roots of the characteristic equation, the characteristic roots. (c) If there are n distinct characteristic roots, a, a,...an, then the general solution of the recurrence relation is S(k) = b₁a₁k + b₂a2k +...+ b₂ªn*. If there are fewer than n characteristic roots, then at least one root is a multiple root. If a, is a double root, then the bjajk term is replaced with (bjo+bjak) a. In general, if a, is a root of multiplicity p, then the bajk term is replaced with (bo+bj.1k+·· + bj.(p-1) k³−¹) at. (d) If n initial conditions are given, we get n linear equations in n unknowns (the by's from Step 3) by substitution. If possible, solve these equations to determine a final form for S(k). Although this algorithm is valid for all values of n, there are limits to the size of n for which the algorithm is feasible. Using just a pencil and paper, we can always solve second-order equations. The quadratic formula for the roots of ar²+bx+c=0 is -b± √b²-4ac 2a The solutions of a² + C₁a + C₂ = 0 are then x= (-C₁+√C₁²-40₂) and (-C₁-√C₁² - 40₂) Although cubic and quartic formulas exist, they are too lengthy to introduce here. For this reason, the only higher-order relations (n ≥ 3) that you could be expected to solve by hand are ones for which there is an easy factorization of the characteristic polynomial.
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