) Practice using Algorithm 8.3.12 by working through the Fibonacci sequence to find a closed formula.
) Practice using Algorithm 8.3.12 by working through the Fibonacci sequence to find a closed formula.
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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Please help solve and explain. Thank you.
![(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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48ceae03-ce82-43df-b816-3449c4a6aa24%2F50c1cc29-e6fc-41ae-933d-8163aa2586b6%2Fd3rorku_processed.png&w=3840&q=75)
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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