Unknown

5.3. Location of Complex Roots 183

(6cosB+ 1)12(cosf?+isin0) - 39. Sketch the Nyquist diagram

for this polynomial (taking M = 6) and verify that it winds once

around the origin. Deduce that the quadratic is not stable.

12. Discuss stability of the following polynomials:

(a) z3 + 2

(b) z3 + 2z2 + 32 + 1

(c) z4 + 3z3 + z2 + z + 8.

Exploration

E.50. Recursion Relations. The Fibonacci sequence was introduced in

Exploration E.14; it is given by the recursion relations

Fl = F2 = 1 F,+l = F,,+F,,-l (n=2,3,4 ,... ).

While any term of the sequence can be found using these equations, the

process is both tedious and sensitive to a miscalculation, since an error at

any point would contaminate the computation from then on. One would

like to have a formula from which the nth term can be found directly.

For some sequences, such a formula is easy to find. For example, let

S,, be the sum of the positive integers up to n. This sequence can be

defined recursively by Si = 1, S,, = S,-i + n. However, one can directly

compute each S,, by the familiar formula S, = (1/2)n(n + 1). For the

Fibonacci sequence, there, too, is a formula, but one would have to be

clever indeed to guess it. However, with the proper approach, it can be

found in a straightforward way.

Often, we can solve a problem by looking to a simpler one for guidance.

Either we might find a stepping stone to the more complex situation or at

least get a better idea of the ingredients of a solution. What would happen

if we had a sequence for which each term depended only on its immediate

predecessor, to wit

u,+l = run (n = 2,3,4,...)

for some number r independent of n? This sequence is geometric and the
nth term is ulr”-l. Can we reduce the Fibonacci sequence to this one?
With r a number to be specified later, we can write the recursion relation
for F,, as:

F,+I + (r - l)Fn = rF, + F,.,-1 = r(F” + r-‘F,+l).

If we pick r so that r - 1 = r-l, then u, = Fn+l + (r - l)F,, will be a
geometric progression. Verify that

F,+I + (r - l)F,, = (F2 + (r - 1)Fl)r n-1 = rn for n > 2.