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184 5. Approximation and Location of Zeros

Verify that there are two possible values of r: r1 = (l/2)(1 + &), r2 =

(l/2)(1 - I/?$. El’ lminate Fn+l from the two equations

Fn+l + (r1 - l)F,, = r;

to obtain that

F,+I + (r2 - l)F,, = ry

F, = (l/&)[rT - rt].

Check that this formula actually works for 1 5 n 5 5.
Use a pocket calculator to work out the values of r1 and r2. Argue that
F,,+JF,, gets closer and closer to rr as n becomes larger (so that F,, be-
comes close to being a geometric progression).
There are a number of striking relations satisfied by the terms of the
Fibonacci sequence which can be established with the aid of the formula
for the general term. For example, show that:

(i) F,f + Fz+l = &+I (n L 1)

(ii) F,,+lF*-l - Fz = (-1)” (n >_ 2)

(iii) F,f+l + F,f - F,fsl = F3,, (n >_ 2)

F2k = i(F4n - hzn) (n 2 1)

(v) if mln, then FmlFn.

Other recursion relations can be handled in a similar way. Work out the
general term for the recursion xn = 3x,-r - 2x,-2 (n 1 3), given that the
initial terms are (a) x1 = 1, 22 = 3; (b) cl = 1, 22 = 2.
Sometimes the recursion gives rise to a polynomial equation with a dou-
ble root. Show that, if xn+l = 2axn - u2x,-1 (n 1 2), the form xn+l -
(2~ - r)x, = r(x,, - (2a - r)x,-1) is possible only for r = a. How
would you modify the technique to deal with this situation? Obtain x,, =
a”-‘21 + (n - l)~“-~(x2 - axr).
Now consider a sequence for which the general term depends on its three
immediate predecessors, such as for example one whose terms satisfies

v,,+l = 6v, - llv,-1 + 6~~2 (n 2 3).

Show that, if we rewrite this in the form

(v,+l + QV, + bun- I) = r(v, + a~,-1 + bv,-2),

then r3 = 6r2 - llr + 6 and ( r, a, b) = (1, -5,6), (2, -4,3), (3, -3,2). Use

this information to set up a system of three linear equations to solve for

v,+r, VA, 21,-l in terms of vr, ~2, 213.