128 4. Equations
(c) Use 2 cos n0 = (cos 0 + i sin 0)” + (COS 0 - i sin t9)n to verify that
2cos2e=(2cose)y-2=y2-2
2cos3e = (2cos2e)y- 2cose = y3 - 3y
and, in general, that
2 cos ne = (2 co+ - i)e)y - 2 CO+ - 2)e
is a polynomial in y for n > 4. \
(d) Verify that
for 3 _< r 5 n/2.
(e> Verify that L(Y) = YLI(Y) - JLZ(Y) for n L 3.
(f) Show that 2 cos n0 = fn(2 COST), and deduce that, if ]a] 5 1,
the equation is satisfied by y = 2 case, where cos nB = a. [See
Exploration E.30 for a similar sequence of polynomials.]
(g) To handle th e case in which Ial 1 1, we go back to (b) and begin
with the observation that 21 and l/u are the roots of the equation
t2 - yt + 1 = 0, where y = u+ l/u. In a similar way, show that
2(u” + IL-” ) = fn(u + u-l), and deduce that the equation is
satisfied by y = u + u-i where u2” - au” + 1 = 0.
(h) How many roots does the equation have?
- One strategy in solving polynomial equations is to make a transfor-
mation of the equations into a special form which will be easier to
handle. For example, the method of completing the square (Exercise
1.2.1) permits the reduction of a general quadratic equation to one
of the form t2 - c = 0.
Suppose that p(t) is a polynomial of degree n. Show that there is a
constant k such that p(t - 1) is a constant multiple of a polynomial
of the form
t” + a,-2tnm2 + an-3tn-3 +...
in which the coefficient of P-l vanishes.
Show that each solution of the equation p(t) = 0 is of the form s - k
where t = s is a solution of the equation p(t - k) = 0. - Because of the transformation of Exercise 4, in order to handle the
general quintic equation, it suffices to deal with quintic polynomials
of the form
t5 + at3 + bt2 + ct + d.