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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?

4. 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.


5. 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.