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Answers to Exercises; Chapter 1 245

Alternatively, the identical vanishing of p(t) implies the same for all of its

derivatives, whence Taylor’s Theorem identifies all the coefficients as 0.

The reader might wish to reflect on the validity of the following proof,

which assumes the Factor Theorem. Let n be the degree of a nonzero poly-

nomial p which vanishes identically. Since p vanishes at 0, 1, 2, 3,... , n, we

can write

p(t) = t(t - 1)(t - 2). f *(t - n)q(t),

for some polynomial q. The degree of the left side is n while that of the

right side is at least n + 1, a contradiction.

1.13. (a) If f(2t) = h(f(t)), then deg f(t) = deg f(2t) = deg h(t) deg f(t),

so that, either f is a constant c and h(c) = c, or else h(t) is linear of the

form ut + 21. If f(t) = untn +... + alt + a0 with a, # 0, we have

2”u,t” + 2n-lun-#+1 +... + 2u1t + uo

= uuntn + UU,-ltn-l +... + uu1t + (uuo + v).
Hence 2”Uk = ‘1LUk (1 5 k 5 n) and 00 = uao + 2). Thus, f(t) = ant” i- uo
and h(t) = 29 + (1 - 2”) a0 where n is a positive integer, and a, and a0
are arbitrary constants.
(b) The relation has the form f(2t) = h(f(t)), with f(t) = sin2 t and
h(t) = 4t(l-t). But, since h is quadratic, by (a), f cannot be a polynomial.
1.14. First solution. If log t is a polynomial, then, by 13.(a), log 2t = log 2+
logt implies logt = at” + b. But then log2 = (2”ut” + b) - (utn + b) =
(2n - l)uP, which is false.
Second solution. Suppose logt = f(t), a polynomial of degree n. Since
f(t2) = 2f(t), comparing degrees yields 2n = n, whence n = 0. But then
logt must be a constant, which is false.
1.15. Suppose that g(t) = ant” +... is periodic with positive period k and
a, # 0. Then g(t + k) - g(t) = knu,t”-’ +... vanishes identically, so its
leading coefficient /man is 0. Hence n = 0 and g must be constant.
1.16. If t113 is a polynomial of degree k, then 3k = degt = 1, which
contradicts k being an integer.
1.17. p(f - g) = 0. If f - g is not the zero polynomial, then the left side
would be a polynomial of nonnegative degree which vanishes identically, a
contradiction.
1.19. (a) The coefficients of all the odd powers oft are 0.
(b) The coefficients of all the even powers oft are 0.
1.20. Suppose p(t) h as more than one term and p(t) = uP+bP +a. ., where
n = degp and m is the next highest nonnegative exponent corresponding
to a nonzero coefficient. Equating p(t2) and [p(t)12 yields ut2n + bt2”’ +... =
u2t2n + 2ubtm+” +.. ., whence 2m = m + n. But this contradicts m < n.