2.3. The Derivative 67
(c) Differentiate both sides of the equation in (b) repeatedly. Set
t = c to obtain
P(C) = co
P’(C) = Cl
p”(C) = 2!c2
and in general
p@‘(c) = 3!c3
P(~)(C) = k!ck for 0 < - k - < n.
Substitute in (b) to obtain the result.
- What does the Taylor Expansion of p(t) about c amount to when
c = O? - Form the Taylor Expansions for p(t) and q(t) about c. Multiply them
together and verify that the result is
(P(l)(C) + (Pd’W - c) + ;(PnYkw - 4” +... *
- (a) Use Taylor’s Theorem to establish the Binomial Expansion
(1 + t)^ = 1 + nt + (;)tz+...+( ;>tk+...+t”.
(b) Prove that
(u + b)” = a” + na”-lb +... + anskbk +... + b”.
- Another approach to use in expanding p(t) in terms of powers of
(t - c) is to make the substitution t = c + s, and compute p(c + s) in
ascending powers of s. Do this for the polynomial t3 - 4t2 + 7t + 2
and check that the coefficient of s’ is p(‘)(c)/r! for each value of r. - For several polynomials and values of c of your choice, find the Tay-
lor expansion and check your answer by (i) Horner’s Method, (ii) a
substitution of the type t = c + s, (iii) multiplying out and adding
the terms of the expansion. - Multiplicity of zeros. In Exercise 2.2, we found that c is a zero of a
polynomial p(t) i f an d only if (t - c) is a factor of p(t). Using this
result as a basis, we can sharpen the idea of a zero.