Calculus: Analytic Geometry and Calculus, with Vectors

(^634) Series
is valid. The right member of (12.56) is called the Taylor expansion of f
in powers of (x - a). Problem 5 gives an example of a function f whose
Taylor expansion in powers of x exists and converges for each x toa
number which differs from f(x) when x 5,5 0. For this and other reasons,
it is sometimes necessary to use (12.54) and other remainder formulas
to obtain numerical estimates of !Rn(x)I.
The fact that (x - t)n is either always positive or always negative
when t lies between a and x enables us to use (12.54) to obtain other
formulas for Rn(x) that are sometimes, but by no means always, more
easily used than (12.54) itself. The simplest and most widely used of
these formulas is obtained by the observation that the value of Rn(x)
lies between the numbers obtained by replacing the factor f(n+l)(t) by
its minimum and maximum values over the interval from a to x. Hence
the intermediate-value theorem implies existence of a number x between
a and x such that
Rn(x) =^1 t)n dt
n.
and hence
(12.51) Rn(x) =
f(n+l)(x)
(x- a)n+l
(n + 1)!
This is the Lagrange form of the remainder.
The binomial formula

(12.58) (1 + x)° = 1 + x +q(q - 1) x2 + q(q - 1) (q - 2)x3

2! 3!

+q(q-1)(q-2)(q-3)x4+q(q-1)(q-2)(q-3)(q-4)x6

4! 5!

in which the exponent q is not necessarily an integer and IxI < 1, is used

so often that many persons find it worthwhile to remember the rule by

which the coefficients can be written. To prove the formula, we let

f(x) = (1 + x)4 and calculate the derivatives

f'(x) = q(1 + x)4-1, f'(x) = q(q- 1)(1 + x)-z,

f"(x)

= q(q - 1)(q - 2)(1 + x)¢"3,

fcn)(x)= q(q - 1)(q - 2) ... (q - n + 1)(1 + x)4-n.

Since f(0) = 1, f'(0) = q, f"(0) = q(q- 1), , the series in (12.58)

is indeed the Taylor expansion of f inpowers of x. To prove that the

series converges when IxI < 1, we can apply the ratio test, but toprove

that the series converges to (1 + x)Q, we must show that lim Rn(x)= 0.

While we could (without being completely unfashionable) find that the

Lagrange form of the remainder will work when 0 < x < 1 and that

another special form will work when -1 <x < 0, we shun these things