SERIES AND LIMITS
x=a+hin the above expression. It then reads
f(x)=f(a)+(x−a)f′(a)+
(x−a)^2
2!
f′′(a)+···+
(x−a)n−^1
(n−1)!
f(n−1)(a)+Rn(x),
(4.18)
where the remainder now takes the form
Rn(x)=
(x−a)n
n!
f(n)(ξ),
andξlies in the range [a, x]. Each of the formulae (4.17), (4.18) gives us the
Taylor expansionof the function about the pointx=a. A special case occurs
whena= 0. Such Taylor expansions, aboutx= 0, are calledMaclaurin series.
Taylor’s theorem is also valid without significant modification for functions
of a complex variable (see chapter 24). The extension of Taylor’s theorem to
functions of more than one variable is given in chapter 5.
For a function to be expressible as an infinite power series we require it to be
infinitely differentiable and the remainder termRnto tend to zero asntends to
infinity, i.e. limn→∞Rn= 0. In this case the infinite power series will represent the
function within the interval of convergence of the series.
Expandf(x)=sinxas a Maclaurin series, i.e. aboutx=0.
We must first verify that sinxmay indeed be represented by an infinite power series. It is
easily shown that thenth derivative off(x)isgivenby
f(n)(x)=sin
(
x+
nπ
2
)
.
Therefore the remainder after expandingf(x)asan(n−1)th-order polynomial about
x= 0 is given by
Rn(x)=
xn
n!
sin
(
ξ+
nπ
2
)
,
whereξlies in the range [0,x]. Since the modulus of the sine term is always less than or
equal to unity, we can write|Rn(x)|<|xn|/n!. For any particular value ofx,sayx=c,
Rn(c)→0asn→∞. Hence limn→∞Rn(x) = 0, and so sinxcan be represented by an
infinite Maclaurin series.
Evaluating the function and its derivatives atx=0weobtain
f(0) = sin 0 = 0,
f′(0) = sin(π/2) = 1,
f′′(0) = sinπ=0,
f′′′(0) = sin(3π/2) =− 1 ,
and so on. Therefore, the Maclaurin series expansion of sinxis given by
sinx=x−
x^3
3!
+
x^5
5!
−···.
Note that, as expected, since sinxis an odd function, its power series expansion contains
only odd powers ofx.