SERIES AND LIMITS
value ofξthat satisfies the expression forRn(x) is not known, an upper limit on
the error may be found by differentiatingRn(x)withrespecttoξand equating
the derivative to zero in the usual way for finding maxima.
Expandf(x)=cosxas a Taylor series aboutx=0and find the error associated with
using the approximation to evaluatecos(0.5)if only the first two non-vanishing terms are
taken. (Note that the Taylor expansions of trigonometric functions are only valid for angles
measured in radians.)Evaluating the function and its derivatives atx= 0, we find
f(0) = cos 0 = 1,
f′(0) =−sin 0 = 0,
f′′(0) =−cos 0 =−^1 ,
f′′′(0) = sin 0 = 0.So, for small|x|, we find from (4.18)
cosx≈ 1 −x^2
2.
Note that since cosxis an even function, its power series expansion contains only even
powers ofx. Therefore, in order to estimate the error in this approximation, we must
consider the term inx^4 , which is the next in the series. The required derivative isf(4)(x)
and this is (by chance) equal to cosx. Thus, adding in the remainder termR 4 (x), we find
cosx=1−x^2
2+
x^4
4!cosξ,whereξlies in the range [0,x]. Thus, the maximum possible error isx^4 /4!, since cosξ
cannot exceed unity. Ifx=0.5, taking just the first two terms yields cos(0.5)≈ 0 .875 with
a predicted error of less than 0.002 60. In fact cos(0.5) = 0.877 58 to 5 decimal places. Thus,
to this accuracy, the true error is 0.002 58, an error of about 0.3%.
4.6.3 Standard Maclaurin seriesIt is often useful to have a readily available table of Maclaurin series for standard
elementary functions, and therefore these are listed below.
sinx=x−x^3
3!+x^5
5!−x^7
7!+··· for−∞<x<∞,cosx=1−x^2
2!+x^4
4!−x^6
6!+··· for−∞<x<∞,tan−^1 x=x−x^3
3+x^5
5−x^7
7+··· for − 1 <x< 1 ,ex=1+x+x^2
2!+x^3
3!+x^4
4!+··· for −∞<x<∞,ln(1 +x)=x−x^2
2+x^3
3−x^4
4+··· for − 1 <x≤ 1 ,(1 +x)n=1+nx+n(n−1)x^2
2!+n(n−1)(n−2)x^3
3!+··· for−∞<x<∞.