08.07.2017 Views

Calculus 2nd Edition Rogawski

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

SECTION 11.7 Taylor Series 601<br />

EXAMPLE 5 Substitution Find the Maclaurin series for e −x2 .<br />

Solution Substitute −x 2 in the Maclaurin series for e x .<br />

e −x2 =<br />

∞∑<br />

n=0<br />

(−x 2 n<br />

)<br />

=<br />

n!<br />

∞∑ (−1) n x 2n<br />

n=0<br />

n!<br />

= 1 − x 2 + x4<br />

2! − x6<br />

3! + x8<br />

4! − ··· 2<br />

The Taylor expansion of e x is valid for all x, so this expansion is also valid for all x.<br />

EXAMPLE 6 Integration Find the Maclaurin series for f (x) = ln(1 + x).<br />

Solution We integrate the geometric series with common ratio −x (valid for |x| < 1):<br />

1<br />

1 + x = 1 − x + x2 − x 3 +···<br />

∫<br />

dx<br />

ln(1 + x) =<br />

1 + x = x − x2<br />

2 + x3<br />

3 − x4<br />

∞ 4 +···= ∑<br />

n=1<br />

n−1 xn<br />

(−1)<br />

The constant of integration on the right is zero because ln(1 + x) = 0 for x = 0. This<br />

expansion is valid for |x| < 1. It also holds for x = 1 (see Exercise 84).<br />

In many cases, there is no convenient general formula for the Taylor coefficients, but<br />

we can still compute as many coefficients as desired.<br />

n<br />

1<br />

y<br />

EXAMPLE 7 Multiplying Taylor Series Write out the terms up to degree five in the<br />

Maclaurin series for f (x) = e x cos x.<br />

Solution We multiply the fifth-order Taylor polynomials of e x and cos x together, dropping<br />

the terms of degree greater than 5:<br />

)<br />

(1 + x +<br />

)(1 x2<br />

2 + x3<br />

6 + x4<br />

24 + x5<br />

− x2<br />

120 2 + x4<br />

24<br />

Distributing the term on the left (and ignoring terms of degree greater than 5), we obtain<br />

)<br />

)( (1 + x + x2<br />

2 + x3<br />

6 + x4<br />

24 + x5<br />

−<br />

(1 + x + x2<br />

120<br />

2 + x3 x<br />

2<br />

) ( x<br />

4<br />

)<br />

+ (1 + x)<br />

6 2<br />

24<br />

= 1 + x − x3<br />

3 − x4<br />

6 − x5<br />

} {{<br />

30<br />

}<br />

Retain terms of degree ≤ 5<br />

x<br />

321<br />

We conclude that the fifth Maclaurin polynomial for f (x) = e x cos x is<br />

−1<br />

FIGURE 1 Graph of T 12 (x) for the power<br />

series expansion of the antiderivative<br />

∫ x<br />

F (x) = sin(t 2 )dt<br />

0<br />

T 5 (x) = 1 + x − x3<br />

3 − x4<br />

6 − x5<br />

30<br />

In the next example, we express the definite integral of sin(x 2 ) as an infinite series.<br />

This is useful because the integral cannot be evaluated explicitly. Figure 1 shows the graph<br />

of the Taylor polynomial T 12 (x) of the Taylor series expansion of the antiderivative.<br />

EXAMPLE 8 Let J =<br />

∫ 1<br />

0<br />

sin(x 2 )dx.<br />

(a) Express J as an infinite series.<br />

(b) Determine J to within an error less than 10 −4 .

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!