1549901369-Elements_of_Real_Analysis__Denlinger_

8.7 Analytic Functions 521

Proof. (a) See Examples 6.5.5 and 6.5.13.

(b) One can easily verify that the Maclaurin polynomials are (Exercise 4)

x3 x5 x2n+1

T2n+i(x) = X - 3f + Sl - · · · + (-l)n-( 2 -n-+-l-)!'

To(x) = 0, and T2n(x) = T2n-1(x).

We use the ratio test to determine convergence of I: ak with

(-l)kx2k+1

ak= (2k+l)!

r lan+il r I x

2

k+3 (2k+l)!I r I x

2

I

k2..1! --;,; = k2..1! (2k + 3)! · x^2 k+1 = k2..1! (2k + 3)(2k + 2) = O.

Thus, this series converges for all x E JR; the interval of convergence is (-oo, +oo).

To prove that the series converges to sin x, we must prove that V x E JR,
Rn(x) ~ 0. Let x E JR. Using Taylor's theorem with the Lagrange form of the
f(n+l)(z)
remainder, we have Rn(x) = ( )' xn+l
n+ 1.
some z between x and 0. Thus,

IRn(x)I <

1

!x!n+l.

-(n+l)!

±sinz or ± coszxn+l

for

(n + 1)!

Now xis fixed in this inequality, so by Corollary 2.3.11, Rn(x) ~ 0.

(c) Exercise 5. D

Our proof of Example 8.7.4 (b) illustrates a common method often useful
in proving that Rn(x) ~ 0. It is summarized in the following theorem.

Theorem 8. 7.5 Suppose f and all its derivatives exist and are bounded by a
single constant on an open interval I containing c. If the Taylor series of f
about c converges on I, then it converges to f(x) for all x EI;

i.e., if :JM> 0 3 \:/x EI and \:/k EN, lf(kl(x)I ::::; M, then

00 f(k)( )

f(x) = L T(x -cl, \:/x EI.

k=O

Proof. See Theorem 6.5.14. •

Example 8.7.6 Taylor series for ex about c.