Calculus of One and Several Variables

11.3 Power Series for Some Known Functions

If x → e^x has a power series, it is ∑ _k=0^∞

. This is because of Corollary 11.2.2 and what was shown earlier that the derivative of this function is itself while e⁰ = 1. Thus the question is whether this series really equals e^x. You can see that this is the case by looking at the Lagrange form of the remainder discussed earlier. In this case, it is

eξ -------xn+1 (n+ 1)!

where ξ is some number between 0 and x. Does this remainder term converge to 0 as n →∞. The answer is yes because

∑∞ eξ n+1 (n-+-1)! |x| n=1

converges by the ratio test. Indeed,

--eξ-|x|n+2 1 (n+e2ξ)!---n+1-= |x|-----→ 0 (n+1)! |x| n +2

and by the n^th term test, it follows lim_n→∞

xⁿ⁺¹ = 0. Thus

x ∞∑ xk- e = k! k=0

Similar considerations show that

∑∞ x2k+1 ∑∞ x2k sin (x) = (− 1)k(2k-+-1)!, cos(x) ≡ (− 1)k (2k)!- k=0 k=0

The details are left for you to do.