James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 11.10 Taylor and Maclaurin Series 759
In the preceding section we were able to find power series representations for a certain
restricted class of functions. Here we investigate more general problems: Which functions
have power series representations? How can we find such representations?
We start by supposing that f is any function that can be represented by a power series
1
f sxd − c 0 1 c 1 sx 2 ad 1 c 2 sx 2 ad 2 1 c 3 sx 2 ad 3 1 c 4 sx 2 ad 4 1 ∙ ∙ ∙
| x 2 a | , R
Let’s try to determine what the coefficients c n must be in terms of f. To begin, notice that
if we put x − a in Equation 1, then all terms after the first one are 0 and we get
f sad − c 0
By Theorem 11.9.2, we can differentiate the series in Equation 1 term by term:
2
f 9sxd − c 1 1 2c 2 sx 2 ad 1 3c 3 sx 2 ad 2 1 4c 4 sx 2 ad 3 1 ∙ ∙ ∙
and substitution of x − a in Equation 2 gives
| x 2 a | , R
f 9sad − c 1
Now we differentiate both sides of Equation 2 and obtain
3
f 0sxd − 2c 2 1 2 ? 3c 3 sx 2 ad 1 3 ? 4c 4 sx 2 ad 2 1 ∙ ∙ ∙
Again we put x − a in Equation 3. The result is
| x 2 a | , R
f 0sad − 2c 2
Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives
4
f -sxd − 2 ? 3c 3 1 2 ? 3 ? 4c 4 sx 2 ad 1 3 ? 4 ? 5c 5 sx 2 ad 2 1 ∙ ∙ ∙
and substitution of x − a in Equation 4 gives
| x 2 a | , R
f - sad − 2 ? 3c 3 − 3!c 3
By now you can see the pattern. If we continue to differentiate and substitute x − a, we
obtain
f snd sad − 2 ? 3 ? 4 ? ∙ ∙ ∙ ? nc n − n!c n
Solving this equation for the nth coefficient c n , we get
c n − f snd sad
n!
This formula remains valid even for n − 0 if we adopt the conventions that 0! − 1 and
f s0d − f. Thus we have proved the following theorem.
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