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Integrate the power series ln ( 1 + x ) = n = 1 ( −1 ) n + 1 x n n term-by-term to evaluate ln ( 1 + x ) d x .

n = 2 ( −1 ) n x n n ( n 1 )

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Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function f and a power series for f at a , is it possible that there is a different power series for f at a that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if

c 0 + c 1 x + c 2 x 2 + = d 0 + d 1 x + d 2 x 2 +

for all values x in some open interval I about zero, then the coefficients c n should equal d n for n 0 . We now state this result formally in [link] .

Uniqueness of power series

Let n = 0 c n ( x a ) n and n = 0 d n ( x a ) n be two convergent power series such that

n = 0 c n ( x a ) n = n = 0 d n ( x a ) n

for all x in an open interval containing a . Then c n = d n for all n 0 .

Proof

Let

f ( x ) = c 0 + c 1 ( x a ) + c 2 ( x a ) 2 + c 3 ( x a ) 3 + = d 0 + d 1 ( x a ) + d 2 ( x a ) 2 + d 3 ( x a ) 3 + .

Then f ( a ) = c 0 = d 0 . By [link] , we can differentiate both series term-by-term. Therefore,

f ( x ) = c 1 + 2 c 2 ( x a ) + 3 c 3 ( x a ) 2 + = d 1 + 2 d 2 ( x a ) + 3 d 3 ( x a ) 2 + ,

and thus, f ( a ) = c 1 = d 1 . Similarly,

f ( x ) = 2 c 2 + 3 · 2 c 3 ( x a ) + = 2 d 2 + 3 · 2 d 3 ( x a ) +

implies that f ( a ) = 2 c 2 = 2 d 2 , and therefore, c 2 = d 2 . More generally, for any integer n 0 , f ( n ) ( a ) = n ! c n = n ! d n , and consequently, c n = d n for all n 0 .

In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.

Key concepts

  • Given two power series n = 0 c n x n and n = 0 d n x n that converge to functions f and g on a common interval I , the sum and difference of the two series converge to f ± g , respectively, on I . In addition, for any real number b and integer m 0 , the series n = 0 b x m c n x n converges to b x m f ( x ) and the series n = 0 c n ( b x m ) n converges to f ( b x m ) whenever bx m is in the interval I .
  • Given two power series that converge on an interval ( R , R ) , the Cauchy product of the two power series converges on the interval ( R , R ) .
  • Given a power series that converges to a function f on an interval ( R , R ) , the series can be differentiated term-by-term and the resulting series converges to f on ( R , R ) . The series can also be integrated term-by-term and the resulting series converges to f ( x ) d x on ( R , R ) .

If f ( x ) = n = 0 x n n ! and g ( x ) = n = 0 ( −1 ) n x n n ! , find the power series of 1 2 ( f ( x ) + g ( x ) ) and of 1 2 ( f ( x ) g ( x ) ) .

1 2 ( f ( x ) + g ( x ) ) = n = 0 x 2 n ( 2 n ) ! and 1 2 ( f ( x ) g ( x ) ) = n = 0 x 2 n + 1 ( 2 n + 1 ) ! .

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If C ( x ) = n = 0 x 2 n ( 2 n ) ! and S ( x ) = n = 0 x 2 n + 1 ( 2 n + 1 ) ! , find the power series of C ( x ) + S ( x ) and of C ( x ) S ( x ) .

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In the following exercises, use partial fractions to find the power series of each function.

4 ( x 3 ) ( x + 1 )

4 ( x 3 ) ( x + 1 ) = 1 x 3 1 x + 1 = 1 3 ( 1 x 3 ) 1 1 ( x ) = 1 3 n = 0 ( x 3 ) n n = 0 ( −1 ) n x n = n = 0 ( ( −1 ) n + 1 1 3 n + 1 ) x n

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5 ( x 2 + 4 ) ( x 2 1 )

5 ( x 2 + 4 ) ( x 2 1 ) = 1 x 2 1 1 4 1 1 + ( x 2 ) 2 = n = 0 x 2 n 1 4 n = 0 ( −1 ) n ( x 2 ) n = n = 0 ( ( −1 ) + ( −1 ) n + 1 1 2 n + 2 ) x 2 n

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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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