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More Challenging Problems
1. Find a power series for 1/(x2 – 1)
Answer
1. The series is -1 – x2 – x4 – x6 – x8-…
Solution
1. First, note that 1/(x2 – 1)= -1/(1 – x2). Now apply the series of Example 3 to conclude 1/(x2 – 1)= -(1 + x2 + x4 + x6 + x8 + …)
2. From the power series for 1/(x + 1) and for 1/(x – 1), use partial fractions to find a power series for 1/(x2 – 1). What assumption are you making in this approach?
Answer
1. Adding the series for (1/2)(1/(x+1)) and for (1/2)(1/(x-1)), and rearranging the terms, we obtain the series from problem 1.
Solution
1. From Example 1 we know a series for 1/(1 + x) is 1 – x + x2 – x3 + x4– … The series for 1/(x – 1)= -(1/(1 – x)) is -1 – x – x2 – x3 – x4 – … Next, note that using a partial fractions expansion we have 1/(x2 – 1) = (1/2)(1/(x-1)) – (1/2)(1/(x+1)) Assuming we can add the series and rearrange them in any way we wish (This will be addressed in the next section), we find 1/(x2 – 1)= (1/2)(-1 – x – x2 – x3 – x4 – …) – (1/2)(1 – x + x2 – x3 + x4 – …) = -1 – x2 – x4; – x2 – … agreeing with the solution to problem 1.
3. Find a power series for 1/(x2 + 5x + 6). On what interval is this series valid?
Answer
1. The series is (1/2 – 1/3) – (1/4 – 1/9)x + (1/8 – 1/27)x2 – (1/16 – 1/81)x3 + …This series is valid for |x| < 2.
Solution
1. Using the method of partial fractions, we find: 1/(x2 + 5x + 6) = 1/((x + 2)(x + 3)) = 1/(x + 2) – 1/(x + 3). Using the method of Example 4 we find these power series: 1/(2 + x) = 1/(2(1 + x/2)) = (1/2)(1 – x/2 + x2/4 – x3/8 + x4/16 – …) 1/(3 + x) = 1/(3(1 + x/3)) = (1/3) (1 – x/3 + x2/9 – x3/27 + x4/81 – …) Subtracting these we find the series for 1/(x2 + 5x + 6) is: (1/2 – 1/3) – (1/4 – 1/9)x + (1/8 – 1/27)x2 – (1/16 – 1/81)x3 + … The series for 1/(2 + x) is valid for: |x/2| < 1, that is, |x| < 2. The series for 1/(3 + x) is valid for |x/3| < 1, that is, |x| < 3. Both expansions are valid for the intersection of these intervals, |x| < 2.
4. Find a power series for (2x + 5)/(x2 + 5x + 6). On what interval is this series valid?
Answer
1. The series is (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x2 – (1/16 + 1/81)x3 + …This expansion is valid for |x| < 2.
Solution
1. Using the method of partial fractions, we find (2x + 5)/(x2 + 5x + 6) = (2x + 5)/((x + 2)(x + 3))= 1/(x + 2) + 1/(x + 3) Recalling the series for 1/(2 + x) and 1/(3 + x) from problem 3, we find the series for (2x + 5)/(x2 + 5x + 6) is the sum of these (1/2 + 1/3) – (1/4 + 1/9)x + (1/8 + 1/27)x2 – (1/16 + 1/81)x3 + … As in problem 3, this expansion is valid for |x| < 2.
5. Using -1/x = 1/(1 – (x+1)), find a power series for -1/x. On what interval is this series valid?
Answer
1. The series is 1 + (x + 1) + (x + 1)2 + (x + 1)3 + (x + 1)4 + … The series is valid for -2 < x < 0.
Solution
1. The expansion for 1/(1 – (x + 1)) is 1 + (x + 1) + (x + 1)2 + (x + 1)3 + (x + 1)4 + … This converges for |x + 1| < 1. That is, -1 < x + 1 < 1, or -2 < x < 0