power-series-notes

power-series-notes - 112 3.2. C HAPTER 3 P OWER S ERIES M...

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112 CHAPTER 3P OWER SERIES 3.2. M ANIPULATION AND D ERIVATION We concluded the previous section by noting that power seriescanbedifferentiatedand integrated term-by-term (within their radii of convergence). This is quite a strong prop- erty of power series (which does not hold for series in general). We begin this section by showing that term-by-term differentiation and integrationcanbeusedto fnd power series. Our starting point will always (in this section, at least) beageometr icser iesofsome kind, the simplest example of such being n 0 x n 1 1 x for x 1 , which we know from our study of geometric series in Section 2.3. By differentiating (term- by-term) both sides of this equation, we obtain our Frst new power series: n 0 nx n 1 d dx 1 1 x 1 1 x 2 for x 1 . This is one of the three basic forms of power series we derive inthissection.Theothertwo are given in Examples 1 and 2. Example 1. ±ind the power series centered at x 0 for ln 1 x and its radius of conver- gence. Solution. Recall that ln 1 x is the antiderivative of 1 1 x : 1 1 x dx ln 1 x . ±urthermore, we can express 1 1 x as a geometric power series (for x 1 ): 1 1 x 1 1 x n 0 x n n 0 1 n x n . Therefore, all we have to do to get the power series for ln 1 x is integrate this series term-by-term, ln 1 x 1 1 x dx n 0 1 n x n n 0 1 n x n 1 n 1 C.
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SECTION 3.2 MANIPULATION AND DERIVATION 113 But what is C , the constant of integration? To Fnd C we substitute x 0 into both sides. We know that ln 1 0 ln 1 0 ,so C 0 .Th isg ives n 0 1 n x n 1 n 1 . The geometric series we integrated had radius of convergence R 1 ,sotherad iuso f convergence of this series for ln 1 x is also R 1 . The power series for ln 1 x that we found in Example 1 is known as the Mercator series , after Nicholas Mercator (1620–1687). Note that by substituting x 1 into this series, we obtain ln 2 n 0 1 n n 1 n 1 1 n 1 n ? This seems to indicate that the sum of the alternating harmonic series is ln 2 .H ow e v e r , there is a problem with this line of reasoning: term-by-term integration is only guaranteed to work inside the interval of convergence, and x 1 is an endpoint of the interval of convergence for the Mercator series. Nevertheless, this computation can be made rigorous, as shown by Abel, see Exercise 33. (Another proof of this result using Euler’s constant γ is given in Exercises 46 and 47 of Section 2.4.) We move on to another example of using integration to derive a power series. Example 2. ±ind the power series centered at x 0 for arctan x and its radius of conver- gence. Solution. ±or this we need to recall that arctan x 1 1 x 2 dx. Again, we can write 1 1 x 2 as a geometric power series (for x 1 ): 1 1 x 2 1 1 x 2 n 0 x 2 n n 0 1 n x 2 n . Now we integrate this series term-by-term: arctan x 1 1 x 2 dx n 0 1 n x 2 n n 0 1 n x 2 n 1 2 n 1 C.
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114 CHAPTER 3P OWER SERIES Finally, we substitute x 0 into both sides of this equation to see that C 0 ,giving arctan x n 0 1 n x 2 n 1 2 n 1 for x 1 .
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power-series-notes - 112 3.2. C HAPTER 3 P OWER S ERIES M...

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