Lecture 2711.7 Power Series 11.8 Differentiation andIntegration of Power SeriesJiwen He1 Power Series1.1 Geometric Series and VariationsGeometric SeriesGeometric Series:P∞k=0xk∞Xk=0xk= 1 + x + x2+ x3+ · · ·11 − x, if |x| < 1,diverges, if |x| ≥ 1.Power SeriesDefine a function f on the interval (−1, 1)f(x) =∞Xk=0xk= 1 + x + x2+ x3+ · · · =11 − xfor |x| < 1As the Limitf can be viewed as the limit of a sequence of polynomials:f(x) = limn→∞pn(x),where pn(x) = 1 + x + x2+ x3+ · · · + xn.Variations on the Geometric Series (I)Closed forms for many power series can be found by relating the series to thegeometric seriesExamples 1.f(x) =∞Xk=0(−1)kxk= 1 − x + x2− x3+ · · ·=∞Xk=0(−x)k=11 − (−x)=11 + x, for |x| < 1.f(x) =∞Xk=02kxk+2= x2+ 2x3+ 4x4+ 8x5+ · · ·= x2∞Xk=0(2x)k=x21 − 2xfor |2x| < 1.1Variations on the Geometric Series (II)Closed forms for many power series can be found by relating the series to thegeometric seriesExamples 2.f(x) =∞Xk=0(−1)kx2k= 1 − x2+ x4− x6+ · · ·=∞Xk=0(−x2)k=11 − (−x2)=11 + x2, for |x| < 1.f(x) =∞Xk=0x2k+13k= x +13x3+19x5+127x7+ · · ·= x∞Xk=0x23k=x1 − (x2/3)=3x3 − x2for |x2/3| < 1.1.2 Radius of Conve rg enceRadius of ConvergenceThere are exactly three possibilities for a power series:Pakxk.Radius of Convergence: Ratio Test (I)The radius of convergence of a power series can usually be found by applyingthe ratio test. In some cases the root test is easier.2Example 3.f(x) =∞Xk=1k2xk= x + 4x2+ 9x3+ · · ·Ratio Test :ak+1ak=(k + 1)2xk+1k2xk=(k + 1)2k2|x| → |x| as k → ∞Thus the series conve rges absolutely when |x| < 1 and diverges w hen |x| > 1.Radius of Convergence: Ratio Test (II)The radius of convergence of a power series can usually be found by applyingthe ratio test. In some cases the root test is easier.Example 4.f(x) =∞Xk=1(−1)kk!xk= 1 − x +12x2−16x3+ · · · = e−xRatio Test :ak+1ak=xk+1/(k + 1)!xk/k!=k!(k + 1)!xk+1xk=1k + 1|x| → 0 < 1 for all xThus the series conve rges absolutely for all x.Radius of Convergence: Ratio Test (III)The radius of convergence of a power series can usually be found by applyingthe ratio test. In some cases the root test is easier.Example 5.f(x) =∞Xk=1k + 1kk2xk= 2x + (3/2)4x2+ (4/3)9x3+ · · ·Ratio Test :(|ak|)1k= k + 1kk2|x|k!1k=k + 1kk|x|=1 +1kk|x| → e|x| < 1 if |x| < 1/eThus the series converges absolutely when |x| < 1/e and diverges when |x| >1/e.3Interval of ConvergenceFor a series with radius of convergence r, the interval of convergence can be[−r, r], (−r, r], [−r, r), or (−r, r).Example 6. In general, the behavior of a power series at −r and at r is notpredictable. For example, the seriesXxk,X(−1)kkxk,X1kxk,X1k2xkall have radius of convergence 1, but the first series converges only on (−1, 1),the second converges on (−1, 1], but the third converges on [−1, 1), the fourthon [−1, 1].Interval of ConvergenceExample 7.f(x) =∞Xk=1(−1)k−1kxkRatio Test :ak+1ak=xk+1/(k + 1)xk/k=kk + 1|x| → |x|Thus the series converges absolutely when |x| < 1 and diverges when |x| > 1.So the radius of convergence is 1x = −1 :∞Xk=1(−1)k−1k(−1)k=∞Xk=1−1kdivergesx = 1 :∞Xk=1(−1)k−1k(1)k=∞Xk=1(−1)k−1kconverges conditionallyThe interval of convergence is (−1, 1].2 Differentiation and Integration2.1 Differentiatio n and IntegrationDifferentiation and IntegrationTheoremLet f(x) =Pakxkbe a power series with a nonzero radius of convergence r.Thenf0(x) =Xakk xk−1for |x| < rZf(x) dx =Xakk + 1xk+1+ C for |x| < r4Geometric series:11 − x=∞Xk=0xkfor |x| < 1Differentiation:1(1 − x)2=∞Xk=0k xk−1∞Xk=0(k + 1) xkfor |x| < 1Integration: − ln(1 − x) =∞Xk=01k + 1xk+1=∞Xk=11kxkfor |x| < 12.2 ExamplesPower Series Expansion of ln(1 + x)Note:ddxln(1 + x) =11 + x=∞Xk=0(−1)kxkfor |x| < 1Integration: ln(1 + x) =∞Xk=0(−1)kk + 1xk+1(+C = 0)=∞Xk=1(−1)kkxk= x −12x2+13x3−14x4+ · · ·The interval of convergence is (−1, 1]. At x = 1,ln 2 =∞Xk=1(−1)kk= 1 −12+13−14+ · · ·Power Series Expansion of tan−1xNote:ddxtan−1x =11 + x2=∞Xk=0(−1)kx2kfor |x| < 1Integration: tan−1x =∞Xk=0(−1)k2k + 1x2k+1(+C = 0)= x −13x3+15x5−17x7+ · · ·The interval of convergence is (−1, 1]. At x = 1,tan−11 =∞Xk=1(−1)k2k + 1= 1 −13+15−17+ · · · =π4Outline5Contents1 Power Series 11.1 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . .11.2 Radius of Convergence . . . . . . . . . . . . . . . . . . . . . . . . 22 Diff and Integ 42.1 Diff and Integ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .