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Lecture 2711.7 Power Series 11.8 Differentiation and
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![Jiwen He 11 Geometric Series and Variations - UH Jiwen He 11 Geometric Series and Variations - UH](https://pdfprof.com/Listes/17/24888-17lecture27_handout.pdf.pdf.jpg)
Integration of Power Series
Jiwen He
1 Power Series
1.1 Geometric Series and Variations
Geometric Series
Geometric Series:?∞
k=0xk k=0x k= 1 +x+x2+x3+···? ?11-x,if|x|<1, diverges, if|x| ≥1.Power Series
Define a functionfon the interval (-1,1)
f(x) =∞? k=0x k= 1 +x+x2+x3+···=11-xfor|x|<1As the Limit
fcan be viewed as the limit of a sequence of polynomials: f(x) = limn→∞pn(x), wherepn(x) = 1 +x+x2+x3+···+xn.Variations on the Geometric Series (I)
Closed formsfor many power series can be found by relating the series to the geometric seriesExamples1.f(x) =∞? k=0(-1)kxk= 1-x+x2-x3+··· k=0(-x)k=11-(-x)=11 +x,for|x|<1. f(x) =∞? k=02 kxk+2=x2+ 2x3+ 4x4+ 8x5+··· =x2∞? k=0(2x)k=x21-2xfor|2x|<1.1Variations on the Geometric Series (II)
Closed formsfor many power series can be found by relating the series to the geometric seriesExamples2.f(x) =∞? k=0(-1)kx2k= 1-x2+x4-x6+··· k=0(-x2)k=11-(-x2)=11 +x2,for|x|<1. f(x) =∞? k=0x 2k+13 k=x+13 x3+19 x5+127 x7+··· =x∞? k=0? x23 k =x1-(x2/3)=3x3-x2for|x2/3|<1.1.2 Radius of Convergence
Radius of Convergence
There are exactly three possibilities for a power series: ?akxk.Radius of Convergence: Ratio Test (I) The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.2Example3.f(x) =∞?
k=1k2xk=x+ 4x2+ 9x3+···
Ratio Test :
????a k+1a k? ???=????(k+ 1)2xk+1k 2xk? (k+ 1)2k2|x| → |x|ask→ ∞
Thus the series converges absolutely when|x|<1 and diverges when|x|>1.Radius of Convergence: Ratio Test (II)
The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.Example4.f(x) =∞? k=1(-1)kk!xk= 1-x+12 x2-16 x3+···=e-xRatio Test :
????a k+1a k? ???=????xk+1/(k+ 1)!x k/k!? k!(k+ 1)!? ???xk+1x k? ???=1k+ 1|x| →0<1 for allxThus the series converges absolutely for allx.
Radius of Convergence: Ratio Test (III)
The radius of convergence of a power series can usually be found by applying the ratio test. In some cases the root test is easier.Example5.f(x) =∞? k=1? k+ 1k k2 x k= 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/e Thus the series converges absolutely when|x|<1/eand diverges when|x|> 1/e.3Interval of Convergence
For a series with radius of convergencer, the interval of convergence can be[-r,r], (-r,r], [-r,r), or (-r,r).Example6.In general,the behavior of a power series at-rand atris not
predictable. For example, the series ?xk,?(-1)kk xk,?1k xk,?1k 2xk all 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 fourth on [-1,1].