Whenx=-11 and 1, the series diverges by divergence test, so the interval of convergence (-11,1).Example 8.5.8.Find the interval and radius of convergence for the following series:∞Xn=1(x-2)nln(n+ 4)Using the ratio test, we havelimn→∞(x-2)n+1ln(n+ 5)·ln(n+ 4)(x-2)n=|x-2|converges when this limit is less than 1, and so|x-2|<1 tells us that the radius of convergence is 1.The open interval of convergence is thus (2-1,2 + 1) = (1,3). Checking the endpoints, whenx= 1the series converges by alternating series test, and whenx= 3 the series diverges by comparing to theseries∑1n+4. Thus, the interval of convergence is [1,3).95
Section 8.5 ExercisesFind the radius and interval of convergence for the given power series.1.∞Xn=0xnn!2.∞Xn=0(x-2)nn2+ 13.∞Xn=1xnn3n4.∞Xn=1n!(2x-1)n5.∞Xn=1xn1·3·5· · ·(2n-1)96
8.6Representing Functions as Power SeriesWe know that, when|x|<1, we have11-x=∞Xn=0xnWhat this tells us is that the functionf(x) =11-xcan be approximated to any degree of accuracy (forvalues ofxwith-1< x <1 anyway) by just looking at polynomials 1 +x+x2+· · ·+xnfor as largenas we require. This is fantastic as polynomials are well-studied and extremely easy to evaluate.Example 8.6.1.Expressf(x) =11+8x3as a power series and find its interval of convergence.Notice that we havef(x) =11 + 8x3=11-(-8x3)=∞Xn=0(-8x3)n=∞Xn=0(-1)n(8x3)n.This converges when|8x3|<1, i.e., when|x|<12. The interval of convergence is thus(-12,12).Theorem 8.6.2.If the power series∞Xn=0cn(x-a)nhas radius of convergenceR >0, then the functionf(x) =∞Xn=0cn(x-a)nis differentiable on the interval(a-R, a+R)and1.ddxf(x) =∞Xn=0ddx[cn(x-a)n] =∞Xn=1ncn(x-a)n-12.Zf(x)dx=∞Xn=0Zcn(x-a)ndx=C+∞Xn=0cnn+ 1(x-a)n+1Example 8.6.3.Given the power series forf(x) =11-x, use differentiation to expressg(x) =1(1-x)2as a power series. Find the radius of convergence of this new power series.We notice thatf0(x) =1(1-x)2=g(x),and sog(x) =ddxf(x) =∞Xn=0ddx[xn] =∞Xn=1nxn-1.Using the Ratio test, we havelimn→∞an+1an= limn→∞(n+ 1)xnnxn-1=|x|,which converges when|x|<1, so the radius of convergence is 1, which is exactly the same as theradius of convergence for the series representation off(x).97
Proposition 8.6.4.Given the series with radius of convergenceR >0and the functionf(x)definedin the premise for Theorem 8.6.2, the new series obtained fromddx[f(x)]andRf(x)dxboth have radiusof convergenceR.Example 8.6.5.Given the power series forf(x) =11 +x, find a power series representation forg(x) = ln(1 +x).Notice thatg0(x) =11 +x=f(x),so we haveg(x) =Zf(x)dx=Z11-(-x)dx=Z"∞Xn=0(-1)nxn#dx=∞Xn=0Z(-1)nxndx=C+∞Xn=0(-1)nn+ 1xn+1.To determineC, we setx= 0 and get thatg(0) = 0 =C, so our power series representation is justg(x) =∞Xn=0(-1)nn+ 1xn+1=∞Xn=1(-1)n-1nxn.