pow_series_for_funct - (12.9 Power Series for Functions Recall the formula for geometric series n=0 arn = a provided |r| < 1(We really 1-r a mean

# pow_series_for_funct - (12.9 Power Series for Functions...

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(12.9) Power Series for FunctionsRecall the formula for geometric series:n=0arn=a1-rprovided|r|<1.(We reallymean,n=0arnconverges toa1-rprovided|r|<1.) So if we leta= 1 andr=x, we haven=0xn= 1 +x+x2+x3+· · ·=11-xprovided|x|<1, or-1< x <1. So we thinkoff(x) =n=0xnas a function ofx.We therefore know is that it is really the functionf(x) =11-x, but only for-1< x <1 (forxoutside that interval, the functionf(x) isundefined).We can use this series to find power series for other functions.ExampleFind a power series for the functionf(x) =2xx-3.We start with11-x= 1 +x+x2+x3+· · ·. Trick: substitutex3forx. We have11-x3= 1 +x3+x32+x33+· · ·or33-x= 1+x3+x29+x327+· · ·. We adjust this a little (by dividing by 3 then multiplyingby-1):1x-3=-13-x9-x227-x381- · · ·.This is1x-3=n=0-xn3n+1. Now multiply by 2xto get the desired result:2xx-3=-2x3-2x29-2x327-2x481- · · ·=n=0-2xn+13n+1=n=0-2xn3n.So we are done, but we should determine the interval of convergence. The geometric series33-x= 1 +x3+x29+x327+· · ·converges ifx3<1 or-3< x <3. Multiplying by 2xdoesn’t affect convergence. So we conclude that