Homework 4.7

Find a power series representation for the function; find the interval of convergence. (Give your power series representation centered at x = 0.)


f(x)=41+xf(x) = \frac {4}{1+x}


∑n=0∞\sum_{n=0}^\infty ( _________ )  provided ∣x∣<|x| < ___________


Write the function as a geometric series:


f(x)=41+x=4⋅11−(−x)f(x) = \frac {4}{1+x} = 4 \cdot \frac {1}{1-(-x)}


For ∣−x∣<1|-x| < 1 (i.e. ∣x∣<1|x| < 1) we have the geometric series expansion


11−r= ∑n=0∞(r)n\frac 1{1- r} = \sum_{n=0}^\infty ( r ) ^n


11−(−x)= ∑n=0∞(−x)n= ∑n=0∞(−1)nxn\frac 1{1-(-x)} = \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty (-1)^n x^n


Multiplying by the original 4 gives 


41+x=∑n=0∞4(−1)nxn\frac 4{1+x} = \sum_{n=0}^\infty 4(-1)^nx^n;      ∣x∣<1|x| < 1


Interval of convergence is (−1,1)(-1,1)

At x=1x=1 the series is 4∑n=0∞(−1)n4\sum_{n=0}^\infty (-1)^n which does not converge (terms do not tend to 0), and at x=−1x=-1 the series is 4∑n=0∞14\sum_{n=0}^\infty 1 which also diverges. Hence endpoints are excluded.

Find a power series representation for the function; find the interval of convergence. (Give your power series representation centered at x=0x=0.)


f(x)=81+x2f(x) = \frac 8 {1 + x²}


∑n=0∞\sum_{n=0}^\infty (___________)    provided  ∣x∣<|x| <  ____________


f(x)=81+x2=8⋅11+x2f(x) = \frac 8 {1 + x²} = 8 \cdot \frac 1{1+x²} 


11−(−x2)=∑n=0∞(−x2)n=∑n=0∞(−1)n(x2)n\frac 1 {1 - (-x²)} = \sum_{n=0}^\infty (-x²)^n = \sum_{n=0}^\infty (-1)^n (x²)^n


∑n=0∞8(−1)n(x2)n\sum_{n=0}^\infty 8 (-1)^n (x²)^n;    ∣x∣<1|x| < 1

Find a power series representation for the function; find the radius of convergence, R. (Give your power series representation centered at x = 0.)


f(x)=ln⁡(1−5x)f(x) = \ln(1-5x)


∑n=0∞\sum_{n=0}^\infty (__________)   provided R=R= __________


ln⁡(1−5x)=∑n=0∞−5n+1n+1xn+1\ln(1-5x) = \sum_{n=0}^\infty -\frac {5^{n+1}}{n+1}x^{n+1}       R=15R = \frac 15


f(x)=11−6xf(x) = \frac {1}{1-6x}


∑n=0∞\sum_{n=0}^\infty (__________)  provided  ∣x∣<|x| < ___________


11−6x⇒11−(6x)⇒r=6x\frac 1{1-6x} \Rightarrow \frac 1{1 - (6x)} \Rightarrow r = 6x 


∑n=0∞(6x)n\sum_{n=0}^\infty (6x)^n 


∣6x∣<1⇒∣x∣<16|6x| < 1 \Rightarrow |x| < \frac 16

Find a power series representation for the function; find the interval of convergence. (Give your power series representation centered at x=0x=0.)


f(x)=x1−x5f(x) = \frac x{1-x^5}


∑n=0∞\sum_{n=0}^\infty (___________)  provided ∣x∣<|x| < ___________


x⋅11−x5⇒r=x5x \cdot \frac 1{1-x^5} \Rightarrow r = x^5


∑n=0∞x(x5)n\sum_{n=0}^\infty x (x^5)^n