Question

Find the power series representation for the function and determine the interval of convergence

f(x)=x/((x^2)+16)

Answer 1

i'll build my solution from this \[\large \frac{ 1 }{ 1+x }=\sum_{n=0}^{+ \infty}\left( -1 \right)^n x^n\]

Answer 2

\[\large \frac{ 1 }{ 16+x^2 }=\frac{ 1 }{ 16 }\frac{ 1 }{ 1+\left( \frac{ x^2 }{ 16 } \right) }\] so use (x^2)/16 instead of x in the series \[\large \frac{ 1 }{ 16+x^2 }=\frac{ 1 }{ 16 }\sum_{n=0}^{+ \infty} \left( -1 \right)^n \left( \frac{ x^2 }{ 16 } \right)^n=\frac{ 1 }{ 16 }\sum_{n=0}^{+ \infty}\left( -1 \right)^n \frac{ x^{2n} }{ 4^{2n} }\]

Answer 3

the radius of convergence is \[\large \left| \frac{ x^2 }{ 4^2 } \right| <1\] or \[\large -4<x<4\]

Answer 4

finally, \[\large \frac{ x }{ 16+x^2 }=x \left( \frac{ 1 }{ 16+x^2 } \right)=\sum_{n=0}^{+ \infty}\left( -1 \right)^n \frac{ x^{2n+1} }{ 4^{2n} }\]