Question

"(b) use part (a) to find a power series for f(x) = 1 (4 + x)3 ."

Answer 1

Assuming


`f(x)=\frac1{(4+x)^3}`

Recall that for
`|x|<1`,


`\displaystyle\sum_(n\ge0)x^n=\frac1{1-x}`

Denote the above by
`s(x)`. Then


`s'(x)=\displaystyle\sum_(n\ge1)nx^(n-1)=\frac1{(1-x)^2}`

`s''(x)=\displaystyle\sum_(n\ge2)n(n-1)x^(n-2)=\frac2{(1-x)^3}`

Now,


`\frac1{(4+x)^3}=\frac1{4^3}\frac1{\left(1-\left(-\frac x4\right)\right)^3}=\frac1{2^7}\frac2{\left(1-\left(-\frac x4\right)\right)^3}`

which means we have


`f(x)=\frac1{2^7}s''(x)=\frac1{128}\displaystyle\sum_(n\ge2)n(n-1)\left(-\frac x4\right)^(n-2)`

which is valid for
`\left|-\frac x4\right|<1`, or
`|x|<4`.