Question

The function f(x) = \ln(3 - x) is represented as a power series \displaystyle f(x) = \sum_{n=0}^\infty c_n x^n . Find the first few coefficients in the power series.

Answer 1

The coefficients
`c_n` are the same ones used in the Taylor series for
`f(x)` centered at
`x=0`. So

`c_n=(f^((n))(0))/(n!)`

We have

`f(x)=\ln(3-x)\implies f(0)=\boxed{c_0=\ln3}`

`f'(x)=\frac1{x-3}\implies f'(0)=\boxed{c_1=-\frac13}`

`f''(x)=-\frac1{(x-3)^2}\implies f''(0)=-\frac19\implies\boxed{c_2=-\frac1{18}}`

`f'''(x)=\frac2{(x-3)^3}\implies f'''(0)=-\frac2{27}\implies\boxed{c_3=-\frac1{81}}`

For
`n\ge1`, we have the pattern

`f^((n))(x)=(-1)^(n-1)((n-1)!)/((x-3)^n)\implies f^((n))(0)=-((n-1)!)/(3^n)`

so that

`c_n=-((n-1)!)/(n!3^n)=-\frac1{n3^n}`