Question

Find the Taylor series expansion formula of `f(x)=\lnx` at `a=3`?

`f'=1/x=x^-1`
`f''=-1/x^2=(-1)x^-2`
`f'''=2/x^3=(-1)(-2)x^-3`
`f^4=(-1)(-2)(-3)x^-4...`

must find `C_n` before finding `f(x)` expansion

Answer 1

`f(x)=ln3+\sum_(n=1)^\infty(-1)^(n+1)1/(n3^n)(x-3)^n`

Explanation:

`f'(3)=(3)^-1`

`f''(3)=(-1)3^-2`

`f'''(3)=(-1)(-2)(3)^-3`

`f^4(3)=(-1)(-2)(-3)(3)^-4`

In general, for `n>=1` we have

`f^n(3)=(-1)^n(n-1)!3^(-n)=((-1)^n(n-1)!)/3^n`

`C_n=f^(n)(3)/(n!)=(1/(n!))*((-1)^(n-1)(n-1)!)/3^n=(-1)^(n-1)1/(3^n(n))`

We also have `\color(red)(f(3) = ln 3)` and so `\color(red)(C_0 = ln(3))`

`\rArrf(x)= sum_{n=0}^oo C_n(x-3)^n`

`qquad qquad = \color(red)(ln(3))+\sum_(n=1)^\infty(-1)^(n+1)1/(n3^n)(x-3)^n`