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

1 Answer

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