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

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#