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

## $f ' = \frac{1}{x} = {x}^{-} 1$ $f ' ' = - \frac{1}{x} ^ 2 = \left(- 1\right) {x}^{-} 2$ $f ' ' ' = \frac{2}{x} ^ 3 = \left(- 1\right) \left(- 2\right) {x}^{-} 3$ ${f}^{4} = \left(- 1\right) \left(- 2\right) \left(- 3\right) {x}^{-} 4. . .$ must find ${C}_{n}$ before finding $f \left(x\right)$ expansion

May 24, 2018

$f \left(x\right) = \ln 3 + \setminus {\sum}_{n = 1}^{\setminus} \infty {\left(- 1\right)}^{n + 1} \frac{1}{n {3}^{n}} {\left(x - 3\right)}^{n}$

#### Explanation:

$f ' \left(3\right) = {\left(3\right)}^{-} 1$

$f ' ' \left(3\right) = \left(- 1\right) {3}^{-} 2$

$f ' ' ' \left(3\right) = \left(- 1\right) \left(- 2\right) {\left(3\right)}^{-} 3$

${f}^{4} \left(3\right) = \left(- 1\right) \left(- 2\right) \left(- 3\right) {\left(3\right)}^{-} 4$

In general, for $n \ge 1$ 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 $\setminus \textcolor{red}{f \left(3\right) = \ln 3}$ and so $\setminus \textcolor{red}{{C}_{0} = \ln \left(3\right)}$

$\setminus \Rightarrow f \left(x\right) = {\sum}_{n = 0}^{\infty} {C}_{n} {\left(x - 3\right)}^{n}$

$q \quad q \quad = \setminus \textcolor{red}{\ln \left(3\right)} + \setminus {\sum}_{n = 1}^{\setminus} \infty {\left(- 1\right)}^{n + 1} \frac{1}{n {3}^{n}} {\left(x - 3\right)}^{n}$