We begin by creating a table of derivatives of \(\ln x\) evaluated at \(x=1\). While this is not as straightforward as it was in the previous example, a pattern does emerge, as shown below.

\[

\begin{array}{lll}

f(x)=\ln x & \Rightarrow & f(1)=0 \\

f^{\prime}(x)=1 / x & \Rightarrow & f^{\prime}(1)=1 \\

f^{\prime \prime}(x)=-1 / x^2 & \Rightarrow & f^{\prime \prime}(1)=-1 \\

f^{\prime \prime \prime}(x)=2 / x^3 & \Rightarrow & f^{\prime \prime \prime}(1)=2 \\

f^{(4)}(x)=-6 / x^4 & \Rightarrow & f^{(4)}(1)=-6 \\

\vdots & & \vdots \\

f^{(n)}(x)= & \Rightarrow & f^{(n)}(1)= \\

\frac{(-1)^{n+1}(n-1) !}{x^n} & & (-1)^{n+1}(n-1) !

\end{array}

\]

we have,

\[

\begin{aligned}

p_n(x) &=f(c)+f^{\prime}(c)(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^2+\frac{f^{\prime \prime \prime}(c)}{3 !}(x-c)^3+\cdots+\frac{f^n(c)}{n !}(x-c)^n \\

&=0+(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\cdots+\frac{(-1)^{n+1}}{n}(x-1)^n .

\end{aligned}

\]

Note how the coefficients of the \((x-1)\) terms turn out to be "nice."