Question

In Problems 17 and 18 use an appropriate series in (2) to find the Taylor series of the given function centered at the indicated value of. Write your answer in summation notation.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{n}\mathit{x}\mathbf{;}\mathit{a}\mathbf{=}\mathbf{2}$

Short answer

$\ln x=\ln 2+\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n{2}^n}(x-2{)}^n$

Step-by-step solution

Definition

Taylor seriesof a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point.

Simplify

Convert the given function in known series expansion.

Consider$f\left(x\right)=\ln x;a=2$

$$\begin{gathered} \ln x=\ln 2\left(1+\frac{x-2}{2}\right) \\ =\ln 2+\ln \left(1+\frac{x-2}{2}\right) \end{gathered}$$

Thus,$f\left(x\right)=\ln 2+\ln \left(1+\frac{x-2}{2}\right)$

Taylor expansion

Maclaurin series of $\ln (1+x)$is,

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

Replace$x$ with$\frac{x-2}{2}$ in the above equation.

$\ln \left(1+\frac{x-2}{2}\right)=\left(\frac{x-2}{2}\right)-\frac{{\left(\frac{x-2}{2}\right)}^2}{2}+\frac{{\left(\frac{x-2}{2}\right)}^3}{3}-\frac{{\left(\frac{x-2}{2}\right)}^4}{4}+\cdots$

Hence, Taylor series representation of $f\left(x\right)=\ln x$function cantered at$a=2$ is

$$\begin{gathered} f\left(x\right)=\ln 2+\ln \left(1+\frac{x-2}{2}\right) \\ =\ln 2+\left(\frac{x-2}{2}\right)-\frac{{\left(\frac{x-2}{2}\right)}^2}{2}+\frac{{\left(\frac{x-2}{2}\right)}^3}{3}-\frac{{\left(\frac{x-2}{2}\right)}^4}{4}+\cdots \\ =\ln 2+\frac{1}{2}(x-2)-\frac{1}{2^{2}2}(x-2{)}^2+\frac{1}{2^{3}3}(x-2{)}^3-\frac{1}{2^{4}4}(x-2{)}^4+\cdots \\ =\ln 2+\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n{2}^n}(x-2{)}^n \end{gathered}$$