Question

Question: In Problems 29–34, determine the Taylor series about the point X₀for the given functions and values of X₀.

32. f(x)=ln(1+x), x₀ =0

Short answer

The required expression is$\sum _{n-1}^{\infty }\frac{{(-1)}^{n=1}}{n}.{\left(x\right)}^n$ .

Step-by-step solution

Taylor series

For a function f(x) the Taylor series expansion about a point x₀ is given by,

$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).(x-x_0)+f\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right).\frac{{(x-x_0)}^3}{3!}+...$

Derivatives of function at x0

We have to calculate the Taylor series expansion for,f (x)= ln (1 + x) at.

Calculating the derivatives of function at x₀,

f (x) =ln (1+x) then f (x₀)=0

f'(x) = $\frac{1}{1+x}$thenf '(x₀)=1

f''(x) = $\frac{-1}{{(1+x)}^2}$ then f''(x₀)= -1

f'''(x) = $\frac{2}{{\left(1+x\right)}^3}$then f'''(x₀) = 2

f''''(x) = $\frac{-6}{{(1+x)}^4}$ then f''''(x₀) = -6

Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at^$x_0=0$, then,

$\ln (1+x)=0+1.(x-0)-1.\frac{(x-0^2}{2!}+2\frac{{\left(x-0\right)}^3}{3!}-6.\frac{{(x-0)}^4}{4!}+....$

= $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$

= role="math" localid="1664200514125" $\sum _{n-1}^{\infty }\frac{{(-1)}^{n+1}}{n}.{\left(x\right)}^n$

Hence, the required expression is$\sum _{n-1}^{\infty }\frac{{(-1)}^{n+1}}{n}.{\left(x\right)}^n$