Q-32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

32. f(x)=ln(1+x), x₀ =0  

Step-by-Step Solution

Answer

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

Step 1: Taylor series

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

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

Step 2: Derivatives of function at x 0

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) = $\frac{1}{1+x}$then f '(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

Step 3: Substitute the derivatives in Taylor series

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

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

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

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

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