Solve the problem.
-Evaluating the Taylor series at 0 for f(x) = ln (1 + x) at x = 0.6 produces the following series. Use four terms in this series to approximate ln 1.6, and then estimate the error in this approximation.

Question

Solve the problem.
-Evaluating the Taylor series at 0 for f(x) = ln (1 + x) at x = 0.6 produces the following series.  Use four terms in this series to approximate ln 1.6, and then estimate the error in this approximation.

Quizplus · Accepted Answer

The Taylor series for ln(1+x) at x=0 is: 
ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...
Substituting x=0.6 gives: 
ln(1.6) = 0.6 - (0.6^2)/2 + (0.6^3)/3 - (0.6^4)/4
= 0.5104 
To estimate the error, we use the remainder term: 
Rn(x) = (f^(n+1)(c))/(n+1)! * (x-a)^(n+1), where c is some number between a and x. 
For ln(1+x), f^(n+1)(x) is (-1)^n * n! / (1+x)^(n+1). 
So, the remainder term for the fourth degree Taylor polynomial is: 
R4(0.6) = (-1)^5 * 5! / (1+c)^5 * (0.6-0)^5 for some c between 0 and 0.6. 
The absolute value of this expression is maximized when c=0 (because 1+c is minimized), giving: 
|R4(0.6)| = 5!/6^5 * 0.6^5 = 0.0004832 
Therefore, the error in the approximation ln(1.6) ≈ 0.5104 is less than 0.0005. 
Choice D is the closest approximation to the actual value and is therefore the best choice.

Solve the Problem. -Evaluating the Taylor Series at 0 for F(x) = Ln