Math Problem Statement

Given the function f(x) = ln(x + 1), determine the Taylor polynomial of second order around a = 0, derive the remainder term, approximate the value at x = 1.5, and calculate the absolute error.

Solution

From the uploaded image, the task involves the Taylor polynomial of order 2 and the corresponding remainder term for the function f(x)=ln(x+1)f(x) = \ln(x+1) around the expansion point a=0a = 0. Let me break down and explain each part of the problem.


a) Construct the Taylor polynomial of second order

The Taylor polynomial of f(x)=ln(x+1)f(x) = \ln(x+1) at a=0a = 0 is given by:

T2(x)=f(0)+f(0)x+f(0)2x2T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2

Step 1: Compute derivatives of f(x)f(x)

  1. f(x)=ln(x+1)f(x) = \ln(x+1)
    f(x)=1x+1f'(x) = \frac{1}{x+1}
  2. f(x)=1(x+1)2f''(x) = -\frac{1}{(x+1)^2}

Step 2: Evaluate at x=0x = 0

  • f(0)=ln(0+1)=0f(0) = \ln(0+1) = 0
  • f(0)=10+1=1f'(0) = \frac{1}{0+1} = 1
  • f(0)=1(0+1)2=1f''(0) = -\frac{1}{(0+1)^2} = -1

Step 3: Substitute into the Taylor polynomial formula

T2(x)=0+1x+12x2=xx22T_2(x) = 0 + 1\cdot x + \frac{-1}{2}x^2 = x - \frac{x^2}{2}

Thus, the second-order Taylor polynomial is: T2(x)=xx22T_2(x) = x - \frac{x^2}{2}


b) Corresponding remainder term

The remainder term (Lagrange form) for f(x)f(x) is: R3(x)=f(3)(c)3!x3R_3(x) = \frac{f^{(3)}(c)}{3!}x^3 where c[0,x]c \in [0, x] and f(3)(x)f^{(3)}(x) is the third derivative.

Step 1: Compute f(3)(x)f^{(3)}(x)

f(3)(x)=2(x+1)3f^{(3)}(x) = \frac{2}{(x+1)^3}

Step 2: Express R3(x)R_3(x)

R3(x)=23!(c+1)3x3=13(c+1)3x3,c[0,x]R_3(x) = \frac{2}{3! (c+1)^3}x^3 = \frac{1}{3(c+1)^3}x^3, \quad c \in [0, x]


c) Approximation at x=1.5x = 1.5 and absolute error

Step 1: Compute T2(1.5)T_2(1.5)

Using T2(x)=xx22T_2(x) = x - \frac{x^2}{2}: T2(1.5)=1.5(1.5)22=1.52.252=1.51.125=0.375T_2(1.5) = 1.5 - \frac{(1.5)^2}{2} = 1.5 - \frac{2.25}{2} = 1.5 - 1.125 = 0.375

Step 2: Compute absolute error

The absolute error is: Error=f(1.5)T2(1.5)\text{Error} = |f(1.5) - T_2(1.5)|

Evaluate f(1.5)=ln(1.5+1)=ln(2.5)0.9163f(1.5) = \ln(1.5+1) = \ln(2.5) \approx 0.9163 (rounded to 4 decimals).

Thus: Error=0.91630.375=0.5413\text{Error} = |0.9163 - 0.375| = 0.5413


Final Answers:

  1. Taylor polynomial: T2(x)=xx22T_2(x) = x - \frac{x^2}{2}

  2. Remainder term: R3(x)=13(c+1)3x3R_3(x) = \frac{1}{3(c+1)^3}x^3

  3. Approximation at x=1.5x = 1.5: T2(1.5)=0.375T_2(1.5) = 0.375

  4. Absolute error: Error=0.5413\text{Error} = 0.5413


Do you need more detailed calculations or additional explanations? Here are some related questions:

  1. How is the Lagrange remainder term used to estimate the error?
  2. Why does the approximation improve as the degree of the Taylor polynomial increases?
  3. Can this method be extended to functions with discontinuities?
  4. What is the significance of rounding in numerical approximations?
  5. How can one verify that c[0,x]c \in [0, x] satisfies the conditions for the remainder term?

Tip: Always compute derivatives carefully and keep track of negative signs when constructing Taylor series expansions!

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Math Problem Analysis

Mathematical Concepts

Taylor series expansion
Logarithmic functions
Numerical approximation
Error analysis

Formulas

Taylor polynomial T_n(x) = f(a) + f'(a)(x-a) + (f''(a)/2)(x-a)^2 + ... + (f^(n)(a)/n!)(x-a)^n
Remainder term R_n(x) = (f^(n+1)(c)/(n+1)!)(x-a)^(n+1), c ∈ [a, x]
Natural logarithm properties: f(x) = ln(x+1), f'(x) = 1/(x+1), f''(x) = -1/(x+1)^2, f'''(x) = 2/(x+1)^3

Theorems

Taylor's theorem with Lagrange remainder

Suitable Grade Level

Undergraduate mathematics or advanced high school (Grades 11-12)