Math Problem Statement

The 3rd degree Taylor polynomial for f(x) = ln(x+1) centered at x = 0 is T_3(x) = x - x^2/2 + x^3/3. Find the remainder for this Taylor polynomial. Use the formula R_n(x) = f^(n+1)(z)/(n+1)! (x-c)^(n+1), where c is the center and z is some point between x and c. Now use the Taylor/Lagrange remainder theorem to find all positive values of x for which this approximation is within 0.001 of f(x).

Solution

To solve this problem systematically, we’ll address the following:

  1. Determine the remainder term R3(x)R_3(x) using the given formula.
  2. Find the interval for xx where the approximation error is within 0.0010.001.

1. Remainder Term R3(x)R_3(x):

The remainder for the Taylor polynomial is given by:

Rn(x)=f(n+1)(z)(n+1)!(xc)n+1,R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}, where cc is the center (here c=0c = 0) and zz lies between xx and 00. For T3(x)T_3(x), we calculate R3(x)R_3(x), the remainder after the 3rd degree term.


Derivatives of f(x)=ln(1+x)f(x) = \ln(1+x):

  • f(x)=ln(1+x)f(x) = \ln(1+x)
  • f(x)=11+xf'(x) = \frac{1}{1+x}
  • f(x)=1(1+x)2f''(x) = -\frac{1}{(1+x)^2}
  • f(x)=2(1+x)3f'''(x) = \frac{2}{(1+x)^3}
  • f(4)(x)=6(1+x)4f^{(4)}(x) = -\frac{6}{(1+x)^4}

Thus, the 4th derivative f(4)(z)=6(1+z)4f^{(4)}(z) = -\frac{6}{(1+z)^4}.


Remainder R3(x)R_3(x) for n=3n = 3:

Substitute into the remainder formula:

R3(x)=f(4)(z)4!x4R_3(x) = \frac{f^{(4)}(z)}{4!} x^4

Since f(4)(z)=6(1+z)4f^{(4)}(z) = -\frac{6}{(1+z)^4} and 4!=244! = 24, we get:

R3(x)=624(1+z)4x4=x44(1+z)4,R_3(x) = \frac{-6}{24 (1+z)^4} x^4 = -\frac{x^4}{4 (1+z)^4}, where zz lies between 00 and xx.


2. Find xx for Approximation Error R3(x)<0.001|R_3(x)| < 0.001:

We need the absolute value of the remainder:

R3(x)=x44(1+z)4.\left| R_3(x) \right| = \frac{x^4}{4 (1+z)^4}.

  • Since zz is between 00 and xx, the maximum value of (1+z)(1+z) occurs at z=0z = 0, giving (1+z)41(1+z)^4 \geq 1.
  • Thus, to bound the remainder, we approximate:

R3(x)x44.\left| R_3(x) \right| \leq \frac{x^4}{4}.

We require R3(x)<0.001\left| R_3(x) \right| < 0.001, so:

x44<0.001.\frac{x^4}{4} < 0.001.

Multiply through by 4:

x4<0.004.x^4 < 0.004.

Take the 4th root:

x<0.0044.x < \sqrt[4]{0.004}.

Calculate 0.0044\sqrt[4]{0.004}:

0.0044=0.004=0.06320.251.\sqrt[4]{0.004} = \sqrt{\sqrt{0.004}} = \sqrt{0.0632} \approx 0.251.


Final Answer:

The Taylor polynomial approximation is accurate within 0.0010.001 for:

0<x0.251.0 < x \leq 0.251.


Summary:

  1. Remainder Term: R3(x)=x44(1+z)4R_3(x) = -\frac{x^4}{4(1+z)^4}, where zz is between 00 and xx.
  2. Approximation Interval: 0<x0.2510 < x \leq 0.251.

Would you like a detailed explanation of any step or further clarifications?


Related Questions:

  1. How do you derive the remainder formula for Taylor polynomials?
  2. Why does the error decrease as the polynomial degree increases?
  3. How do higher-order derivatives influence the Taylor remainder?
  4. What is the geometric significance of the remainder term?
  5. How can we improve the accuracy of a Taylor polynomial approximation?

Tip:

When approximating with Taylor polynomials, always use the remainder term to ensure the desired level of accuracy!

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Remainder Term
Approximation Error
Lagrange Remainder Theorem

Formulas

R_n(x) = f^(n+1)(z) / (n+1)! * (x-c)^(n+1)
f(x) = ln(1 + x)
R_3(x) = -x^4 / 4(1+z)^4

Theorems

Taylor/Lagrange Remainder Theorem

Suitable Grade Level

University Calculus I