Math Problem Statement

Use the Remainder Estimation Theorem to find an interval containing x = 0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval. Check your answer by graphing |f(x) − p(x)| over the interval you obtained. f(x) = sin(x), p(x) = x − x^3/3! Round your answer to two decimal places.

Solution

The given problem involves using the Remainder Estimation Theorem to find an interval containing x=0x = 0 over which the function f(x)=sin(x)f(x) = \sin(x) can be approximated by the polynomial p(x)=xx33!p(x) = x - \frac{x^3}{3!} with a three-decimal-place accuracy.

Let's go step by step:

Step 1: Taylor Series Approximation

The Taylor series for sin(x)\sin(x) around x=0x = 0 is: sin(x)=xx33!+x55!\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots The polynomial p(x)=xx36p(x) = x - \frac{x^3}{6} is the Taylor polynomial of degree 3 for sin(x)\sin(x).

Step 2: Remainder Estimation Theorem

The remainder after the third term in the Taylor series for sin(x)\sin(x) is given by: Rn(x)=f(n+1)(c)(n+1)!xn+1R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} For p(x)p(x), we are using the third-degree polynomial, so the remainder R3(x)R_3(x) is the next term in the Taylor series: R3(x)=x55!=x5120R_3(x) = \frac{x^5}{5!} = \frac{x^5}{120} We need to find the maximum value of xx such that this remainder is less than or equal to 0.00050.0005 (since three decimal places means the error must be less than 0.00050.0005).

Step 3: Solve for xx

We want: x51200.0005\left| \frac{x^5}{120} \right| \leq 0.0005 Solving for xx: x51200.0005\frac{x^5}{120} \leq 0.0005 x50.0005×120=0.06x^5 \leq 0.0005 \times 120 = 0.06 x(0.06)1/5x \leq (0.06)^{1/5} Now, I'll compute the fifth root of 0.060.06.The value of xx such that the remainder is less than or equal to 0.00050.0005 is approximately x0.57x \leq 0.57.

Thus, the interval over which f(x)=sin(x)f(x) = \sin(x) can be approximated by p(x)=xx33!p(x) = x - \frac{x^3}{3!} with three-decimal accuracy is x0.57|x| \leq 0.57.

Would you like further details or have any other questions?

Here are 5 related questions:

  1. How does the Remainder Estimation Theorem apply to higher-degree polynomials?
  2. What happens to the interval if we include more terms in the Taylor expansion?
  3. How does the error bound change if we require four decimal places of accuracy?
  4. What are other common functions approximated using Taylor series?
  5. How do we handle the error in approximations for functions that are not smooth?

Tip: When approximating functions using Taylor series, increasing the degree of the polynomial generally increases the accuracy over a larger interval.

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Math Problem Analysis

Mathematical Concepts

Taylor Series
Remainder Estimation Theorem
Polynomial Approximation

Formulas

f(x) = sin(x)
p(x) = x − x^3/3!
R_3(x) = x^5/120

Theorems

Remainder Estimation Theorem

Suitable Grade Level

College-level Calculus