Math Problem Statement

Determine the truncation in given that 0.98\sqrt{0.98} = (1 - 0.02)0.5^{0.5}

Solution

To determine the truncation error when approximating 0.98\sqrt{0.98} using the expression (10.02)0.5(1 - 0.02)^{0.5}, we can use a Taylor series expansion around x=0x = 0 for the function (1x)0.5(1 - x)^{0.5}.

Taylor Series Expansion

The Taylor series expansion for (1x)0.5(1 - x)^{0.5} around x=0x = 0 is given by:

(1x)0.5=112x18x2116x3+(1 - x)^{0.5} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 + \cdots

For x=0.02x = 0.02:

(10.02)0.5=10.0220.02280.02316+(1 - 0.02)^{0.5} = 1 - \frac{0.02}{2} - \frac{0.02^2}{8} - \frac{0.02^3}{16} + \cdots

Let's expand this up to the quadratic term to see the truncation error when truncating after the linear term:

(10.02)0.510.022(0.02)28(1 - 0.02)^{0.5} \approx 1 - \frac{0.02}{2} - \frac{(0.02)^2}{8}

Calculating each term separately:

  1. First term: 11

  2. Linear term: 0.022=0.01- \frac{0.02}{2} = -0.01

  3. Quadratic term: (0.02)28=0.00048=0.00005- \frac{(0.02)^2}{8} = - \frac{0.0004}{8} = -0.00005

Including only the linear term, we have:

(10.02)0.510.01=0.99(1 - 0.02)^{0.5} \approx 1 - 0.01 = 0.99

Including up to the quadratic term, we have:

(10.02)0.510.010.00005=0.98995(1 - 0.02)^{0.5} \approx 1 - 0.01 - 0.00005 = 0.98995

Exact Value Calculation

The exact value of 0.98\sqrt{0.98} is:

0.980.9899494937\sqrt{0.98} \approx 0.9899494937

Truncation Error

  1. When truncating after the linear term: Truncation Error (Linear Term Only)=0.980.99=0.98994949370.99=0.0000505063\text{Truncation Error (Linear Term Only)} = \sqrt{0.98} - 0.99 = 0.9899494937 - 0.99 = -0.0000505063

  2. When truncating after the quadratic term: Truncation Error (Quadratic Term)=0.980.98995=0.98994949370.98995=0.0000005063\text{Truncation Error (Quadratic Term)} = \sqrt{0.98} - 0.98995 = 0.9899494937 - 0.98995 = -0.0000005063

Relative Error

  1. Relative Error (Linear Term Only): Relative Error (Linear Term)=0.00005050630.98994949370.000051\text{Relative Error (Linear Term)} = \frac{-0.0000505063}{0.9899494937} \approx -0.000051

  2. Relative Error (Quadratic Term): Relative Error (Quadratic Term)=0.00000050630.98994949370.000000512\text{Relative Error (Quadratic Term)} = \frac{-0.0000005063}{0.9899494937} \approx -0.000000512

Summary

  • Truncated value (linear term only): 0.990.99
  • Truncated value (up to quadratic term): 0.989950.98995
  • Exact value: 0.98994949370.9899494937
  • Truncation Error (linear term only): 0.0000505063-0.0000505063
  • Truncation Error (quadratic term): 0.0000005063-0.0000005063
  • Relative Error (linear term only): 0.000051-0.000051
  • Relative Error (quadratic term): 0.000000512-0.000000512

Would you like more details or have any questions? Here are some related questions for further exploration:

  1. What is a Taylor series expansion and how is it used in approximations?
  2. How does the Taylor series apply to fractional exponents?
  3. Why do higher-order terms in the expansion improve accuracy?
  4. What are the applications of truncation error analysis in numerical methods?
  5. How can we determine the optimal number of terms to include in a series expansion for a given accuracy?
  6. What is the difference between truncation error and round-off error?
  7. How does the choice of xx affect the accuracy of Taylor expansions?
  8. What are alternative methods for approximating square roots and their respective errors?

Tip: When using series expansions for approximations, consider including higher-order terms for better accuracy, especially when higher precision is required.

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Math Problem Analysis

Mathematical Concepts

Taylor Series Expansion
Approximations
Truncation Error
Relative Error

Formulas

Taylor series expansion for (1 - x)^{0.5}

Theorems

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Suitable Grade Level

Advanced High School