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Unit 1.4.2 Loss of digits of accuracy

Homework 1.4.2.1.

Let \(\alpha = -14.24123 \) and \(\widehat \alpha = -14.24723 \text{.}\) Compute

  • \(\displaystyle \vert \alpha\vert = \)

  • \(\displaystyle \vert \alpha - \widehat \alpha \vert = \)

  • \(\displaystyle \frac{\vert \alpha - \widehat \alpha \vert}{\vert \alpha \vert} = \)

  • \(\log_{10} \left( \frac{\vert \alpha - \widehat \alpha \vert}{\vert \alpha \vert} \right) = \)
Solution

Let \(\alpha = -14.24123 \) and \(\widehat \alpha = -14.24723 \text{.}\) Compute

  • \(\displaystyle \vert \alpha\vert = 14.24123 \)

  • \(\displaystyle \vert \alpha - \widehat \alpha \vert = 0.006 \)

  • \(\displaystyle \frac{\vert \alpha - \widehat \alpha \vert}{\vert \alpha \vert} \approx 0.00042\)

  • \(\log_{10} \left( \frac{\vert \alpha - \widehat \alpha \vert}{\vert \alpha \vert} \right) \approx -3.4 \)

The point of this exercise is as follows:

  • If you compare \(\alpha = -14.24123 \)
    \(\hat \alpha = -14.24723 \) and you consider \(\hat \alpha \) to be an approximation of \(\alpha \text{,}\) then \(\hat \alpha\) is accurate to four digits: \(-14.24 \) is accurate.

  • Computing \(\log_{10} \left( \frac{\vert \alpha - \hat \alpha \vert}{\vert \alpha \vert} \right) \) tells you approximately how many decimal digits are accurate: \(3.4 \) digits.

Be sure to read the solution to the last homework!