## Unit1.4.2Loss of digits of accuracy

###### Homework1.4.2.1.

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!