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If we measure a length and is measured as $12.5$ meters long, accurate to $0.1$ of a meter this means the absolute error is $0.05$m.
The relative error is: $\frac{0.05}{12.5} = 0.004$. This means that the measurement is accurate to $\frac{4}{1000}$ or $0.4$%. or equivalently that for each unit we measure we introduce an error of $0.004$.

Now if we compare the $10^{-50}$ with $10^{-6}$ then the absolute error is: $|10^{-50} - 10^{-6}| = 10^{-6}$.
The relative error then is $\frac{absolute\space error}{measured\space value} = \frac{10^{-6}}{10^{-6}} = 1$

What is the meaning of $1$ here? How is it interpreted and how from a small absolute difference we get something that seems to indicate $100$% error?

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  • $\begingroup$ What are the $10^{-50}$ and $10^{-6}$? If the measured value is $10^{-6}$ but the lowest possible value is $10^{-50}$, then the relative error may really be that large. $\endgroup$
    – peterwhy
    Commented Nov 7, 2021 at 0:16
  • $\begingroup$ @peterwhy: the $10^{-50}$ should be the actual value and $10^{-6}$ the measured/estimated value $\endgroup$
    – Jim
    Commented Nov 7, 2021 at 0:17
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    $\begingroup$ Then the absolute error, or that absolute difference between the actual and measured values, are almost $100\%$ of the measured value and not small relative to the measured value. $\endgroup$
    – peterwhy
    Commented Nov 7, 2021 at 0:21
  • $\begingroup$ @peterwhy: that helps thank you $\endgroup$
    – Jim
    Commented Nov 8, 2021 at 17:38

1 Answer 1

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Briefly, when the absolute value of the relative error exceeds unity, you can no longer trust the sign. If $T$ is the target value and $A$ is the approximation, then the absolute error is $$E = T-A$$ and the relative error is $$R=\frac{E}{T}.$$ It follows that $$A = T - (T-A) = T - \frac{T-A}{T}T = T-RT= T(1-R).$$ Typically, we do not know the relative error, but we have an upper bound for the absolute value. Now if $$|R|<1,$$ then $A$ and $T$ have the same sign. If $|R|\geq1$, then it is entirely possible that $A$ and $T$ have different sign.

Computing the correct sign is critical in root finding applications. If $f : \mathbb{R} \rightarrow\mathbb{R}$ is continuous and $f(x_1)$ and $f(x_2)$ have different signs, then $f$ has a zero in the interval between $x_1$ and $x_2$. If we cannot trust the computed value of the sign of $f(x_i)$, then we cannot make this determination with certainty.

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  • $\begingroup$ In this case though $R= 0$ right? $\endgroup$
    – Jim
    Commented Nov 7, 2021 at 10:27
  • $\begingroup$ @Jim: To what case are you referring? $\endgroup$ Commented Nov 7, 2021 at 13:12
  • $\begingroup$ In my post, the relative error is $1$. So $A = T$ in your formula right? $\endgroup$
    – Jim
    Commented Nov 7, 2021 at 15:59
  • $\begingroup$ @Jim. No, in your post you do not have $R=0$ because $10^{-50} \not =0$ and in general $T \not =A$. $\endgroup$ Commented Nov 7, 2021 at 20:23
  • $\begingroup$ The relative error in the post is $1$. So based on your formula $A = T(1 - 1) = T\cdot 0 = A$. What does this mean? $\endgroup$
    – Jim
    Commented Nov 7, 2021 at 22:04

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