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Let's say I have two numbers $x$ and $y$ which I'd like to compare to see whether they are equal up to their first $N$ digits.

For instance $ 1.002 \approx 1.001$ and $1002 \approx 1001$ up to 3 digits.

Intuitively, this means that the difference of $x$ and $y$ should be $N$ orders of magnitude smaller than (say) $x$. In symbols, the digits of accuracy, $D$ is $$ D = \log_{10}(|x|) -\log_{10} (|x-y|) = - \log_{10} \left| \frac{y}{x} - 1\right| $$

To make the definition symmetric, it seems to make sense to replace the $\frac{y}{x} - 1$ by $\ln(\frac{y}{x})$.

So a final definition would be

$$ D = \begin{cases} -\log_{10} \left| \ln |x| - \ln |y| \right| & \text{ if } xy > 0 \\ -\infty & \text{ otherwise } \end{cases}$$

Is this a reasonable definition? If so, does it have a name? Where is it referenced?

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    $\begingroup$ This is essentially an extension of the definition of relative error, using the log base 10 to obtain a notion of "digits" that extends to non-integer values. I don't think it has a name, other than "the log of the relative error". $\endgroup$ – Paul Dec 20 '17 at 21:38
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    $\begingroup$ So $1.004\approx 1.001$ up to 3 digits, but does $3.012 \approx 3.003$ up to 3 digits (same relative error)? $\endgroup$ – Kirill Dec 20 '17 at 22:17
  • $\begingroup$ 1100 ≈ 1099 giving D=3.04 makes more sense if you take it to mean "x and y will match once they are both rounded to N digits". 3.012≈3.003 actually gives D=2.52 which brings up the question: Should D be rounded up or down before using it as N? $\endgroup$ – candied_orange Dec 21 '17 at 2:41
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Short version

In scientific computing, the notion of relative error is way more popular than accuracy to $N$ digits. Whenever we present the results, we usually plot the obtained (scaled) data and relative/absolute error in multiple different ways. Reporting the number of correct obtained digits (in addition to the relative error) is usually limited to the following scenarios:

  1. In a table form, for the column "number of digits".
  2. In a sentence of the type: "...this method allowed to obtain a $K$-digit accurate solution...".

For both cases, I can hardly imagine the value of providing a counter-intuitive, non-integer value.


Moreover (since we usually prefer to think in base 10), one finds the quantity $D$ as: $$ D = -\log_{10}\left(\frac{|x-y|}{|x|}\right)=-\log_{10}(\eta) $$

where $x$ is the test value, $y$ is the reference value, and $\eta$ is the relative error between the two. Without loss of generality, let's not touch cases when $x=0$ and $xy<0$ ($x$ and $y$ have different signs).

However, knowing $D$ does not give me an immediate answer to the question of how many correct digits $x$ has. Simply, because $D$ is not an integer, and the answer to this question (usually) implies an integer answer. $D$ can tell me that $x$ has at least $\lfloor D \rfloor$ and at most $\lceil D\rceil$ correct digits, reducing the uncertainty to two possible answers.

For a more elaborate derivation & discussion on this part and how this is justified, see this question on Math SE.

Is it possible to decide solely on $D$ (or $\eta$), how many digits are correct? No.

The comments provide counterexamples (explained by the above paragraph): $$ x_1 = 1.004,\quad y_1 = 1.001,\quad \eta_1 \approx 2.988\cdot10^{-3},\quad D_1 = 2.5246...\\ x_2 = 3.012,\quad y_2 = 3.003,\quad \eta_2 \approx 2.988\cdot10^{-3},\quad D_2 = 2.5246... $$ Here, $D_1=D_2$, but

  • In case of $(x_1,y_1)$, $D_1$ has to be ceiled to get the correct answer of 3-digit match
  • In case of $(x_2,y_2)$, $D_2$ has to be floored to get the correct answer of 2-digit match

So, the definition of $D$ in extended to a non-integer number of digits sense is counterintuitive and hard to rely upon.

Now, to always be correct one may answer that: $x$ is at least $\lfloor D \rfloor$-digit accurate wrt $y$. That implies the following conclusion: if one has a method that provides a $10^{-3}$-level relative error, that leaves us with a 2-digit accurate solution. Even though, in some cases, it might be 3-digit accurate.

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