Hacker News Reader

Comment by Buttons840 | Hacker News Reader

I don't understand the percentage. What, for example, would an approximation that was only 20% accurate look like?

1.7 is about 20% off of accurate. 0.3 to 7.0 seem to be within the range of what 20% accurate would mean (1/5th to 5x the actual value).

The thread shows what they mean a few tweets in.

I don’t think ‘Accuracy’ is used much in math. https://mathworld.wolfram.com/Accuracy.html exists, but I think it would be better to report relative error (https://en.wikipedia.org/wiki/Approximation_error, https://mathworld.wolfram.com/RelativeError.html), which is less than one in a million here.

The math, in general, isn’t as good as it, IMO, should be. The claim

“Thus, the interval [x₀, 2/x₀] envelopes √2.

From this, it follows that the mid-point of the interval [x₀, 2/x₀] is a better approximation to √2”

isn’t correct. The midpoint is a better approximation, but that doesn’t follow from “the interval [x₀, 2/x₀] envelopes √2.”. [1,999] envelopes √2, but 500 isn’t a better approximation of √2 than 1.

(Aside: https://en.wikipedia.org/wiki/Accuracy_and_precision is useless for finding a definition of accuracy. The chapter “Common technical definition” doesn’t define anything, and the picture in that section seems to imply that, for both precision and accuracy, lower is better)

The math as explained in Twitter is correct, your example is not relevant because [1,999] is not in the form [x,2/x].

I'm sure there's a better way to see the conclusion stated in the tweet, but you can convince yourself with the brute-force way of graphing:

- `abs(x-sqrt(2))` (the distance between x and sqrt(x))

- `abs(2/x-sqrt(2))` (the distance between 2/x and sqrt(x))

- `abs((x/2+1/x)-sqrt(2))` (the distance between the midpoint and sqrt(2))

and seeing that the last one is always smaller than the first two between 1 and sqrt(2), the values for x stated in a previous tweet.

To simplify your other responses, 20% accurate means you're 80% inaccurate. So +/- 80% of the correct value.

It would mean their error would be less than it'd look like 1.2*sqrt(2).

Read the thread he explains it very well. The percentage is 1 - (absolute error divided by the true value) * 100

Note that the parent asked for 20% accurate, not 20% error (80% accurate).

Accuracy is generally defined as (mean - true)/true

But it's a common note that that's really the inaccuracy.

approx_20pct * 0.2 <= sqrt(2) <= approx_20pct / 0.2; so it might look like 0.3, or 4.5.

Reply on news.ycombinator.com