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September 5

Does somone ever tried to research rounding economy related to different number bases?

As some example, at base 10, if you want to round to nearest number with last X digits being 0, you will need to add or remove at average a value of (10^X)/4, where X is the amount of digits you want to be 0.

Does someone ever tried to find what is the number base, where you would need at average add/remove the minimum amount of numbers to round the value you want?191.250.232.243 (talk) 17:55, 5 September 2021 (UTC)[reply]

I suspect that there are no published articles on this concept of rounding economy, determined by the average value of ｜i − round_n(i)｜, where rounding to n digits 0 takes place in some base b ≥ 2. We can take the arithmetic mean over the range 0 ≤ i < bⁿ. The above formula for base 10 generalizes for even bases to:

1⁄4 bⁿ.

For odd bases, this is slightly more complicated:

1⁄4 (bⁿ − b⁻ⁿ).

As n → ∞, this is asymptotically equivalent to the formula for even bases. But how to compare two bases, say 2 and 10? As shown by 2¹⁰ ≈ 10³, 10 binary digits have about the same information as 3 decimal digits. For a fair comparison, we must replace n by something that depends on b. The information in a base-b digit is proportional to log b, so we should replace n by an expression that is inversely proportional to log b, such as λ(log b)⁻¹. But observe that

b^{λ(log b)⁻¹} = e^λ,

regardless of the value of b. This means the two expressions above become

1⁄4 e^λ and

1⁄4 (e^λ − e^−λ).

So it really does not make a difference, although odd bases have a slight edge for small values of λ. Consider 2⁸ ≈ 3⁵. To level the field, we compare 2^7.9624 = 249.41... with 3^5.0237 = 249.40... Then

1⁄4 2^7.9624 = 62.354..., while

1⁄4 (3^5.0237 − 3^−5.0237) = 62.352...

Somewhat unimpressive. --Lambiam 21:15, 5 September 2021 (UTC)[reply]