Efficient Integer Representations of Irrationals

[Raman]

Originally Posted by jasonp

Yes, there is a standard estimate for the error term between a continued fraction and the decimal number it is supposed to represent, based only on the size of p and q (it's been ~15 years, I don't remember it). You can also prove that to get a better approximation of the error, you must use numbers larger than p and q.

For example, to getting with this following sequence
2⁴⁶ ≈ 7.10¹³
2¹³⁹ ≈ 7.10⁴¹
2³³⁵ ≈ 7.10¹⁰⁰
2²⁴⁷¹ ≈ 7.10⁷⁴³
2¹⁵⁷⁷² ≈ 7.10⁴⁷⁴⁷
2¹⁵⁷³²⁶ ≈ 7.10⁴⁷³⁵⁹

[R.D. Silverman]

Originally Posted by jasonp

Yes, there is a standard estimate for the error term between a continued fraction and the decimal number it is supposed to represent, based only on the size of p and q (it's been ~15 years, I don't remember it). You can also prove that to get a better approximation of the error, you must use numbers larger than p and q.

Look up Thue's Thm. or Liouville's Thm. --> Diophantine approximation

BTW, I does everyone know the theorem that in a well-defined sense
(I'll let people look it up if they are interested), the "Golden Ratio" is the
"most irrational" number that there is because it is the hardest to
represent as the ratio of integers. (in terms of the heights of the numerator
and denominator). Consider its CF expansion...

[Raman]

Originally Posted by Raman

The fun is that
73/153, when both the numerator, denominator are being multiplied by using 2232, yields 162936/341496
but that the value for 162935/341496 is always being a better closer approximation to the value for log 3

2098 = 2251 - 153
17855 = 7*2251 + 2098
17855 = 8*2251 - 153
17855 + 153 = 8*2251

17855*19 + 2251
= (8*2251 - 153)*19 + 2251
= 152*2251 - 19*153 + 2251
= 153*2251 - 19*153
= 153*(2251 - 19)
= 153*2232
= 341496 ≡ 0 (mod 153)

162936 = 73*2232 ≡ 0 (mod 73)
162935 = 8519*19+1074
= 153 * 1074 - 73 * 19
≡ 72 (mod 73)

Thereby 44, 109, 153, 2098, 2251, 17855 are all being the denominator values for the continued fraction expansion for the real number log 3 itself

as such, again repeatedly, similar to the following repetitions as well
as follows
109 = 153 - 44
2098 = 13*153 + 109
2098 = 14*153 - 44
2098 + 44 = 14*153
2098 + 153 + 44 = 15*153
2251 + 44 = 15*153