4 ms·
The powers of 2 and 3 getting further away bit is not correct as you mention but it isn't actually all that important. My Github page is a WIP so bear with me..
by dankearney 4y ago
The powers of 2 and 3 getting further away bit is not correct as you mention but it isn't actually all that important. My Github page is a WIP so bear with me..
If a cycle does exist for some large-ish cycle length, let’s say 100 odd numbers, then it means that (3+1/x1)(3+1/x2)…(3+1/x100) exactly equals some power of 2.
Since (3+1/xn) rapidly approaches 3.0, for any large x1, the product of (3+1/x1)(3+1/x2) etc is going to be pretty close to a power of 3, but it needs to be much larger than that to equal a power of 2, since powers of 2 and 3 are always somewhat far apart. That means that x1, x2 etc must be relatively small so that the product of the (3+1/x) can actually be substantially larger than a power of 3. That puts a hard floor on the possible cycle values, and that floor actually trends smaller as the cycle gets larger since the distance between a power of 3 and the nearest power of 2 also trends larger.
- LegionMammal978 4y ago> since powers of 2 and 3 are always somewhat far apart. This is the part where your argument fails. On a logarithmic scale, powers of 2 and powers of 3 can become arbitrarily close together. To see this, notice that for any ε > 0, we can create a rational approximation x/y such that 0 < ln 2/ln 3 - x/y ≤ ε. Multiplying by y ln 3, we have 0 < y ln 2 - x ln 3 ≤ εy ln 3, or equivalently, x ln 3 < y ln 2 ≤ (x + εy) ln 3. Finally, we take the exponential to get 3^x < 2^y ≤ 3^(x + εy) = 3^(x(1 + εy/x)). As a consequence of Dirichlet's approximation theorem, εy/x can become arbitrarily small, so our upper bound 3^(x(1 + εy/x)) can become arbitrarily close to 3^x on a logarithmic scale.