10/19/2025, 04:00 PM
my definition of tetration goes like this:
1. i construct the xth super root of y by lagrange interpolation.
2. i evaluate my constructed super root function at a non-integer x value.
3. i finally construct my tetration function by using the y value i got for my super root function.
more details are given in this desmos graph: https://www.desmos.com/calculator/ooxd3v6twb
as can be seen in the desmos graph, i constructed the xth super root of 2 and i chose 3.5 as my x value.
that's how i constructed the 3.5th super root of 2 tetrated to x.
if you want to know the value for the 3.5th super root of 2 using my method and kneser's, here it is:
my method: \( \sqrt[3.5]{2}_{s}=1.45849906738\)
kneser's method: \( \sqrt[3.5]{2}_{s}=1.4584946676580992\)
my method and kneser's method give slightly different results. that might be because the lagrange interpolation from my method does not approximate the super root function as well as kneser's method. but maybe the approximation problem can be solved using limits.
here's what i mean: we take the limit of the xth super root as x goes to infinity.
we interpolate the function using the limit values and we apply recursive formulas to construct the xth super root.
it doesn't seem like it's easy to find recursive formulas for the xth super root.
like i said in the title of this thread, my tetration method SEEMS reasonable. but i don't yet know if it IS reasonable.
maybe someone finds a quick algorithm in desmos to compute super roots. the algorithm i found required a lot of iterations of newton's method. too many iterations of newton's method would be bad for my computer.
and another point i would like to make: it seems like i can continue on a similar path for pentation, hexation, etc.
and by continuing on a similar path, i mean adapting my tetration method for higher hyperoperations.
1. i construct the xth super root of y by lagrange interpolation.
2. i evaluate my constructed super root function at a non-integer x value.
3. i finally construct my tetration function by using the y value i got for my super root function.
more details are given in this desmos graph: https://www.desmos.com/calculator/ooxd3v6twb
as can be seen in the desmos graph, i constructed the xth super root of 2 and i chose 3.5 as my x value.
that's how i constructed the 3.5th super root of 2 tetrated to x.
if you want to know the value for the 3.5th super root of 2 using my method and kneser's, here it is:
my method: \( \sqrt[3.5]{2}_{s}=1.45849906738\)
kneser's method: \( \sqrt[3.5]{2}_{s}=1.4584946676580992\)
my method and kneser's method give slightly different results. that might be because the lagrange interpolation from my method does not approximate the super root function as well as kneser's method. but maybe the approximation problem can be solved using limits.
here's what i mean: we take the limit of the xth super root as x goes to infinity.
we interpolate the function using the limit values and we apply recursive formulas to construct the xth super root.
it doesn't seem like it's easy to find recursive formulas for the xth super root.
like i said in the title of this thread, my tetration method SEEMS reasonable. but i don't yet know if it IS reasonable.
maybe someone finds a quick algorithm in desmos to compute super roots. the algorithm i found required a lot of iterations of newton's method. too many iterations of newton's method would be bad for my computer.
and another point i would like to make: it seems like i can continue on a similar path for pentation, hexation, etc.
and by continuing on a similar path, i mean adapting my tetration method for higher hyperoperations.