10/06/2011, 12:15 AM
Oh very very interesting!
So I guess, the definitive property for exponentiation would be:
\( a\cdot a^x =a^{x+1} \) and no other function satisfies this requirement.
I wonder, what exactly differentiates tetration such that there are multiple solutions given for \( ^b a \) when \( a^{^b a} = \,\,^{b+1} a \) is the definitive property.
I think it's because square roots, or root functions in general, are exponentiation by the multiplicative inverse. They are binded by multiplication, and there is only one number b (besides its negative sqrt) that when multiplied by itself gives a.
However, super square roots (which there is only one value for every x) are not \( ^{\frac{1}{2}} a \) so this eliminates uniqueness by relation to super square roots.
Instead if we want a different sort of parallel to squaring a squareroot involving (1/2) for tetration we can have \( \exp_a^{\circ \frac{1}{2}}(\exp_a^{\circ \frac{1}{2}}(1)) = a \) but since \( \exp_a^{\circ \frac{1}{2}}(x) \) is determined by the extension of tetration, we see there is no external reason limiting half values to one specific value.
I guess this is a lot like how the one periodic function creates a separate super function for exponentiation. I think I'm beginning to understand why we have different solutions for tetration; I guess the real difficulty is defining what extra parameter makes it the solution.
So I guess, the definitive property for exponentiation would be:
\( a\cdot a^x =a^{x+1} \) and no other function satisfies this requirement.
I wonder, what exactly differentiates tetration such that there are multiple solutions given for \( ^b a \) when \( a^{^b a} = \,\,^{b+1} a \) is the definitive property.
I think it's because square roots, or root functions in general, are exponentiation by the multiplicative inverse. They are binded by multiplication, and there is only one number b (besides its negative sqrt) that when multiplied by itself gives a.
However, super square roots (which there is only one value for every x) are not \( ^{\frac{1}{2}} a \) so this eliminates uniqueness by relation to super square roots.
Instead if we want a different sort of parallel to squaring a squareroot involving (1/2) for tetration we can have \( \exp_a^{\circ \frac{1}{2}}(\exp_a^{\circ \frac{1}{2}}(1)) = a \) but since \( \exp_a^{\circ \frac{1}{2}}(x) \) is determined by the extension of tetration, we see there is no external reason limiting half values to one specific value.
I guess this is a lot like how the one periodic function creates a separate super function for exponentiation. I think I'm beginning to understand why we have different solutions for tetration; I guess the real difficulty is defining what extra parameter makes it the solution.