Hi.

I was wondering about a way by which one could theoretically "rescue" the continuum sum formula of Ansus.

The idea is this. Let be a function that decays to 0 as . Then, there exists a "natural" notion of continuum sum, namely,

.

provided the sum converges.

We can modify this for a function that decays to a fixed point as well. If decays to a fixed point , then will decay to 0, and so we have

By linearity,

so

and the sum of a constant, of course, has a natural definition, and so we get

as the Ls cancel out when we plug into our sum formula.

Now take Ansus' continuum sum formula, here for base e,

and we reverse to its negative, yielding

Now, we can apply the identity

.

Then we get

and if the tetrational decays exponentially to a fixed point in the left half of the complex plane, the continuum sum is well defined.

The only problem I see here is that cannot be 0, otherwise there will be singularities at the integers and then the sum formula won't work. If we use a displacement along the imaginary axis then it might work but how then do you evaluate that derivative at , or integrate to -1 to obtain the iterating formula? Hmm...

What do you think about this?

I was wondering about a way by which one could theoretically "rescue" the continuum sum formula of Ansus.

The idea is this. Let be a function that decays to 0 as . Then, there exists a "natural" notion of continuum sum, namely,

.

provided the sum converges.

We can modify this for a function that decays to a fixed point as well. If decays to a fixed point , then will decay to 0, and so we have

By linearity,

so

and the sum of a constant, of course, has a natural definition, and so we get

as the Ls cancel out when we plug into our sum formula.

Now take Ansus' continuum sum formula, here for base e,

and we reverse to its negative, yielding

Now, we can apply the identity

.

Then we get

and if the tetrational decays exponentially to a fixed point in the left half of the complex plane, the continuum sum is well defined.

The only problem I see here is that cannot be 0, otherwise there will be singularities at the integers and then the sum formula won't work. If we use a displacement along the imaginary axis then it might work but how then do you evaluate that derivative at , or integrate to -1 to obtain the iterating formula? Hmm...

What do you think about this?