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
 - L = \sum_{n=1}^{\infty} f(n) - f(n + x))
By linearity,
 - L = \sum_{n=1}^{x} f(n) - \sum_{n=1}^{x} L)
so
 = \left(\sum_{n=1}^{x} f(n) - L\right) + \sum_{n=1}^{x} L)
and the sum of a constant, of course, has a natural definition, and so we get
 = Lx + \sum_{n=1}^{\infty} f(n) - f(n + x))
as the Ls cancel out when we plug
into our sum formula.
Now take Ansus' continuum sum formula, here for base e,
}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{z-1} \mathrm{tet}(n + z_0)\right))
and we reverse
to its negative, yielding
}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{-z-1} \mathrm{tet}(n + z_0)\right))
Now, we can apply the identity
.
Then we get
}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=1}^{z} -\mathrm{tet}(-n + z_0)\right))
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
provided the sum converges.
We can modify this for a function that decays to a fixed point as well. If
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
Now take Ansus' continuum sum formula, here for base e,
and we reverse
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
What do you think about this?