Hi.
I was wondering about a way by which one could theoretically "rescue" the continuum sum formula of Ansus.
The idea is this. Let \( f(x) \) be a function that decays to 0 as \( x \rightarrow \infty \). Then, there exists a "natural" notion of continuum sum, namely,
\( \sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x) \).
provided the sum converges.
We can modify this for a function that decays to a fixed point as well. If \( f(x) \) decays to a fixed point \( L \), then \( f(x) - L \) will decay to 0, and so we have
\( \sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{\infty} f(n) - f(n + x) \)
By linearity,
\( \sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{x} f(n) - \sum_{n=1}^{x} L \)
so
\( \sum_{n=1}^{x} f(n) = \left(\sum_{n=1}^{x} f(n) - L\right) + \sum_{n=1}^{x} L \)
and the sum of a constant, of course, has a natural definition, and so we get
\( \sum_{n=1}^{x} f(n) = Lx + \sum_{n=1}^{\infty} f(n) - f(n + x) \)
as the Ls cancel out when we plug \( f(n) - L \) into our sum formula.
Now take Ansus' continuum sum formula, here for base e,
\( \frac{\mathrm{tet}'(z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{z-1} \mathrm{tet}(n + z_0)\right) \)
and we reverse \( z \) to its negative, yielding
\( \frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{-z-1} \mathrm{tet}(n + z_0)\right) \)
Now, we can apply the identity
\( \sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n) \).
Then we get
\( \frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=1}^{z} -\mathrm{tet}(-n + z_0)\right) \)
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 \( z_0 \) 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 \( z_0 \), 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 \( f(x) \) be a function that decays to 0 as \( x \rightarrow \infty \). Then, there exists a "natural" notion of continuum sum, namely,
\( \sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x) \).
provided the sum converges.
We can modify this for a function that decays to a fixed point as well. If \( f(x) \) decays to a fixed point \( L \), then \( f(x) - L \) will decay to 0, and so we have
\( \sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{\infty} f(n) - f(n + x) \)
By linearity,
\( \sum_{n=1}^{x} f(n) - L = \sum_{n=1}^{x} f(n) - \sum_{n=1}^{x} L \)
so
\( \sum_{n=1}^{x} f(n) = \left(\sum_{n=1}^{x} f(n) - L\right) + \sum_{n=1}^{x} L \)
and the sum of a constant, of course, has a natural definition, and so we get
\( \sum_{n=1}^{x} f(n) = Lx + \sum_{n=1}^{\infty} f(n) - f(n + x) \)
as the Ls cancel out when we plug \( f(n) - L \) into our sum formula.
Now take Ansus' continuum sum formula, here for base e,
\( \frac{\mathrm{tet}'(z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{z-1} \mathrm{tet}(n + z_0)\right) \)
and we reverse \( z \) to its negative, yielding
\( \frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{-z-1} \mathrm{tet}(n + z_0)\right) \)
Now, we can apply the identity
\( \sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n) \).
Then we get
\( \frac{\mathrm{tet}'(-z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=1}^{z} -\mathrm{tet}(-n + z_0)\right) \)
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 \( z_0 \) 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 \( z_0 \), or integrate to -1 to obtain the iterating formula? Hmm...
What do you think about this?
