I really have to put my foot down on this one. The lower limit is is not zero as it appears in the Tetration_Summary page. It is 1. I have re-derived a more general formula for this that accentuates this lower index:
\(
P(a, x_0, x_1)
= \prod_{k=x_0}^{x_1} \text{sexp}_a(k) \ln(a)
= \ln(a)^{(x_1 - x_0 + 1)}\prod_{k=x_0}^{x_1} \text{sexp}_a(k)
= \frac{{\text{sexp}_a}'(x_1)}{{\text{sexp}_a}'(x_0 - 1)}
\)
as you can see from this, if the final derivative in the denominator is evaluated at (\( 0 = x_0 - 1 \)), then this means \( x_0 = 1 \), which means the lower index of the product is (k=1), not (k=0).
@Ansus
Your derivations are based on the (k=0) formula (which is wrong), but other than that, they are quite clever! I never thought to do that. I think there would be less room for error if we use the "P" function to simplify things. Starting with the basic derivatives:
\( {\text{sexp}_a}'(x) = {\text{sexp}_a}'{(x_0)} P(a, x_0 + 1, x) \)
\( {\text{spow}_x}'(a) = \frac{1}{a \ln a} \sum_{k=1}^{x} P(a, x - k, x) \)
combining them gives:
\( \frac{{\text{spow}_x}'(a)}{{\text{sexp}_a}'(x)} = \frac{1}{a \ln a} \sum_{k=1}^{x} \frac{1}{{\text{sexp}_a}'(x - k - 1)} \)
which is about as rigorous as I can make it, so that should be right.
Andrew Robbins
\(
P(a, x_0, x_1)
= \prod_{k=x_0}^{x_1} \text{sexp}_a(k) \ln(a)
= \ln(a)^{(x_1 - x_0 + 1)}\prod_{k=x_0}^{x_1} \text{sexp}_a(k)
= \frac{{\text{sexp}_a}'(x_1)}{{\text{sexp}_a}'(x_0 - 1)}
\)
as you can see from this, if the final derivative in the denominator is evaluated at (\( 0 = x_0 - 1 \)), then this means \( x_0 = 1 \), which means the lower index of the product is (k=1), not (k=0).
@Ansus
Your derivations are based on the (k=0) formula (which is wrong), but other than that, they are quite clever! I never thought to do that. I think there would be less room for error if we use the "P" function to simplify things. Starting with the basic derivatives:
\( {\text{sexp}_a}'(x) = {\text{sexp}_a}'{(x_0)} P(a, x_0 + 1, x) \)
\( {\text{spow}_x}'(a) = \frac{1}{a \ln a} \sum_{k=1}^{x} P(a, x - k, x) \)
combining them gives:
\( \frac{{\text{spow}_x}'(a)}{{\text{sexp}_a}'(x)} = \frac{1}{a \ln a} \sum_{k=1}^{x} \frac{1}{{\text{sexp}_a}'(x - k - 1)} \)
which is about as rigorous as I can make it, so that should be right.
Andrew Robbins
