(09/02/2009, 10:09 AM)Ansus Wrote: Note that this formula ... works both with lower limit 0 and 1 because \( f_a(0)=1 \).
True, but it does have unfortunate misunderstandings later on, like the extra \( \ln(a)^2 \) in the final formula, which is incorrect. Using index substitution, the right formula is:
\( \frac{{\text{spow}_x}'(a)}{{\text{sexp}_a}'(x)} = \frac{1}{a \ln a} \sum_{k=0}^{x-1} \frac{1}{{\text{sexp}_a}'(k - 1)} \)
Our formulas are identical except for the \( \ln(a)^2 \), which should be \( \ln(a) \). This extra logarithm comes from the wrong index.
