Closed-form derivatives

[andydude]

Looking around the forum, I can't find good references for these, although there are lots of references to these formulas, I think it would be good to have these together, so here they are:

\(
\begin{tabular}{rl}
\frac{\partial}{\partial a} ({}^{x}{a}) & = \frac{1}{a} \sum_{k=1}^{x} \ln(a)^{k-1} \prod_{j=0}^{k} {}^{{(x-j)}}{a} \\
\frac{\partial}{\partial x} ({}^{x}{a}) & = T(a) \ln(a)^x \prod_{k=1}^{x} {}^{k}{a}
\end{tabular}
\)
where
\( T(a) = \left[\frac{\partial}{\partial x} ({}^{x}{a})\right]_{x=0} \)

I also found a descent approximation to T(a):

\( T(a) \approx \log_{2.55} \log_{2.55} (2.29 + 2.58 a) \)

Andrew Robbins

[bo198214]

The I use this thread for giving the general formula for a superfunction \( \sigma \) of \( f \):
\( \sigma(x+1)=f(\sigma(x)) \)
\( \sigma'(x+1)=f'(\sigma(x))\sigma'(x) \)
\( \sigma'(x+n) = \sigma'(x)\prod_{k=0}^{n-1} f'(\sigma(x+k)) \)

[bo198214]

\(
\frac{\partial}{\partial x} ({}^{x}{a}) = T(a) \ln(a)^x \prod_{k=0}^{x - 1} {}^{k}{a}
\)

Doesnt it need to read
\(
\frac{\partial}{\partial x} ({}^{x}{a}) = T(a) \ln(a)^x \prod_{k=0}^{x - 1} {}^{k+1}{a}
\)
?

[andydude]

@Ansus
Thank you so much for finding that mistake! I fixed it in my original post.

Also, I noticed in the MathFacts page that it is a very complete overview, except for two things: the base-sqrt(2) tetration approximation: \( {}^{x}{(\sqrt{2})} \approx 2 \frac{x+1}{x+2} \), and intuitive/natural tetration, for which i would say that the matrix encoding of

\( \alpha(b^x) = \alpha(x) + 1 \)

is

\( (C[b^x]^T - I)D[\alpha] = D[1] \)

which can be solved for \( \alpha(x) \) "intuitively" despite the fact that \( (C[b^x]^T - I) \) is a noninvertible matrix.

[bo198214]

@Ansus
Thank you so much for finding that mistake! I fixed it in my original post.

Hey, I also discovered that mistake! Only my solution suggestion was different from Ansus'.

[andydude]

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

[andydude]

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.

[andydude]

Only because it gives extra ln(a)?

Yes, but wait. I'm wrong now. It is \( \ln(a)^2 \) in the denominator... I'm sorry. You're right. All of this off-by-one stuff is hard to keep track of.