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 Closed-form derivatives andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/23/2009, 01:12 AM (This post was last modified: 08/24/2009, 01:40 AM by 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/23/2009, 10:01 AM (This post was last modified: 08/23/2009, 10:04 AM by 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/23/2009, 10:14 AM (08/23/2009, 01:12 AM)andydude Wrote: $ \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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/24/2009, 05:37 AM (This post was last modified: 08/24/2009, 05:39 AM by 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/25/2009, 09:49 AM (08/24/2009, 05:37 AM)andydude Wrote: @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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 09/02/2009, 01:52 AM (This post was last modified: 09/03/2009, 07:55 AM by 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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 09/03/2009, 01:32 AM (This post was last modified: 09/03/2009, 01:33 AM by andydude.) (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. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 09/03/2009, 04:02 AM (This post was last modified: 09/03/2009, 07:58 AM by andydude.) (09/03/2009, 01:51 AM)Ansus Wrote: 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. « Next Oldest | Next Newest »

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