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 What am I doing wrong here? JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/12/2011, 07:53 PM (This post was last modified: 05/12/2011, 11:31 PM by JmsNxn.) I was looking at tetration and considering its little taylor series, given by: (1) $sexp(x) = \bigtriangleup \sum_{n=0}^{\infty} a_n + n(x \bigtriangledown l) - ln(n!)$ where: $\bigtriangleup \sum_{n=0}^{r} f(n)= f(0) \bigtriangleup f(1) \bigtriangleup...{f®}$ and: $x \bigtriangleup y= \ln(e^x + e^y)$ $x \bigtriangledown y = \ln(e^x - e^y)$ specifically, what I was trying to do is find a relationship between the coefficients of tetrations' little taylor series and normal taylor series. We'll say: $sexp(x) = \sum_{n=0}^{\infty} b_n \frac{(x-l)^n}{n!}$ so I'm looking for $b_n$ as an expression of $a_n$ or vice versa. So I started off by looking at (1): $sexp(x) = \bigtriangleup \sum_{n=0}^{\infty} a_n + n(x \bigtriangledown l) - ln(n!)$ $sexp(x) = ln(\sum_{n=0}^{\infty} e^{a_n + n(x \bigtriangledown l) - ln(n!)})$ $sexp(x+1) = \sum_{n=0}^{\infty} e^{a_n} \frac{(e^x - e^l)^n}{n!}$ and now let this equal our formula for sexp(x+1) using $b_n$ or the normal taylor series. $\sum_{n=0}^{\infty} e^{a_n} \frac{(e^x - e^l)^n}{n!} = \sum_{d=0}^{\infty} b_d \frac{(x+1-l)^d}{d!}$ and subtract the right hand side from the left hand side $\sum_{n=0}^{\infty} \frac{e^{a_n}(e^x - e^l)^n - b_n (x+1-l)^n}{n!} = 0$ and now, since x is essentially arbitrary, let x = l to give: $\sum_{n=0}^{\infty} \frac{b_n}{n!} = 0$ but this contradicts the original Taylor series expansion $sexp(x) = \sum_{n=0}^{\infty} b_n \frac{(x-l)^n}{n!}$ which states: $\sum_{n=0}^{\infty} \frac{b_n}{n!} = sexp(l+1)$ Any help would be greatly appreciated, thanks. The only solution I have is $l = -2$ bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/21/2011, 08:04 PM (This post was last modified: 05/21/2011, 08:04 PM by bo198214.) (05/12/2011, 07:53 PM)JmsNxn Wrote: $x \bigtriangledown y = \ln(e^x - e^y)$ One problem could be that if you let x=l then $x \bigtriangledown l = -\infty$. And calculating with infinities is always somewhat risky. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/26/2011, 02:21 AM (This post was last modified: 05/26/2011, 02:29 AM by JmsNxn.) oh, that's obvious there in the first equation. I hate missing stuff that's right in front of me. yeah, I've really started to notice that infinity is difficult to work with. It frustrated me at first but then I realize it comes with the territory. « Next Oldest | Next Newest »

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