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 \( l = -2 \)
(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 \( l = -2 \)