Notes on A091736 from Enrico T. Federighi (rico125162(AT)aol.com), Feb 02 2004 and Nov 12 2007
By the canonical half-exponential function hexp(x) I mean a power series
similar to the exponential function but half way. If r^f(x)=f(x+1) with f(-1)=0,
then f(-1/2)=hexp(sqrt(ln(r)). There are several ways of defining f(-1/2)
but the most natural way is the one that comes closest to (1-sqrt(ln(r))/(1-(ln(r))
which this gives. If one considers in Maple6 for ln(r) the following:
hf := x->a+b*x+c*x^2+d*x^3; ff := x->hf(hf(x)); series(ff(x),x,4); c0 := coeff(ff(x),x,0);
eq1 := c0=1; c1 := coeff(ff(x),x); eq2 := c1=ln(r); c2 := coeff(ff(x),x,2); eq3 := c2=1/2*ln(r)^2;
c3 := coeff(ff(x),x,3); eq4 := c3=1/6*ln(r)^3; fsolve({eq1,eq2,eq3,eq4},{a,b,c,d},-.01..1);
The solution of a will be close to the f(-1/2) which we seek. If more terms are taken the answer will converge.
To clarify my last statement with r^f(x)=f(x+1), f(-1)=0. Define f(-1/2) by the series given and then f(1)=x, and f(2)=exp(x) and f(3/2)=hexp(x) which indeed is half-way.
Let b be a tower base and h a tower height where ^^ is defined as follows:
b^^(-1)=0 and b^^(h+1)=b^(b^^h). A091736 gives the
values for h=-1/2. Suppose b=1.20. Then x would be
sqrt(ln(b))=sqrt(ln(1.20)). Using A091736 and A091737
we obtain 1.20^^(-1/2)=.70065. The true number is
.700654152555. This series can only be used when
1.01