Math time!

No there isn't, but for most people \varepsilon and \varnothing look better than \epsilon and \emptyset

The Taylor series centered on 'a' will converge for all points in the open ball B_a(a). Do you even complex variables?

[math] \ln(x) = \ln(a) - \sum^{ \infty } _ {1} (-1)^{n} \frac{ (x-a)^n } { n a^n } [/math]

[math] radius = \lim_{n \to \infty} \frac { \frac{ 1 } { n a^n } } { \frac{ 1 } { (n+1) a^{n+1} } } = a [/math]

1. ln(x) has a vertical asymptote at x=0
2. The slope of ln(x) decreases as x->∞
A Taylor expansion can't satisfy the first requirement and can only satisfy the second over a limited range.

f(x)=sqrt(x)-1/sqrt(x) satisfies both of those properties, and provides a reasonable approximation for small x (and f(1)=0).

No, you're off by 1 and forgot the (x-1)^n term

You failed it.

This is taught in calc 2. Its first applicatiom is in diffeq with eulers equation for solving linear, second degree, constant coefficient, differential equations when the factorization of the auxillary equation is r=a+/-bi. Even then, you just produce another equation to memorize. You should know this or not comment about your ignorance.

I have wondered about the
following logarithmic functions:
L(x, 1) = ln(x)
L(x, 2) = ln(ln(x)+1)
L(x, 3) = ln(ln(ln(x)+1)+1)
...
L(x, n) = ln(L(x, n-1)+1)
... and how to represent
an intermediate function
L(x, r) where r = a/b
for whole number _a_ and _b_

should have stated
"posint _a_ and _b_"
late night, sorry

interpolation.