Taylor Product?

We learned, so long ago we've probably forgotten when, about the Taylor Series. $$\begin{equation} f(x) = f(x_0) + (1/2)(x-x_0) f^{'}(x_0) + (1/6)(x-x_0)^2 f^{''}(x_0) + ... \end{equation}$$ If the derivatives are large, this might not converge very quickly (if at all). If the function $f$ is reasonably well behaved, and positive, we can try looking at products instead. Never mind the complex logarithms for now. $$\begin{eqnarray} f(x) = e^{log(f(x))} \\ g(x) \equiv log(f(x)) \\ g(x) = g(x_0) + (x-x_0)g^{'}(x_0) + (1/2)(x-x_0)^2 g^{''}(x_0) + \dots \\ f(x) = f(x_0) e^{(x-x_0)g^{'}(x_0)} e^{(1/2)(x-x_0)^2 g^{''}(x_0)} \dots \end{eqnarray}$$ Will this converge any faster? For a distance use the difference between the approximation so far and the true value, divided by the true value. Pick a couple of simple examples: $f(x)= e^{x}$ and $f(x) = x^2$. The first one converges much faster with a "Taylor Product" $$\begin{eqnarray} f(x) = e^x \\ g(x) = x \\ g(x) = x_0 + (x-x_0) 1 + 0 + 0 + 0 \dots \\ f(x) = f(x_0) e^{(x-x_0)} \times 1 \times 1 \dots \end{eqnarray}$$ The "distances" for the approximations are $$\begin{eqnarray} (f(x)-f(x_0))/f(x) = 1 - e^{x-x0} \\ (f(x)-f(x_0) e^{(x-x_0)})/f(x) = 0 \\ 0 \\ \dots \end{eqnarray}$$ The second function example is, of course, much easier to approximate with a Taylor Series; you only need three terms for it to be exact. $$\begin{equation} f(x) = x_0^2 + (x-x_0)\times 2x_0 + (1/2) (x-x_0)^2 \times 2 = x^2 \end{equation}$$ But never mind that; let's use the "Taylor Product" anyway. Here $g(x) = 2\log(x)$ If we let $x=x_0+1/2$, and let $x_0 = 1$

order	g deriv	$g^{(n)}$ at $x_0=1$	scale term	f cumulative
		at $x_0=1$		error frac
0	$2\log(x)$	0	1	.555
1	$2/x$	2	2.718	-.208
2	$-2/x^2$	-2	.7788	.059
3	$4/x^3$	4	1.0869	-.023
4	$-12/x^4$	-12	.9692	.009
...	...	...	...	...

Suppose instead that $x=x_0 + 1/2$ but $x_0 = 1000$. It probably won't surprise you to see that it converges faster, using the given distance measure. \hline

order	g deriv	$g^{(n)}$ at $x_0=1$	scale term	f cumulative
		at $x_0=1$		error frac
0	$2\log(x)$	0	1000000	.001
1	$2/x$	.002	1.0010005	-2.5 E-7
2	$-2/x^2$	-.000002	.999999	8.3 E-11
3	$4/x^3$	$4\times 10^{-9}$	1.00000	-3.1 E-14
...	...	...	...	...

I don't know what this is actually called, and search engines turned up reams of irrelevancies. On a related note, MathJax in Blogger doesn't understand tabular mode.