# What is the correct terminology to say that $\small f(x)=a+bx+cx^2+…$ can be expressed by $\small g(x)=A(1-x)+B(1-x)(2-x)+C(1-x)(2-x)(3-x)+…$

Hm, I do not even know the best formulation for my question in the header. It is not for the math but for the proper writing/terminology. I've come across the term "base change" recently but the context was a bit different, so I don't know whether this is possibly correct/incorrect here.

As I've said in the subject, I have two expressions of the same function; one time I express it as a power series in x, say $\small f(x) = a + bx + cx^2+dx^3 + ...$, and in a certain article I find the same problem handled, but with a formula like $\small g(x)=A + B(1-x) + C(1-x)(2-x) + D(1-x)(2-x)(3-x) + ...$ (the actual coefficients don't matter here)

If I expand $\small g(x)$ and collect like powers of x to make a power series of it, it is expected, that that power series has the same coefficients as f(x) (or: "they are identical"). (There is a problem in it, that the expansion leads to divergent sums for x but that need not be discussed here). Let's assume, I'm correct and the series come out to be identical. Btw, I know that the transformation behind this involves the Stirling numbers 1st kind.

My question is: how do I write in a small article, that the series f(x) is expressible by g(x) and vice versa? Perhaps "g(x) is a Stirling-transformation on f(x)" ? or "We do a change-of-basis from f(x) to g(x)" ?

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For "vice versa" I worry a little: Suppose $1=A=B=C=\ldots$, then $g(0) = \sum_{n=0}^\infty n!$ or so, but $f(0) = a$. Not all $g$ can be expressed as $f$, I think. – Jack Schmidt Oct 17 '12 at 19:38
I think the divided difference formula mentioned by lhf may mean that every $f$ that converges at all positive integers has a $g$, but that non-convergent $f$ (like $f(x)=1/(1.5-x)$) will have trouble. – Jack Schmidt Oct 17 '12 at 19:45

$f$ is expressing a polynomial in the monomial basis.
$g$ is expressing the same polynomial in the Newton basis using data points in $\mathbb N$.
The Newton basis is actually $(x-1)(x-2)\cdots$, so $g$ is that except for a change of sign in the coefficients. – lhf Mar 14 '12 at 12:13
Ah, thanks. Can I use this also for the case, that I have not a polynomial but a series? And also, is the reference to "Newton" replacable by some other reference, since I've actually $\small (1-x),(1-x)(a-x),(1-x)(a-x)(a^2-x),...$ where a is some additional problemspecific fixed parameter ? – Gottfried Helms Mar 14 '12 at 12:24