Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)" ?

share|improve this question
    
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

1 Answer 1

up vote 5 down vote accepted

$f$ is expressing a polynomial in the monomial basis.

$g$ is expressing the same polynomial in the Newton basis using data points in $\mathbb N$.

share|improve this answer
    
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
    
But he doesn't describe a polynomial, he describes a power series. While any polynomial can be written in the Newton basis, that's not true for any power series, I think. –  Thomas Andrews Mar 14 '12 at 12:26
    
Oh, I missed that it was a power series. Sorry for the noise. –  lhf Mar 14 '12 at 12:29
    
@ThomasAndrews, I'd be very interested in seeing that result about power series. –  lhf Mar 14 '12 at 12:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.