3

I'm reading My Numbers, My Friends by Paulo Ribenboim and I've encountered this:

Thus $U_n = f_n(P,Q)$, where $f_n(X,Y) \in \mathbb{Z}[X, Y]$. The function $f_n$ is isobaric of weight $n-1$, where $X$ has weight 1 and $Y$ has weight 2.

The topic of the paragraph is the Fibonacci numbers, $P$ and $Q$ are taken from the definition of discriminant: $D = P^2 - 4Q$ and $U_n$ is the first Lucas sequence.

Can you please explain what the author means when he says $f_n$ is isobaric?

Gerry Myerson
  • 179,216
wvxvw
  • 1,011
  • A function used at constant pressure is called isobaric? xD – AvZ Feb 28 '15 at 10:43
  • @AvZ sounds pretty close, since X and Y are said to have weights :) but I'm still not sure whether this was a typo, and if it was, then what was originally meant? – wvxvw Feb 28 '15 at 10:49

1 Answers1

3

$f(X,Y)$ is a polynomial, so it's a sum of terms of the form $c_{ij}X^iY^j$. Isobaric of weight $n-1$ means $i+2j=n-1$ for each of these terms.

Gerry Myerson
  • 179,216
  • 2
    You may wish to write "Isobaric of weight n−1 where $X$ has weight $1$ and $Y$ has weight $2$ means", instead. Otherwise it is not clear why you are taking $i+2j$ instead of, say $i+j$. – A.P. Feb 28 '15 at 11:46
  • I think I'm getting closer, but, can you extend your answer please by saying what $n$ would be in this case? In other words, are you saying that the number of terms of the polynomial is related to the index in the sequence? (that is my understanding now). – wvxvw Feb 28 '15 at 11:47
  • @A.P, of course you are correct. I am relying on the reader to remember that condition from the question statement. – Gerry Myerson Feb 28 '15 at 11:47
  • I'm not saying anything about the number of terms in the polynomial. The weight of the polynomial is one less than its index in the sequence. – Gerry Myerson Feb 28 '15 at 11:50
  • Sorry, the weight of a polynomial is a new concept for me. Is it the highest power of a polynomial? – wvxvw Feb 28 '15 at 11:52
  • @wvxvw: Not just that. He's saying that the monomials in $f_n$ have a very special form. In particular, $f_n$ does indeed have degree $n-1$. – A.P. Feb 28 '15 at 11:53
  • No. The weight, in this case, is $i+2j$, which is $n-1$. $X$ has weight 1, $Y$ has weight 2, so $c_{ij}X^iY^j$ has weight $i+2j$. – Gerry Myerson Feb 28 '15 at 11:54
  • @A.P, no, the polynomial doesn't (necessarily) have degree $n-1$. – Gerry Myerson Feb 28 '15 at 11:55
  • Sorry, @GerryMyerson that's not really explaining it. I don't know what property of a polynomial you are referring to. Are there perhaps references online / in the literature that I can read about it? – wvxvw Feb 28 '15 at 11:56
  • Not explaining what? It's a bloody polynomial. It has monomials of the form $c_{ij}X^iY^j$. And the restriction on the pairs $i,j$ that it's allowed to have is that $i+2j$ must equal $n-1$. What exactly don't you get? – Gerry Myerson Feb 28 '15 at 11:58
  • 1
    @GerryMyerson Ah, yes... I left out an "at most": $\deg(f_n) \leq n-1$. – A.P. Feb 28 '15 at 12:01
  • 1
    Oh, sorry, now I figured it. Took me a while... thanks both of you, A.P and Gerry Myerson! – wvxvw Feb 28 '15 at 12:09