Is there a standard name for multivariate polynomials wherein each term consists of only one coordinate? That is, polynomials of this form: $p(x_1, \ldots, x_n)= \sum_{i=1}^n \sum_{j=1}^n a_{i,j}x_i^{j}$ where all of the $a_{i,j}$ are constants.
Multivariate polynomial with no mixed terms
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0Good point, you capture a subtlety I've missed. I've changed it accordingly. – 2012-06-01
3 Answers
Why not just call it a "sum of univariate polynomials"? [I see G.H. already suggested this.]
If the polynomial were a form (that is, every monomial had the same total degree), then it would be called "diagonal" if it had no fixed terms. So you could also called it a sum of diagonal forms.
A Google search says:
- no hits for "polynomial(s) with no mixed terms"
- 5 hits for "polynomial(s) without mixed terms"
- ~30 hits for "sum(s) of univariate polynomials"
This isn't very high, so maybe there is a better term.
EDIT: since no one else is making suggestions, here is what I think of the terms.
Multivariate polynomial without mixed terms is good but I'm not sure everyone has in mind what a "mixed term" is, so you might need to recall the definition.
Sum of univariate polynomials is shorter and only uses very well-known definitions. The only non-obvious part is that it mentally requires you to embed $R[x_i]$ into $R[x_1,\dots,x_n]$. However since $R[x]$ is already stable under addition, it should be fairly clear that we're not looking for a univariate sum, so there is no ambiguity.
I think this is called the normal form for the polynomial $p$. Suppose $p=\sum a_{ij}x_i x_j$ is a homogeneous degree 2 polynomial in $k[x_1,\ldots, x_n]$. Then there is a systematic procedure to rewrite the polynomial so that $p=\sum_{i=1}^n c_i x_i^2$. The procedure is basically using "completing the square" multiple times. See page 402 in Ideals, Varieties, and Algorithms in Cox, Little, and O'Shea.
As of the moment, I believe that you could rearrange the terms so that $p=\sum_{i=1}^k p_i$, where each $p_i$ is homogeneous of degree $i$ and apply an analogous concept to each $p_i$ to obtain
$p(x_1, \ldots, x_n)= \sum_{i=1}^n \sum_{j=1}^n a_{i,j}x_i^{j}.$
I hope this helped.
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1I think li$f$e is much more complicated when the de$g$ree is larger than 2. – 2012-06-06