# Why the expansion of functions of 2 variables in legendre polynomials does not take into account all products Pi (x) Pj (y)?

My question arises from this reading link (In this article, in equation 6 on page 3, also expands a function of 2 variables in this way Parametric estimate of intensity inhomogeneities applied to MRI), where on page 10 expand a function of 2 variables in a sum of two-dimensional legendre polynomials:

My question is why in this case they do not use all the orthogonal combinations Pi (x) Pj (y) in the summation as is usual in other cases such as for example the discrete expansion of fourier:

• In English, no need for an interrogation point in front of a sentence. – Jean Marie Sep 30 '17 at 21:11
• This approach results in a polynomial of maximal degree $n$, i.e. the powers of $x$ and $y$ add to at most $n$. If one uses all combinations $P_i(x)P_j(x)$ with $i, j \le n$, then there could be terms such as $x^4y^4$ if $n = 4$, but no terms $x^6y^2$. – Hans Engler Sep 30 '17 at 21:22
• yes, it is true what you mention, in fact in the expansion in series of taylos in 2 variables for example, the polynomials that appear in each of the summands have the same degree, for example in the component of second order appear the polynomials $x^{2}$,$xy$ and $y^{2}$. – Roger Figueroa Quintero Sep 30 '17 at 22:15
• I will have to study a little more about why this way of combining the polynomials results in a base generating the vector space of the real functions. – Roger Figueroa Quintero Sep 30 '17 at 22:23