Consider the bivariate function f(x,y) with Taylor series development

f(x,y)=\sum _{i,j=0}^{\mathrm{\infty}}{c}_{ij}{x}^{i}{y}^{j}

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around the origin. We know that a solution of the univariate Padé approximation problem for

f(x)=\sum _{i=0}^{\mathrm{\infty}}{c}_{i}{x}^{i}

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is given by

p(x)=\left|\begin{array}{cccc}{\sum}_{i=0}^{m}{c}_{i}{x}^{i}& x{\sum}_{i=0}^{m-1}{c}_{i}{x}^{i}& \cdots & {x}^{n}{\sum}_{i=0}^{m-n}{c}_{i}{x}^{i}\\ {c}_{m+1}& {c}_{m}& \cdots & {c}_{m+1-n}\\ \vdots & \vdots & \ddots & \vdots \\ {c}_{m+n}& {c}_{m+n-1}& \cdots & {c}_{m}\end{array}\right|

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and

q(x)=\left|\begin{array}{cccc}1& x& \cdots & {x}^{n}\\ {c}_{m+1}& {c}_{m}& \cdots & {c}_{m+1-n}\\ \vdots & \vdots & \ddots & \vdots \\ {c}_{m+n}& {c}_{m+n-1}& \cdots & {c}_{m}\end{array}\right|.

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Let us now multiply the *j* th row in p(x) and q(x) by {x}^{j+m-1} (j=2,\dots ,n+1) and afterwards divide the *j* th column in p(x) and q(x) by {x}^{j-1} (j=2,\dots ,n+1). This results in a multiplication of numerator and denominator by {x}^{mn}. Having done so, we get

\frac{p(x)}{q(x)}=\frac{\left|\begin{array}{cccc}{\sum}_{i=0}^{m}{c}_{i}{x}^{i}& {\sum}_{i=0}^{m-1}{c}_{i}{x}^{i}& \cdots & {\sum}_{i=0}^{m-n}{c}_{i}{x}^{i}\\ {c}_{m+1}{x}^{m+1}& {c}_{m}{x}^{m}& \cdots & {c}_{m+1-n}{x}^{m+1-n}\\ \vdots & \vdots & \ddots & \vdots \\ {c}_{m+n}{x}^{m+n}& {c}_{m+n-1}{x}^{m+n-1}& \cdots & {c}_{m}{x}^{m}\end{array}\right|}{\left|\begin{array}{cccc}1& 1& \cdots & 1\\ {c}_{m+1}{x}^{m+1}& {c}_{m}{x}^{m}& \cdots & {c}_{m+1-n}{x}^{m+1-n}\\ \vdots & \vdots & \ddots & \vdots \\ {c}_{m+n}{x}^{m+n}& {c}_{m+n-1}{x}^{m+n-1}& \cdots & {c}_{m}{x}^{m}\end{array}\right|}

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if (D=det{D}_{m,n}\ne 0).

This quotient of determinants can also immediately be written down for a bivariate function f(x,y). The sum {\sum}_{i=0}^{k}{c}_{i}{x}^{i} will be replaced by the *k* th partial sum of the Taylor series development of f(x,y) and the expression {c}_{k}{x}^{k} by an expression that contains all the terms of degree *k* in f(x,y). Here a bivariate term {c}_{ij}{x}^{i}{y}^{j} is said to be of degree i+j.

If we define

p(x,y)=\left|\begin{array}{cccc}{\sum}_{i+j=0}^{m}{c}_{ij}{x}^{i}{y}^{j}& {\sum}_{i+j=0}^{m-1}{c}_{ij}{x}^{i}{y}^{j}& \cdots & {\sum}_{i+j=0}^{m-n}{c}_{ij}{x}^{i}{y}^{j}\\ {\sum}_{i+j=m+1}{c}_{ij}{x}^{i}{y}^{j}& {\sum}_{i+j=m}{c}_{ij}{x}^{i}{y}^{j}& \cdots & {\sum}_{i+j=m+1-n}{c}_{ij}{x}^{i}{y}^{j}\\ \vdots & \vdots & \ddots & \vdots \\ {\sum}_{i+j=m+n}{c}_{ij}{x}^{i}{y}^{j}& {\sum}_{i+j=m+n-1}^{m}{c}_{ij}{x}^{i}{y}^{j}& \cdots & {\sum}_{i+j=m}^{m}{c}_{ij}{x}^{i}{y}^{j}\end{array}\right|

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and

q(x,y)=\left|\begin{array}{cccc}1& 1& \cdots & 1\\ {\sum}_{i+j=m+1}{c}_{ij}{x}^{i}{y}^{j}& {\sum}_{i+j=m}{c}_{ij}{x}^{i}{y}^{j}& \cdots & {\sum}_{i+j=m+1-n}{c}_{ij}{x}^{i}{y}^{j}\\ \vdots & \vdots & \ddots & \vdots \\ {\sum}_{i+j=m+n}{c}_{ij}{x}^{i}{y}^{j}& {\sum}_{i+j=m+n-1}^{m}{c}_{ij}{x}^{i}{y}^{j}& \cdots & {\sum}_{i+j=m}^{m}{c}_{ij}{x}^{i}{y}^{j}\end{array}\right|,

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then it is easy to see that p(x,y) and q(x,y) are of the form

\begin{array}{r}p(x,y)=\sum _{i+j=mn}^{mn+m}{a}_{ij}{x}^{i}{y}^{j},\\ q(x,y)=\sum _{i+j=mn}^{mn+n}{b}_{ij}{x}^{i}{y}^{j}.\end{array}

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We know that p(x,y) and q(x,y) are called Padé equations [3, 14]. So, the multivariate Padé approximant of order (m,n) for f(x,y) is defined as

{r}_{m,n}(x,y)=\frac{p(x,y)}{q(x,y)}.

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