 Solution

Clearing the denominators in the rational interpolation conditions givesThese are $n+1$ linear equations for $(m+1)+\ell =n+1$ unknowns. The $i$th row of the coefficient matrix, in the order $a_0,\dots ,a_m,b_0,\dots ,b_{\ell -1}$, isThis rational interpolation linearization is equivalent to the original fraction conditions only if $r(x_i)\ne 0$ at every interpolation node. If $r(x_i)=0$, the cleared equation merely forces $q(x_i)=0$, leaving the fraction undefined.

For $f_i=x_i^{m+1}$, the coefficient row becomesThis is the Vandermonde matrix, whose determinant is ${\prod }_{0\le i<j\le n}(x_j-x_i)\ne 0$. Equivalently, a vector in its null space would define a degree-at-most-$n$ polynomial vanishing at all $n+1$ distinct interpolation nodes, which is zero by the preceding uniqueness proof. Thus the cleared linear system has exactly one solution.

There is also a constructive description of that solution. Define the monic polynomialThe polynomial $x^{m+1}r(x)-q(x)$ is monic of degree $n+1$ and vanishes at every interpolation node. It must therefore equal $W(x)$. By polynomial division, write uniquelyThe quotient $R$ is monic of degree $\ell$, andis the unique coefficient solution. This does not assert that $R$ is nonzero at every interpolation node.

For a counterexample with $m=0$, $\ell =1$, choose $x_0=0,x_1=1$ and the specified data $f_i=x_i$, so $f_0=0,f_1=1$. The cleared equations reduce togiving the unique solution $q(x)=0$, $r(x)=x-1$. But at $x_1=1$ the quotient is $0/0$. Indeed, any genuine rational interpolant of this prescribed form would need $r(0)\ne 0$; then $q(0)/r(0)=0$ forces the constant numerator to be zero. At a nonzero denominator its value at $1$ would also be zero, never one. Cancelling the common factor in the cleared candidate would merely give the identically zero rational function, which still fails the data. Hence