# Transposing the coefficients of a bivariate generating function

Consider a triangular of integers¢ $a_{1,0}$ ¢ $a_{2,1}, \, a_{2,0}$ ¢ $a_{3,2}, \, a_{3,1}, \, a_{3,0}$ ¢ $\quad \vdots$ ¢ and also its "transpose" ¢ $a_{1,0}$ ¢ $a_{2,0}, \, a_{2,1}$ ¢ $a_{3,0}, \, a_{3,1}, \, a_{3,2}$ ¢ $\quad \vdots$ ¢ Given a bivariate getting function $f(x,y)$ with collection $$ a_{1,0}*y + (a_{2,1}x + a_{2,0})y^2 + (a_{3,2}x^2 + a_{3,1}*x + a_{3,0})y^3 + \cdots $$ what is the creating function which offers the coefficients of the polynomials of $x$ as the shifted triangular, i.e. just how can $g(x,y)$ be gotten with collection $$ a_{1,0}y + (a_{2,1} + a_{2,0}x)y^2 + (a_{3,2} + a_{3,1}x + a_{3,0}x^2)y^3 + \cdots $$

$g(x, y) = \frac{1}{x} f \left( \frac{1}{x}, xy \right)$. First you need to recognize that there is an extra standard method below where you turn around the coefficients of a polynomial by replacing $x \mapsto \frac{1}{x}$ and also clearing up , and also afterwards it is not tough to fine-tune every little thing else to exercise.

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