The generating function is given by, $$\phi(x,h)= (1-2hx+h^2)^{-\frac 12}$$ where $|h|\le1$. Why is it that, $$\phi(1,h)= (1-2h+h^2)^{-\frac 12}=\frac 1{1-h}$$ and not, $$\frac 1{h-1} ?$$
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When $|h|$ is small and $h$ is real, $1-h$ is positive while $h - 1$ is negative. By convention, $x^{1/2}$ (for $x > 0$) refers to the positive square root of $x$, and $x^{-1/2}$ is the reciprocal of that.
Robert Israel
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Since $1 - 2h + h^2 = (1 - h)^2$, $\phi(1,h) = |1 - h|^{-1}$. If $|h| < 1$, in particular, $h < 1$. Therefore, $1 - h > 0$ and $|1 - h|^{-1} = (1 - h)^{-1}$, or $\frac{1}{1 - h}$.
kobe
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