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The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ as series does not converge, and the $g(x)$ expression goes to infinity as $x\to 1^-$. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ and $g(1)$ is infinity and the series does not converge.

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ as series does not converge, and the $g(x)$ expression goes to infinity as $x\to 1^-$. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ so $g(x)\to -\infty$ as $x\to 1$ and the series does not converge.

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ does not converge. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ and $g(1)$ is infinity so the series does not converge.

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ as series does not converge, and the $g(x)$ expression goes to infinity as $x\to 1^-$. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ and $g(1)$ is infinity and the series does not converge.

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ does not converge.

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ does not converge. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ and $g(1)$ is infinity so the series does not converge.