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We have the following functions, $g(x) = \sum_{n=0}^\infty {k \choose n}x^n$ and $h(x) = (1+x)^{-k}g(x)$. We have to deduce that $g(x) = (1+x)^k$ (in previous exercises we showed that $h'(x)=0$).

So what I assumed to be a logical course of action was saying that $g(x) = h(x)(1+x)^k$, where $h(x) = c$ for a certain constant $c$ because $h'(x) = 0$. In the case that $x=0$, we'd have $h(0) = 1^k g(0)$. Since $g(0) = 1$, we'd have $h(0) = h(x) = 1$, and we'd arrive to our solution.

However, here we used $0^0=1$ to conclude that $g(0)=0$, which is controversial. I know that $0^0=1$ in the context of Newton's binomium, but is there a way of reaching the same conclusion without using this fact?

2 Answers 2

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$g(x) = \sum_{n=0}^\infty {k \choose n}x^n = \sum_{n=0}^\infty {k \choose n}x^n 1^{k-n} = (1+x)^k$ where the last equality is just the binomial formula.

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In the definition of $g(x):=\sum_0^n{k\choose n}x^n$ there is the question of how to interpret $x^n$ when $n=0$. It is safe to assume that $x^0=1$ for all $x$, including for $x=0$; otherwise we won't have continuity of $g(x)$, and if $g(x)$ isn't continuous then $h(x)$ won't be differentiable.