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In the proof of Taylor's theorem, 5.15 in Rudin, it is stated that for continuously differentiable function $f:[a,b]\to\mathbb{R}$ and $P(\beta) = \sum_{k=0}^{n-1}\frac{f^{(k)}(\alpha)}{k!}(\beta - \alpha)^k$, we have $$f^{(k)}(\alpha) = P^{(k)}(\alpha),\quad k=0,1,\dots$$

This seems to be based on the fact that $P(\alpha) = f(\alpha)0^0$ (assuming I got it correctly), which confuses me a lot, Zero to the zero power - is $0^0=1$?

Community
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Aad
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    In this context, yes, by convention. Or if you prefer, you could write $$P(\beta) = f(\alpha) + \sum_{k=1}^{n-1} \frac{f^{(k)}}{k!} (\beta-\alpha)^k$$ – Tryss Oct 10 '15 at 15:07

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In general when working with polynomials and power series using sigma notation, one adopts the convention that $0^0 = 1$ for notation convenience.

Dan
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