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I'm trying to prove the following proposition:

Let $U\in R^n$ be open, and $f,g\colon U\to R$ be $C^k$ functions, then the Taylor polynomial of $fg$ is computed as $P_{f,a}^k(a+\vec{h})\cdot P_{g,a}^k(a+\vec{h})$ and discarding the terms of degree > $k$, where $P_{f,a}^k(a+\vec{h})$ denotes the degree $k$ Taylor polynomial of $f$ at $a$.

And here's what I've got so far:

$$P_{fg,a}^k=\sum_{m=0}^k\sum_{I\in I_n^m}\frac{1}{I!}D_I(fg)(a)$$ using the definition of Taylor polynomial of multi-variable functions. ($I=(i_1,i_2,...,i_n)$ and $D_If=D_1^{i_1}D_2^{i_2}...D_n^{i_n}f)$.

Then I think $D_I(fg)$ can be written as $D_I(f)g+fD_I(g)$ using the product rule so that $$P_{fg,a}^k(a+\vec{h})=P_{f,a}^k(a+\vec{h})\cdot g+f\cdot P_{g,a}^k(a+\vec{h})$$

but I can't figure out how this leads to the final result.

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I think each $D_I(fg)$ can be represented using the multinomial theorem. –  bgins Dec 12 '11 at 17:32
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up vote 1 down vote accepted

You're trying to do things the hard way. To save myself some typing, I'll write $ a+h = x$, $P^k_{f,a}(a+h) = T_f(x)$ and similarly for $g$ and $fg$. We have $f(x) = T_f(x) + o(|h|^k)$ and similarly for $g$ and $fg$. So $$T_f(x) T_g(x) - T_{fg}(x) = (f(x) + o(|h|^k))(g(x) + o(|h|^k)) - (f(x) g(x) + o(|h|^k)) = o(|h|^k)$$

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thank you for this simple and clear proof. Let me ask one more question, what you did above shows that $T_f(x)T_g(x)$ is the best approximation to $T_{fg}(x)$, but not necessarily equal to each other, correct? Is my question a valid question? (I'm not a math major and trying to teach myself some advanced calculus) –  user11869 Dec 12 '11 at 19:53
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