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I've been told that we can write a taylor series for functions $f:R^n\rightarrow R$ but we can't write one for $f:R^n\rightarrow R^n$. I'm not quite sure why this not possible, but I suspect it have something to do with the mean value theorem. Could anyone shed some light on this?

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  • $\begingroup$ Actually you can write Taylor series for mappings between arbitrary Banach spaces. You need to be careful in estimating the remainder term, though. A readable introduction to this is S. Lang Undergraduate analysis, chapter "Derivatives in Vector Spaces". $\endgroup$ Commented Jan 15, 2013 at 21:38
  • $\begingroup$ By the way, the subtlety I was referring to is exactly the fact that you do not have the following form of the mean value theorem when $f$ is not real-valued: $$f(x+h)-f(x)=f'(x+\theta h)h,\quad \text{some}\ \theta\in (0, 1).$$ In this sense your suspicion hits the point. $\endgroup$ Commented Jan 15, 2013 at 21:41
  • $\begingroup$ @GiuseppeNegro: What form does it take then? $\endgroup$
    – Paul
    Commented Jan 15, 2013 at 23:00
  • $\begingroup$ The integral form of the remainder is true even for vector-valued functions. $\endgroup$ Commented Jan 15, 2013 at 23:26

3 Answers 3

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If $f : \mathbb{R}^n \to \mathbb{R}^n$, then you can split $f = (f_1, f_2, \ldots, f_n)$, and (under the usual smoothness assumptions) each component can be expanded in a Taylor series.

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For each $f_j:\mathbb{R}^n\to\mathbb{R}$ in $f=(f_1,\ldots,f_j,\ldots,f_n)$ use a notation for multi index $\alpha=(\alpha_1,\ldots,\alpha_i,\ldots,\alpha_n)$ in Higher-order partial derivatives, norms, factorials $!$ and coordinates of vectors $x,v\in\mathbb{R}^n$. Then for $f:\mathbb{R}^n\to\mathbb{R}^n$ we have Taylor series of $f(x+v)=$ \begin{align} \bigg(\sum_{k=0}^{\infty}\sum_{|\alpha|=k}\frac{1}{\alpha!}\partial^\alpha f_1(x)\cdot v^\alpha,\ldots,\sum_{k=0}^{\infty}\sum_{|\alpha|=k}\frac{1}{\alpha!}\partial^\alpha f_j(x)\cdot v^\alpha,\ldots, \sum_{k=0}^{\infty}\sum_{|\alpha|=k}\frac{1}{\alpha!}\partial^\alpha f_n(x)\cdot v^\alpha\bigg) \end{align}

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  • $\begingroup$ coodenadas $\mapsto$ coordinates $\endgroup$
    – Pedro
    Commented Jan 16, 2013 at 0:23
  • $\begingroup$ @PeterTamaroff Tank's. $\endgroup$ Commented Jan 17, 2013 at 0:38
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The same formula that applies for scalar fields applies for vector fields.

$$f(x) = f(a) + [(x-a) \cdot \nabla] f \Big|_a + \frac{1}{2!} [(x-a) \cdot \nabla]^2 f \Big |_a + \ldots$$

$f$ can be scalar or vector.

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