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Let us suppose the $i^{th}$ partial derivative of $f:\Bbb{R}^n\to \Bbb{R}$ exists at $P$; i.e. if $P=(x_1,x_2,\dots,x^n)$, $$\frac{f(x_1,x_2,\dots,x_n+\Delta x_n)-f(x_1,x_2,\dots,x_n)}{\Delta x_n}=f'_n (P)$$

My book says this implies that $$f(x_1,x_2,\dots,x_n+\Delta x_n)-f(x_1,x_2,\dots,x_n)=f'_n(P)\Delta x_n + \epsilon_n \Delta x_n$$

such that $\lim\limits_{\Delta x_n\to 0} \epsilon_n=0$.

I don't understand where $\epsilon_n$ comes into the picture. Why can't we just have $f(x_1,x_2,\dots,x_n+\Delta x_n)-f(x_1,x_2,\dots,x_n)=f'_n(P)\Delta x_n$, considering we're anyway using $\Delta x_n$ as a real number rather than an operator.

Justification for asking on overflow- I'm doing research on multi-variable calculus..?

Thanks!

user92664
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  • But since your research is not about research level mathematics I fear the question is going to be closed as well... –  Sep 04 '13 at 19:41
  • I was scared this would happen. But before this question is closed, any helpful comments would be just great. I've been stuck on this for a very long time. – user92664 Sep 04 '13 at 19:43
  • A hint: look up the definition of partial derivative. –  Sep 04 '13 at 20:24
  • There is a "lim" symbol missing in your first displayed formula. – Etienne Sep 04 '13 at 21:28

2 Answers2

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It would be more useful to understand why the $\varepsilon$ appears in dimension one. We define $\varepsilon(h):=\frac{f(x+h)-f(x)}h-f'(x)$ for $h\neq 0$. Then by definition of the derivative, $\lim_{h\to 0}\varepsilon(h)=0$, and $f(x+h)=f(x)+hf'(x)+h\varepsilon(h)$ but in genereal $\varepsilon(\cdot)$ is not identically $0$.

Davide Giraudo
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Compare the definition of the derivative to your ratio when $f:\mathbb R\to\mathbb R$, $x\mapsto x^2$.

Did
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