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There is a formula that describes this expansion using big O notation, I'm very curious on how this is derived. I also understand that the order term may very depending on what $\Delta x$ approaches to in limit ?

$(x + \Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + O((\Delta x)^{2})$

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marked as duplicate by Antonio Vargas, achille hui, T. Bongers, Shuhao Cao, Pedro Tamaroff Aug 25 '13 at 19:07

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Could you state what the formula is? –  T. Bongers Aug 25 '13 at 18:26
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You'll use $O$ depending on what you want to "discard". Maybe a little detail is needed to really understand what you want. –  Pedro Tamaroff Aug 25 '13 at 18:26
    
Hint : It's easy. Just use Newton expansion. –  Zia Aug 25 '13 at 18:31

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up vote 2 down vote accepted

You use the binomial theorem, that states that $$(x+\Delta x)^n=\sum_{k=0}^n\binom nk x^{n-k}(\Delta x)^{k}$$

When $k\geq 2$, you'll always get terms $\Delta x$ in powers of two, three, four, etc, so you can write $$(x+\Delta x)^n=x^n+nx^{n-1}\Delta x+O((\Delta x)^2)$$

The term being "omitted" is

$${(\Delta x)^2}\sum\limits_{k = 2}^n \binom nk{x^{n - k}}(\Delta x)^{k - 2} = {(\Delta x)^2} \left(\binom n2{x^{n - 2}} + \binom n3{x^{n - 3}}\Delta x + \cdots + {(\Delta x)}^{n - 2}\right) $$

It follows $$(x+\Delta)^2-x^n-nx^{n-1}\Delta x=(\Delta x)^2\cdot (\text{ something bounded as } (\Delta x)^2\to 0)$$

This is what the big $O$ means here. One should be careful, since the $O$ is context dependent: here we're taking $\Delta x\to 0$; but in other cases the limiting process might change, leading to different asymptotics.

One can write this more appropriately in order to take $\Delta x\to 0$ as $$(x+\Delta x)^n=x^n+nx^{n-1}\Delta x+o(\Delta x)$$

Where $o(\Delta x)$ means this is a term such that $$\lim_{\Delta x\to 0}\frac{o(\Delta x)}{\Delta x}= 0$$

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