Question regarding notation when omitting terms

[kent davidge]

If we want to expand a function ##f(x)## up to first order around ##x = 0## say, we usually write ##f(x) = f(0) + (df/dx)|_0 x + \mathcal O(x^2)##.

But what if I want to expand ##f(x)## in the whole series, and showing only the first order term in x? What notation do you use for that? (Aside from ##f(x) = f(0) + (df/dx)|_0 x + \sum \frac{d^{k-2}f}{dx^{k-2}}x^{k-2} / (k-2)!##.)

My thought: ##f(x) = f(0) + (df/dx)|_0 x + \mathcal O(x^{\geq 2})##

 

[BvU]

Hi,

if I want to expand ##f(x)## in the whole series

You don't want that, because it doesn't make sense. The only term of interest around zero is the ##x^2## term. All higher order terms in ##f(x) - f(0) - (df/dx)\Big |_0 \; x \ ## vanish for ##x\downarrow 0##.

In other words, ##\mathcal O(x^{2}) ## is the same as your ##\mathcal O(x^{\geq 2})## (or: implies it).

 

[mfb]

x³ < x² for small x, therefore ##ax^3 \in O(x^2)##, same for all higher orders.

 

[kent davidge]

x³ < x² for small x, therefore ##ax^3 \in O(x^2)##, same for all higher orders.

I'm not considering only small ##x## though

 

[BvU]

Doesn't matter. The meaning doesn't change:
$$f(x) - f(0) - (df/dx)\Big |_0 \; x \ = \mathcal O (x^2) \equiv \lim_{x \downarrow 0 }{f(x) - f(0) - (df/dx)\Big |_0 \; x \over x^2} \ \ \text {is finite}$$no more, no less. If you want to consider huge ##x## is upon you -- it probably won't be a good approximation, though.

( from wikipedia sin )

 

[mfb]

For large x you have to choose between physics and computer science convention of the notation, but in the latter case considering the constant and linear term is irrelevant and you'll likely need a completely new approach.