Undergrad 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.