What is the assumption for making approximations in expressions?

[weetabixharry]

If I have some expression such as: y = 1 + \epsilon^2 - 5\epsilon^4 and then make this approximation:y \approx 1 + \epsilon^2 then, if I understand correctly, my specific assumption is that 5\epsilon^4 \ll 1 + \epsilon^2 which, for example, would always be satisified if I happened to know that 5\epsilon^4 \ll 1.
However, if I have something like a Taylor series, I'm not sure exactly what my assumption is. For example:\sqrt{1 - x} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \frac{5}{128}x^4 - \dots \\ \ \ \ \ \ \ \ \ \ \ \ \ \approx 1 - \frac{1}{2}x It seems that my exact assumption here involves an infinite series that might be tricky to evaluate.
So, more loosely speaking, is the approximation satisfied, for example, if \frac{1}{8}x^2 \ll 1? Is there a sensible/rigorous way of dealing with things like this?

 

[haruspex]

It is certainly a good idea to compute a bound on the error that arises by curtailing the expansion.
In the example you give, you have the signs wrong in the expansion. All except the first term should be negative. The general term has magnitude (x/4)^{r 2r}C_r. Asymptotically, that approximates x^r/√r, which is clearly less than x^r. If you curtail the expansion at r = n, the sum of the remaining terms cannot exceed xⁿ/(1-x).
Faced with a series which does alternate in sign, you can usually do better than this by combining pairs of consecutive terms and putting a bound on those.

 

[Nabeshin]

Well when x \ll 1, x^2 \gg x^4 \gg x^6 \cdots, So each next term in the series is much smaller than the one which preceded it.

And since x \ll 1 you know that the 1 out front is important, so you simply keep the leading order term in x, namely the linear one.

(This does assume something about the coefficients, however. But from Taylor we know that they're decreasing as well so this specific example is OK. If the coefficients were, for some reason, increasing, then a more detailed analysis would be in order.)

 

[chiro]

Typically what happens is that you have the independent variables in the form of x + epsilon and then you expand it out in the context of the model to see how the error propagates in the model.

One common and powerful way to do this is to use the norm and metric identities like the triangle inequality since you can get a bound on on the error and convert the x + epsilon in terms of x and epsilon separately.

 

[weetabixharry]

In the example you give, you have the signs wrong in the expansion.

You're right - thanks for spotting my error. I'll edit the OP to avoid confusion.

If you curtail the expansion at r = n, the sum of the remaining terms cannot exceed xⁿ/(1-x).

I'll have to have a think about this, but I expect this is exactly what I'm looking for.

Faced with a series which does alternate in sign, you can usually do better than this by combining pairs of consecutive terms and putting a bound on those.

Yes - this makes intuitive sense. Many thanks for your clear explanations.

 

[awkward]

In the specific case of Taylor series, there are formulas for the remainder term which can be used to bound the error. See the section titled "Explicit formulae for the remainder" in

http://en.wikipedia.org/wiki/Taylor's_theorem