1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to make approximations?

  1. Jul 15, 2012 #1
    If I have some expression such as: [tex]y = 1 + \epsilon^2 - 5\epsilon^4[/tex] and then make this approximation:[tex]y \approx 1 + \epsilon^2[/tex] then, if I understand correctly, my specific assumption is that [itex]5\epsilon^4 \ll 1 + \epsilon^2[/itex] which, for example, would always be satisified if I happened to know that [itex]5\epsilon^4 \ll 1[/itex].
    However, if I have something like a Taylor series, I'm not sure exactly what my assumption is. For example:[tex]\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[/tex] 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 [itex]\frac{1}{8}x^2 \ll 1[/itex]? Is there a sensible/rigorous way of dealing with things like this?
    Last edited: Jul 16, 2012
  2. jcsd
  3. Jul 15, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 2rCr. Asymptotically, that approximates xr/√r, which is clearly less than xr. 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.
  4. Jul 15, 2012 #3


    User Avatar
    Science Advisor

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

    And since [itex] x \ll 1 [/itex] 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.)
  5. Jul 15, 2012 #4


    User Avatar
    Science Advisor

    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.
  6. Jul 16, 2012 #5
    You're right - thanks for spotting my error. I'll edit the OP to avoid confusion.

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

    Yes - this makes intuitive sense. Many thanks for your clear explanations.
  7. Jul 16, 2012 #6
    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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook