- #1
weetabixharry
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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?
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?
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