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Math Help - little o notation

  1. #1
    Junior Member
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    little o notation

    I'm on pg 288 of Apostol's Calculus volume 1. A part of the theorem at the top says that 1/(1 + g(x)) = 1 - g(x) + o(g(x)) if g(x) -> 0 as x -> a. The first subsequent example is to show tan x = x +(1/3)x ^3 + o(x^3) as x -> 0. Using the theorem and the taylor approximation to cosine, we have 1/cos x = 1/(1 - (1/2)x^2 + o(x^3)) = 1 +(1/2)x^2 +o(x^2) as x ->0. I can't see how this follows from the theorem or the definition of o notation.
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  2. #2
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    A Taylor approximation for cosine is \cos x = 1 - {x^2\over2!} + o(x^3). Letting g(x) = - {x^2\over2!} + o(x^3) and using the theorem you get the approximation \sec x = 1 - g(x) + o(g(x)) = 1 + {x^2\over2!} + o(x^3) + o(-{x^2\over2!} + o(x^3)). Can you see that o(x^3) + o(-{x^2\over2!} + o(x^3)) = o(x^2)?
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  3. #3
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    It's the last identity that i can't figure out how to justify. but since g(x) = -(x^2)/2! + o(x^3) shouldn't we have sec x = 1 + (x^2)/2! - o(x^3) + o((-x^2)/2! + o(x^3)) (with a minus sign preceding the o(x^3) term)? Either way, if you can explain how to justify the last sentence you wrote, it would be extremely helpful. i'm not good with the algebraic manipulations involving o-signs i guess.
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  4. #4
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    Oh crap, yes, I lost the sign. It actually doesn't matter since the little o term represents only a magnitude. Anyway, basically the x^2 term is largest, so it dominates. Formally, let f and g be o(x^3) and let h be o(x^2/2! + g). We want to show that f+h is o(x^2). So compute {f+h\over x^2} = {f\over x^2}+{h\over x^2} = x{f\over x^3} + {h\over x^2/2! + g}({1\over2!} + x{g \over x^3}). The first term goes to zero as x does because f is o(x^3) and the second goes to zero because h is o(x^2/2! + g). (I should really have {|f+h|\over x^2} I guess but I'm lazy. You get the idea.) Its easier if you have more theorems about how to manipulate little o's.
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