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Thread: Analysis, Power Series

  1. #1
    Nov 2008

    Analysis, Power Series


    Here's the question: Consider the function f defined uniquely by the relation


    Show that f has a series expansion in powers of x with strictly positive radius. (Hint: consider g(h(x)) where g(y)=(1+y)^\frac{1}{3} and h(x)=\frac{sin(x)-x}{x})

    So far I figured out that f(x)=x g(h(x)), but that's about it. I tried "Taylor-expanding" these different functions, but can't do it about 0 and so it gets complicated. Also i am not too sure about composing the functions. It seems you can do it with formal power series, but i don't know for this particular problem.

    I could possibly argue the existence of the series using some theorems, but the next question ask to compute the series for f as far as the term in x^5, so I guess I really have to find a series expansion explicitely.

    I would be thankful if anyone could tell me how I should start working on this...
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  2. #2
    Nov 2008
    I think I got it. It turns out that the series expansion of h doesn't have any constant term, so by some theorem g(h(x)) has a series expansion with strictly positive radius. To calculate the terms up to x^5 i only need to consider the first 3 terms of f(x) = x(b_0+b_1h(x)+b_2(h(x))^2+...) (where the b_i are the coefficient of the series expansion of g) since all other terms have powers of x greater than 5 and then collect terms...

    The next question is showing that the radius of convergence cannot exceed Pi. I showed that f(x) is not differentiable at x=Pi using the definition of the derivative.

    Does any of this make sense to you?
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