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A Newton's Generalized Binomial Theorem

  1. Sep 29, 2017 #1

    JBD

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    I'm trying to expand the following using Newton's Generalized Binomial Theorem.
    $$[f_1(x)+f_2(x)]^\delta = (f_1(x))^\delta + \delta (f_1(x))^{\delta-1}f_2(x) + \frac{\delta(\delta-1)}{2!}(f_1(x))^{\delta-2}(f_2(x))^2 + ...$$
    where $$0<\delta<<1$$

    But the condition for this formula is that $$\lvert f_1(x)\rvert > \lvert f_2(x)\rvert$$

    And that's where my problem is. Since both functions are sinusoidal, there are times when indeed $$\lvert f_1(x)\rvert > \lvert f_2(x)\rvert$$ but there are also values of x such that $$\lvert f_2(x)\rvert > \lvert f_1(x)\rvert$$. Take for example the graphs of cos^2 x and sin^2x.

    In other words, since the condition is violated, the expansion is not true for all x.

    I'm thinking of separating the two instances. At x's where $$\lvert f_1(x)\rvert > \lvert f_2(x)\rvert$$ then I can use the above expansion. If $$\lvert f_2(x)\rvert > \lvert f_1(x)\rvert$$, then:

    $$[f_2(x)+f_1(x)]^\delta = (f_2(x))^\delta + \delta (f_2(x))^{\delta-1}f_1(x) + \frac{\delta(\delta-1)}{2!}(f_2(x))^{\delta-2}(f_1(x))^2 + ...$$

    But, how can I separate the two instances? Or is there another way to solve this problem?
     
  2. jcsd
  3. Sep 29, 2017 #2

    mfb

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    Staff: Mentor

    You need the order for convergence of the series, and I don't see a way to avoid using two cases.
     
  4. Sep 29, 2017 #3

    JBD

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    How do you find the order of convergence? Do you mean use convergence tests? Most of the time f1 > f2 so I think the series converges. Does this mean that I'm allowed to use my first expansion? Thanks.
     
  5. Sep 29, 2017 #4

    mfb

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    It is a power series in ##\frac{f_2}{f_1}##, this fraction has to be smaller than 1 to make the power series converge. There is nothing to test.
    If it is larger than 1, you can swap the two values as you did in post 1 and then use the formula with swapped values.
     
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