# A Newton's Generalized Binomial Theorem

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1. Sep 29, 2017

### JBD

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. Sep 29, 2017

### Staff: Mentor

You need the order for convergence of the series, and I don't see a way to avoid using two cases.

3. Sep 29, 2017

### JBD

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.

4. Sep 29, 2017

### Staff: Mentor

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.