Graduate Newton's Generalized Binomial Theorem

Click For Summary
The discussion focuses on applying Newton's Generalized Binomial Theorem to expand the expression \([f_1(x)+f_2(x)]^\delta\) under the condition \(|f_1(x)| > |f_2(x)|\). The user encounters issues because both functions are sinusoidal, leading to instances where the condition is violated. They propose separating cases based on the values of \(x\) to apply the theorem correctly. The conversation also touches on the order of convergence for the series, emphasizing that the power series converges when \(\frac{f_2}{f_1} < 1\). Ultimately, the approach involves using the appropriate expansion based on which function is dominant at given values of \(x\).
JBD
Messages
15
Reaction score
1
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?
 
Mathematics news on Phys.org
You need the order for convergence of the series, and I don't see a way to avoid using two cases.
 
  • Like
Likes JBD
mfb said:
You need the order for convergence of the series, and I don't see a way to avoid using two cases.
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.
 
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.
 
  • Like
Likes JBD
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
795
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K