Binomial Expansion (Arfken/Weber/Harris 1.3.9)

[CJ2116]

Hi everyone,

I'm currently working through Mathematical Methods for Physicists 7th ed. by Arfken/Weber/Harris and there's one question that's been giving me some difficulty. I would appreciate any feedback if possible.

Thanks!

Chris

Homework Statement

The relativistic sum w of two velocities u and v in the same direction is given by
$$\frac{w}{c}= \frac{u/c+v/c}{1+uv/c^2}$$
If
$$\frac{v}{c}=\frac{u}{c}=1-\alpha,$$
where ##0\le\alpha\le 1##, find w/c in powers through terms in ##\alpha^3##

Homework Equations

Binomial Expansion:
$$\left(1+x\right)^m=1+mx+\frac{m(m-1)x^2}{2!}+\frac{m(m-1)(m-2)x^3}{3!}$$

The Attempt at a Solution

This seems like a straightforward substitution and expansion:
$$\frac{w}{c}= \frac{u/c+v/c}{1+uv/c^2}=\frac{(1-\alpha)+(1-\alpha)}{1+(1-\alpha)^2}=2(1-\alpha)\left(1+(1-\alpha)^2\right)^{-1}$$
Now, expanding the last term I'm getting
$$\frac{w}{c}=2(1-\alpha)\left(1-(1-\alpha)^2+\frac{-1(-1-1)(1-\alpha)^4}{2!}+\frac{-1(-1-1)(-1-2)(1-\alpha)^6}{3!}+...\right)$$
$$\Rightarrow \frac{w}{c}=2(1-\alpha)\left(1-(1-\alpha)^2+(1-\alpha)^4-(1-\alpha)^6+...\right)$$

I'm noticing two things with this that make me think I'm doing something stupid. The first is that ##0\le\alpha\le 1## and at ##\alpha=0## this series diverges.

The second is that this is giving an infinite number of polynomials to expand and I'm not seeing how to get something of the form ##1+a_1\alpha+a_2\alpha^2+a_3\alpha^3...##, which is what I assume the problem is asking.

Can anybody see something wrong with my assumptions or calculations?

 

[fresh_42]

Why do you have ##1+\alpha## in the denominator and not ##1-\alpha## ?

 

[Orodruin]

Adjusting for what fresh_42 said:

You seem to be using the expansion for ##1/(1+x)## for ##x = (1-\alpha)^2##, indicating that you are expanding around ##(1-\alpha)^2 = 0##, i.e., ##\alpha = 1##. The radius of convergence for the expansion of ##1/(1+x)## is 1 and so you actually should expect something that does not converge at ##\alpha = 0##. Try expanding around ##\alpha = 0## instead.

 

[CJ2116]

Sorry, fixed it. That's what I get for copying and pasting a typo!

 

[Charles Link]

The denominator should be ## 1+(1-\alpha)^2 =2-2 \alpha+\alpha^2=2(1-\alpha+\frac{\alpha^2}{2}) ##. You want to do the expansions with ## x=- \alpha +\frac{\alpha^2}{2} ##. ## \\ ## If you expand ## \frac{1}{1+x} =1-x+x^2-x^3+... ## with ## x>1 ##, the series diverges.## \\ ## (If you use ## x=(1-\alpha)^2 ##, the series is so close to diverging (radius of convergence =1), that it basically will not give any useful result. You can plug in a small value for ## \alpha ##, but the expression for ## \frac{1}{1+x} ## with the ## x=(1-\alpha)^2 ## converges far too slowly=it almost diverges=so that you might need a couple hundred terms to get anything close to the answer. If you choose ## x ## properly, the series converges rapidly for small ## \alpha ##).

 

[CJ2116]

Excellent, thanks for the responses!

With this I'm now getting:
$$\frac{w}{c}= \frac{2(1-\alpha)}{\left(1+(1-\alpha)^2\right)}=\frac{2(1-\alpha)}{\left(2-2\alpha+\alpha^2\right)}=\frac{(1-\alpha)}{\left(1-\alpha+\alpha^2/2\right)}$$
$$\Rightarrow \frac{w}{c}=(1-\alpha)\left(1-(-\alpha+\alpha^2/2)+(-\alpha+\alpha^2/2)^2-(-\alpha+\alpha^2/2)^3+O(\alpha^4)\right)$$
Ignoring anything above order 4:
$$\frac{w}{c}=(1-\alpha)\left(1+\alpha-\alpha^2/2+\alpha^2-\alpha^3+\alpha^3\right)=(1-\alpha)\left(1+\alpha+\alpha^2/2\right)$$
$$\frac{w}{c}=1+\alpha+\alpha^2/2-\alpha-\alpha^2-\alpha^3/2=1-\alpha^2/2-\alpha^3/2$$

Looks like that did it! Thanks Again!

Chris