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