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CJ2116

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Hi everyone,

I'm currently working through

Thanks!

Chris

The relativistic sum

$$\frac{w}{c}= \frac{u/c+v/c}{1+uv/c^2}$$

If

$$\frac{v}{c}=\frac{u}{c}=1-\alpha,$$

where ##0\le\alpha\le 1##, find

Binomial Expansion:

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

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?

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?

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