Binomial Expansion (Arfken/Weber/Harris 1.3.9)

In summary, the conversation discusses a question involving the relativistic sum of two velocities and the use of binomial expansion to find the solution. After correcting a typo in the original equation, the correct expansion is found and the solution is determined to be w/c = 1 - α^2/2 - α^3/2.
  • #1
CJ2116
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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?
 
Last edited:
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  • #2
Why do you have ##1+\alpha## in the denominator and not ##1-\alpha## ?
 
  • #3
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.
 
  • #4
Sorry, fixed it. That's what I get for copying and pasting a typo!
 
  • #5
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 ##).
 
Last edited:
  • #6
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
 
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Related to Binomial Expansion (Arfken/Weber/Harris 1.3.9)

1. What is binomial expansion?

Binomial expansion is a mathematical technique used to expand a binomial expression to a higher power. It is based on the binomial theorem, which states that any binomial expression raised to a positive integer power can be written as a sum of terms with coefficients determined by Pascal's triangle.

2. How is binomial expansion calculated?

Binomial expansion is calculated using the binomial theorem or by using the binomial coefficient formula. The binomial coefficient formula is nCr = n! / (r! * (n-r)!) where n is the power of the binomial expression and r is the term number.

3. What is the significance of binomial expansion?

Binomial expansion is a useful tool in mathematics and has applications in various fields such as probability, statistics, and engineering. It allows for the simplification of complex expressions and the calculation of binomial probabilities.

4. Can binomial expansion be used for non-integer powers?

No, binomial expansion can only be used for positive integer powers. For non-integer powers, other techniques such as the binomial series expansion or Taylor series expansion must be used.

5. Are there any limitations to binomial expansion?

Binomial expansion is limited to binomial expressions, meaning it can only be applied to expressions with two terms. It also has a limited range of convergence, meaning it may not accurately approximate the expression for certain values of its variables.

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