Simplifying Binomial Expansion for Relativistic Velocity Sum

[ognik]

Hi, given the sum of 2 relativistic velocities $ \frac{w}{c} = \frac{\frac{u}{c}+\frac{v}{c}}{1+\frac{uv}{c^2}}$ and setting $ \frac{u}{c}= \frac{v}{c} = (1-\alpha)$, find $ \frac{w}{c} $ in powers of $ \alpha $ through $ {\alpha}^{3} $

I used a binomial expansion to get a tidy intermediate series of $ \frac{w}{c} = 2\left(\left(1-\alpha\right) -2{\left(1-\alpha\right)}^{2} + 3{\left(1-\alpha\right)}^{3} -4{\left(1-\alpha\right)}^{4} + ... \right) $

Now the final series I should arrive at is given as $ 1 - \frac{{\alpha}^{2}}{2} - \frac{{\alpha}^{3}}{2} + ...$
However when I multiply out my intermediate series, I get a messy set of terms of powers of $ \alpha $ that don't simplify to anything remotely like this ($ -2 + 154\alpha - 186{\alpha}^{2} + ...$)
I've redone the laborious expansion a couple of times without success. I'm pretty sure my intermediate series is right? Is there a better way of simplifying it?

I can write the intermediate series as $ 2\sum_{n=0}^{\infty}\left(n+1\right)\left(1-\alpha\right)^{n+1}\left(-1\right)^{n} $. It seems to me that it might simplify easily - if I could expand the $ \left(1-\alpha\right)^{n+1} $ term, again using binomial, but I don't know how to treat the (n+1) and (-1) aspects, or how to expand when already within a series?

 

[GJA]

Hi ognik,

Given that we're using the relativistic addition of velocities, consider $u=v\approx c$. Then $1-\alpha\approx 1.$ Now write the power series expansion for $\frac{x}{1+x^{2}}$ centered about $a=1$ to third order. Replacing $x$ with $1-\alpha$ and multiplying by 2 should give you what you're looking for.

Let me know if anything is unclear/not quite right.

 

[I like Serena]

Hi, given the sum of 2 relativistic velocities $ \frac{w}{c} = \frac{\frac{u}{c}+\frac{v}{c}}{1+\frac{uv}{c^2}}$ and setting $ \frac{u}{c}= \frac{v}{c} = (1-\alpha)$, find $ \frac{w}{c} $ in powers of $ \alpha $ through $ {\alpha}^{3} $

I used a binomial expansion to get a tidy intermediate series of $ \frac{w}{c} = 2\left(\left(1-\alpha\right) -2{\left(1-\alpha\right)}^{2} + 3{\left(1-\alpha\right)}^{3} -4{\left(1-\alpha\right)}^{4} + ... \right) $

Now the final series I should arrive at is given as $ 1 - \frac{{\alpha}^{2}}{2} - \frac{{\alpha}^{3}}{2} + ...$
However when I multiply out my intermediate series, I get a messy set of terms of powers of $ \alpha $ that don't simplify to anything remotely like this ($ -2 + 154\alpha - 186{\alpha}^{2} + ...$)
I've redone the laborious expansion a couple of times without success. I'm pretty sure my intermediate series is right? Is there a better way of simplifying it?

I can write the intermediate series as $ 2\sum_{n=0}^{\infty}\left(n+1\right)\left(1-\alpha\right)^{n+1}\left(-1\right)^{n} $. It seems to me that it might simplify easily - if I could expand the $ \left(1-\alpha\right)^{n+1} $ term, again using binomial, but I don't know how to treat the (n+1) and (-1) aspects, or how to expand when already within a series?

Hi ognik,

If we look only at the terms without $\alpha$ in your intermediate series, we get $2 \cdot (1-2+3-4+...)$, which diverges.
How did you find $-2$?
Anyway, I'm not sure how you got to your intermediate series, but it doesn't look right.
And even if it would be right, it won't help you, since it diverges.

Perhaps we can start with rewriting $\frac wc$ in a form that is ready to be expanded? (Wondering)
Then we can see which approaches are feasible and that will lead to a convergent series.

For the expansion itself, I propose to use the geometric series expansion \(\displaystyle \frac 1{1-y}=1+y+y^2+y^3+...\) that applies if $|y|<1$ and preferably if $|y|<<1$.

 

[ognik]

Hi - here's how I got my intermediate series:
$ \frac{w}{c} = \frac{2\left(1-\alpha\right)}{1+\left(1-\alpha\right)^2} $
Binomial: $ n = -2, x=(1-\alpha) $
$ \frac{1}{1+\left(1-\alpha\right)^2} = 1 - 2\left(1-\alpha\right) + \frac{\left(-2\right)\left(-3\right) }{2!}\left(1-\alpha\right) ^2 + \frac{\left(-2\right)\left(-3\right)\left(-4\right) }{3!}\left(1-\alpha\right)^3 - ... $
$ \therefore \frac{w}{c} = 2\left(1-\alpha\right) [ 1 - 2\left(1-\alpha\right) + 3\left(1-\alpha\right) ^2 -4\left(1-\alpha\right)^3 + ...] $
$ = 2 [ \left(1-\alpha\right) - 2\left(1-\alpha\right)^2 + 3 \left(1-\alpha\right) ^3 -4\left(1-\alpha\right)^4 + ...] $
I am interested if this is right/wrong, even if it's not the correct approach?
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I don't think we can assume u=v≈c , from what I understand even as low as v/c = 0.1 would be considered relativistic?

With the techniques I am aware of, I can't see how we can use a geometric series, there is a power of 2 in the divisor?

 

[I like Serena]

That's the binomial expansion of \(\displaystyle \frac{1}{\Big(1+\left(1-\alpha\right)\Big)^2}\), which is a different expression.

Instead of a binomial expansion we could do for instance a geometric series expansion of \(\displaystyle \frac{1}{1-y}\) with \(\displaystyle y=-(1-\alpha)^2\).

Apparently we are assuming that $u=v$, which are "close" to $c$, so that we write them as $u=v=(1-\alpha)c$, where $\alpha$ is considered to be "small", although it could indeed still be 0.9, meaning convergence of the power series will be somewhat slow.

 

[ognik]

I think I am missing something. The binomial expansion of the 'top', IE $ \left(1-\alpha\right) $ (with n=1 and x=-\alpha) happens to be $ \left(1-\alpha\right) $
So I am multiplying all terms of the 'bottom' series by the 'top' series, is that wrong?
------------------
Although I can see the approach is right, I am also struggling with $ \frac{1}{\left(1-y\right)}$ :
$ \frac{1}{\left(1-y\right)} = \frac{1}{1+\left(1-\alpha\right)^2}= \frac{1}{2\left(1-\alpha\right)+{\alpha}^{2}}
=\frac{\left(1-\alpha\right)}{2+\frac{{\alpha}^{2}}{\left(1-\alpha\right)}} \ne\ \frac{2\left(1-\alpha\right)}{1+\left(1-\alpha\right)^2} $ ?
--------------------
Going with GJA's approach:
$ \frac{x}{1+{x}^{2}} = x-{x}^{3}+{x}^{5}-{x}^{7}+...\: (for \left| x \right| \lt 1) $. We are only interested in 0 < x <1
Being impatient, I threw this at Mathematica, and got $ 4\alpha -14{\alpha}^{2} + 26{\alpha}^{3} - ...$

 

[GJA]

Hi ognik,

What you wrote in your last response was the power series centered at $a = 0.$ Like I mentioned before, you want to use the power series centered at $a=1.$

 

[ognik]

You're right thanks, Taylor around 1 gives $ \frac{1}{2} - \frac{1}{4}(x-1)^2 - \frac{1}{4}(x-1)^3 + ... $
so $ \frac{w}{c} = 1 - \frac{\alpha^2 }{2}- \frac{\alpha^3}{2} - ... $
Perfect!

About the meaning: $ \alpha $ is the percentage that v is below c, right? So if I put $ \alpha $ = 0.1 then v is 90% of c...
Then I could also put $ \alpha $ = 0.9 and have v at 10% of c, right? But then $ 1 - \alpha $ would not be close to 1?
So I can see that w/c cannot be > 1, but why did you chose to expand about 1?
------------------------------------
I would still like to clear up the approach I used for my 'intermediate series' - please explain why I can't do that?

Thanks guys :-)

 

[I like Serena]

I think I am missing something. The binomial expansion of the 'top', IE $ \left(1-\alpha\right) $ (with n=1 and x=-\alpha) happens to be $ \left(1-\alpha\right) $
So I am multiplying all terms of the 'bottom' series by the 'top' series, is that wrong?

Nothing wrong with that.
The problem is that the expansion for the 'bottom' is wrong.
Your expansion is for $\frac{1}{\Big(1+(1-\alpha)\Big)^2}$ instead of for $\frac{1}{\Big(1+(1-\alpha)^2\Big)}$.
Note the location of the 'square'.

Although I can see the approach is right, I am also struggling with $ \frac{1}{\left(1-y\right)}$ :
$ \frac{1}{\left(1-y\right)} = \frac{1}{1+\left(1-\alpha\right)^2}= \frac{1}{2\left(1-\alpha\right)+{\alpha}^{2}}
=\frac{\left(1-\alpha\right)}{2+\frac{{\alpha}^{2}}{\left(1-\alpha\right)}} \ne\ \frac{2\left(1-\alpha\right)}{1+\left(1-\alpha\right)^2} $ ?

The idea is that we substitute $y=-(1-\alpha)^2$ in $\frac 1{1-y}=1+y+y^2+y^3+...$.
Doing so, gives us:
$$\frac 1{1-(-(1-\alpha)^2)} = 1 + (-(1-\alpha)^2) + (-(1-\alpha)^2)^2 + (-(1-\alpha)^2)^3 + ...$$
Simplifying, we get your 'bottom':
$$\frac 1{1+(1-\alpha)^2} = 1 - (1-\alpha)^2 + (1-\alpha)^4 - (1-\alpha)^6 + ...
= (1-1+1-1+...) + \alpha(2 - 4 + 6 - ...) + ... $$
As you can see, we still have a problem though, since the series for the coefficients do not converge.

You're right thanks, Taylor around 1 gives $ \frac{1}{2} - \frac{1}{4}(x-1)^2 - \frac{1}{4}(x-1)^3 + ... $
so $ \frac{w}{c} = 1 - \frac{\alpha^2 }{2}- \frac{\alpha^3}{2} - ... $
Perfect!

Hmm... how did you get the Taylor expansion around 1?

 

[GJA]

Hi ognik,

The reason I chose to center the power series at $a=1$ was because this gave the answer you were looking for. It's not unlike solving an algebra problem. Many times people I work with will solve them correctly, but want their answer to match the back of the book. In turn I need to show them something to get things in the "correct" form.

I think that ultimately this problem is not stated in a mathematically precise way. If we were told $v/c$ was small (i.e. close to zero), the correct approach would be to expand around zero. Essentially, given the form of the answer you posted, it became evident that centering the power series around 0 would not work, so I tried what came next in my mind: see what happens for $v/c\approx 1$ (the case where relativity must surely be in play).

All that said, we can make an argument that our power series will hold for $v/c$ small as well using the radius of convergence. Unfortunately, as far as I can tell, there is not a clean formula for the coefficients in the power series to use the Ratio Test to determine the radius of convergence. Thus, I will resort to the use of the "theoretical" radius of convergence (https://en.wikipedia.org/wiki/Radius_of_convergence)

$r=\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}}$

and the numerical coefficients provided by Wolfram.

Use Wolfram to expand the power series centered at 1 to a decent number of terms (I went out to 32). Now, assuming the pattern we observe continues, we see that $|a_{n}|\leq 1/2$ for all $n.$ Thus,

$\sqrt[n]{|a_{n}|}\leq\frac{1}{2^{1/n}}$

so that

$\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}\leq 1.$

This implies

$1\leq r,$

so that the radius of convergence (of the full power series centered at 1) is at least 1.

Furthermore, looking at the graphs provided by Wolfram of the function $\frac{x}{1+x^{2}}$ and the third order polynomial, we see a good correlation near $x=0.$ This means that even if we do chose $\alpha$ close to $1$, we should expect a numerically acceptable error. For example, choosing $\alpha=.9$ gives an error of approximately $0.032480$

 

[ognik]

Thanks GJA, that was an excellent reply for me.
ILS, I can only say - aaargh! I must look careless, even though I'm trying not to be. So in conclusion, the function doesn't lend itself to binomial expansion, so I'd have to use Taylor, which leaves me back at GJAs approach.
Thanks again both.