Applications of the Binomial Theorem

[courtrigrad]

How would you quickly derive the binomial series? Would you have to use Taylor's Theorem/ Taylor Series? And does the Binomial Theorem follow from the binomial series? Are there any applications at all of the binomial series/ Binomial Theorem to special relativity? I know the binomial series is (x+a)^{v} = \sum_{k=0}^{\infty}\binom{v}{k} x^{k}a^{v-k}. v is a real number. But I guess when v is a positive integer n we get (x+a)^{n} = \sum_{k=0}^{\infty}\binom{n}{k} x^{k}a^{n-k}. Why does the series terminate at n = v?

Thanks

 

[quasar987]

That's a lot of question! Take your breath!

How would you quickly derive the binomial series?

It is not quick and painless but it is simply a result of applying Taylor's expansion theorem to the function of one variable f(\epsilon)=(1+\epsilon)^v. The resulting series is

(1+\epsilon)^{v} = \sum_{k=0}^{\infty}\binom{v}{k} \epsilon^{k}

and its radius of convergence is found to be 1. One can then decide to set \epsilon = x/a and multiply both sides of the equation by a^v to get

(x+a)^{v} = \sum_{k=0}^{\infty}\binom{v}{k} x^{k}a^{v-k}

of radius of convergence 'a'.

And does the Binomial Theorem follow from the binomial series?

Yes; in the particular case v=n, a positive integer, the series is finite (the series "terminate" at k=n) and equals the binomial sum. More on that later.

Are there any applications at all of the binomial series/ Binomial Theorem to special relativity?

The binomial series is useful for approximations. When we have an expression of the form (1+\epsilon)^{v} where |\epsilon |<1, we can write it as its Taylor series (which is in this case, the binomial series) and make the approximation that (1+\epsilon)^{v} is equal to only the first few terms of the series. You should be able to see why this is reasonable in the scope of 0<|\epsilon|<1.

In special relativity, the factor \gamma is of the above form, with \epsilon =-(v/c)^2 and v=-½, so a series expansion is possible and we can make the approximation of keeping only the first few terms in in order to ease our calculations or get insights into the equations.

I know the binomial series is (x+a)^{v} = \sum_{k=0}^{\infty}\binom{v}{k} x^{k}a^{v-k}. v is a real number. But I guess when v is a positive integer n we get (x+a)^{n} = \sum_{k=0}^{\infty}\binom{n}{k} x^{k}a^{n-k}. Why does the series terminate at n = v?

Watch what happens to \binom{n}{k} for k>n*. It vanishes. So all terms of the binomial series past k=n are zero.

*The definition of \binom{n}{k} used is this: http://en.wikipedia.org/wiki/Binomial_series