How Can Approximations by Expansion Simplify Scientific Calculations?

[eep]

Hi,
I've never really studied various ways of expanding expressions in order to obtain an approximation that can make calculations easier. For example,

<br /> p^t = \frac{m}{\sqrt{1 - v^2}}<br />

reduces to

<br /> p^t = m + \frac{1}{2}mv^2 + ...<br />

for v << 1.

How does one arrive at something like this? What other expansions are useful? I used to think that I'd just calculate everything exactly but I now realize these sorts of expansions are extremeley important.

 

[quasar987]

Are you familiar with Taylor series? It says that if a function can be expanded in a power series, then it will be of the form

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

Wheter this series converges towards the original function is another matter. (Actually, it is a matter of calculating the radius of convergence of a power series)

For exemple, the function f(x) = (1+x)^r where r is any real number has as its Taylor expansion

\sum_{n=0}^{\infty} \frac{r(r-1)...(r-n+1)}{n!}x^n

and the series converges to f(x) = (1+x)^r for all |x|<1 but not for any other value of x.

This is exactly what has been done in your post. p(v)=m(1+v^2)^{-1/2} is of the form (1+x)^r with x=-v² and r=-½, so it converges to the series expansion you wrote for all values of v such that |v²|<1.

N.B. in the context of special relativity, v is always lesser than 1 since it has been assumed that c=1, so the formula really is valid for all velocity.The only other expansion I know of is the expansion in a Fourier series.

 

[quasar987]

Btw, wouldn't it be
<br /> p = m - \frac{1}{2}mv^2 + ...<br />
instead? (with a "-" sign instead of a "+" sign on odd terms)

 

[mathman]

The sign is +. It results from multiplying two - signs, -1/2 and -v².