Aproximations by Expansion

  • Thread starter eep
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  • #1
eep
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Hi,
I've never really studied various ways of expanding expressions in order to obtain an approximation that can make calculations easier. For example,

[tex]
p^t = \frac{m}{\sqrt{1 - v^2}}
[/tex]

reduces to

[tex]
p^t = m + \frac{1}{2}mv^2 + ...
[/tex]

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.
 

Answers and Replies

  • #2
quasar987
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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

[tex]f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n[/tex]

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 [itex]f(x) = (1+x)^r[/itex] where r is any real number has as its Taylor expansion

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

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

This is exactly what has been done in your post. [itex]p(v)=m(1+v^2)^{-1/2}[/itex] is of the form [itex](1+x)^r[/itex] 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.
 
Last edited:
  • #3
quasar987
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Btw, wouldn't it be
[tex]
p = m - \frac{1}{2}mv^2 + ...
[/tex]
instead? (with a "-" sign instead of a "+" sign on odd terms)
 
  • #4
mathman
Science Advisor
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The sign is +. It results from multiplying two - signs, -1/2 and -v2.
 

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