0
$\begingroup$

I don't seem to get the answer the book "The Geometry of Spacetime" by Callahan does e.g. $1 + 1/2(v^2) + O(v^4)$ on Pg. $100$ and it is rather crucial for the ensuing discussion

$\endgroup$
  • 2
    $\begingroup$ Does this mean that we need to read p.100 of the book? $\endgroup$ – Wuestenfux Jul 13 at 15:10
  • 1
    $\begingroup$ Please use [MathJax](math.meta.stackexchange.com/questions/5020/… $\endgroup$ – saulspatz Jul 13 at 15:12
  • 1
    $\begingroup$ Use the definition: $f(v) = f(0) + f'(0) v + f''(0) v^2/2 + \ldots$. Compute $f(0)$, $f'(0)$ and $f''(0)$ and you have your result. $\endgroup$ – Winther Jul 13 at 15:12
  • $\begingroup$ Thanks -- evaluating the results at v = 0 makes all the horrible stuff go away $\endgroup$ – luysii Jul 13 at 15:20
2
$\begingroup$

The Taylor series of $(1+x)^\alpha$ (for all $\alpha\in\Bbb R$ and $x\in(-1,1)$ ) is the so called binomial series.

$$(1+x)^\alpha=\sum_{k=0}^\infty \binom \alpha kx^k$$

Where the generalized binomial is $\binom\alpha k=\frac{\alpha\cdot(\alpha-1)\cdots(\alpha-k+1)}{k!}$. In your case you are looking at $(1+x)^{-1/2}$ with $x=-v^2$.

$\endgroup$
  • 1
    $\begingroup$ If the OP cannot understand the expression in his/her text book, it is doubtful that this answer will be of much use. $\endgroup$ – Mark Viola Jul 13 at 15:16
  • $\begingroup$ Sometimes one works in vain. $\endgroup$ – Gae. S. Jul 13 at 15:17
  • $\begingroup$ @MarkViola , OP doesn't even know how to use MathJax. How is it that they got such a high reputation? $\endgroup$ – evaristegd Jul 13 at 15:26
  • $\begingroup$ @evaristegd You, the OP and I have by no means a high reputation. Also, sound questions sometimes get upvoted. $\endgroup$ – Gae. S. Jul 13 at 15:31
1
$\begingroup$

The square root is locally Lipschitz (even differentiable) for positive arguments. Thus $\sqrt{a+\epsilon}=\sqrt{a}+O(\epsilon)$ for $a>0$. With that transform $$ \frac1{\sqrt{1-v^2}}=\frac{\sqrt{1+v^2}}{1+O(v^4)}=\sqrt{1+v^2+\frac{v^4}4}+O(v^4)=1+\frac{v^2}2+O(v^4) $$

This could be extended to higher orders, $$ \frac1{\sqrt{1-v^2}}=\frac{\sqrt{1+v^2+v^4}}{\sqrt{1-v^6}}=\sqrt{(1+\frac{v^2}2)^2+\frac{3v^4}4}+O(v^6)=...=1+\frac{v^2}2+\frac{3v^4}8+O(v^6) $$ etc.

$\endgroup$
0
$\begingroup$

It is simpler to consider the Taylor series $$(1-v)^{-1/2}=(1-0)^{-1/2}+\frac12(1-0)^{3/2}v+o(v)$$

and plug $v^2$, giving

$$(1-v^2)^{-1/2}=1+\frac12v^2+o(v^2).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.