I've read in a book that this function: $$b(r) = \sqrt{\frac{r^3}{r - 2M}}$$ behaves approximately as $$ b(r) \approx r + M $$ for $r \gg M$.

I checked it numerically, by plotting a graph of this function, and found that it is true. How to show this mathematically?


Let $R >0.$


Binomial expansion for rational exponent $\alpha$:


$ 1+\alpha x+(\alpha (\alpha-1)/2!)x^2+ O(x^3)$

converges for $|x|<1.$

Hence for large $R$:

$b(R)=R(1+M/R +O((M/R)^2))$ https://brilliant.org/wiki/fractional-binomial-theorem/


This is just a Taylor expansion $$b(r) = \sqrt{\frac{r^3}{r - 2M}}=r \sqrt{\frac{r}{r - 2M}}$$ Use long division or Taylor series $$\frac{r}{r - 2M}=1+\frac{2 M}{r}+\frac{4 M^2}{r^2}+O\left(\frac{1}{r^3}\right)$$ $$\sqrt{\frac{r}{r - 2M}}=1+\frac{M}{r}+\frac{3 M^2}{2 r^2}+O\left(\frac{1}{r^3}\right)$$ $$b(r)=r+M+\frac{3 M^2}{2 r}+O\left(\frac{1}{r^2}\right)=r+M+O\left(\frac{1}{r}\right)$$


$$\frac{\sqrt{\frac{r^3}{r - 2M}}}{r+M}=\sqrt{\frac{r^3}{(r-2M)(r+M)^2}}=$$ $$\sqrt{\frac{r^3}{r^3\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}=\sqrt{\frac{1}{\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}\to 1$$ as $r\to\infty$.

  • 1
    $\begingroup$ While this is true, it works for $r+K$ for any constant $K$, so I don't think it fundamentally addresses OP's question. $\endgroup$ – Mees de Vries Jul 19 at 8:22

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