I want to prove that $$ |x -y|^{-2} \leq \frac{1}{|x|^2} + O(|x|^{-3}) $$ when $| y| \leq R$, and $|x| \geq 2R$, $R>0$. The hint is to use that $$ |x -y|^{-2}= |x|^{-2}(1-2\frac{x \cdot y}{|x|^{2}}+ \frac{|y|^2}{|x|^2})^{-1} $$

I tired different attempts, but I am not able to show it. I would really appreciate any hint. Thanks!

EDIT: I found this estimate in the book (Vorticity and incompressible flow ) by Majda and Bertozzi. Here is a screen shot

  • $\begingroup$ The dot product can be written as $|x||y|\cos\theta$ and you can make further approximations from there. $\endgroup$ – Everiana Jul 30 at 1:27
  • $\begingroup$ Are you trying to prove the hint, or to use it to prove the problem? $\endgroup$ – Ross Millikan Jul 30 at 2:40
  • $\begingroup$ Are $x$ and $y$ real numbers here or are they vectors? If they're vectors, the problem needs to be restated with $|x|^2$ and so on. $\endgroup$ – Ted Shifrin Jul 30 at 2:41
  • $\begingroup$ They are vectors. Yes you are right! $\endgroup$ – Demha Jul 30 at 14:24

The statement is not correct, even if $x,y$ are reals. If the are vectors, let them both be along the same axis. Then if $|x|=2R, |y|=R$ and they are in the same direction $|x-y|^{-2}=\frac 1{R^2}$ while $\frac 1{|x|^2}=\frac 1{4R^2}$ and the difference is $\frac 3{4R^2} \not \in O(x^{-3})$

  • $\begingroup$ The statement is about vectors. There is a typo in what I wrote. There should absolute value signs on the right-hand side. The statement should be $|x-y|^{-2} \leq \frac{1}{|x|^2} +O(\frac{1}{|x|^3})$. $\endgroup$ – Demha Jul 30 at 14:15
  • $\begingroup$ I found this estimate in the book (Vorticity and incompressible flow) by Majda and Bertozzi. $\endgroup$ – Demha Jul 30 at 14:19
  • $\begingroup$ The statement is true if $y$ is limited to a fixed size, but if it is allowed to grow with $x$ it is not. You can just count powers of $R$ in the term $\frac {x\cdot y}{|x|^2}$. They all cancel, so the term is the same size as the $\frac 1{|x|^2}$ out front. $\endgroup$ – Ross Millikan Jul 30 at 14:36

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