I have a question regarding this question here.


if I set $ x+2=c/c-v$ , will I have
$U_n = M\left(c-\frac{c}{n+2}\right)-M\left(c-\frac{c}{n+1}\right), \label{1}\tag{$U$} $

and will $M(c)=M(c-1)$, or $M(c)=-M(c-c)=-M(0)=-m_0$?

  • $\begingroup$ What question? Your link just tries to go to the general list mathoverflow.net/questions. $\endgroup$
    – LSpice
    Jul 26, 2023 at 1:49
  • $\begingroup$ Sorry, it's okay now $\endgroup$
    – z.10.46
    Jul 26, 2023 at 1:58

1 Answer 1


Are the $U_n$ supposed to have a physical meaning or is this a purely mathematical exercises? if we directly interpret:

$$ U_n = M\left(c-\frac{1}{n+2}\right)-M\left(c-\frac{1}{n+1}\right)\label{2}\tag{$U$} = \frac{m_0}{\sqrt{1 - 1 + \frac{2}{c(n+2)} - \frac{1}{c^2(n+2)^2}}} - \frac{m_0}{\sqrt{1 - 1 + \frac{2}{c(n+1)} - \frac{1}{c^2(n+1)^2}}} = m_0 \left( \frac{1}{\sqrt{\frac{2}{c(n+2)} - \frac{1}{c^2(n+2)^2}}} - \frac{1}{\sqrt{\frac{2}{c(n+1)} - \frac{1}{c^2(n+1)^2}}} \right) = m_0 c \left( \frac{n+2}{\sqrt{2c(n+2) -1 }} - \frac{n+1}{\sqrt{2c(n+1)-1}} \right) = $$

It's not clear to me how to assign units to the last object. It seems like the $n$ and $1$ and $2$ need to be given units for the whole thing to remain physical with dimensional analysis.

Anyways... I believe that $U_n = O(n^{-\frac{1}{2}}) $ so one could use the Euler Maclaurin formula on this and see if that suggests a constant term in its asymptotic expansion. This will be very tedious and I'll get around to it later (unless someone beats me to it :) )

  • $\begingroup$ Thank you very much. It would be a physically sound theory formulated mathematically in that case. Have you considered that $x+2=\frac{1}{1-\frac{v^2}{c^2}}$? $\endgroup$
    – z.10.46
    Jul 25, 2023 at 20:38
  • $\begingroup$ im sorry but your comment doesn't make sense to me. What does " It would be a physically sound theory formulated mathematically in that case" mean? do you mean to say "if we gave n,1,2 appropriate units then the expression is physically sound?" $\endgroup$ Jul 25, 2023 at 20:41
  • $\begingroup$ Sorry i edit question $\endgroup$
    – z.10.46
    Jul 25, 2023 at 22:55

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