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I came across the relation in the title in a physics textbook and wondered how I get to it.

$$n^2-1 \approx (n-1)2$$

for $$n-1\ll 1$$

Could anybody maybe help me out?

Thanks!

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  • $\begingroup$ An asymptotic relation as $n\to\infty$, maybe? $\endgroup$ May 18, 2020 at 10:07
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    $\begingroup$ Maybe your book assumes $n\approx 1$? In that case, $n^2-1=(n-1)(n+1) \approx (n-1)2$ $\endgroup$
    – Tavish
    May 18, 2020 at 10:08
  • $\begingroup$ I think it is an asymptotic for $n\to 1$ $\endgroup$
    – Exodd
    May 18, 2020 at 10:08

4 Answers 4

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$$n^2-1=(n-1+2)(n-1)\approx2(n-1)$$ because $n-1$ is negligible compared to $2$.

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You can also work this out in terms of $n-1$,

$$n^2-1=(n-1+1)^2-1=(n-1)^2+2(n-1)$$ and the first term is negligible.

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Write $n=1+\epsilon,\,\epsilon\ll1$ so $\frac{n^2-1}{(n-1)2}=1+\frac12\epsilon\sim1$.

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I think the point is that if you consider the function $f(x) = x^2 -1$ around $x_0=1$, you can expand it using Taylor series, up to the second term (because the function itself is quadratic, that means just $f(x_0)$ and the linear term. Since $f(x_0)=0$, you have $$ f(x) \approx 2(x-1) $$ in the vicinity of $x_0=1$, which is what @YvesDaoust plotted.

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$n^2-1=(n+1)(n-1)$ so if $n$ is very close to $1$, then $n+1$ is very close to $2$ and $n^2-1$ is very close to $2(n-1)$.

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