# Where does this relation come from? $n^2-1 \approx (n-1)2$ for $n-1 \ll 1$

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!

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

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

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

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.

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