Magnetic moment μ approximation

Question

I've been reading a bit about the magnetic moment (spin-only) $\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons

$$\mu_{s.o}=\sqrt{n(n+2)}$$

where $n$ is the number of unpaired electrons.

However in our lecture today we were using the approximation $\mu_{s.o} \approx n+1$. Is this an acceptable approximation for the magnetic moment or should I stick to using the previous one.

Obviously using $\mu_{s.o} \approx n+1 $ is easier to use for calculations but I would like someone's opinion on this.

Answer 1

I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.

This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$ By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.

Answer 2

To me this seems like a blind usage of the Taylor expansion.

Let's say we want to get a Taylor approximation of $\mu (n) = \sqrt{n(n+2)}$. Then we know we can expand any function $f(x) = f(x_0 +h)$ if $h/x_0 \ll 1$ into $f(x_0 + h) \approx f(x_0) + \frac{\partial f}{\partial x}(x_0) \cdot h $.

If I apply this to the function $\mu(n) = \mu(0 + n)$ then I get the approximate result $\mu(n) \approx \frac{n+1}{\sqrt 2}$. But here of course it is not true that $n/0 \ll 1$, so the assumption needed for using Taylor is violated.

However, if I plot both those functions in Wolframalpha they seem to agree OK enough for large n. So I guess a different approximation technique might have been used deriving this result, and it depends if you're working at large n or small n.

Answer 3

Using $\sqrt{x}-\sqrt{y} = { x -y \over \sqrt{x} + \sqrt{y}} $ we have $\sqrt{n(n+2)} - (n+1) = -{ 1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )} $, so $|\sqrt{n(n+2)} - (n+1)| \le {1 \over 2n}$.

The approximation is reasonable for large $n$.

Addendum: To see where the approximation comes from, note that $\sqrt{1+ {2 \over n}}+1 + {1\over n} \ge 2$, hence $ {1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )} \le {1 \over 2n}$.