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


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.


3 Answers 3


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 $x=\infty$. Laurent Series of the function

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 $n=10$.

Plot of true function and approximation

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    $\begingroup$ In fact, using algebra, one can show that the error of approximation is bounded by ${1 \over 2n}$. $\endgroup$
    – copper.hat
    Oct 24, 2017 at 16:07
  • $\begingroup$ I would hesitate to call this a great approximation because it's really quite bad for small n which is the only part of the plot that makes physical sense since you're rarely going to have close to 10 unpaired electrons. Good answer though mathematically. $\endgroup$
    – jheindel
    Oct 24, 2017 at 18:32
  • $\begingroup$ @jheindel true though I can see it be useful for back of the envelope/order of magnitude type calculations. And sure 0 and 1 aren't great, but for those its hardly worth making the approximation in the first place because you could probably just memorize those. By n=2, you only have about 6% error. $\endgroup$
    – Tyberius
    Oct 24, 2017 at 18:54
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    $\begingroup$ Ya true. I suspect this is not very useful beyond a coarse approximation anyways so approximating it more can't be too bad. $\endgroup$
    – jheindel
    Oct 24, 2017 at 18:55

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.

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    $\begingroup$ It's not really a Taylor expansion, so much as using the arithmetic mean of two numbers ($n$ and $n+1$) as an approximation to their geometric mean. $\endgroup$
    – psmears
    Oct 24, 2017 at 16:05
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    $\begingroup$ @psmears: You mean $n$ and $n + 2$, right? $\endgroup$ Oct 24, 2017 at 17:00
  • $\begingroup$ Why do you assume that it's a blind usage of anything? $\endgroup$ Oct 24, 2017 at 17:03
  • $\begingroup$ @DavidRicherby: The approximation apparently transforms a nonlinear function into a linear one. The simplest and most common way to do this is to do a Taylor approximation. $\endgroup$ Oct 24, 2017 at 17:18
  • $\begingroup$ My (literal) emphasis was on "blind". Saying it's "a blind usage of Taylor expansions" means that you think that the teacher used Taylor expansions without giving any thought to whether they were suitable. Why are you accusing them of that? $\endgroup$ Oct 24, 2017 at 18:25

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}$.

  • $\begingroup$ @Tyberius: Good suggestion. Done. $\endgroup$
    – copper.hat
    Oct 24, 2017 at 16:20

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