I wrote in my analysis notes the following: $\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right)$.

I'm guessing it comes from Taylor's formula but I don't understand how they got that result.

Also, it said: $\sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$

Do we simply assume that since $ \omicron\left(\left| \dfrac{x}{n} \right|\right) \rightarrow 0 \implies$ $(\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right))^2 = \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$

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    $\begingroup$ Shouldn't it be $\sin\left(\frac{x}{n}\right) = \frac{x}{n} - ...$? $\endgroup$ Commented May 26, 2016 at 5:53
  • $\begingroup$ I'm quite sure $\sin \left( \dfrac x n \right)^2$ means the sine of $\left( \dfrac x n \right)^2$. The square of $\sin \left( \dfrac x n \right)$ is usually written $\sin^2 \left( \dfrac x n \right)$. $\endgroup$
    – M. Vinay
    Commented May 26, 2016 at 6:01
  • $\begingroup$ Pay attention to the answer and @ArjitSeth's comment. The sign you have used is incorrect. It doesn't get "squared away" in the second formula, as you seem to be thinking. There never is a negative sign there to begin with, and the whole expression is not being squared. Only the $\dfrac x n$ is being replaced by its square. That would not affect the sign outside it (if there were one). $\endgroup$
    – M. Vinay
    Commented May 26, 2016 at 6:03
  • $\begingroup$ @ArjitSeth Yes I think $\sin\left(\dfrac{x}{n}\right) = \dfrac{x}{n} - \omicron\left(\left| \dfrac{x}{n} \right|\right)$ too $\endgroup$
    – aribaldi
    Commented May 26, 2016 at 13:14

1 Answer 1


The taylor series for $\sin y$ around $y=0$ is $$ \sin y=y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}+\cdots. $$ Plug in $y=x/n$ to get $$ \sin\frac{x}{n}=\frac{x}{n}\underbrace{-\frac{x^{3}}{n^{3}3!}+\frac{x^{5}}{n^{5}5!}+\cdots.}_{o(|\frac{x}{n}|)} $$

  • $\begingroup$ (the Taylor expansion of order $k$ being true for every $C^k$ function, not only analytic functions. in particular it is very possible to prove the formula above before you even proved that $\cos x = (\sin x)'$) $\endgroup$
    – reuns
    Commented May 26, 2016 at 7:30

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