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Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$

In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it is easy.

However how to prove that $r(\sin x -x)=o(x^5)$?
I tried to do it but then I have: $$\frac{r(\sin x -x)}{\sin x -x}\cdot \frac{\sin x -x}{x^5}=\frac{r(\sin x -x)}{\sin x -x}\cdot(\frac{\sin x}{x}-1)\cdot \frac{1}{x^4} \rightarrow 0\cdot(1-1)\cdot(-\infty)$$Can you help me how can I prove it?

MP3129
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    $\sin(\sin x - x) = \sin x - x + \mathcal{O}(\sin x - x)^3) = \sin x - x + \mathcal{O}(\mathcal{O}(x^3)^3) = \sin x - x + \mathcal{O}(x^6)$ – Brevan Ellefsen Apr 16 '19 at 20:54
  • @BrevanEllefsen your $\mathcal{O}$ is the same as my small $o$? Why these equality are real? – MP3129 Apr 16 '19 at 21:51
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    there's actually a typo in my comment, it should be $\mathcal{O}(x^9)$, which trivially implies your result. You are using O-notation, so you should know what it means if you look it up :) – Brevan Ellefsen Apr 16 '19 at 21:59
  • @BrevanEllefsen Ok, I understand your sollution. However have you got idea how to prove it using my $r(\sin x - x)$? – MP3129 Apr 16 '19 at 22:42

2 Answers2

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I would be stupid about it and work formally, in the ring $\Bbb Q[[x]]/(x^6)$.

Now, in that ring, $\sin(x)=x-\frac{x^3}6+\frac{x^5}{120}$. Let me write $g(x)=\sin(x)-x$, which is divisible by $x^3$. Then $\sin(g(x))=g(x)+g(g(x))$, but since $g(g(x))$ is divisible by $x^9$, we can ignore it. Thus in $\Bbb Q[[x]]/(x^6)$, we get $\sin(\sin(x)-x)=-\frac{x^3}6+\frac{x^5}{120}$, just the same as $\sin(x)-x$.

Lubin
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You can note that $$\sin x - x=-\frac{x^3}{6}+o(x^3)\tag{1}$$ and replacing $x$ with $\sin x - x$ we get $$\sin(\sin x - x) - (\sin x - x) =-\frac{(\sin x - x)^3}{6}+o((\sin x - x) ^3)$$ Using $(1)$ we can see that both the terms in right hand side of above equation are $o(x^5) $ and we are done.

Paramanand Singh
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