# Composed series expansion understanding

Recently I'm studying series expansion and there is something I don't understand but it's hard to tell what... For example, imagine I need to calculate the series expansion of $$\sqrt{1+\sin(x)}$$ at $$a=0$$.

I want to do it as fast as possible, so my idea is to have the series expansion of $$\sqrt{1+x}$$ and then plug the series expansion of $$\sin(x)$$ in it. Since when $$x = 0$$, $$\sin(x) = 0$$ too...

So for $$\sqrt{1+x}$$ I find $$1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4)$$ for example.

For $$\sin(x)$$ it's $$x-\frac{x^3}{3!}+O(x^5)$$ Which are both correct but when I replace the $$x$$ of the first expansion by the $$\sin(x)$$'s one I find the result don't change from $$1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4)$$ while wolfram tells me the $$x^3$$ term should be $$\frac{-1}{48}$$ And it's where it doesn't make sense to me, why would it be the only term to change? Computers are always right so I do believe it's a misunderstanding but where could it be? Here's the thing I have for $$x^3$$ : $$\frac{3}{8}\cdot\frac{1}{3!}\cdot\frac{1}{(1+x^3-\frac{x^3}{3!})^{5/2}}\cdot x^3 = \frac{1}{16}\cdot\frac{1}{(1+x^3-\frac{x^3}{3!})^{5/2}}\cdot x^3$$ and for me the last term is equal to $$1$$ since $$x$$ is close to $$0$$.

Your approach is correct. Until you've got it mastered I'd use two different variables: so $$\sqrt{1+y} = 1 + \frac{1}{2}y - \frac{1}{8}y^2 + \frac{1}{16}y^3 + O(y^4)$$ Then you want to substitute $$y = x - \frac{1}{3!}x^3 + O(x^5)$$ So $$\sqrt{1+y} = 1 + \frac{1}{2}\left(x - \frac{1}{3!}x^3 + O(x^5)\right) - \frac{1}{8}\left(x + O(x^3)\right)^2 + \frac{1}{16}\left(x + O(x^3)\right)^3 + O(x^4)$$ Note that there's no need to expand the third and fourth terms beyond $$O(x)$$ because they're squared/cubed and will give terms $$O(x^4)$$. This is why it's only the cubic term that changes.
So the extra term in $$x^3$$ you're looking for comes from collecting all terms in $$x^3$$ above.
• That's it thanks a lot, I wasn't properly reading the equation so I forgot to include the $\frac{1}{3!}x^3$ in 1/2. Now I get the correct result! Dec 7, 2021 at 12:58