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I'm trying to calculate the Taylor series of $arcsin(1-x)$ about $x=0$. I'm having trouble because I can't compute the derivative there.

I can see the correct solution on WolframAlpha (http://www.wolframalpha.com/input/?i=taylor+series+arcsin(1-x)), but I'd like to know how to derive it. I found a similar question regarding the Taylor series for arccos(1-x) (Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms), but the derivation takes advantage of the fact that $arccos(1-x) << 1$, which is not the case for arcsin.

Any help would be greatly appreciated.

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  • $\begingroup$ Notice that the function is defined for $0<x<2$ $\endgroup$ – user84413 Jun 24 '16 at 2:26
  • $\begingroup$ Why can't you compute the derivative there? $\endgroup$ – GFauxPas Jun 24 '16 at 2:30
  • $\begingroup$ I got as far as $\frac{d}{dx}arcsin(1-x) = -\frac{1}{\sqrt{1-(1-x)^2}} = -\frac{1}{\sqrt{2x-x^2}}$ and didn't know how to proceed from there because the denominator is $0$ at $x=0$. $\endgroup$ – user349852 Jun 24 '16 at 3:21
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Let $f(x)=\arcsin(1-x)$ for $x\in[0,2]$.

Since the derivative of $f(x)=O\left( x^{-1/2}\right)$ for $x\sim 0$, we let $t=x^{1/2}$ and $g(t)=\arcsin(1-t^2)$.

We will now develop the first few terms of the Taylor series for $g(t)$ around $t=0$.


We have for the first derivative $g^{(1)}(t)$

$$\begin{align} g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\ &=-\frac{2}{\sqrt{2-t^2}}\tag 1 \end{align}$$


Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$

$$\begin{align} g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2 \end{align}$$


Continuing, we have for $g^{(3)}(t)$

$$\begin{align} g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3 \end{align}$$


And finally, we have for $g^{(4)}(t)$

$$\begin{align} g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4 \end{align}$$


We evaluate $(1)-(4)$ at $t=0$ and form the expansion

$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$

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