Doing a normal Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms by taking derivatives doesn't work because of division by zero.
I've put this into wolfram alpha: . It's nice but I need to show that I can do it myself. This isn't an analysis class so I have never seen square roots in a series expansion before.
I found this in my search: Some approximations for $\arccos(1/(1+x))$ but I need the expansion to two terms.
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$\begingroup$Let $y=f(x)=\arccos(1-x)$. Then $$1-x=\cos y=1-\frac{y^2}2+\frac{y^4}{24}+O(y^6),$$ so $$x=\frac{y^2}2-\frac{y^4}{24}+O(y^6).$$ Now clearly there is no actual Taylor series for $y$ about $x=0$ because $f'(0)$ does not exist. However, a generalized power series solution can be written down, known variously as the Frobenius method or the asymptotic expansion of $y=f(x)$ near $x=0$. Solving this equation formally:
$$2x=y^2\left(1-\frac{y^2}{12}+O(y^4)\right)\Rightarrow y=\sqrt{2x}\left(1-\frac{y^2}{12}+O(y^4)\right)^{-1/2}$$
Since $0<y\ll1$, $\frac{y^2}{12}\ll1$ so that we can use the binomial theorem $(1+x)^p=1+px+\cdots$ to get the next-leading order term:
$$y=\sqrt{2x}+\sqrt{2x}\frac{y^2}{24}+O(y^4)=\sqrt{2x}+\frac{\sqrt{2x}}{24}\left(\sqrt{2x}+\frac{\sqrt{2x}}{24}y^2+O(y^4)\right)^2+O(y^4)$$ $$=\sqrt{2x}+\frac{(2x)^{3/2}}{24}\left(1+\frac{1}{12}y^2+O(y^4)\right)+O(y^4)=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+\frac{(2x)^{3/2}}{24\cdot 12}y^2+O(x^{3/2}y^4)+O(y^4)$$
Now, since $y=O(\sqrt x)$ (which follows from the leading order term), we can simplify all that to get $$y=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+O(x^2).$$
$\endgroup$ $\begingroup$Start from the derivative of the function
$$\frac{1}{\sqrt{x}\sqrt{2-x}}= \frac{1}{\sqrt{2x}}\frac{1}{\sqrt{1-x/2}} = \frac{1}{\sqrt{2x}}(1+\frac{1}{4}x+\frac{3}{32}x^2+\dots )=\dots\,. $$
Now, integrate the above series to get your Taylor series.
$\endgroup$ $\begingroup$Although addressed indirectly in the other answers here, it seems relevant to point out that, strictly speaking, the resulting series expansion in all cases (so far) is -not- a Taylor series.
The Taylor series of a function $f(x)$ about $x = c$ is given by:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n, $$
i.e. it is an expansion in terms of positive, integer powers of $\mathbf{(x-c)}$, which the series given thus far are not. On the WolframAlpha link in the OP, the result is called a Puiseux series.
Personally, I don’t see any way of getting a Taylor series for this function, about $x = 0$. WolframAlpha certainly doesn’t list any such form.
$\endgroup$ $\begingroup$The same problem happens if you try to expand $\sqrt x$ as a Taylor series. The vertical slope kills you, because polynomials can't do that. The solution is what Alpha does-divide out the "singular part", which in your case is $ \sqrt x.\ \ \frac {\arccos (1-x)}{\sqrt x}$ is nicely behaved at the origin and can give you a Taylor series.
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