# Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms

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: http://www.wolframalpha.com/input/?i=taylor+series+arccos%281-x%29. 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.

• What value are you expanding "about"? – icurays1 Dec 11 '12 at 0:15
• "Taylor series about $x=0$" means a series in nonnegative integer powers of $x$. This isn't. It is still a series expansion, just not a Taylor series. And that explains why the derivative formulas for Taylor series don't work here. – GEdgar Dec 11 '12 at 2:02

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 $$y=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+O(x^2).$$

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.

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.

• Why $2\sqrt x$? If you want the leading term, it's $\sqrt{2x}$. If you are just dividing out the singular part, $\sqrt{x}$ will do. (Not that you are wrong, it just seems a bit arbitrary.) – Mario Carneiro Dec 11 '12 at 0:52
• @MarioCarneiro: true. I misremembered it typing up. Have deleted the $2$ – Ross Millikan Dec 11 '12 at 1:29
• When I try to take the derivative of $\frac {\arccos(1-x)}{\sqrt x}$ I get $\frac{ \frac {-\sqrt x}{\sqrt{2x-x^2}} + \frac {\arccos(1-x) }{2 \sqrt {x}^3}}{x}$ = $-\frac{1}{\sqrt{2x^2-x^3}}$ which is still bad at $x=0$. Is it just that my algebra is wrong? – 無色受想行識 Dec 11 '12 at 1:48
• @RossMillikan What do you mean by “The solution is what Alpha does-...” a) The linked solution gives a series solution for $cos^{(-1)}(1-x)$ and b) doesn’t adding a factor of $\frac{1}{\sqrt{x}}$ make the behaviour as $x \to 0$ even worse?! – Rax Adaam Jun 27 at 19:15
• @RaxAdaam: The Alpha series is a polynomial times $\sqrt x$. It is not a Taylor series because a Taylor series is a polynomial, where all the exponents are nonnegative integers. – Ross Millikan Jun 27 at 19:54

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.