# 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.

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.

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 '19 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 '19 at 19:54