Recently in these pages we have had occasion to see the trigonometric identity $$ 2\arcsin\sqrt x = \frac \pi 2 + \arcsin(2x-1). $$ From this we can immediately deduce that $$ \sqrt x = \sin\left( \frac \pi 4 + \frac 1 2 \arcsin(2x-1) \right) $$ This suggests that the arcsine function near the left endpoint of its graph is shaped very similarly to the square root function. Part of this is obvious: it has a vertical tangent and is concave downward.

Sorry ‒ there was a typo in the question. What I intended is what the FIRST paragraph of the question was about; the second paragraph failed to be consistent with that since I left out the π/2 term. I've fixed it now.

But it makes me wonder about $x\mapsto \left(\frac \pi 2 + \arcsin x\right)^2$ near $x=-1.$ Is there anything interesting to say about an expansion of that function in powers of $(x+1)\text{ ?}$

  • $\begingroup$ I'm unclear on what you would like to see. As far as I can tell, it'll look like $\frac{\pi^2}{4} - \sqrt{2}\pi\sqrt{x+1}+2(x+1)+o(x+1)$ -- what qualifies as "interesting"? $\endgroup$ – Clement C. Jul 28 '17 at 23:43
  • $\begingroup$ @ClementC. : Isn't that for $\arcsin$ rather than for $(\arcsin)^2\text{ ?} \qquad$ $\endgroup$ – Michael Hardy Jul 28 '17 at 23:53
  • $\begingroup$ I don't think so (here what WolframAlpha confirms)... unless I misunderstood the square -- you do not mean composition, do you? $\endgroup$ – Clement C. Jul 28 '17 at 23:55
  • $\begingroup$ math.stackexchange.com/questions/515412/… $\endgroup$ – lab bhattacharjee Jul 29 '17 at 2:00

Apply an integral representation and a binomial expansion:

$$\begin{align}\arcsin(x)&=-\frac\pi2+\int_{-1}^x\frac1{\sqrt{1-t^2}}~\mathrm dt\\&=-\frac\pi2+\int_{-1}^x\frac1{\sqrt{(1-t)(1+t)}}~\mathrm dt\\&=-\frac\pi2+\int_{-1}^x\frac1{\sqrt{(1+t)}}\sum_{n=0}^\infty\binom{-1/2}n2^{-\frac12-n}(-1)^n(1+t)^n~\mathrm dt\\&=-\frac\pi2+\sum_{n=0}^\infty\binom{-1/2}n2^{-\frac12-n}(-1)^n\int_{-1}^x(1+t)^{n-\frac12}~\mathrm dt\\&=-\frac\pi2+\sum_{n=0}^\infty\binom{-1/2}n2^{\frac12-n}(-1)^n\frac{(1+x)^{n+\frac12}}{2n+1}\\&=-\frac\pi2+\sqrt{2(x+1)}+\frac{(x+1)^{3/2}}{6\sqrt2}+\mathcal O((x+1)^{5/2})\end{align}$$

One can multiply a few of these terms out to get

$$\small(\arcsin(x))^2=\frac{\pi^2}4-\pi\sqrt{2(x+1)}+2(x+1)-\frac\pi{6\sqrt2}(x+1)^{3/2}+\frac13(x+1)^2+\mathcal O((x+1)^{5/2})$$

Of course, for the newly edited question, simply use the above expansion to see that

$$\left(\frac\pi2+\arcsin(x)\right)^2=2(x+1)+\frac13(x+1)^2+\mathcal O((x+1)^3)$$

And of course, the entire Cauchy product:


  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. The exchange was pleasant, but it sounded like it had converged. Site hygiene then suggests relocating it to a chat room. If there is something that you think will help future viewers, please @-ping me or flag this post with an explanation. Diamond moderators can undelete selected comments. $\endgroup$ – Jyrki Lahtonen Jul 30 '17 at 7:27

Let $f(x)=\arcsin(x)$ for $x\in[-1,1]$.

Since $f'(x)=O\left( (x+1)^{-1/2}\right)$ for $x\sim -1$, we let $t=(x+1)^{1/2}$ and $g(t)=\arcsin(t^2-1)$.

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

$$g(t)=-\frac\pi2+\sqrt 2 t+\frac{\sqrt 2}{12}t^3+O(t^5)$$

which reveals that the arcsine function has the expansion

$$\arcsin(x)=-\frac{\pi}{2}+\sqrt{2}(x+1)^{1/2}+\frac{\sqrt{2}}{12}(x+1)^{3/2}+O\left((x+1)^{5/2}\right) \tag5$$

And finally, squaring $(5)$ yields the coveted expansion

$$\begin{align}\arcsin^2(x)=\frac{\pi}{4}-\sqrt 2\pi(x+1)^{1/2}+2(x+1)+\frac{\sqrt 2\pi}{12}(x+1)^{3/2}+\frac13(x+1)^2+O\left((x+1)^{5/2}\right) \end{align}$$

which agrees with the result reported by @SimplyBeautifulArt.


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