2
$\begingroup$

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 $(x+1)\text{ ?}$

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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:

$$\left(\frac\pi2+\arcsin(x)\right)^2=2\sum_{n=0}^\infty\sum_{k=0}^n\binom{-1/2}k^2\frac{(-2)^{-n}(1+x)^{n+1}}{(2k+1)(2n-2k+1)}$$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.