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$r'(\theta)^2 + r(\theta)^2 = \theta^2,\quad r(t=0)=0\tag{1}$

There is an interesting approach to prove that the solutions of the equation $(1)$ have power series representations of form $r(\theta) = \pm \ \left(\frac{\theta^2}{2}-\frac{\theta^4}{32}+\frac{\theta^6}{768}+\cdots\right)$, and the power series convergence. (More details

However, I am still unable to compute the radius of convergence analytically. Any areas of math to solve the problem? Any help, thoughts are highly welcomed.

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    $\begingroup$ Doesn't @RobertBryant give all the relevant references in his very nice answer to the original question? (mathoverflow.net/questions/121402/…) $\endgroup$
    – Igor Rivin
    Aug 7 '13 at 17:44
  • $\begingroup$ @Igor Revin I understand that this is a nice prove, but what about the radius? $\endgroup$
    – Mikhail Gaichenkov
    Aug 7 '13 at 18:03
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    $\begingroup$ I don't know who this Igor Revin is, but as for @RobertBryant's proof, if you read it to the end, he gives a reference addressing the very question you asks. $\endgroup$
    – Igor Rivin
    Aug 8 '13 at 2:02
  • $\begingroup$ @Igor Rivin. The reference just explains the power series convegence, but I cannot see that the radius is, I guess, 7/2? $\endgroup$
    – Mikhail Gaichenkov
    Aug 8 '13 at 4:15
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    $\begingroup$ Well, the kind of mathematics can prove the convergence, but I cannot see any way to compute the radius of convergence. For eg how to prove it is 7/2? $\endgroup$ Aug 8 '13 at 18:06
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The reason you didn't get a satisfactory answer so far is that your problem is not a bona fide initial problem and cannot be turned into one using suitable substitutions and the like. When you introduce, e.g., a new independent variable $u$ by means of $t:=\sqrt{2u}$ you get an "initial value problem" of the form $$\dot r(u)=\sqrt{1-{r^2(u)\over u}},\quad r(0)=0\ .$$ Changing the initial condition to $r(0)=10^{-100}$ lets the solution disappear, contrasting the case of "ordinary" IVP's, where the solutions depend continuously on the initial conditions.

Nevertheless, the solution you are interested in can be written as $$r(t)=\sum_{k=1}^\infty a_k t^{2k}\ ,$$ whereby the $a_k$ satisfy the following recursion: $$a_1={1\over2},\quad a_n=-{1\over 4n}\left(\sum_{k=1}^{n-1}a_k a_{n-k}+4\sum_{k=2}^{n-1}k(n+1-k)a_ka_{n+1-k}\right)\qquad(n\geq2)\ .$$ Using this recursion and putting $b_n:=|a_n|^{-1/2n}$ one obtains the following plot (note that $\lim\inf_{n\to\infty} b_n$ is the radius of convergence you are after):

enter image description here

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Have you tried the Cauchy-Hadamard theorem?

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An asymptotic formula for the coefficients of the power series would help if explicit expression is not available. The derivation of the asymptotic approximation depends on the way you obtained that power series.

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