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Which real values of $\theta$ allows us to represent $|\cos\theta|^{1/2}+|\sin\theta|^{1/2}$ as a series of the form $$1+a_1\sin 2\theta+a_2\sin^22\theta+\cdots+a_k\sin^k2\theta+a_{k+1}\sin^{k+1}2\theta+\cdots ?$$

In that case how can we find constants $a_1, a_2, a_3, \cdots ?$

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  • $\begingroup$ Do you want absolute values around the sine and cosine under the square roots...? $\endgroup$ – Andrew D. Hwang Dec 23 '16 at 1:38
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    $\begingroup$ The domain will be $\theta\in\left\{z\ |\ 2\pi n\le z\le2\pi n+\pi/2\right\}$ for integral $n$. $\endgroup$ – Will Sherwood Dec 23 '16 at 1:40
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At $\theta = 0$, the $|\sin(\theta)|^{1/2}$ gives you a singularity like $|\theta|^{1/2}$. But if your series converges on some interval, it is $f(\sin(2\theta))$ where $f$ is an analytic function in a neighbourhood of $0$, and this can't have that square root singularity at $0$. So the answer is: nowhere.

EDIT: If you let $\theta = x^2$, then for $0 < \theta < \pi$ you have $$\eqalign{&\sin(\theta)^{1/2} = \sin(x^2)^{1/2}\cr &= x-{\frac{1}{12}}{x}^{5}+{\frac{1}{1440}}{x}^{9}-{\frac{1}{24192}}{x}^ {13}-{\frac{67}{29030400}}{x}^{17}-{\frac{1}{5677056}}{x}^{21}-{\frac{ 64397}{4649508864000}}{x}^{25}-{\frac{113249}{100429391462400}}{x}^{29 }+\ldots} $$ while for $0 < \theta < \pi/2$ $$ \eqalign{&\cos(\theta)^{1/2} = \cos(x^2)^{1/2} \cr &= 1-{\frac{1}{4}}{x}^{4}-{\frac{1}{96}}{x}^{8}-{\frac{19}{5760}}{x}^{12 }-{\frac{559}{645120}}{x}^{16}-{\frac{29161}{116121600}}{x}^{20}-{ \frac{2368081}{30656102400}}{x}^{24}-{\frac{276580459}{11158821273600} }{x}^{28}+\ldots}$$

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  • $\begingroup$ Thank you sir. According to you and Doug M, there is a problem at $\theta=0.$ But I could not find any gap of the solution given by Doug M. Could you explain little bit more. Since we are looking for real values of $\theta,$ I think we do not think along singularities? $\endgroup$ – Bumblebee Dec 23 '16 at 3:29
  • $\begingroup$ If a series in powers of $z$ converges at $z=a$, then it converges for all $z$ (real or complex) with $|z| < a$, and the sum is an analytic function of $z$. $\endgroup$ – Robert Israel Dec 23 '16 at 16:34
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$\sin^2 \theta = \frac {1}{2} (1 - \cos 2\theta)\\ |\sin \theta|^\frac 12 = (\frac {1}{2})^\frac 14 (1-\cos 2\theta)^\frac 14\\ |\cos \theta|^\frac 12 = (\frac {1}{2})^\frac 14 (1+\cos 2\theta)^\frac 14$

We can do a binomial expansion:

$|\sin \theta|^\frac 12 = (\frac {1}{2})^\frac 14(1^\frac 14 - \frac 14 (1^{-\frac {3}{4}}) \cos 2\theta - \frac {3}{2!4^2}(1^{-\frac74}) \cos^2 2\theta -\frac{3\cdot7}{3!4^3}\cos^3 2\theta - \frac{3\cdot7\cdot\ 11}{4!4^4}\cos^4 2\theta\cdots)$

$|\cos \theta|^\frac 12$ will look very similar, with the exception of some sign switching. When you add them together.

$|\sin \theta|^\frac 12 + |\cos \theta|^\frac12= 2(\frac {1}{2})^\frac 14(1 - \frac {3}{2\cdot4^2} \cos^2 2\theta - \frac{3\cdot7\cdot\ 11}{4!4^4}\cos^4 2\theta- \frac{3\cdot7\cdot\ 11\cdot 15\cdot 19}{6!4^6}\cos^6 2\theta\cdots)$

$\cos^2 2\theta = 1 - \sin^2 2\theta$ And I am feeling a little too exhausted to go through all of those substitutions. But it will work.

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  • $\begingroup$ Your series is in powers of $\cos(2\theta)$, not $\sin(2\theta)$. The substitution $\cos^2(2\theta) = 1 - \sin^2(2\theta)$, followed by expansion of powers to rearrange terms, is not likely to give you a convergent series. And indeed, as I wrote in my answer, it can't converge. $\endgroup$ – Robert Israel Dec 23 '16 at 16:37
  • $\begingroup$ @RobertIsrael yes I realized a little bit late in the game that I needed to get from a series in terms of $\cos 2\theta$ to a series in terms of $\sin 2\theta$. The series above does converge, and it seems like it it is likely to converge have going through the $\cos^2 2\theta = 1-sin^22\theta$ substitution. But, haven't worked it far enough to prove by any of the standard tests. I am following your rationale as to why it can't converge. $\endgroup$ – Doug M Dec 23 '16 at 20:08
  • $\begingroup$ By using Raabe's test I found that the rearranging series of $\sin^22\theta$ does not convergent. Anyway thank you for your nice solution. $\endgroup$ – Bumblebee Dec 27 '16 at 18:09

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