I am trying to prove that dnθdxn where x=1−cosθ2 exists ∀n>0. I found that dθdx=2sinθ2 and based on this answer I think I only need to prove that sinθ2 is smooth (notice is a function of θ and its firsts derivatives looks like fun).
I also found that cos−1(x)=π2+∑∞k=0x1+2k(12)kk!+2kk! (at wolfram alpha). But I am not sure that the existence of its Taylor Series implies its smoothness. I thought that by linearity of differentiation and because all terms in the series are polynomials (smooth funcions) there could be a way to make it work.
Is there a simple (elegant) way to prove its smoothness? Until now my only approach is to find a pattern of the n-th derivative, however I am actively trying to avoid that because of this.
edit:
I am trying to prove that θ=2cos−1(1−x) is smooth (in other words, has infinitely many derivatives ∃dnθdxn∀n>0)
I tried different approaches, but based on the Taylor Expansion of cos−1 I found that θ=2cos−1(1−x) can be expressed by polynomials (which are smooth functions) and therefore its derivatives.
I have never worked with Taylor Series, so I am not really confident about this last statement.
Side note: My original expression was x=1−cosθ2,−1≤x≤1 (which implies θ=2cos−1(1−x)) and because of that I derived dθdx=2sinθ2 applying ddx to both sides.
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